BIO 10/e Lab Manual

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Mader: Biology 11/e Lab Resource Guide

Part I The Cell

1 Scientific Method

2 Metric Measurement and Microscopy

3 Chemical Composition of Cells

4 Cell Structure and Function

5 How Enzymes Function

6 Photosynthesis

7 Cellular Respiration

Part II The Genetic Basis of Life

8 Mitosis and Meiosis

9 Mendelian Genetics

10 Human Genetics

11 DNA Biology and Technology

12 Evidences of Evolution

13 Natural Selection

Part III Microbiology and Evolution

14 Bacteria and Protists

15 Fungi

Part IV Plant Evolution and Biology

16 Nonvascular Plants and Seedless Vascular Plants

17 Seed Plants

18 Organization of Flowering Plants

19 Water Absorption and Transport in Plants

20 Control of Plant Growth and Responses

21 Reproduction in Flowering Plants

Part VAnimal Evolution and Diversity

22 Introduction to Invertebrates

23 Invertebrate Coelomates

24 The Vertebrates

Part VI Comparative Animal Biology

25 Animal Organization

26 Basic Mammalian Anatomy I

27 Basic Mammalian Anatomy II

28 Chemical Aspects of Digestion

29 Homeostasis

30 Nervous System and Senses

31 Musculoskeletal System

32 Animal Development

Part VII Ecology

33 Sampling Ecosystems

34 Effects of Pollution on Ecosystems

Laboratory 1 Scientific Method

(LM pages 1–8)

Eleventh Edition Changes

This lab has been much improved in a number of ways. In Section 1.1, examples are provided for each step of the scientific method and in Section 1.3 and 1.4, students use plus and minus signs to hypothesize and then record a pillbug's reaction to test substances. Table 1.4 has an improved design for collecting class data. A virtual lab, described on page 2, concerns mealworm behavior and can be used as an addition to or replace Sections 1.3 and 1.4.

New/Revised Figures: 1.1 Pillbugs on leaf

New/Revised Tables: 1.1 Pillbug Speed; 1.2 Hypotheses About Pillbug's Response to Potential Foods; 1.3 Pillbug's Response to Potential Foods; 1.4 Class Results

MATERIALS AND PREPARATIONS

Instructions are grouped by procedure. Some materials may be used in more than one procedure.

Special Requirements

Living material. Live pillbugs, Armadillidium vulgare, for all sections of lab. Earthworm alternative. See appendix below if you wish to use earthworms instead of pillbugs in all sections of lab.

Fresh material. Substances for instructor to feed pillbugs and substances for students to test pillbug behavior (see Section 1.4 below).

1.2 Observing the Pillbug (LM pages 4–5)

_____ pillbugs, Armadillidium vulgare, live (Carolina 14-3082)

_____ pen, white (or correction fluid, white) or tape tags

_____ magnifying lenses or stereomicroscopes

_____ small glass or plastic dishes, such as disposable petri dishes

_____ graduated cylinders or small beakers for observing pillbug movement

_____ rulers, metric, 30 cm plastic

_____ stopwatch

Live pillbugs (LM pages 1–7). Obtain 50 pillbugs for a class of 20 to 35 or more students. Order pillbugs so that they arrive as close as possible to the date they will be needed. Use one container of fresh pillbugs for each lab.

Care and feeding of pillbugs: Follow care and feeding instructions provided with the pillbug order. Withdraw food 1–2 days prior to the experiment.

Use white correction fluid or tape tabs to number the pillbugs for identification.

Collecting pillbugs (LM pages 1–7). Pillbugs like moisture, and avoid sunlight. They can be found next to brick buildings along the grass line or next to sidewalks, or under logs and planks of wood. They are attracted to wet grass covered with a cardboard box or plastic tarp. Encourage students to collect their own pillbugs and give them lab participation points. Collect pillbugs in the spring, summer, and fall as they are hard to find in the winter.

Maintaining pillbugs in the lab (LM pages 1–7). After collecting, pillbugs can be easily maintained in a terrarium to keep a fresh supply all year long. They feed primarily on decaying organic matter; they like moisture and avoid sunlight. They like carrots and cucumbers. Change the food daily to prevent mold growth.

1.3 Formulating Hypotheses (LM page 6)

1.4 Performing the Experiment and Coming to a Conclusion (LM pages 6–7)

_____ pillbugs, Armadillidium vulgare, live (Carolina 14-3082)

_____ small beakers, 35-mm film cans, watch glasses, or small petri dishes for distributing test substances

_____ petri dishes, preferably 150 mm (or else 100 mm) for testing the pillbugs

_____ small plastic bottle for spritzing

_____ distilled water

_____ cotton balls

Suggested test substances:

_____ flour

_____ cornstarch or bran flakes

_____ coffee creamer

_____ baking soda

_____ fine sand (control)

_____ milk

_____ orange juice or apple juice

_____ ketchup

_____ applesauce

_____ carbonated beverage

_____ water (control)

Do not use salt, vinegar, or honey, as these substances are harmful to pillbugs.

Plain water is used as a control for liquids. Fine sand is used as a control for powders.

Experimental design (LM pages 6–7). These methods are recommended: For a dry substance, make a circle of the test substance in a petri dish and put the pillbug in the center of the circle. For a liquid, put a cotton ball soaked with the test substance in the pillbug's path. Rinse pillbugs between testing procedures by spritzing with distilled water and then placing them on a paper towel to dry.

Cleanup (LM pages 6–7). Cleanup is easier and the experiment goes well if there is a limited number of test substances and each student chooses only two dry and two liquid test substances. Substances can be distributed to several stations in small beakers, 35-mm film cans, watch glasses, or small petri dishes. Testing pillbugs in 150 mm petri dishes works well.

EXERCISE QUESTIONS

1.1 Using the Scientific Method (LM pages 2–3)

Why does the scientific method begin with observations? To study the natural world, scientists have to observe natural phenomena.

What is the benefit of formulating a hypothesis? The hypothesis tells what is to be tested by experiment or further observations.

Why must a scientist keep complete records of an experiment? So others can repeat the experiment and can check that the data are valid.

What is the purpose of the conclusion? The conclusion tells what has been learned from the experiment (or further observations).

How is a scientific theory different from a conclusion? Each experiment has a conclusion. A scientific theory is based on many conclusions from various experiments in related fields.

1.2 Observing the Pillbug (LM pages 3–5)

Observation: Pillbug’s External Anatomy (LM pages 3–4)

1. Examine the exterior of a pillbug.

• How can you recognize the head END of a pillbug? The head bears antennae and eyes.

• How many segments and pair of walking legs are in the thorax? There are 7 segments and 7 pairs of legs.

Observation: Pillbug’s Motion (LM page 5)

1. Watch a pillbug's underside.

a. Describe the action of the feet and any other motion you see.

The seven pairs of legs move with the front pair leading, and each pair moves in succession thereafter.

b. Allow a pillbug to crawl on your hand. Describe how it feels. It tickles the skin as it moves.

c. Does a pillbug have the ability to move directly forward? yes

d. Do you see evidence of mouthparts on the underside of the pillbug? A pillbug has four pairs of mouthparts.

2. As you watch the pillbug, identify

a. the anatomical parts that allow a pillbug to identify and take in food. Antennae, eyes, and mouthparts

b. behaviors that will allow a pillbug to acquire food. For example, is the ability of a pillbug to move directly forward a help in acquiring food? Explain. Yes because it is the most efficient way to reach food. What other behaviors allow a pillbug to acquire food? A pillbug has the ability to eat food.

c. a behavior that helps a pillbug avoid dangerous situations The pillbug rolls into a ball when it is threatened.

3. Measure the speed of three pillbugs. See Table 1.1 Pillbug Speed

Table 1.1 Pillbug Speed*

Pillbug Millimeters Time Speed

Traveled (mm) (sec) (mm/sec)

1 71 30 2.36

2 132 60 2.20

3 64 30 2.13

Average speed 2.23 mm/sec

*Answers will vary. The answers provided here are examples.

1.3 Formulating Hypotheses (LM page 6)

2. Hypothesize in Table 1.2 how you expect the pillbug to respond, and offer an explanation for your reasoning. The following is an example of three possible student hypotheses regarding flour. Students uses "0" for no response, "—" for moves away from the substance and "+" for moving toward the substance and eating it.

Table 1.2 Hypotheses About Pillbug’s Response to Potential Foods

Substance Hypothesis Reason for Hypothesis

Flour 0 Flour is a bland substance.

Flour — Flour is a dry substance.

Flour + Flour is a food substance.

1.4 Performing the Experiment and Coming to a Conclusion (LM pages 6–7)

Experimental Procedure: Pillbug’s Response to Potential Foods (LM pages 6–7)

5. Do your results support your hypotheses?

Table 1.3 Pillbug’s Response to Potential Foods

Substance Pillbug’s Response Hypothesis supported?

Flour + Depends on hypotheses

Cornstarch +

Coffee creamer +

Baking soda —

Fine sand 0*

Milk +

Orange juice —

Ketchup —

Applesauce +

Carbonated beverage +

Water 0*

*pillbugs may move toward these substances but do not eat them.

6. Are there any hypotheses that were not supported by the experimental results (data)? Answer depends on student hypotheses.

7. Compare your results with those of other students who tested the same substances. Complete Table 1.4.

Table 1.4 Pillbug's Response to Potential Foods: Class Results

Answers will vary depending on class data.

8. On the basis of the class data do you need to revise your conclusion for any particular pillbug response? Depends on class data

9. Did the pillbugs respond as expected to the controls, i.e., did not eat them? Depends on experimental results.

LABORATORY REVIEW 1 (LM page 8)

1. What are the essential steps of the scientific method? The scientific method usually includes :new observations, formulating a hypothesis, testing the hypothesis through experimentation and further observations, and reaching a conclusion.

2. What is a hypothesis? A hypothesis is a tentative explanation of observed phenomena.

3. Is it sufficient to do a single experiment to test a hypothesis—why or why not? No, multiple experiments are needed to assure accuracy of the results.

4. What do you call a sample that goes through all the steps of an experiment but does not contain the factor being tested? a control

5. What part of a pillbug is for protection, and what does it do to protect itself?

The exoskeleton is for protection; it rolls into a ball to protect itself.

6. Name one observation that you used to formulate your hypotheses regarding pillbug reactions toward various substances. Personal experience with the test substances.

7. Why is it important to test one substance at a time when doing an experiment? Allows a response of attraction or avoidance to be recorded accurately per substance.

Indicate whether statements 8 and 9 are hypotheses, conclusions, or scientific theories.

8. The data show that various vaccines protect people from disease. conclusion

9. All living things are made of cells. scientific theory

10. How should an affirmative conclusion always be worded? The results of this study support the hypothesis that … (hypothesis should be restated).

Earthworm Alternative

Earthworms can be used instead of pillbugs for all of the exercises in this laboratory.

Place earthworms in large rectangular plastic storage containers and let them roam around for approximately 15 min. Can also be used to keep earthworms between experiments. Plexiglass is also needed to place test substances on while holding earthworms above to see behavior towards substances.

Earthworms want to move rapidly to escape. They are inclined to move away from light, move under things, and seem to want to move downward. They are expected to move away from heat source. They also move toward each other and pile up on each other. They can move up and down on glass at a 45 degree angle.

With regard to what student already knows about earthworm activity, they might predict certain behaviors. Earthworms live (or hide) in the soil, so they would move down and through soil. Soil prevents desiccation and keeps them cool and moist. By moving under things, they could stay cooler, stay moist, and stay hidden in the dark. Perhaps light bothers them also.

Earthworms can move backward and forward from both ends. When they are investigating a substance, they make a long, skinny point out of the end they are investigating with, and if they are repelled by a substance, they pull back and the end becomes thick and round.

When testing with liquids, if earthworm gets even close to the substance, the substance will be pulled along the earthworm’s body without the earthworm doing anything. Capillary action or cohesion tension? To prevent this, hold the earthworm above the substance, in case the substance (especially lemon juice) might harm the earthworm. Just let the worm move its pointed end into or near the substance. You can tell when it is repelled as it will pull away. Rinse the earthworm right away if it touches a substance (especially lemon juice).

WHEN FINISHED WITH EARTHWORMS, mix damp potting soil with some oatmeal, potato peels, lettuce, or other organic matter from the test—not too much, just enough to give the earthworms something to eat. Add earthworms. Cover container with newspaper. Keep soil damp. When completely finished, release earthworms into garden or greenhouse soil.

Laboratory 2 Metric Measurement and Microscopy

(LM pages 9–26)

Eleventh Edition Changes

Section 2.3 has a better design in that the stereomicroscope is now opposite to the parts to be identified. The design of Section 2.5 has also been improved and now includes an examination of pond water with accompanying illustration.

New/Revised Figures: 2.6 Stereomicroscope; 2.7 Compound Light Microscope; 2.12 Microorganisms found in pond water

New/Revised Tables: Table 2.1. Metric Units of Length Measurement; Table 2.5 Differences Between Onion Epidermal and Human Epithelial Cells

MATERIALS AND PREPARATIONS

Instructions are grouped by procedure. Some materials may be used in more than one procedure.

Special Requirements

Living material. Euglena.

Fresh material. Onion, pond water (order if not available locally).

2.1 The Metric System (LM pages 10–13)

Length

_____ rulers, plastic millimeter

_____ meterstick, metric and English

_____ long bones from disarticulated human skeleton

_____ cardboard (10 cm x 30 cm), two pieces

Weight

_____ balance scale

_____ wooden block, small enough to hold in hand

_____ object, such as a penny, a piece of granite, or a trilobite fossil, small enough to fit through the opening of a small graduated cylinder

Volume

_____ wooden block and object from above

_____ graduated cylinders, 50 mL or 100 mL

_____ test tubes (large enough to hold 20 mL of water)

_____ dropper bottles containing water

_____ index card, blank white (20 cm or 30 cm)

_____ beaker, 50 mL

_____ graduated pipette (for demonstration)

Temperature

_____ thermometer, Celsius

_____ cold water, hot water, ice water (conveniently available for temperature measurement

2.3 Stereomicroscope (Dissecting Microscope) (LM pages 16–17)

_____ microscope, stereomicroscope with illuminator

_____ lens paper

_____ an assortment of objects for viewing (e.g., coins, plastomount)

2.4 Use of the Compound Light Microscope (LM pages 18–22)

_____ microscopes, compound light

_____ lens paper

_____ slide, prepared: letter e ; or newspaper, scissors, slides and coverslips

_____ rulers, clear plastic millimeter from above

_____ slide, prepared: colored threads; or to prepare your own, you will need slides and coverslips, three or four colors of sewing thread (or hairs), scissors, and a dropper bottle of water

2.5 Microscopic Observations (LM pages 23–25)

All exercises:

_____ microscope slides (glass or plastic)

_____ coverslips

_____ lens paper

_____ microscopes, compound light

_____ methylene blue solution, or iodine-potassium-iodide (IKI) solution

(premade) in dropper bottle

Onion Epidermal Cells

_____ onion, fresh

_____ scalpel

Human Epithelial Cells

_____ toothpicks, prepackaged flat for student to obtain cells from mouth

_____ ethyl alcohol (ethanol), 70% ; or alcohol swabs (if toothpicks are not individually prepackaged)

_____ biohazard waste container for toothpicks

_____ container of 10% bleach solution for slides and coverslips (to be washed directly or autoclaved and washed at lab technician’s discretion)

or

_____ prepared slide: human stratified squamous epithelium, cheek

Euglena

_____ Protoslo® (Carolina 88-5141) or methyl cellulose solution

_____ Live Euglena culture

and/or

_____ pond water, obtain locally or order from supply house

_____ pictorial guides such as:

Jahn, T. L., et al. 1979. How to Know the Protozoa, 2nd ed. Wm. C. Brown Publishers, ISBN: 0697047598 (Carolina 45-4100).

Needham, J. G., and P.R. Needham. A Guide to the Study of Freshwater Biology: With Special Reference to Aquatic Insects and Other Invertebrate Animals, 5th ed. Charles C. Thomas Publishers, ISBN: 0070461376.

Patterson, D. J. and Hedley, S. 1996. Free0Living Freshwater Protozoa: A Color Guide. John Wiley & Sons, ISBN: 0470235675.

Prescott, G. W. 1978. How to Know the Freshwater Algae, McGraw-Hill Higher Education, ISBN: 0697047547

Rainis, K. G., and Russell, B. J. 1997. Guide to Microlife, Franklin Watts, Inc., ISBN: 053112667

Microscope supplies. Set aside an area in the laboratory for storage of clean microscope slides, coverslips, and lens paper. Post a notice in this area, outlining the established procedures for handling dirty slides. Possible procedures include:

1. Wash, rinse, and dry all slides, and return them to their boxes; discard plastic coverslips.

2. Wash and rinse all slides, and place them in the drying rack.

3. Place dirty slides in the detergent solution provided; discard plastic coverslips. Some laboratories prefer that the laboratory assistant wash all slides in an ultrasonic cleaner, rinse the slides in distilled water, and allow the slides to drain dry.

4. Discard plastic coverslips. Glass coverslips should be placed in detergent solution in a beaker.

5. To eliminate the possibility of contact with pathogens, the epithelial slide exercise (page 24) can be done as a demonstration using a flexscope or videoscope for students to view from their seats. Otherwise, use a biohazardous waste container for toothpick disposal, and wash slides and coverslips in a 10% bleach solution. Microscopes should also be wiped with a disinfecting solution.

Order solutions/reagents or prepare your own.

Methylene blue solution (LM page 23). Make up a 1.5% stock solution, using 1.5 g methylene blue stain (dye powder) in 100 mL of 95% ethyl alcohol . Dilute one part stock solution with nine parts water for laboratory use, or use iodine (IKI) solution. Methylene blue staining solution can also be purchased premade.

Iodine (IKI) solution (LM page 23). Iodine-potassium-iodide (IKI) solution can be purchased premade, or the ingredients can be purchased separately as potassium iodide (KI) and iodine (I). These dry ingredients have a long shelf life and can be mixed as needed according to the following recipe:

To make a liter of stock solution, add 20 g of potassium iodide (KI) to 1 liter of distilled water, and stir to dissolve. Then add 4 g of iodine crystals, and stir on a stir plate; dissolution will take a few hours or more. Keep the stock reagent in dark, stoppered bottles. For student use, place in dropper bottles. Label as “iodine (IKI) solution.”

Iodine solution stored in clear bottles loses potency over time. If the solution lightens significantly, replace it. Small dropper bottles can be stored for about a month, and they are used in other exercises. A screw-capped, brown bottle of stock iodine can be stored for about six months. Dispose of it if the solution turns light in color.

Pond water (LM page 25). A good culture of pond water can be maintained to provide algae and protozoans during any season. Collect pond water during an active growing season from any local pond or stream. Include some algae and a small amount of organic debris and living aquatic (aquarium) plants, such as Elodea. Place the collected pond water and other items in a transparent container with a large surface area. Both container and lid should be transparent. A large culture dish covered with another culture dish or a small aquarium is suitable as a container. If kept in diffuse window light or under artificial illumination, the culture will grow and provide material for future labs, even in the middle of winter. If live cultures of pond water organisms or Euglena are purchased for a particular laboratory, they can be added to the maintained culture once they are no longer in use.

Protoslo® (or methyl cellulose solution) (LM page24). You can also use glycerol (Carolina 86-5530) and water as a substitute for Protoslo®. Note: Thickened Protoslo® can be reconstituted with distilled water.

EXERCISE QUESTIONS

2.1 The Metric System (LM pages 10–13)

Length (LM pages 10–1)

Experimental Procedure: Length (LM pages 10–11)

1. How many centimeters are represented? usually 15 One centimeter equals how many millimeters? 10 According to Table 2.1, 1 µm = 0.001 mm, and

1 nm = 0.000001 mm. Therefore, 1 mm = 1,000 µm = 1,000,000 nm.

2. Measure the diameter of the circle shown to the nearest millimeter. This circle is 38 mm = 38,000 µm = 38,000,000 nm.

For example because there are 1,000 µm in one mm, 38 mm = 38,000 µm 

3. How many centimeters are in a meter? 100 How many millimeters are in a meter? 1,000 The prefix milli means thousandth.

4. For example, if the bone measures from the 22 cm mark to the 50 cm mark, the length of the bone is 28 cm. If the bone measures from the 22 cm mark to midway between the 50 cm and 51 cm marks, its length 28.5 cm = 285 mm.

5. Record the length of two bones. Recorded lengths will vary.

Weight (LM page 11)

2g = 2,000 mg; 0.2 g = 200 mg; and 2 mg = 0.002 g

Experimental Procedure: Weight (LM page 11)

2. Measure the weight of the block to the tenth of a gram. Answers will vary.

3. Measure the weight of an item small enough to fit inside the opening of a 50 mL graduated cylinder. Answers will vary.

Volume (LM pages 12–13)

Experimental Procedure: Volume (LM pages 12–13)

1. For example, use a millimeter ruler to measure the wooden block used in the previous Experimental Procedure to get its length, width, and depth. Answers will vary according to the size of the block used. Computations of volume will also vary.

3. Hypothesize how you could find the total volume of the test tube. Fill the test tube with water, and pour the water into the graduated cylinder. Read the volume in milliliters. What is the test tube’s total volume? Answers will vary.

4. Hypothesize how you could use this setup to calculate the volume of an object. Fill the cylinder with water to the 20 mL mark. Drop the object into the cylinder, and read the new elevated volume. The difference between the two readings is the volume of the object alone. Now perform the operation you suggested. Answers will vary.

5. Hypothesize how you could determine how many drops from the pipette of the dropper bottle equal 1 mL. Using a 10 mL graduated cylinder, count the number of drops it takes to get to 1 mL.

How many drops from the pipette of the dropper bottle equal 1 mL? approximately 10 (Answers will vary with student’s technique and with the type of pipette.) Are pipettes customarily used to measure large or small volumes? small

Temperature (LM page 13)

Experimental Procedure: Temperature (LM page 13)

1a. Water freezes at either 32°F = 0°C.

1b. Water boils at either 212°F =100°C.

2. Human body temperature of 98°F is what temperature on the Celsius scale?

37°C

3. Record any two of the following temperatures in your lab environment. Answers will vary.

2.2 Microscopy (LM pages 14–15)

Electron Microscopes (LM page 15)

Conclusions: Microscopy (LM page 15)

• Which two types of microscopes view the surface of an object?

(1) stereomicroscope; (2) scanning electron microscope

• Which two types of microscopes view objects that have been sliced and treated to improve contrast? compound light microscope and transmission electron microscope

• Of the microscopes just mentioned, which one resolves the greater amount of detail? transmission electron microscope

2.3  Stereomicroscope (Dissecting Microscope) (LM pages 16–17)

Identifying the Parts (LM page 16)

2. What is the magnification of your eyepieces? 10x or 20x

Locate each of these parts on your stereomicroscope, and label them on Figure 2.6.

Figure 2.6, top to bottom: eyepiece lens, magnification changing knob, binocular head; illuminator, focusing knob

Focusing the Stereomicroscope (LM page 17)

4. Does your microscope have an independent focusing eyepiece? yes (most likely) Is the image inverted? no

5. What kind of mechanism is on your microscope? Answers will vary.

2.4 Use of the Compound Light Microscope (LM pages 18–22)

Identifying the Parts (LM pages 18–19)

1. What is the magnifying power of the ocular lenses on your microscope? The magnifying power of the ocular lenses is marked on the lens barrel (usually 4x, 10x and 40x).

5. Objectives (objective lenses)

a. What is the magnifying power of the scanning objective lens on your microscope? (usually 4x).

b. What is the magnifying power of the low-power objective lens on your microscope? The magnifying power of the low-power objective lens is marked on the lens barrel (usually 10x).

c. What is the magnifying power of the high-power objective lens on your microscope? The magnifying power of the high-power objective lens is marked on the lens barrel (usually 40x).

d. Does your microscope have an oil immersion objective? depends on microscope

6. Does your microscope have a mechanical stage? depends on microscope

Figure 2.7. Left side, top to bottom: ocular lens or lenses, viewing head, nosepiece, objective lens or lenses, condenser, diaphragm/diaphragm control lever, light source

Right side, top to bottom: arm, stage clips, stage, coarse-adjustment knob, fine-adjustment knob, base

Inversion (LM page 20)

Observation: Inversion (LM page 20)

1. Draw the letter e as it appears on the slide (with the unaided eye, not looking through the eyepiece). The letter should be in the normal position.

2. Draw the letter e as it appears when you look through the eyepiece. The letter should be upside down and reversed.

3. What differences do you notice? The letter is inverted—that is, it appears to be upside down and reversed compared to its appearance when viewed by the unaided eye.

4. Which way does the image appear to move? When moved to the right, the slide appears to move to the left. Explain. This is due to inversion.

Focusing the Compound Light Microscope—Higher Powers (LM page 20)

5. On a drawing of the letter e, draw a circle around the portion of the letter that you are now seeing with high-power magnification. A small portion will be circled.

Total Magnification (LM pages 21)

Observation: Total Magnification (LM page 21)

Calculate total magnification figures for your microscope, and record your findings in Table 2.3.

Table 2.3 Total Magnification*

Objective Ocular Objective Total

Lens Lens Magnification

Scanning power (if present) 10x 4x 40x

Low power 10x 10x 100x

High power 10x 40x 400x

Oil immersion (if present) 10x 100x 1,000x

*Answers may vary with equipment.

Field of View (LM pages 21–22)

Observation: Field of View (LM pages 21–22)

Low-power (10x) Diameter of Field (LM page 21)

2. Estimate the number of millimeters, to tenths, that you see along the field: approx. 1.6 mm. Convert the figure to micrometers: approximately 1,600 µm.

High-power (40x) Diameter of Field (LM page 21)

1. To compute the high-power diameter of field (HPD), substitute these data into the formula given:

a. LPD = low-power diameter of field (in micrometers) = 1,600 µm

b. LPM = low-power total magnification (from Table 2.3) = 100x

c. HPM = high-power total magnification (from Table 2.3) = 400x

HPD = (1,600 µm) x (100x) = 400µm

(400x)

Conclusions: Total Magnification and Field of View (LM page 22)

• Does low power or high power have a larger field of view (one that allows you to see more of the object)? low power

• Which has a smaller field but magnifies to a greater extent? high power

• To locate small objects on a slide, first find them under low power; then place them in the center of the field before rotating to high power.

Depth of Field (LM page 22)

Observation: Depth of Field (LM page 22)

2. Determine the order of the threads or hairs, and complete Table 2.4.

Table 2.4 Order of Threads (or Hairs)*

Depth Thread (or Hair) Color

Top Red

Middle Blue

Bottom Yellow

*The order of threads given is that of Carolina Biological Supply Company slide 29-1418. The order of threads in other slides may be different.

2.5 Microscopic Observations (LM pages 23–25)

Observation: Onion Epidermal Cells (LM page 23)

. Label Figure 2.10. 1. nucleus; 2. cell wall

5. stet 5Count the number of onion cells that line up end to end in a single line across the diameter of the high-power (40x) field. for example, five cells

What is your high-power diameter of field (HPD) in micrometers? 400 µm.

Calculate the length of each onion cell (HPD / number of cells): for example, 80 µm.

Observation: Human Epithelial Cells (LM page 24)

3. stet 3Label Figure 2.11. 1. plasma membrane; 2. nucleus; 3. cytoplasm

Table 2.5 Differences Between Onion Epidermal and Human Epithelial Cells Differences Onion Epidermal Cells Human Epithelial Cells (Cheek)

Shape Square or rectangular Rounded

Orientation Regular (in rows) Random

Boundary Thick Thin

Observation: Euglena (LM page 24)

5.stet 5 Compare your Euglena specimens with Figure 2.12. List the labeled features that you can actually see: Answers will vary.

LABORATORY REVIEW 2 (LM page 26)

1. Make the following conversions:

a. 1 mm = 1,000 µm = 0.1 cm

b. 15 mm = 1.5 cm = 15,000 µm

c. 50 mL = 0.05 liter

d. 5 g = 5,000 mg

2. Explain the designation “compound light” microscope:

a. compound There are two sets of lenses—objective and ocular.

b. light Light is used to view the object.

3. What function is performed by the diaphragm of a microscope? The diaphragm regulates the amount of light coming through the lenses.

4. Why is it helpful for a microscope to be parfocal? Little, if any, adjustment is needed when switching from low to high power.

5. Why is locating an object more difficult if you start with the high-power objective than with the low-power objective. The diameter of field is smaller in high power than in low power.

6. How much larger than actual size does an object appear with a low-power objective lens? 100x

7. A virus is 50 nm in size.

a. Would you recommend using a stereomicroscope, compound light microscope, or an electron microscope to see it? electron microscope Why? Only an electron microscope has the capability of observing an object this small because it magnifies more and has greater resolving power.

b. How many micrometers is the virus? 0.05 µm

8. If the diameter of a field is 1.6 mm, and you count forty consecutive cells from one end of the field to the other, how wide is each cell of micrometers? 40 µm

9. What type of microscope, aside from the compound light microscope, might you use to observe the organisms found in pond water? stereomicroscope

10. Briefly describe the necessary steps for observing a slide at low-power under the compound light microscope. Center the slide on the stage. Looking from the side, decrease the distance between the slide and the objective lens until the lens comes to a stop. Looking through the ocular lens(es), use the coarse-adjustment knob to increase the distance between the slide and the lens until the object comes into view.

Laboratory 3 Chemical Composition of Cells

(LM pages 27–39)

Eleventh Edition Changes

Introducing both positive and negative controls, this lab was reorganized to place carbohydrates before proteins for better flow. Testing for the composition of starch was removed and Figure 3.4 is now close to its reference in Section 3.3. Section 3.4 Testing Foods and Unknowns was revised and has new conclusion questions.

New/Revised Figures. 3.7 Carbohydrate foods; 3.8 Protein foods;

3.9 Lipid foods

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Fresh material. Potato and onion (3.1 Carbohydrates) and the test materials (3.4. Testing the Chemical Composition of Everyday Materials and an Unknown)

All Exercises

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other clothing protection

_____ wax pencils

_____ rulers, plastic millimeter

_____ dropper bottles

_____ test tubes and racks

_____ water, distilled

_____ boiling chips, pumice

_____ microscopes, compound light

_____ lens paper

_____ slides and cover slips

_____ paper towels

Test tubes. The exercises in this laboratory require students to add solutions to test tubes. Disposable 12 x 75 test tubes are less expensive than reusable test tubes and students achieve better results than when they wash and re-use test tubes. Disposable test tubes can be marked with sharpies—the marks do not come off in a hot water bath as do wax pencil marks. Measuring: As an expedient, students are asked to mark off the tubes at various centimeter levels with a ruler and then to fill to these marks. You may prefer to have students use a dropper and count the number of drops as per your instructions, or you may prefer to have students use a standard method of measuring volume, such as with a graduated cylinder or a pipette. Most experiments use the standard size test tube. A few experiments require the large size test tube. Mini test tubes can be substituted for most laboratory exercises as long as the total volume in a given tube does not exceed 9 cm. Use of mini test tubes will reduce the volume of reagents used by approximately one-third.

Test tube sizes/volumes are as follows:

Mini 13 x 100 mm 1 cm = 1.0 mL

Standard 16 x 150 mm 1 cm = 1.5 mL

Medium large 20 x 150 mm 1 cm = 2.4 mL

Large 25 x 150 mm 1 cm = 4.0 mL

3.1 Carbohydrates (LM pages 28–32)

_____ 1% starch suspension (solution)

_____ iodine (IKI) solution

_____ dropper bottles, one per student group

_____ potato, fresh

_____ potato juice from fresh peeled potato

_____ onion, fresh

_____ onion juice from fresh onion

_____ mortar and pestle

_____ scalpels or razor blades, single-edged

_____ cutting board for potato, onion

_____ paper towels

_____ boiling water bath:

_____ hot plate (See Carolina’s Apparatus: Laboratory Equipment

and Supplies section.)

_____ large beaker

_____ pair beaker tongs

_____ test tube clamps

_____ boiling chips, pumice

_____ thermometer, Celsius 50–150°C range

_____ 1% glucose solution

_____ Benedict’s reagent powder or Benedict’s reagent solution

Order solutions/reagents or prepare your own:

Iodine (IKI) solution (LM page 29). Use one dropper bottle per student group. Pre-made iodine-potassium-iodide solution can be purchased, or the ingredients can be purchased separately as potassium iodide (KI) and iodine (I). These dry ingredients have a long shelf life and can be mixed as needed, according to the instructions in Laboratory 2.

1% glucose solution (LM page 29). This makes enough for all procedures for 20 student groups. Place 1 g of glucose in 50 mL of distilled water. Stir to dissolve, and bring the volume up to 100 mL.

Onion and potato juice (LM page 29). Onion may be mashed the day before, then stored in a refrigerator overnight. Potato juice should be extracted as close to time of use as possible. Good results come from blending a peeled potato with water, 1:1. After blending, add a pinch of calcium chloride to keep the juice from turning brown and obscuring the test results.

Benedict’s reagent (LM page 31). 50 mL per student group is sufficient. Benedict’s reagent can be purchased as a powder to make 1 liter. Or to make 1 liter, mix 173 g of sodium citrate and 100 g of sodium carbonate, anhydrous (Na2CO3) (Carolina 88-8770) with 800 mL of distilled water. Warm this mixture to dissolve; then cool and filter it. Add distilled water to make 850 mL. Then dissolve 17.3 g of copper sulfate (cupric sulfate, pentahydrate) in 100 mL of distilled water, and stir slowly into the first solution. Add distilled water to make 1 liter. When testing, Benedict’s reagent should be boiled approximately 5 minutes or longer.

Boiling water bath (LM page 31). Place the large beaker of water on a hot plate. Adjust the dial on the hot plate so that the water is maintained at a gentle rolling boil during the experiment. Thermometers are optional since students should know that boiling water is 100°C.

3.2 Proteins (LM pages 32–34)

_____ biuret reagent

_____ 1% albumin solution

_____ 1–2% pepsin solution

_____ 1–2% starch solution

_____ Parafilm® (Carolina 21–5600)

Order solutions/reagents or prepare your own:

Biuret reagent (LM page 33). 30 mL per student group should be sufficient (using standard test tubes for all procedures). If you buy prepared biuret, use only ten to fifteen drops; otherwise, the solution will be too dark, or dilute to a 10% solution (10 mL biuret with 90 mL distilled water). To prepare your own biuret reagent, maintain separate stock solutions of 3% copper sulfate—3 g of copper sulfate per 100 mL of distilled water and 10% potassium hydroxide or sodium hydroxide—100 g of potassium hydroxide or sodium hydroxide pellets (per 1,000 mL of distilled water). Adding five drops of copper sulfate solution and ten drops of potassium hydroxide solution to each experimental tube produces more consistent results. Biuret reagent should be prepared fresh for each lab.

1% albumin solution (LM page 33). 20 mL per student group should be sufficient (using standard test tubes for all procedures). Prepare pH 7 buffer solution, as per directions on the vial, and dissolve 1 g of albumin per 100 mL of buffer. Allow time for precipitation to occur, and then decant. Also, check pH with indicator paper, and adjust to pH 7 with dilute acid or base. Swirl the stock, before distributing it to students.

1–2% pepsin solution (LM page 33). 20 mL per student group should be sufficient (using standard test tubes for all procedures). To make a 1% solution, dissolve 1 g of pepsin in 100 mL of distilled water. Stir to dissolve.

1–2% starch solution (LM page 33). 20 mL per student group should be sufficient (using standard test tubes for all procedures). Care must be taken in preparing this solution. To make a 1% solution, dissolve 1 g of starch in a small amount of cold water to form a pourable paste. Add this to 100 mL of boiling distilled water, while stirring, and

mix a few minutes. Cool. Add a pinch of sodium chloride (NaCl). If refrigerated, this solution will last for several weeks; otherwise, a fresh supply should be prepared each day.

3.3 Lipids (LM pages 34–37)

_____ paper, brown of loose-leaf (squares 4 cm 3 4 cm)

_____ droppers

_____ water

_____ vegetable oil

_____ whole milk

_____ test tubes

_____ Tween® (Wards Biology 38W2179)

_____ adipose tissue slide

Order solutions/reagents or make your own:

2% Tween® or bile solution (LM page 36). Tween® is a standard wetting agent. Measure 1 mL of Tween® of 1 g of bile salts, and dissolve in 100 mL of distilled water. As an inexpensive alternative, liquid dishwashing detergent can be substituted for the Tween® of bile salts.

3.4 Testing Foods and Unknowns (LM pages 37–8)

_____ everyday materials for unknowns, suggestions: bread crumbs, flour and water mixture, squeeze bottle of margarine, skim milk, orange juice, syrup from canned fruit.

_____ Benedict’s reagent (see 3.1 Carbohydrates)

_____ iodine solution (see 3.1 Carbohydrates)

_____ biuret reagent (see 3.2 Proteins)

_____ paper, brown or loose-leaf (squares 4 cm x 4 cm) (see 3.3 Lipids)

EXERCISE QUESTIONS

3.1Carbohydrates (LM pages 28–32)

Test for Starch (LM pages 29–30)

Experimental Procedure: Test for Starch (LM page 29)

Table 3.2 Iodine Test for Starch

Tube Contents Color Conclusions

1 Water Brown (no change) No starch is present.

2 Starch suspension Black Starch is present.

3 Onion juice Brown (no change) No starch is present.

4 Potato juice Black Starch is present.

5 Glucose solution Brown (no change) No starch is present.

Conclusions: Starch (LM page 30)

• Does the potato or the onion store glucose as starch? potato How do you know? Potato juice tests positive for starch.

Experimental Procedure: Microscopic Study of Potato and Onion (LM page 30)

Potato (LM page 30)

3. What is the color of the small oval bodies? black

What is the chemical composition of these oval bodies? starch What is the name of these oval bodies? starch granules

Onion (LM page 30)

3. Does onion contain starch? no

4. Are these results consistent with those you recorded for onions in Table 3.2? yes

Test for Sugars (LM pages 31–34)

Experimental Procedure: Test for Sugars (LM pages 31–32)

Table 3.4 Benedict’s Test For Sugars

Tube Contents Color (After Heating) Conclusions

1 Water Blue No sugar

2 Glucose solution Orange to red A very high amount of sugar

3 Starch suspension Usually blue; may be yellow No sugar to a low amount of

sugar

4 Onion juice Yellow to orange A moderate to high amount

of sugar is present.

5 Potato juice Yellow Low amount of sugar

Conclusions: Sugars (LM pages 31–32)

• Which tubes served as (negative, positive) controls? tube one was negative control and tube two was positive control.

• In what form do potatoes store carbohydrate? potatoes store carbohydrate as starch In what form do onions store carbohydrate? sugars Does this explain your results in Table 3.4? yes Why? Because potatoes convert sugar to starch, they contain a low amount of sugar. Because onions do not convert sugar to starch, they do contain sugar.

3.2 Proteins (LM pages 32–34)

Test for Proteins (LM pages 33)

Experimental Procedure: Test for Proteins (LM page 33)

Table 3.6 Biuret Test for Protein

Tube Contents Final Color Conclusions

1 Distilled water Blue No protein is present.

2 Albumin Purple Protein is present.

3 Pepsin Purple Protein is present.

4 Starch Blue No protein is present.

Conclusions: Proteins (LM page 34)

• Enzymes are composed of what type organic molecule? protein

• According to your results, is starch a protein? No, starch is not a protein.

• If your test results are not as expected, inform your instructor, who will advise you how to proceed. Answers will vary; the reagent may be faulty.

• Which of the four tubes is the negative control sample? tube 1

• Why do experimental procedures include control samples? The control samples usually validate the experiment. If positive control sample gives a negative result and/or the negative control sample gives a positive result, the experiment is invalid—in this case, the reagents may be contaminated or the procedure may need improvement.

3.3 Lipids (LM pages 34–37)

Test for Fat (LM page 35)

Experimental Procedure: Test for Fat (LM page 35)

1. Place a small drop of water on a square of brown or loose-leaf paper. Describe the immediate effect. The water sits on top.

2. Place a small drop of vegetable oil on the paper. Describe the immediate effect. Oil soaks into the paper and creates an oil spot.

Table 3.7 Paper Test for Fat

Sample Observations Conclusions

Water spot Evaporates and does Substance is not an oil.

not penetrate paper

Oil spot Penetrates paper and Substance is an oil

does not evaporate

Emulsification of Oil (LM pages 35–36)

Experimental Procedure: Emulsification of Oil (LM page 36) stet 36

Tube 1: 2. Is vegetable oil soluble in water? no

Tube 2: 2. Describe how the distribution of oil in tube 2 compares with the distribution in tube 1. In Tube 1 the oil forms a heavy layer. In Tube 2 the heavy layer of oil is no longer present.

5. Record your observations in Table 3.8.

Table 3.8 Emulsification of Oil

Tube Contents Observations Conclusions

1 Oil Oil droplets are large. Oil does not mix with water

Water

2 Oil Oil droplets are numerous and Emulsifier causes oil to mix Water small with water

Emulsifier

3. Milk Fat droplets are numerous and Milk contains fat

Water

Conclusions: Emulsification of Oil (LM page 36)

• Explain the correlation between your macroscopic observations (how the tubes look to your unaided eye) and your microscopic observations. Tube 1 has two layers (a heavy oil layer and a water layer) because the oil droplets are large. Tube 2 has no heavy top layer of oil because the oil droplets are small. Tube 3 has no heavy top layer of fat because milk is homogenized.

Adipose Tissue (LM page 37)

Observation: Adipose Tissue (LM page 37)

2. Explain why fat does not disperse in cytoplasm. Fat is not soluble in water and an emulsifier is not present.

3.4 Testing Foods and Unknowns (LM pages 37–38)

Experimental Procedure: Testing Foods and Unknowns (LM page 38)

1. Develop instructions to test foods for carbohydrate, protein and fat. Liquefy the substance and put 1 cm in each of two test tubes. (a) To one add two drops iodine (black color is positive test for starch) and add Benedicts reagent before boiling (green to orange-red test for sugar).

(b) To test for protein add 5 drops biuret reagent to 1 cm of liquefied substance. Look for purple color. (c) To test for fat place a small quantity of the substance to brown or loose leaf paper and examine for an oily spot that does not evaporate.

3. Record your results in Table 3.9. Results will vary depending on the substances provided by the instructor.

Conclusions: Testing Foods and Unknowns (LM page 38)

• What type foods tested positive for only one of the organic compounds? For example, oil is purely lipid and egg white is purely protein.

• What type foods tested positive for more than one organic compound? For example bread or milk are complex substances composed of several organic compounds.

• What type carbohydrate would give a positive iodine test but a negative Benedict's test? Starch

LABORATORY REVIEW 3 (LM page 39)

1. What three major classes of organic molecules studied today are present in cells? carbohydrate (sugars and starch), protein and fat

2. You have been assigned the task of constructing a protein. What type of building block would you use? amino acids

3. A hydrolytic digestive enzyme breaks down starch to maltose molecules. Maltose is a disaccharide.

4. Why is it necessary to shake an oil and vinegar salad dressing before adding it to a salad? The oil in salad dressing is not soluble in vinegar which is mostly water, and shaking mixes the otherwise separate layers.

5. How would you test for each of the following substances?

a. Sugar Benedict’s reagent

b. Fat apply to brown paper

c. Starch iodine (IKI) solution

d. Protein biuret reagent

6. Assume that you have tested an unknown sample with both biuret reagent and Benedict’s reagent and that both tests result in a blue color. What have you learned? The sample does not contain protein or sugar.

7. What purpose is served when a test is done using water instead of a sample substance? The test done with water is a negative control that is expected to give a negative result. If a positive result occurs, the experiment is invalid.

8. A test tube contains starch, a digestive enzyme for starch, and water. The biuret test is negative. After thirty minutes, the Benedict’s test is positive. What substance is present? sugar

9. A test tube contains either albumin or pepsin. The test for protein is positive. Explain. Albumin (and pepsin) are proteins.

10. A student adds iodine solution to egg white and waits for color change. How long will the student have to wait? Forever.

Laboratory 4 Cell Structure and Function

(LM pages 41–56)

Eleventh Edition Changes

This lab is essentially the same as it was in the 10e. Note, however, the call out for a virtual lab (LM page 45) that can help students learn the structure and function of organelles.

MATERIALS AND PREPARATIONS

Instructions are grouped by procedure. Some materials may be used in more than one procedure.

Special Requirements

Living material. Elodea is needed for Section 4.2; whole sheep blood for Section 4.4.

Fresh material. Potato is needed for Section 4.4.

4.1 Prokaryotic Versus Eukaryotic Cells (LM page 42)

_____ slide, prepared: a prokaryote, such as Anabaena

_____ slide, prepared: human kidney

_____ microscope, compound light

_____ lens paper

4.2 Animal Cell and Plant Cell Structure (LM pages 43–46)

_____ Elodea, living

_____ dropper bottle of water

_____ slides

_____ coverslips

_____ microscopes, compound light

_____ lens paper

Elodea (LM page 46). Live Elodea can be purchased locally at aquarium stores or through Carolina Biological Supply Company. Place Elodea in distilled water in an aquarium with a continuous air supply from an aquarium air stone and pump. Place in indirect window light or under artificial illumination.

4.3 Diffusion (LM pages 47–49)

Solute Diffusion Through a Semisolid (LM page 47)

To prepare petri dish for student examination:

_____ petri dish

_____ gelatin powder or agar powder for 1.5% solution

_____ potassium permanganate (KMnO4) crystals (Carolina 88-4130)

_____ rulers, plastic millimeter (preferably transparent)

Diffusion demonstration through gelatin or agar (LM page 47). (Note: Agar allows faster diffusion than gelatin.) Prepare one dish per student group. At least a day ahead, prepare a 1.5% gelatin solution in a beaker or flask by dissolving 1.5 g of gelatin powder or agar in 100 mL of boiling water; stir thoroughly until dissolved. Allow to cool until the glassware can be handled with a hot mitt. Fill a petri dish 3–5 mm deep with gelatin solution. Put a lid on dish until cool. After cooling, store the dish in a refrigerator. After gelling, make a small depression in the center of the dish. Using a forceps, drop a crystal of potassium permanganate into the depression.

Solute Diffusion Through a Liquid (LM page 47)

_____ potassium permanganate (KMnO4) crystals

_____ petri dish (one per student group)

_____ water

_____ forceps

_____ rulers, plastic millimeter

Potassium permanganate (LM page 47). Only 1–2 crystals are needed per student group. While wearing gloves, dispense several crystals of potassium permanganate into a shallow, wide-mouth, screw-top container appropriately labeled. (Note: Potassium permanganate diffuses very quickly.)

Diffusion Through Air (LM pages 48)

_____ perfume or cologne

Solute Diffusion Across the Plasma Membrane (LM pages 48–49)

_____ dialysis tubing, approximately 15 cm per setup

_____ 1% glucose solution

_____ plastic droppers or Pasteur pipettes

_____ 1–2% starch solution

_____ rubber bands to close off the end of dialysis tubing

_____ rubber bands that fit snugly around brim of 250 mL beaker

_____ beaker , 250 mL (one per group)

_____ water, distilled

_____ iodine (IKI) solution

_____ rubber band that goes around top of beaker

_____ test tube (one per group)

_____ test-tube rack

_____ wax pencils

_____ Benedict’s reagent or glucose test strips, optional

_____ boiling water bath:

_____ hot plate (See Carolina’s Apparatus: Laboratory Equipment

and Supplies section.)

_____ large beaker

_____ beaker tongs

_____ test tube clamps

_____ boiling chips, pumice

Boiling water bath (LM page 49). Place a large beaker of water on a hot plate. Adjust the dial on the hot plate so that the water is maintained at a gentle rolling boil during the experiment. Thermometers are optional since students should know that boiling water is 100°C.

Order solutions/reagents or prepare your own:

Glucose solution (LM page 48). Prepare as described in the resource guide instructions for Laboratory 3.

Starch solution (LM page 48). Prepare as described in the resource guide instructions for Laboratory 3.

Iodine (IKI) solution (LM page 48). Prepare as described in the resource guide instructions for Laboratory 2.

Benedict’s reagent (LM page 49). Prepare as described in the resource guide instructions for Laboratory 3. Glucose test strips can be used in place of Benedict’s reagent to test for glucose in bag and beaker.

4.4 Osmosis: Diffusion of Water Across Plasma Membrane (LM pages 49–53)

Experimental Procedure: Speed of Osmosis (LM pages 49–50)

_____ Osmosis Demonstration Unit (Carolina 68-4100)

_____ 20% -50% starch solution

_____ plastic syringe for filling thistle tube (Carolina 19-9606)

Osmosis Demonstration Alternative

_____ dialysis tubing

_____ beaker

_____ 10–20% sucrose solution

_____ rubber bands to close off the bottom of dialysis tubing

_____ plastic clamps to close off the top of dialysis tubing

Osmosis demonstration (LM page 50). The Osmosis Demonstration Unit is particularly easy to fill and empty. Partially fill the thistle tube with 50% starch (or similar) solution. Place the apparatus in a beaker containing distilled water. (The standard animal membrane disc (Carolina 68-4030) can be replaced with 1” dialysis tubing and a dialysis tubing clamp. Opening the clamp makes emptying easy). Other osmometers can be found in Carolina’s Osmosis and Diffusion: Physiology section. Alternative: This demonstration can also be done using dialysis tubing and a beaker. See Experimental Procedure: Diffusion Across the Plasma Membrane for set up. Tie off one end of the tubing, then fill with 10–20% sucrose solution. Clamp or tie it off at the open end. Pat the bag dry and weigh. Place the bag in a beaker of water for 45 minutes to 1 hour. Remove, pat dry weigh immediately.

Experimental Procedure: Effect of Tonicity in Red Blood Cells

(LM pages 51–52)

To prepare stoppered test tubes for students:

_____ test tubes, Pyrex 16 mm X 150 mm with stoppers

_____ stoppers, rubber laboratory, solid, size 1

_____ sheep blood, pooled, citrated

_____ water, distilled

_____ 0.9% and 10% sodium chloride (NaCl) solutions

_____ dropper

_____ whole blood demonstration slides (optional)

_____ microscopes, compound light

Whole blood (LM page 51). Blood should not be human blood. Use any available animal blood, other than human, to remove the risk of transmission of the HIV virus. Use caution with any animal blood as it may contain pathogens. Blood is shipped in iced, insulated containers and should be stored in the refrigerator. If kept refrigerated, sheep blood may be stored for up to 2 weeks. Prepare the test tubes as follows:

Tube 1 5 mL 0.9% NaCl plus three drops of sheep blood

Tube 2 5 mL 10% NaCl plus three drops of sheep blood

Tube 3 5 mL 0.9% NaCl plus distilled water and three drops of sheep blood

Stopper the tubes.

NaCl solutions:

0.9% NaCl: Add 9 g of NaCl to 1 liter of distilled water. Smaller volumes may be prepared.

10% NaCl: Add 100 g of NaCl to 1 liter of distilled water. Smaller volumes may be prepared.

Slides of whole blood (optional). Prepare a demonstration slide of the 0.9% sheep blood solution (Tube 1) and the 10% sheep blood solution (Tube 2) for student observation.

Experimental Procedure: Elodea Cells (LM pages 52–53)

_____ see materials listed previously in Section 4.2

_____ 10% NaCl from the whole blood demonstration

Experimental Procedure: Potato Strips (LM page 53)

_____ potato, fresh

_____ cutting board for potato

_____ scalpels or razor blades, single-edged

_____ rulers, plastic millimeter

_____ test tubes (2) per group

_____ wax pencil

_____ water

_____ 10% sodium chloride (NaCl) in wash bottles

_____ paper towel

4.5 pH and Cells (LM pages 54–55)

Experimental Procedure: pH and Cells (LM page 54–55)

supplies

_____ test tubes (3 per group) labeled water, inorganic buffer, and cytoplasm

_____ test tube rack

_____ pH paper (range pH 0–14)

_____ rods, glass stirring

_____ 0.1 N hydrochloric acid (HCl) in dropper bottles

To prepare test tubes

_____ water

_____ pH 7 buffer (inorganic) solution

_____ protein (e.g., albumin) solution, buffered

Order buffer or prepare your own:

pH 7 buffer (LM page 54). 50 mL per student group is sufficient. If you wish to make it yourself, combine 50 mL 0.1 M potassium dihydrogen phosphate (1.36 g per 100 mL distilled water) with 29.1 mL 0.1 M NaOH (0.4 g per 100 mL distilled water). Dilute this mixture to 100 mL with distilled water.

Buffered “cytoplasm” (e.g., albumin solution) (LM page 54). 50 mL per student group should be sufficient. Mix 1 g of albumin with 100 mL of pH 7.0 buffer

0.1 N HCl solution (LM page 54). Mix 0.83 mL concentrated HCl with 100 mL distilled water. Place in dropper bottles.

Experimental Procedure: Effectiveness of Antacids (LM page 55)

_____ mortar and pestle

_____ antacids: Alka-Seltzer, Rolaids, Tums, or other antacid tablet

_____ 0.04% phenol red solution

_____ beakers, plastic 250 mL (one per antacid tested)

_____ rods, glass stirring

_____ 0.1 N hydrochloric acid (HCl) in dropper bottles

Order solutions/reagents or prepare your own:

0.04% phenol red solution (LM page 55). Prepare 100 mL per student group. Dissolve 0.04 g of phenol red in 100 mL of distilled water.

EXERCISE QUESTIONS

4.1 Prokaryotic Versus Eukaryotic Cells (LM page42)

Observation: Prokaryotic/Eukaryotic Cells

1. Do you agree that there are no nuclei in these cells? Yes

2. Do you agree that there are nuclei in these cells? Yes

4.2 Animal Cell and Plant Cell Structure (LM pages 43–46)

Study Table 4.1 to determine structures that are unique to plant cells and unique to animal cells, and write them below the examples given:

Plant Cells Animal Cells

1. Large central vacuole 1. Small vacuoles

2. Cell wall 2. Centrioles

3. Chloroplast

Animal Cell Structure (LM page 44)

With the help of Table 4.1, give a function for each of these structures, and label Figure 4.2. (For Figure 4.2 labels, see figure in Biology 11/e text.)

Structure Function

Plasma membrane Selective passage of molecules

into and out of cell

Nucleus Storage of genetic information

Nucleolus Ribosomal formation

Ribosome Protein synthesis

Endoplasmic Synthesis and/or modification;

reticulum transport by vesicle formation

Rough ER Protein synthesis

Smooth ER Various functions; lipid synthesis in some cells

Golgi apparatus Processing, packaging, and distributing molecules

Vesicles Storage and transport of substances

Lysosome Intracellular digestion

Mitochondrion Cellular respiration

Centrioles in Formation of basal bodies

centrosome

Cytoskeleton Cell shape and movement

Plant Cell Structure (LM page 45)

With the help of Table 4.1, give a function for these structures unique to plant cells, and label Figure 4.3. (For Figure 4.3 labels, see figure in Biology 11/e text.)

Structure Function

Cell wall Supports and protects plant

Central vacuole, large Gives added support to cell; depository of wastes

Chloroplasts Photosynthesis

Observation: Plant Cell Structure (LM page 46)

6. Can you locate the cell nucleus? Answers will vary, but usually yes.

7. Why can't you see the other organelles featured in Figure 4.3? They are too small to be seen by light microscope.

8. Can you detect the movement of chloroplasts in this cell or any other cell? Answers will vary, but usually yes.

4.3 Diffusion (LM pages 47–49)

Experimental Procedure:Speed of Diffusion (LM pages 47–48)

Record all data in Table 4.2

Table 4.2 Speed of Diffusion

Diffusion data will depend on room temperature, gelatin consistency, and the molecular weight of the dye used.

Conclusions: Speed of Diffusion (LM page 48)

• In which experiment was diffusion the fastest? diffusion through air

• What accounts for the difference in speed? Friction resists the movement of molecules. Air offers the least amount of friction; also air currents aid diffusion. A solid offers the most amount of friction.

Solute Diffusion Across Plasma Membrane (LM pages 48–49)

Experimental Procedure: Solute Diffusion Across Plasma Membrane (LM pages 48–9)

Table 4.3 Solute Diffusion Across Plasma Membrane

At Start of Experiment At End of Experiment

Contents Color Color Benedict’s Test Conclusion

Bag Glucose Colorless Black _______ Iodine diffused

Starch into bag.

Beaker Water Yellowish Colorless Positive (+) Glucose diffused

Iodine out of bag.

Conclusions: Solute Diffusion Across Plasma Membrane (LM page 49)

• Which solute did not diffuse across the dialysis membrane from the bag to the beaker? starch How do you know? Starch molecules are too large to diffuse across the dialysis membrane. Contents of beaker did not turn black.

4.4 Osmosis: Diffusion of Water Across Plasma Membrane (LM pages 49–53)

Experimental Procedure: Speed of Osmosis (LM pages 49–50)

1. Note the level of liquid in the thistle tube, and measure how far it travels in 10 minutes:

1 mm

2. Calculate the speed of osmosis under these conditions: 6 mm/hr

Conclusions: Osmosis (LM page 50)

• In which direction was there a net movement of water? from beaker to thistle tube Explain what is meant by “net movement” after examining the arrows in Figure 4.7b. Water moves in and out of thistle tube, but more water moves in than moves out of tube.

• If the starch molecules in corn syrup moved from the thistle tube to the beaker, would there have been a net movement of water into the thistle tube? No Why wouldn’t large starch molecules be able to move across the membrane from the thistle tube to the beaker? They are too large to cross a membrane.

• Explain why the water level in the thistle tube rose: In terms of solvent concentration, water moved from the area of higher water concentration to the area of lower water concentration across a differentially permeable membrane.

Tonicity (LM pages 50–53)

Experimental Procedure: Effect of Tonicity on Red Blood Cells (LM pages 51–52)

Table 4.4 Effect of Tonicity on Red Blood Cells

Tube Tonicity Print Visibility Explanation

1 Isotonic No Cells are intact.

2 Hypertonic No Cells are intact.

3 Hypotonic Yes Cells have burst.

Experimental Procedure: Effect of Tonicity on Elodea Cells (LM page 52–3)

As depicted in Figure 4.9a, where do you find the chloroplasts? Next to the plasma membrane and cell wall. As depicted in Figure 4.9b, where do you find the chloroplasts? Bunched up in the middle of the cell.

Hypotonic Solution

2. Compare your slide with Figure 4.9a. Because the cells are in a hypotonic solution, the chloroplasts are next to the plasma membrane and cell wall.

Hypertonic Solution

2. Compare your slide with Figure 4.9b. Because the cells are in a hypertonic solution, the chloroplasts are bunched up in the middle.

Table 4.5 Effect of Tonicity on Elodea Cells

Tonicity Appearance of Cells Due to (scientific term)

Hypotonic Normal Turgor pressure

Hypertonic Shriveled center Plasmolysis

Conclusions: Hypotonic and Hypertonic Solutions (LM page 53)

• In a hypotonic solution, the large central vacuole of plant cells exerts turgor pressure, and the chloroplasts are seen up against the cell wall.

• In a hypertonic solution, the central vacuole loses water, and the cytoplasm including the chloroplasts have moved away from the cell wall. This is called plasmolysis.

Experimental Procedure: Potato Strips (LM page 53)

5. Which tube has the limp potato strip? tube 2 Use tonicity, to explain why water diffused out of the potato strip in this tube? The solution in tube 2 was hypertonic.

Which tube has the stiff potato strip? Tube 1 Use tonicity to explain why water diffused into the potato strip in this tube? The solution in tube 1 was hypotonic.

6. Create a table to display your results.

Table Effect of Tonicity on Potato Strips (This table is an example and student tables can vary.)

Contents Tonicity Results Explanation

Tube 1 Potato Hypotonic Stiff potato Water diffused into potato

Water

Tube 2 Potato Hypertonic Limp potato Water diffused out of potato 10% NaCl

4.5 pH and Cells (LM pages 54-55)

Why are cells and organisms buffered? To maintain pH of the cells and body fluids. Only then do they remain healthy.

Experimental Procedure: pH and Cells (LM pages 54–55)

Table 4.6 pH and Cells*

Tube Contents pH Before Acid pH After Acid Explanation

1  Water 6–6.5 2–3 Not buffered

2  Buffer 7 7 Buffered

3  Cytoplasm 7 7 Buffered

*These results are based on 1 mL of test solution.

Conclusions: pH and Cells (LM page 55)

• Why would you expect cytoplasm to be as effective as the buffer in maintaining pH? Cytoplasm and all body fluids are buffered.

Experimental Procedure: Effectiveness of Antacids (LM page 55)

Table 4.7 Effectiveness of Antacids

Data will depend on the antacids used.

Conclusions: Effectiveness of Antacids (LM page 55)

• Did dosage in mg have any effect on the results? depends on antacid used

• Which of the substances on the label could be a buffer? depends on antacid

LABORATORY REVIEW 4 (LM page 56)

1. Contrast the location of DNA in a eukaryotic cell with that of a prokaryotic cell. DNA is located in the nucleus of a eukaryotic cell and in a nucleoid of a eukaryotic cell.

2. What characteristics do all eukaryotic cells have in common? the presence of a nucleus and membrane-bounded organelles

3. Why would you predict that an animal cell, but not a plant cell, might burst when placed in a hypotonic solution? Animal cells do not have cell walls. The cell wall of plant cells keeps them from bursting.

4. Which of the cellular organelles would be included in a category called:

a. Membranous canals and vacuoles endoplasmic reticulum, Golgi apparatus, vesicles, vacuoles, lysosomes, peroxisomes

b. Energy-related organelles mitochondria and chloroplasts

5. How do you distinguish between rough endoplasmic reticulum and smooth endoplasmic reticulum?

a. Structure Rough endoplasmic reticulum has ribosomes; smooth endoplasmic reticulum does not.

b. Function Rough endoplasmic reticulum is the site of protein synthesis; smooth endoplasmic reticulum has various functions.

6. If a dialysis bag filled with water is placed in a molasses solution, what do you predict will happen to the weight of the bag over time? The bag will lose weight. Why? Water would diffuse out of the bag and enter the molasses solution.

7. How does plant cell structure aid the ability of plants to stand upright? Strong cell walls and water-filled vacuoles that maintain turgor pressure help plants to stand upright.

8. The police are trying to determine if some nondescript material removed from a crime scene was plant matter. What would you suggest they look for? To determine if it was plant matter, the police should microscopically look for cell walls and chloroplasts, and they should test for starch.

9. A test tube contains red blood cells and a salt solution. When the tube is held up to a page, you can see the print. With reference to a concentration of 0.9% sodium chloride (NaCl), how concentrated is the salt solution? The solution has a lower concentration than 0.9% NaCl. Since it is hypotonic to them, the cells have burst.

10. Predict the microscopic appearance of cells in the leaf tissue of a wilted plant. The central vacuole has pulled away from the cell wall, and the chloroplasts have moved to the center of the cell.

Laboratory 5 How Enzymes Function

(LM pages 57–66)

Eleventh Edition Changes

New to this edition, also, each section asks students to hypothesize the outcome of the experimental procedure before the experiment is done. In this way students can get a sense of the scientific process. An analogy has been added to the introduction for section 5.3. A modified virtual lab concerning Enzyme-Controlled Reaction (LM pages 64--65) can be an addition to this lab or can replace Sections 5.3 and 5.4.

MATERIALS AND PREPARATIONS

Instructions are grouped by procedure. Some materials may be used in more than one procedure.

Special Requirements

Fresh material. A potato is needed per lab to prepare catalase for Sections 5.1--5.4.

Equipment. Incubator (or water bath) and refrigerator (or ice bath) for Section 5.2 Effect of Temperature on Enzyme Activity. 15 minute incubation required.

All Exercises

_____ water, distilled

_____ wax pencils

_____ rulers, plastic centimeter

_____ test tubes and racks

_____ beakers

_____ droppers

Test tubes. Disposable 12 x 75 test tubes are less expensive than reusable test tubes and students achieve better results than when they wash and re-use test tubes. Disposable test tubes can be marked with sharpies—the marks do not come off in a hot water bath as do wax pencil marks. Measuring: As an expedient, students are asked to mark off the tubes at various centimeter levels with a ruler and then to fill to these marks. You may prefer to have students use a dropper and count the number of drops as per your instructions, or you may prefer to have students use a standard method of measuring volume, such as with a graduated cylinder or a pipette. Most experiments use the standard size test tube. A few experiments require the large size test tube. To minimize reagents, mini test tubes can usually be substituted as long as the total volume in a given tube does not exceed 9 cm.

Test tube sizes/volumes are as follows:

Mini 13 x 100 mm (Carolina 73-0008) 1 cm3 = 1.0 mL

Standard 16 x 150 mm (Carolina 73-0014) 1 cm3 = 1.5 mL

Medium large 20 x 150 mm (Carolina 73-0018) 1 cm3 = 2.4 mL

Large 25 x 150 mm (Carolina 73-0024) 1 cm3 = 4.0 mL

5.1 Catalase Activity (LM pages 58–59)

_____ catalase, buffered

_____ hydrogen peroxide, purchased locally

_____ 5% sucrose solution

_____ potassium phosphate, dibasic

_____ potassium phosphate, monobasic

_____ pre-mixed buffer, pH7

Order solutions/agents or prepare your own.

Buffered catalase (LM page 59). Make potato catalase fresh for each lab by grinding one small potato or half of a large potato with 50 mL water in a blender. Strain the potato mixture through a sieve to remove any large pieces of potato. Sections 5.1--5.3 requires buffered catalase. Section 5.4 requires nonbuffered catalase. You may put the buffered/nonbuffered catalase in a beaker on the supply bench, and students can use dropper pipettes to dispense the enzyme into test tubes.

Phosphate buffer (LM page 59). Add 8.70 g potassium phosphate, dibasic, K2HPO4, and 6.80 g potassium phosphate monobasic, KH2PO4, to one liter distilled water. Mix, check pH, and use to dilute catalase as needed. Premixed buffer may be used, as well.

Hydrogen peroxide (LM page 59). The hydrogen peroxide used in this experiment can be purchased from a local store.

5% sucrose solution (LM page 59). Dissolve 5 g sucrose in 100 mL distilled water. Dispense from a beaker with dropper pipettes.

Note: Caution the students that they should swirl the enzyme and substrate to mix, then allow the tube to sit for 20 seconds before measuring the height of the bubble column. The bubbles produced by the reaction are very small, and resemble shaving cream foam. If the catalase/sucrose mixture is swirled for 20 seconds, the catalase will produce large bubbles, which some students confuse for the enzyme reaction.

5.2 Effect of Temperature on Enzyme Activity (LM pages 59–60)

_____ catalase (see section 5.1)

_____ hydrogen peroxide (purchased locally)

_____ incubator

_____ refrigerator or ice bath

_____ boiling water bath:

_____ hot plate

_____ large beakers

_____ beaker tongs

_____ thermometer

_____ test tube holders

5.3 Effect of Concentration on Enzyme Activity (LM page 61)

_____ catalase (see section 5.1)

_____ hydrogen peroxide (purchased locally)

5.4 Effect of pH on Enzyme Activity (LM pages 62–63)

_____ catalase, nonbuffered (see section 5.1)

_____ 5 M HCl

_____ hydrogen peroxide (purchased locally)

_____ 5 M NaOH

Order solutions/agents or prepare your own.

5 M HCl. CAUTION—This solution will get HOT (LM page 62). Add 400 mL distilled water to a 1-liter graduated beaker. Place beaker with magnetic spinbar on a stirring plate. While stirring, slowly pour in 416 mL concentrated HCl. Add distilled water to bring the volume up to 1,000 mL.

5 M NaOH. CAUTION—This solution will get very HOT (LM page 62). In a 1-liter beaker with a magnetic spinbar, gradually add a total of 200 grams NaOH pellets to

750 mL distilled water, allowing the heat to dissipate between additions of NaOH. After the solution cools, add distilled water to bring the volume up to 1,000 mL.

EXERCISE QUESTION

5.1 Catalase Activity (LM pages 58–59)

What is the reactant in this reaction? H2O2 What is the substrate for catalase? H2O2

What are the products in this reaction? H2O and O2 Bubbling occurs as the reaction proceeds. Why? O2 production

Hypothesize which tube 1,2,or 3 in Table 5.1 will have the greater bubble height. Tube 1 Explain your answer. Only this tube contains both catalase and its substrate.

Experimental Procedure: Catalase Activity (LM pages 59)

Table 5.1 Catalase Activity

Tube Contents Bubble Column Height Explanation

1 Catalase 20 mm Substrate and enzyme are Hydrogen peroxide both present.

2 Water 0 mm Tube lacks enzyme. Hydrogen peroxide

3 Catalase 0 mm Tube lacks correct substrate. Sucrose solution

Conclusions: Catalase Activity (LM page 59)

• Which tube showed the bubbling you expected? Tube 1 Was your hypothesis supported? Depends on hypothesis.

• Which tube is a negative control? tube 2 If this tube showed bubbling, what could you conclude about your procedure? Results are not due to catalase; therefore, experiment is invalid.

• Enzymes are specific. Which tube exemplifies this characteristic of an enzyme?

tube 3.

5.2 Effect of Temperature on Enzyme Activity (LM pages 59–60)

With this information in mind, examine Table 5.2 below and hypothesize which tube (1, 2 or 3) will have more product per unit time as judged by bubble height. Tube 2 Explain your answer. The enzyme exposed to normal body temperature will perform best because the molecules will be moving about but the temperature is not high enough to denature the enzyme.

Experimental Procedure: Effect of Temperature (LM pages 60)

5. Record your results in Table 5.2. Plot your results in Figure 5.3.

Table 5.2 Effect of Temperature

Tube Temperature (°C) Bubble Column Explanation

Height (mm)

1 Refrigerator 5°C 9 mm Temperature below optimum.

2 Incubator 37°C 23 mm Optimum temperature enzyme activity

3 Boiling water 100°C 0 mm Temperature too hot. Denaturation occurred.

[pic]

Conclusions: Effect of Temperature (LM page 60)

• Was your hypothesis supported. Depends on hypothesis.

• What is your conclusion concerning the effect of temperature on enzyme activity? A warm temperature speeds an enzymatic reaction, but a hot temperature denatures an enzyme.

5.3 Effect of Concentration on Enzyme Activity (LM page 61)

With this in mind, examine Table 5.3 and hypothesize which tube (1, 2 or 3) will have more product per unit time as judged by bubble height. Tube 3 Explain your answer. The more enzyme molecules, the more active sites per substrate, and the more product within a limited time frame.

Experimental Procedure: Effect of Enzyme Concentration (LM page 61)

Table 5.3 Effect of Enzyme Concentration

Tube Amount of Bubble Column Explanation

Enzyme Height (mm)

1 1 cm 20 mm Explanation for all tubes: The

2 2 cm 30 mm greater the enzyme concentration,

3 3 cm 40 mm the more O2 is produced during the

20 second time period.

Conclusions: Effect of Concentration (LM page 61)

• Was your hypothesis supported. Depends on hypothesis.

• If unlimited time was allotted, would the final results be the same in all tubes? Yes. Explain why or why not. All tubes have the same amount of substrate and enzymes can be used over and over again.

• Would you expect similar results if the substrate concentration were varied in the same manner as the enzyme concentration? yes Why or why not? It would take less time for the substrate to encounter an active site.

• What is your conclusion concerning the effect of concentration on enzyme activity? Increased amount of enzyme or substrate will increase the rate of enzyme activity.

5.4 Effect of pH on Enzyme Activity (LM page 62)

With this information in mind, examine Table 5.4 and hypothesize which tube (1, 2 or 3) will have more product per unit time as judged by bubble height. Tube 2 Explain your answer. pH 7 is optimum pH for Catalase

Experimental Procedure: Effect of pH (LM page 62–63)

1-3. Measure the height of the bubble column and record your results in Table 5.4.

Table 5.4 Effect of pH

Tube pH Bubble Column Height (mm) Explanation

1 3 17 mm pH too acidic for catalase

2 7 35 mm optimum pH for catalase

3 11 12 mm pH too basic for catalase

Plot your results from Table 5.4 here.

[pic]

Conclusions: Effect of pH (LM page 63)

• Was your hypothesis supported. depends on hypothesis.

• What is your conclusion concerning the effect of pH on enzyme activity? Any pH other than the optimum pH will decrease the activity of an enzyme.

5.5Comparison of Factors That Affect Enzyme Activity (LM page 63)

Table 5.5 Factors that Affect Enzyme Activity

Factors Promote Enzyme Activity Inhibit Enzyme Activity

Enzyme Specificity Active site available Active site not available

Temperature Intermediate Extreme

Enzyme or substrate High Low

concentration

pH Optimum pH Too acidic or basic for the enzyme

Conclusions: Factors That Affect Enzyme Activity (LM page 63)

• Why does enzyme specificity promote enzyme activity? The active site of an enzyme has a shape that is complementary to the shape of its substrate. Only then can enzyme and substrate interact and the reaction occur.

• Why does a warm temperature promote enzyme activity? It increases the motion of molecules and therefore the number of times enzyme and substrate interact.

• Why does increasing enzyme concentration promote enzyme activity? It increases the number of active sites available.

• Why does optimum pH promote enzyme activity? Optimum pH is required to maintain the active site of an enzyme

Virtual Lab Enzyme-Controlled Reactions

Experiment 1 Determining an Enzyme's Optimum pH

Hypothesis The optimum pH allows an enzyme to function at a faster rate.

Steps of the Experiment

Which of these factors should you hold constant during this experiment for determining an enzyme's optimum pH? amount of substrate

Knowing how enzymes work why might you decide to use the maximum concentration of substrate available? To increase the chance of enzymatic reactions.

Table 1 Product/min at substrate concentration 8 g

pH pH 3 pH 5 pH 7 pH 9 pH 11 Product/min 96 198 350 223 121

Did the enzyme perform best (more product/min) at a particular pH? yes What pH? 7 Use this graph to show your results.

[pic]

Conclusion

Results of this experiment (support or do not support) the hypothesis? Support

Experiment 2 Determining the Optimum Substrate Concentration

Knowing that enzymes perform at a faster rate when their active site is always filled with substrate, hypothesize how increasing the substrate concentration could affect the product/min. Increasing the substrate concentration will increase product/min until a maximum is reached.

Steps of the Experiment

In this experiment what experimental factor will you hold constant? pH Which pH would you use and why? pH 7 because that was optimum pH in the first experiment.

Table 2 Product/min at pH 7

Amount of Substrate 0.5 g 1.0 g 2.0 g 4.0 g 8.0 g

Product/min 72 145 300 350 350

Use this graph to show your results.

[pic]

Did the amount of product/min continue to increase as substrate concentration increased to 8.0 g? No Explain. There was no further increase after 4.0 g.

Conclusion

Results of this experiment (support or do not support) the hypothesis. Support

Explain your results on the basis of the active site model. The active sites of the enzyme molecules were full and therefore increasing the concentration to 8.0 g resulted in no further increase in the rate.

LABORATORY REVIEW 5 (LM page 66)

1. What happens at the active site of an enzyme? Substrates are oriented to bring about the reaction.

2. On the basis of the active site, explain why the following conditions speed a chemical reaction:

a. More enzyme There are more active sites available for substrates.

b. More substrate It is more likely that a substrate molecule will encounter an active site.

3. Name three other conditions (other than the ones mentioned in question 2) that maximize enzymatic reactions.

a. specific enzyme for a given substrate

b. optimum pH

c. optimum temperature

4. Explain the necessity for each of the three conditions you listed in question 3.

a. Correct enzyme/substrate Need this enzyme for the substrate.

b. Optimum pH Enzyme shape is maintained.

c. Optimum temperature Movement of molecules increases as temperature rises. Beyond optimum temperature, the enzyme may have the wrong shape for the substrate; at boiling enzyme is denatured.

5. Lipase is a digestive enzyme that digests fat droplets in the basic conditions (NaHCO3 is present) of the small intestine. Indicate which of the following test tubes would show digestion following incubation at 37°C, and explain why the others would not.

Tube 1: Water, fat droplets no enzyme

Tube 2: Water, fat droplets, lipase wrong pH

Tube 3: Water, fat droplets, lipase, NaHCO3 digestion occurs

Tube 4: Water, lipase, NaHCO3 no substrate

6. Fats are digested to fatty acids and glycerol. As the reaction described in question 5 proceeds, the solution will become what type pH? acidic Why? Fatty acids are released.

Given the following reaction:   Catalase

2 H2O2 ——————> 2 H2O + O2

hydrogen water oxygen

peroxide

7. Which substance is the substrate? hydrogen peroxide

8. Which substance is the enzyme? catalase

9. Which substances are the end products? water and oxygen

10. Is this a synthetic or degradative reaction? degradative 

Laboratory 6 Photosynthesis

(LM pages 67–76)

Eleventh Edition Changes

This lab remains the same as the previous edition: students are asked to formulate hypotheses before doing the experiments. The directions were improved for the experimental procedure involving green light (LM pages 72--3). At adopter request, a carbon cycle diagram was restored

New/ Revised Figures: 6.1 Paper Chromatography (This should be Figure 6.2?)

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Fresh material. Fresh or frozen spinach, depending on preparation alternative chosen, for 6.1 Plant Pigments.

Living material. Elodea (Anacharis) for 6.2 Solar Energy and 6.3 Carbon Dioxide Uptake.

Equipment pre-assembly required. Volumeter for 6.2 Solar Energy and 6.3 Carbon Dioxide Uptake.

Fume hood for 6.1 Photosynthetic Pigments.

All Exercises

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other clothing protection

_____ distilled water

_____ wax pencils

_____ rulers, plastic centimeter

6.1 Photosynthetic Pigments (LM pages 68–69)

_____ fresh spinach pigment extract:

_____ spinach, fresh

_____ blender, glass or stainless steel

_____ jar, wide-mouth, screw-cap

_____ cheesecloth and funnel, or Buchner funnel (Carolina 73-4624)

_____ filter paper

_____ acetone

_____ frozen spinach pigment extract alternative:

_____ spinach, frozen, 40 g.

_____ blender, glass or stainless steel

_____ amber jars, wide-mouth, screw-cap

_____ acetone

_____ ethanol

_____ sodium chloride

_____ filter, paper

_____ test tubes, large culture 25 X 150 mm

_____ cork, #11 with paper-clip hook for large test tubes

_____ chromatography paper, Whatman #1

_____ glass capillary tube

_____ paper towels

_____ scissors

_____ fume hood

_____ chromatography solution:

_____ petroleum ether

_____ jar, wide-mouth, screw-cap

_____ acetone

_____ test tube rack, 25 mm holes

_____ pencils

Whatman #1 chromatography paper (LM page 68). Use sheets of 12 x 12 cm Whatman #1 chromatography paper. Cut the sheets to fit the chromatography apparatus, rounding or pointing one end.

Fume hood and cautions (LM page 69). For the chromatography exercise, direct students’ attention to the fume hood and ether cautions in the lab manual.

Fresh spinach pigment extract (LM page 69). If a fume hood is available, prepare the extract there. Wash and thoroughly drain the spinach. Cut the veins and petioles from the leaves. Put the spinach in a glass or stainless steel blender, add acetone, and blend to form a thick slurry. Extract should be filtered, using a cheesecloth plug in a funnel or a small Buchner funnel with aspiration. Refrigerate the slurry in a tightly stoppered container labeled “Pigment Extract.” Extract exposed to light and room temperature begins decomposing within an hour, while refrigerated extract may last overnight. Have an empty wide-mouth, screw-capped jar, labeled “Used Pigment Extract,” available in which to place used pigment. Keep the jar tightly closed.

An alternate method involves drying spinach leaves slowly in a dry oven and then pulverizing them in a blender or with a mortar and pestle. Leaf powder is useful for weeks if stored in a sealed container and placed in a cool, dark area. Pulverization reduces leaf volume considerably. The dry leaf powder can be added to a small amount of acetone to form a thick slurry.

Frozen spinach pigment extract alternative (LM page 69). Partially defrost and divide a package of frozen spinach into 40 g portions. Combine 40 g frozen spinach with 200 mL acetone in a blender. Blend 2–3 minutes on high. Let stand 3 minutes. Decant supernatant, save as 1. Add 100 mL ethanol to solids remaining in blender. Blend 2–3 minutes on high. Decant supernatant, save as 2. Combine 1 and 2, and filter to remove any remaining solids. Add a pinch of sodium chloride. Refrigerate in amber jar.

Chromatography solution (LM page 69). 100 mL is sufficient for five student groups. Combine forty-five parts petroleum ether with five parts acetone, and store in a screw-capped container. Label as “Chromatography Solution.” Keep the container tightly closed, since this solution is volatile and extremely flammable. (If a fume hood is available, prepare the solution there.) Have a wide-mouth, screw-capped jar, labeled “Used Chromatography Solution,’’ available in which to place used solution. Keep the jar tightly closed.

Disposal (LM page 69). Organic solvents should be recycled or disposed of according to local procedures and regulations.

6.2 Solar Energy (LM pages 70–71)

_____ Elodea, fresh

_____ scalpels or razor blades, single-edged to make fresh cuts

_____ volumeter:

_____ test tubes, large culture, 25 x 150 mm

_____ rubber stoppers, #5.5, single-holed

_____ pipette, graduated

_____ lamp, 150 watt or aquarium light (full-spectrum bulb)

_____ beaker of water to serve as heat absorber

_____ Sodium bicarbonate solution

_____ sodium bicarbonate (NaHCO3) 3% solution

_____ water, distilled

_____ aquarium aerator for sodium bicarbonate

_____ beaker, 1,000 mL

_____ aluminum foil

_____ green dye or food coloring, or green filter for use with the light, or

green plastic (Wrap tube with green plastic for Green Light procedure.)

White and green light experiments (LM pages 70–73). These photosynthesis experiments work best when other lights are minimized. For the green light experiment, green plastic may be used instead of the dye or filter. Simply wrap the tube with the green plastic.

Volumeter (LM page 71). Prepare one volumeter per student group ahead of time. Insert a graduated pipette into a single-holed rubber stopper that fits into a large culture test tube, as shown in Figure 6.4. When the rubber stopper is in place during the experiments, a continuous column of liquid will form between the test tube and the pipette. Adjust the placement of the leading edge of the liquid by applying pressure to the stopper. The oxygen emitted by the Elodea will displace the liquid in the test tube, thus moving the edge of the liquid in the pipette. The student will read the change in millimeters.

3% sodium bicarbonate (NaHCO3) solution (LM page 70). Prepare 125 mL per student group. Dissolve 30 g of NaHCO3 in 1,000 mL of distilled water. Aerate the solution with an aquarium aerator for 30 minutes before the laboratory exercise to saturate with carbon dioxide. Discard the solution after use. This paragraph should be moved before Volumeter (LM page 71)?

Elodea (LM page 70). Use fresh Elodea (one healthy sprig per student group is sufficient) that has been maintained in continuously aerated distilled water. Change the water at least every two days.

6.3 Carbon Dioxide Uptake (LM page 74)

_____ 0.04% phenol red solution

_____ straws, individually packaged

_____ Elodea, volumeter, lamp, beaker as described in Section 6.2.

0.04% phenol red solution (LM page 74). Prepare 100 mL per student group. Dissolve 0.04 g of phenol red in 100 mL of distilled water. Have students use caution when blowing through the straw into the test tube of phenol red. Overzealous students may blow the phenol red out of the tubes and onto themselves. Students need only blow on the surface of the liquid to get a color change.

EXERCISE QUESTIONS

6.1 Photosynthetic Pigments (LM pages 68–69)

Restate this hypothesis here: Leaves contain various pigments and this allows them to absorb the energy of different colored light .

Experimental Procedure: Photosynthetic Pigments (LM pages 68–69)

4. Use this formula to calculate the Rf (ratio-factor) values for each pigment, and record these values in Table 6.1. Carotenes will have the largest values and chlorophyll b will have the smallest.

Conclusions: Photosynthetic Pigments (LM page 69)

• Do your results support the hypothesis that plant leaves contain various pigments? yes Explain. Photosynthetic chromatography separated the various pigments.

6.2 Solar Energy (LM pages 70–73)

Role of White Light (LM pages 70–71)

Experimental Procedure: White Light (LM pages70–71)

4. Record your results in Table 6.2. Why did the edge move upward? The edge moved upward in response to oxygen production, which forced the liquid outward in the tubing.

5. Record this value in Table 6.2. Why does cellular respiration, which occurs in a plant all the time, cause the edge to move downward? Oxygen, which was produced during photosynthesis, was being used by the plant during cellular respiration. As the volume of oxygen decreased (because photosynthesis is not occurring when the tube is wrapped by foil), less water was forced into the tubing, and the edge receded.

6. If the Elodea had not been respiring in step 4, how far would the edge have moved upward? Add the distance the edge moved downward to the distance the edge moved upward during the initial experiment with the white light. Record this number in Table 6.2.

7. Calculate the rate of photosynthesis. 201 mm/hr (Rates will vary with plant condition, distance from the lamp, and room temperature.) Record value in Table 6.2.

Table 6.2 Rate of Photosynthesis (White Light)

Data

Net photosynthesis (white light) 32 mm/10 min

Cellular respiration (no light) 1.5 mm/10 min

Gross photosynthesis (net + cellular respiration) 33.5 mm/10 min

Rate of photosynthesis 201 mm/hr

Role of Green Light (LM pages 72–73)

According to Figure 6.5, what color light do the chlorophylls absorb best? violet, blue, and red  Least? green, the reflected color

What color light do the carotenoids (carotenes and xanthophylls) absorb best? green-yellow Least? yellow, orange, the reflected colors

Hypothesize which color light is minimally utilized for photosynthesis. green

Experimental Procedure: Green Light (LM pages 72–73)

8. This percentage, based on your data in Table 6.3, = 37%

This percentage, based on class data in Table 6.3, = Compute from class data.

Table 6.3 Rate of Photosynthesis (White and Green Light)*

Your Data Class Data

Gross Photosynthesis (mm/10 min)

White (from Table 6.2) 33.5 mm/10 min

Green 12.5 mm/10 min

Rate of Photosynthesis (mm/hr)

White (from Table 6.2) 201 mm/hr

Green 75 mm/hr

*The results presented in this table are sample data. Actual results will vary.

Conclusions: Rate of Photosynthesis (LM page 73)

• Do your results support the hypothesis that green light is minimally used by a land plant for photosynthesis? no Explain with reference to Figure 6.5. Figure 6.5 suggests that green light will be minimally used but this data suggests that it does play a role.

• How does the percentage based on your data differ from that based on class data? Explanation will vary according to particular student.

6.3 Carbon Dioxide Uptake (LM page 74)

Experimental Procedure: Carbon Dioxide Uptake (LM page 74)

2. Blowing onto the solution adds what gas to the test tube? primarily carbon dioxide  When carbon dioxide combines with water, it forms carbonic acid. What causes the color change? Carbonic acid releases hydrogen ions. As the pH decreases, the color of the indicator changes from red to yellow.

5. How long did the color change take. Answers will vary.

6. Hypothesize why the solution in the test sample eventually turned red. As the plant uses carbon dioxide, carbonic acid disappears and the solution becomes basic again.

6.4 The Light Reactions and the Calvin Cycle Reactions (LM page 75)

Light Reactions (LM page 75)

1. a. What is the function of the photosynthetic pigments in photosynthesis? to absorb solar energy

b. How does it benefit a plant to have a variety of photosynthetic pigments? A variety of pigments allows plants to absorb different portions of white light.

c. Account for this observation with reference to the photosynthetic pigments. None of the photosynthetic pigments absorb green light.

d. Does this explain why leaves are green? Yes How so? Leaves are green because they reflect green light.

2. a. What happens to water during the light reaction? Water is split and oxygen is released.

b. What happens to the released oxygen? Much goes into the atmosphere and some is used by the plants and other organisms for cellular respiration.

3. The light reactions take place in the thylakoid membranes.

4. In your own words summarize the light reactions based on this laboratory. During the light reactions, solar energy is absorbed and water is split releasing oxygen.

Calvin Cycle Reactions (LM page 75)

1. What happens to carbon dioxide after it is taken up during the Calvin cycle reactions? It is reduced to a carbohydrate.

2. The Calvin cycle reactions take place in the stroma of chloroplasts.

3. In your own words, summarize the Calvin cycle reactions based on this laboratory. During the Calvin cycle reactions, carbon dioxide is taken up and reduced to a carbohydrate,

Light Reactions and Calvin Cycle Reactions (LM page 75)

1. Examine the overall equation for photosynthesis and show that there is a relationship between the light reactions and the Calvin cycle reactions by drawing an arrow between the hydrogen atoms in water and the hydrogen atoms in the carbohydrate.

[pic]

LABORATORY REVIEW 6 (LM page 76)

1. How are plant pigments involved in photosynthesis? They absorb solar energy.

2. Why is it beneficial to have several different plant pigments involved in photosynthesis? Each absorbs a different color light, to maximize the amount of solar energy that is absorbed.

3. On what basis does chromatography separate substances? solubility in nonpolar solvents

4. Some types of red algae carry on photosynthesis 70 meters beneath the ocean surface. What color light do you predict does not penetrate to this depth? red light

Consider the following reaction:

CO2  +  H2O  —> H2CO3  —> H+  +  HCO3-

        carbonic acid

5. Phenol red, a pH indicator, turns yellow (indicating acid) when you breathe into a solution. How does the reaction explain why the solution turned acidic? Carbon dioxide reacts with water to produce carbonic acid. As carbonic acid dissociates, the solution becomes more yellow, indicating a lower pH.

6. When light is available and a plant is added, the solution returns to its original red color. Why does this occur? The plant uses the CO2 for photosynthesis, causing the reaction to move to the left.

Gas exchange occurs in both photosynthesis and cellular respiration. Contrast these two processes by completing the following table:

Organelle Gas given off Gas Taken Up

7. Photosynthesis Chloroplast O2 CO2

8. Cellular Mitochondrion CO2 O2

respiration

9. What experimental conditions were used in this laboratory to test for cellular respiration in plant cells? no light and the plant then took up oxygen instead of releasing it.

10. Suppose you replaced Elodea with animal cells in the experimental test tube. Would the results differ according to the use of a white light or no light? no Explain. Animal cells do not photosynthesize; therefore, it makes no difference whether or not there is light.

Laboratory 7 Cellular Respiration

(LM pages 77–83)

Eleventh Edition Changes

This laboratory now begins with an introduction that includes the function of the mitochondrion and better stresses the role of the mitochondrion in cellular respiration. The lab was reorganized to place cellular respiration before fermentation and student directives have been clarified. A virtual lab entitled Energy in a Cell, is available on the Biology website and tests students understanding of both photosynthesis and cellular respiration.

New/Revised Figures: 7.1 Cellular Respiration; 7.4 Volumeter .

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Living material. Soybean seeds for Section 7.1 Cellular Respiration. Begin germinating soybean seeds 48 hours prior to the laboratory.

Fresh material. Baker’s yeast cake or active dry yeast for Section 7.2 Fermentation.

All Exercises

_____ water, distilled

_____ wax pencils

7.1 Cellular Respiration (LM pages 78–80)

_____ volumeter (Carolina 68-2200) or optional setup described below

_____ absorbent cotton

_____ 15% potassium hydroxide (KOH) solution

_____ dropper bottle for KOH

_____ soybean seeds, dry

_____ soybean seeds, germinating

_____ glass beads

_____ Brodie manometer fluid

Germinating soybean seeds (LM page 79). Begin germinating soybean seeds 48 hours prior to the laboratory. First, cover the dry soybean seeds with water, and allow them to soak for 24 hours. Then layer them in petri dishes with wet paper towels for another 24 hours.

Volumeter containing three respirometers (LM page 79). Purchase a volumeter that resembles Figure 7.3. Alternative: Prepare three vials, each with an attached stopper and calibrate pipette. according to the instructions in the manual (KOH-soaked cotton, soybeans, and glass beads). Insert the stoppers fitted with pipettes(#4 one-holed stoppers and pipettes calibrated to 0.01 mL). Weigh the vials so that you can lay them in a room temperature water bath with pipettes resting on the masking tape sling. Wait 7 minutes for temperature acclimation. Now move these respirometers so that they lay entirely in water with the calibrations visible from above. Water will enter the pipettes for a short distance. Allow 3 minutes for equilibration. Take the initial reading followed by two readings at 10 minute intervals. Make any volume corrections as indicated in the manual. Students should keep their hands out of the water bath during the experiment.

15% potassium hydroxide (KOH) solution (LM page78). Prepare 30 mL per student group. To make 15% KOH solution, dissolve 15 g of KOH in 100 mL of distilled water.

7.2 Fermentation (LM pages 80–82)

_____ Respirometers

_____ test tubes (15 x 125 mm) four per group

_____ test tubes (20 x 150 mm) four per group

_____ test tube racks

_____ 2% glucose (dextrose) solution

_____ 2% fructose solution

_____ 2% sucrose solution

_____ yeast suspension (five baker’s yeast cakes or five packets active dry yeast per laboratory)

_____ rulers, plastic millimeter

_____ water bath (or incubator), 37°C

_____ hot plate

_____ beaker, large enough to hold four assembled respirometers

_____ beaker tongs

_____ thermometer

Yeast suspension (LM page 78). Prepare 40 mL per student group. Crumble five cakes of baker’s yeast or pour five packets of active dry yeast into 1 liter of water. Shake until well suspended. Re-suspend the yeast before each use, since settling of the yeast cells may cause heterogenous sampling and thus inconsistent results. Some instructors report that they get better results when students are given a portion of a thick yeast paste.

2% solutions of glucose, fructose, and sucrose (LM page 78). Prepare 25 mL of each solution per student group. For 2% solutions, dissolve 2 g of glucose, fructose, or sucrose in 100 mL of distilled water.

EXERCISE QUESTIONS

7.1 Cellular Respiration (LM pages 78–80)

Hypothesis: Germinating soybeans that are growing will be carrying on cellular respiration, whereas nongerminating soybeans that are not growing will not be carrying on cellular respiration.

Experimental Procedure: Cellular Respiration (LM pages 78–80)

Table 7.1 Cellular Respiration*

Vial Contents Initial Reading after Reading after Net Correction Net

Reading 10 Minutes 20 Minutes Change Change

1 Germinating +++

soybeans

2 Dry (nongerminating) +

soybeans

Glass beads

3 Glass beads

*Students will provide actual readings, which will vary considerably. The pluses in this table indicate relative readings for net change in water position in the pipette.

Conclusions: Cellular Respiration (LM page 80)

• Do your results support or fail to support the hypothesis? support Explain. O2 is evidence that the plants are carrying on cellular respiration and lack of O2 uptake is evidence that plants are not carrying on cellular respiration.

• Why was it necessary to absorb the carbon dioxide? The carbon dioxide given off would not allow you to detect that oxygen was taken up.

• Which respirometer in the soybean experiment was the control? vial 3 was the control because if oxygen was taken up by glass beads the experiment is faulty.

7.2 Fermentation (LM pages 80–82)

Ethanol Fermentation (LM pages 81–82)

Hypothesis: Yeast formation of the monosaccharides glucose and fructose will yield better results than fermentation of sucrose, a disaccharide.

Experimental Procedure: Yeast Fermentation (LM pages 82)

Table 7.2 Fermentation by Yeast*

Tube Sugar Initial Gas Final Gas Net Change Ease of Fermentation Height Height

1 Yeast Depends upon +++ Depends upon Yeast can and glucose conditions conditions utilize glucose

in fermentation

2 Yeast Depends upon +++ Depends upon Fructose is and fructose conditions conditions readily changed to glucose, the

molecule that

yeast can act

upon in

fermentation

3 Yeast Depends upon ++ Depends upon Sucrose is a and sucrose conditions conditions disaccharide

and is used less

effectively by

yeast.

4 Yeast Depends upon No change — — — —

and water conditions

* The pluses in this table indicate relative readings for gas bubble height. Students will provide actual measurements.

Conclusions: Yeast Fermentation (LM page 82)

• Do your data support or fail to support you hypothesis. usually support Explain. The gas height for glucose and fructose was higher than for sucrose.

• Can your results be correlated with the comparative structure of the sugars? yes Explain. Glucose and sucrose are monosaccharides and sucrose is a disaccharide.

• Which respirometer was the control? The control was tube 4, containing yeast and water, and no sugar.

LABORATORY REVIEW 7 (LM page 83)

1. In Experimental Procedure: Cellular Respiration, what gas was taken up by the soybeans? oxygen

2. Explain the role of each of the following components in the cellular respiration experiment:

a. KOH  Takes up carbon dioxide given off by cellular respiration.

b. Germinating soybeans  These beans are metabolically active; therefore, any change signifies the extent of cellular respiration

c. Dry (nongerminating) soybeans  These beans are not metabolically; therefore, any change in this tube is used as a correction due to atmospheric pressure temperature changes.

3. If you performed the cellular respiration experiment without soaking the cotton with KOH, what results would you predict? No net change in both vial 1 in addition to 2. Why? The carbon dioxide given off in vial 1 will obscure the uptake of oxygen in that vial.

4. Soybeans don't move as animals do. For what purpose do they need energy? To maintain their bodies and to grow.

5. Both cellular respiration and fermentation ordinarily begin with what molecule? glucose

6. How do the overall equations for these processes indicate that fermentation is anaerobic and that cellular respiration is aerobic? Fermentation has no oxygen in the equation; cellular respiration does.

7. Glucose breakdown results in the breaking of C—H bonds and stored energy is released. Contrast the end products of fermentation and cellular respiration in terms of their energy content. The end products of fermentation, such as ethanol, have C—H bonds, and, therefore, more energy content than carbon dioxide without C—H bonds and water, the end products of cellular respiration.

8. Fermentation results in the net production of only 2 ATP, while cellular respiration results in many more ATP. Explain these results with reference to the end products of both of these processes. Ethanol, but not carbon dioxide and water, contains energy that could have been used for ATP production.

9. In Experimental Procedure: Yeast Fermentation, the gas bubble got larger. What gas was causing this increase in bubble size? carbon dioxide

10. Why might you hypothesize that, of the three sugars (glucose, fructose, and sucrose), glucose would result in the most activity during the fermentation experiment? Glucose is the sugar organisms usually metabolize.

Laboratory 8 Mitosis and Meiosis

(LM pages 85–104)

Eleventh Edition Changes

This lab remains essentially as it was in the previous edition. Note the suggestion for a virtual lab on page 91.

New and Revised Figures: 8.4 Cytokinesis in animal cells; 8.8 Meiosis I compared to mitosis; 8.10 Spermatogenesis and oogenesis in mammals

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

All Exercises

_____ microscopes, compound light

_____ lens paper

8.1 The Cell Cycle (LM pages 86–92)

Animal Mitosis Models (LM page 87)

_____ mitosis models, animal

_____ slide, prepared: whitefish mitosis

Plant Mitosis Models (LM page 90)

_____ mitosis models, plant

_____ slide, prepared: onion (Allium) root tip

8.2 Meiosis (LM pages 93–97)

Building Chromosomes to Simulate Meiosis (LM page 93)

_____ Chromosome Simulation Biokit® (Carolina 17-1100)

8.5 Gametogenesis in Animals (LM pages101–103)

Gametogenesis Models (LM page 102)

_____ gametogenesis model

_____ slide, prepared: testis

_____ slide, prepared: mammalian ovary

EXERCISE QUESTIONS

8.1 The Cell Cycle (LM pages 86–92)

State the event of each stage on the line provided.

G1 Organelles begin to double in number

S Replication of DNA

G2 Synthesis of proteins

M Mitosis

Explain why the term “cell cycle” is appropriate. In dividing cells, the stages repeat.

Mitosis (LM page 86)

Consult Table 8.1 and label the sister chromatids, centromere, and kinetochore in the drawing of a duplicated chromosome in Figure 8.2b. 1. sister chromatids; 2. centromere;

3. kinetochore

Spindle (LM page 87)

Observation: Animal Mitosis (LM page 87)

2. What is the number of chromosomes in each of the cells in this model series? Answers may vary, depending on what model is being used.

Mitosis Phases (LM pages 88–90)

Observation: Plant Mitosis (LM page 90)

Plant Mitosis Models

3. What is the number of chromosomes in each of the cells in this model series? Answers may vary, depending on what model is being used.

Onion Root Tip Slide

3. In the boxes provided, draw and label the stages of mitosis as observed in the onion root tip slide. Students are to draw all phases.

Cytokinesis (LM pages 91–92)

Cytokinesis in Animal Cells (LM page 91)

Were any of the cells of the whitefish blastula slide undergoing cytokinesis? Yes, some cells should be undergoing cytokinesis How do you know? Indentations of the cell membrane are seen along the medial line.

Cytokinesis in Plant Cells (LM page 92)

Were any of the cells of the onion root tip slide undergoing cytokinesis? yes How do you know? The formation of a cell plate along the medial aspect of some cells is apparent.

Summary of Mitotic Cell Division (LM page 92)

1. The nuclei in the daughter cells have the same number of chromosomes as the parent cell had.

2. Mitosis is cell division in which the chromosome number remains the same in the daughter cells as it was in the parent cell.

3. If a parent cell has 16 chromosomes, how many chromosomes do the daughter cells have following mitosis? 16

8.2 Meiosis (LM pages 93–97)

Experimental Procedure: Meiosis (LM pages 93–95)

Prophase I (LM page 94)

4. Why use nonsister chromatids and not sister chromatids? Nonsister chromatids have different genetic material. Sister chromatids have the same genetic material. If sister chromatids were involved in crossing-over, there would be no exchange of different genetic material.

Telophase I (LM page 94)

7. What combinations of chromosomes are at the poles?

For example: Pole A: red-long and blue-short

Pole B: blue-long and red-short

8. What other combinations would have been possible?

For example: Pole A: red-long and red-short

Pole B: blue-long and blue-short

Conclusions: Meiosis I (LM page 94)

• Do the chromosomes inherited from the mother or father have to remain together following meiosis I? no Explain Independent assortment occurs.

• Name two ways that meiosis contributes to genetic variation among offspring:

a. crossing-over

b. independent assortment of homologues

Interkinesis (LM page 94)

Does DNA replication occur during interkinesis? no Explain. Chromosomes are already duplicated.

Metaphase II (LM page 95)

How many chromosomes are at the metaphase plate? two

Anaphase II (LM page 95)

Pull the two magnets of each duplicated chromosome apart. What does this action represent? This action represents centromeres dividing and daughter chromosomes moving to opposite poles.

Conclusions: Meiosis II (LM page 95)

• Therefore, how many nuclei are usually present when meiosis II was complete? four

• In this exercise, how many chromosomes were in the parent cell nucleus undergoing meiosis II? two

• How many chromosomes are in the daughter nuclei? two Explain. When the chromatids of the chromosomes in the parent cell separate, they become daughter chromosomes, which segregate into the daughter cells.

Summary of Meiotic Cell Division (LM page 95)

1. The parent cell has the diploid (2n) number of chromosomes, and the daughter cells have the haploid (n) number of chromosomes.

2. Meiosis is cell division in which the chromosome number has been reduced by half.

3. If a parent cell has 16 chromosomes, the daughter cells will have how many chromosomes following meiosis? 8

4. A zygote contains the same number of chromosomes as the parent, but are these exactly the same chromosomes as either parent had? no Why not? Half the chromosomes came from each parent and the parents have different chromosomes.

5. Fertilization is another way that sexual reproduction results in genetic variation. During fertilization, union of gametes recombines chromosomes.

8.3 Mitosis Versus Meiosis (LM pages 98–99)

Table 8.2 Differences between Mitosis and Meiosis

Mitosis Meiosis

1. Number of divisions One Two

2. Chromosome number in daughter cells Diploid Haploid

3. Number of daughter cells Two Four

Table 8.3 Mitosis Compared with Meiosis I

Mitosis Meiosis I

Prophase: No pairing of chromosomes Prophase I: Pairing of homologues

Metaphase: Duplicated chromosomes at Metaphase I: Homologues at

metaphase plate metaphase plate

Anaphase: Sister chromatids separate Anaphase I: Homologues separate

Telophase: Chromosomes have Telophase I: Chromosomes have

one chromatid two chromatids

8.5 Gametogenesis in Animals (LM pages 101–103)

Observation: Gametogenesis in Mammals (LM pages 102–103) stet 102

Slide of Ovary (LM page 102)

4. How many secondary follicles can you find on your slide? Answers will vary depending on slide. How many vesicular follicles can you find? Answers will vary depending on slide. How does this number compare with the number of sperm cells in the testis cross section (see Fig. 8.12)? Many more sperm than follicles are seen.

Summary of Gametogenesis (LM page 103)

1. What is gametogenesis? formation of gametes In general, how many chromosomes are in a gamete? the haploid number

2. What is spermatogenesis? formation of sperm How many chromosomes does a human sperm have? haploid

3. What is oogenesis? formation of an egg How many chromosomes does a human egg have? haploid

4. Following fertilization, how many chromosomes does the zygote, the first cell of the new individual, have? diploid

LABORATORY REVIEW 8 (LM page 104)

1. During anaphase of mitosis in human or other 2n organisms, do the chromosomes have one or two chromatids as they move toward the poles? The chromosomes have one chromatid each.

2. During anaphase of meiosis I, do the chromosomes have one or two chromatids as they move toward the poles? The chromosomes have two chromatids.

3. During anaphase of meiosis II, do the chromosomes have one or two chromatids as they move toward the poles? The chromosomes have one chromatid.

4. Asexual reproduction of a haploid protozoan can be described as n —> n. Explain. A haploid cell undergoing mitosis produces haploid daughter cells.

5. Explain why furrowing is a suitable mechanism for cytokinesis of animal cells but not plant cells. Animal cells do not have a cell wall; plant cells do. Furrowing only indents the plasma membrane and not the cell wall.

6. A student is simulating meiosis I with chromosomes that are red-long and yellow-long; red-short and yellow-short. Why would you not expect to find both red-long and yellow-long in one resulting daughter cell? They represent a homologous pair, and these separate during meiosis I.

7. If there are 13 pairs of homologous chromosomes in a primary spermatocyte, how many chromosomes are there in a sperm? 13

8. Assume that you have built a homologous pair of chromosomes, each having two chromatids. One homologue contains red beads, while the other contains yellow beads. Describe the appearance of two nonsister chromatids following crossing-over. Each has some red beads and some yellow beads.

9. What are the major differences between mitosis and meiosis? number of divisions (one versus two); number of chromosomes in daughter cells (same number versus half the number); number of daughter cells (two versus four); and occurrence

10. A person with Down syndrome has what type of chromosome anomaly? abnormal chromosome number (as extra chromosome 21)

Laboratory 9 Mendelian Genetics

(LM pages 105–120)

Eleventh Edition Changes

The laboratory was rewritten to streamline the experiments and improve their clarity. A box about the Punnett square has been added to this lab and students are given the opportunity to do Punnett squares throughout the lab. The instructor is given the choice to have students do the fruit-fly experimental procedures or just examine the experimental results. Note the two virtual labs (LM pages 106 and 117) available on the Biology web site for this laboratory.

New/ Revised Figures: Punnett Squares are provided whenever students are to calculate the results of a cross.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Living material. Tobacco seedlings (9.1 One-Trait Crosses) and Drosophila cultures (9.1 One-Trait Crosses) (Note: Dated orders for Drosophila cultures and tobacco seeds or seedlings must be received by Carolina Biological Supply Company at least two weeks in advance of the use date.)

9.1 One-Trait Crosses (LM pages 106–11)

Color of Tobacco Seedlings (LM pages 107–108)

_____ tobacco seedlings (Albino Tobacco Biokit®, Carolina 17-8300)

_____ hand lens or stereomicroscope

_____ lens paper

Tobacco seedling purchase options (LM page 107). The Albino Tobacco Biokit® contains seed, growth chambers, and germination papers for a class of thirty students. Sow seeds approximately ten days before use. The seedlings can be maintained for about a week. (The albino ones will die shortly thereafter.) Directions come with the biokit. Replacement components (Carolina 17-8302) are available. Or green:albino tobacco seeds (Carolina 17-8400) can be germinated on black agar plates (Carolina 17-8442) for easier viewing. Or green:albino tobacco seedlings (Carolina 17-8402) can be ordered. Orders must be dated and will be shipped to arrive three to four days before intended use date. Order at least two weeks in advance of use date. Or corn seedlings germinated from green:albino genetic corn seeds (Carolina 17-7130) can be used instead of tobacco seedlings, eliminating any need for a dissecting microscope to count the seedlings.

Drosophila melanogaster Characteristics (LM pages 108–111)

Instructors should choose between these alternatives:

1. _____ Drosophila melanogaster slides (optional) are available at Wards.

2. _____ Drosophila culture vials one per group or class

_____ FlyNap® anesthetic and anesthetic wands (optional)

_____ hand lens or stereomicroscope

_____ card, white index

_____ brush, camel-hair

_____ morgue (filled with xylol—could use ethyl alcohol, 80%, or

isopropyl alcohol)

or

freeze flies overnight for examination the next day. This does away with the need for FlyNap® and reduces escaped flies.

Wing Length in Drosophila (LM pages 110–111)

Instructors should choose between these alternatives.

1. Three week experiment: Students have their own culture bottle and examine both F1 and F2 flies.

_____ vestigial X wild, Parental flies with these traits can be purchased from Carolina

_____ Also see supplies needed for examination of live flies.

2. One week experiment: Sample F2 flies are provided for the student to examine or the instructor provides the data only.

_____ vestigial X wild, F1 flies can be purchased from Carolina

_____ Also see supplies needed for examination of live flies.

Drosophila crosses (LM page 109). For the three-week experiment, each student or group has their own culture bottle. Culture vials need only be clean, not sterile, for use with the flies. The use of instant medium (Formula 4-24 Plain) significantly reduces preparation time. For the monohybrid cross (vestigial X wild), use F1 Custom Drosophila Cross (Carolina 17-2055). Transfer a few F1 flies to each culture bottle and make sure that each bottle receives at least one female that has been inseminated. Note that at week two, students remove the F1 flies; at week three, they count the F2 flies. Save the culture bottles sent by the supply house to give a continual supply of F2 flies, which can be frozen overnight for examination the next day. The Carolina Drosophila manual that comes with every order gives excellent directions on all aspects of culturing flies.

For the one-week experiment, you can remove the F1 flies from F1 Custom Drosophila Cross (Carolina 17-2055) when you receive it, and then use the vial(s) to provide sample F2 results. Allow one week for F2 flies to be available. If you freeze the F2 flies overnight, you will not need to use FlyNap®.

9.2 Two-Trait Crosses (LM pages 112–115)

Color and Texture of Corn (LM pages 112–113)

_____ Corn Dihybrid Genetics Biokit® (Carolina 17-6380)

Corn Dihybrid Genetics Biokit (LM page 111). The corn biokit comes with fifteen ears, marker pins, a teacher’s manual, and thirty student guides. A variety of other genetic corns and student guides are available.

Wing Length and Body Color in Drosophila (LM page 114)

_____ Vestigial X ebony flies

See instructions above for 3 week or 1 week experiments. See instructions above for choices in examining flies. Frozen flies do not require the use of FlyNap.

9.3 X-Linked Crosses (LM pages 116--117)

Red/White Eye Color in Drosophila (LM pages 114–115)

_____ white eyed X wild flies

See instructions above for 3 week or 1 week experiments. See instructions above for choices in examining flies. Frozen flies do not require the use of FlyNap.

9.4 Chi-Square Analysis (LM pages 118–119)

_____ calculators, and/or

_____ computer with a spreadsheet program, such as Quattro Pro or Excel

Chi-Square Analysis (LM pages 118--119). Students can pool their results so that only one chi-square test is needed. Students can use either calculators or a computer program (or a combination of both) to do the chi-square tests. Instructors can go through the chi-square test with the class data for the whole class and discuss results. Once students understand the concept of chi-square, they can use the computer with one of the spreadsheet programs to do the rest of the calculations.

EXERCISE QUESTIONS

9.1 One-Trait Crosses (LM pages 106–108)

Color of Tobacco Seedlings (LM pages 107--108)

Experimental Procedure: Color of Tobacco Seedlings (LM pages 107--108)

1. What is the expected phenotypic ratio? 3:1 (3 green plants to 1 white plant)

[pic]

Table 9.1 Color of Tobacco Seedlings

Number of Offspring

Green Color White Color

Plate #

Plate #

Plate #

Totals 90 30

Class Data

*These data are ideal; individual and class data will vary. Questions can be answered using these data if students do not do the experiment.

Conclusions: Color of Tobacco Seedlings (LM page 108)

• Calculate the actual phenotypic ratio you observed. If a student counts 120 green seedlings to 40 white seedlings, the actual ratio is 120/40 or three green seedlings to one white seedling or 3:1. Round off the actual numbers, and use only whole numbers for the actual ratio. Does your ratio differ from the expected ratio? yes Explain.

Counting small numbers of offspring is more likely to result in a variation from the expected ratio.

• Chi-square value: Results will depend on the student’s data. See 9.4 Chi-Square Analysis in this guide for possible results.

• Do your class data give a ratio that is closer to the expected ratio and is the chi-square deviation insignificant? Depends on data Explain. The more offspring that are counted, the greater is the probability of achieving the expected ratio.

Drosophila Melanogaster Culture (LM pages 108–109)

Observation: Drosophila melanogaster Culture(LM pages 108–109)

Culture

1. Where in the vial are the adult flies? The adult flies are flying around in the vial and are walking along the surface of the food in the bottom of the vial. the eggs? The eggs are buried near the surface of the food at the bottom of the vial.

2. Where are the larva? The larva look like maggots. They initially wriggle around in the food. They eventually move up out of the food and onto the sides of the vial, where they can be clearly seen.

3. Where are the pupae? The pupae are found on the wall of the vial. They are easily distinguishable from the larvae because they are encased in a light brown pupa case.

Observation: Drosophila Flies (LM pages 109--110)

Table 9.2 Characteristics of Wild-Type and Mutant Flies

Wild-Type Ebony Body Vestigial-Wing Sepia-Eye White-Eye

Wing length Long Long Short Long Long

Color of eyes Red Red Red Brown White

Color of body Gray Black Gray Gray Gray

Wing Length in Drosophila (LM pages 110–111)

Complete this Punnett square:

[pic]

What is the expected phenotypic ratio among the offspring? 3:1 (3 long-wings to 1 short-wings)

Experimental Procedure: Wing Length in Drosophila (LM page 111)

Table 9.3 Wing Length in Drosophila*

    Number of Offspring

Long Wings Vestigial Wings (Short wings)

Your data 91 29

Class data

*These data are possible, however individual and class data will vary. Questions can be answered using these data if students do not do the experiment.

Conclusions: Wing Length in Drosophila (LM page 111)

• Calculate the phenotypic ratio based on the data. If the data is 120 long-wings to 40 short wings, the ratio is 3:1. Do your results differ from the expected ratio? probably yes Explain. If the data involves only a small number of offspring there can be a variation from the expected ratio.

• Chi-square value: depends on the data. See Section 9.4 Chi-Square Analysis, in this Lab Resource Guide for possible results.

• What is the ratio based on class data? depends on class data Is the chi-square deviation now insignificant? most likely yes Explain. The more offspring counted, the greater is the probability of achieving the expected ratio.

• Using the Punnett square provided, determine the phenotypic ratio expected for the cross Ll x ll? 1:1 (1long wings to 1 short wings)

[pic]

9.2 Two-Trait Crosses (LM pages 112–15)

Color and Texture of Corn (LM pages 112–113)

Experimental Procedure: Color and Texture of Corn (LM pages 112–113)

1. Complete this Punnett square:

[pic]

2. What is the expected phenotypic ratio among the offspring? 9:3:3:1 (9 purple smooth to 3 purple rough to 3 yellow smooth to 1 yellow rough)

Table 9.4 Color and Texture of Corn*

Number of Kernels

Purple Smooth Purple Rough Yellow Smooth Yellow Rough

Sample # ____ 162 52 55 18

Sample # ____ 78 29 20 9

Sample # ____ 51 16 18 6

Totals 291 97 93 33

Class Data

*These data are possible, however individual and class data will vary.

Conclusions: Color and Texture of Corn (LM page 113)

• From your data, which two traits seem dominant? purple and smooth Which two traits seem recessive? yellow and rough

• Calculate the actual phenotypic ratio you observed. Divide all quantities in Table 9.4 by the number of seeds that are yellow rough. For example, if the results are twenty-seven purple smooth, nine purple rough, nine yellow smooth, and three yellow rough, the phenotypic ratio is 9:3:3:1. Does your ratio differ from the expected ratio? most likely

• Chi-square value: depends on data. See 9.4 Chi-Square Analysis in this Lab Resource Guide for possible results.

• What is the phenotypic ratio per the class? Answer will vary Do your class data give a ratio that is closer to the expected ratio, and is the chi-square deviation insignificant? probably yes Explain. The more offspring that are counted, the greater is the probability of achieving the expected ratio.

Wing Length and Body Color in Drosophila (LM pages 114–115)

Complete this Punnett square:

[pic]

What is the expected phenotypic ratio for this cross? 9:3:3:1 (9 long-wings, gray body to 3 long wings, black body to 3 short wings, gray body to 1 short wings, black body).

Experimental Procedure: Wing Length and Body Color in Drosophila (LM page 114)

Table 9.5 Wing Length and Body Color in Drosophila*

Phenotypes

Long Wings Long Wings Vestigial Wings Vestigial Wings Gray Body Ebony Body Gray Body Ebony Body

Number of Offspring 78 29 26 9

Class data

*These data are possible, however individual and class data will vary. Questions can be answered using these data if students do not do the experiment.

Conclusions: Wing Length and Body Color in Drosophila (LM page 115)

• Calculate the phenotypic ratio based on your data. A student will most likely observe a variation of the 9:3:3:1 ratio. Do your results differ from the expected ratio? yes Explain. Counting small numbers of offspring will result in a variation from the expected ratio.

• Chi-square value: depends on student’s data. See Section 9.4 Chi-Square Analysis, page 118 of the Lab Manual, for possible results.

• What is the phenotypic ratio per the class? Probably closer to 9:3:31 than individual data Do you expect the class data to give a ratio that is closer to the expected ratio and an insignificant chi-square deviation? yes Explain. The more offspring that are counted, the greater is the probability of achieving the expected ratio.

• Use the Punnett square provided to calculate the expected phenotypic results for the cross LlGg x llgg.

[pic]

What is the expected phenotypic ratio for this cross? 1:1:1:1(1long wings ,gray body: 1 long wings, black body: 1 short wings, gray body: 1 short wings,black body )

Summary (LM page 115)

Using the alleles Aa, and if necessary Bb, make a list of the four types of crosses you have studied and the expected phenotypic results.

Aa x Aa = 3:1

Aa x aa = 1:1

AaBb x AABb = 9:3:3:1

AaBb x aabb = 1:1:1:1

9.3 X-linked Crosses (LM pages 116–117)

Red/White Eye Color in Drosophila (LM pages 116–117)

Complete this Punnett square and state the expected phenotypic ratio for this cross

Females: The phenotypic ratio for females is all red eyes. Males: The results are 1:1; that is, for every red-eyed male, there will be a male that has white eyes.

[pic]

Experimental Procedure: Red/White Eye Color in Drosophila (LM page 117)

Table 9.6 Red/White Eye Color in Drosophila*

Number of Offspring

Your Data: Red Eyes White Eyes

Males 16 17

Females 63 0

Class Data:

Males

Females

*These data are possible, however individual and class data will vary. Questions can be answered using these data if students do not do the experiment.

Conclusions: Red/White Eye Color in Drosophila (LM page 117)

• Calculate the phenotypic ratio based on your data for males and females separately.

Males: Depends on student but 1:1 is expected for males

Females: All red eyes.

• Does this ratio differ from the expected ratio? depends on student

• Chi-square value: depends on data. See 9.4 Chi-Square Analysis of the Lab Resource Guide for possible results.

• Would you expect the class data to give a ratio that is closer to the expected ratio and an insignificant chi-square deviation? Yes What is the ratio? Depends on data. Explain. The more offspring that are counted, the greater is the probability of achieving the expected ratio.

• Using the Punnett square provided, calculate the expected phenotypic results for the cross XRY x XrXr

What is the expected phenotypic ratio among the offspring? females:all red eyes. males: white eyes.

9.4 Chi-square Analysis (LM pages 118-119)

Use Table 9.9 for performing a chi-square analysis of your results from a previous Experimental Procedure in this laboratory. Results will depend on the student’s data. Possible results are provided below for each Experimental Procedure.

Experimental Procedure: Color of Tobacco Seedlings

Possible Chi-Square Results for Color of Tobacco Seedlings

|Phenotype |Observed |Expected |Difference | |Partial Chi-Square |

| |Number |Results (e) |(d) |(d2) |(d2/e) |

|Green |85 |86.25 |1.25 |1.5625 |0.0181159 |

|White |30 |28.75 |-1.25 |1.5625 |0.0543478 |

|Chi-Square=x2=E (d2/e) = | | | | | |

| | | | | | |

Experimental Procedure: Wing Length in Drosophila

Possible Chi-Square Results for Wing Length in Drosophila

|Phenotype |Observed |Expected |Difference | |Partial Chi-Square |

| |Number |Results (e) |(d) |(d2) |(d2/e) |

|Green |91 |90 |-1 |1 |0.0111111 |

|White |29 |30 |1 |1 |0.0333333 |

|Chi-Square=x2=E (d2/e) = | | | | | |

| | | | | | |

Experimental Procedure: Color and Texture of Corn

Possible Chi-Square Results for Color and Texture of Corn

|Phenotype |Observed |Expected |Difference | |Partial Chi-Square |

| |Number |Results (e) |(d) |d2 |(d2/e) |

|Purple, smooth |78 |61 |-17 |289 |4.7377049 |

|Purple, rough |26 |20 |6 |36 |1.8 |

|Yellow, smooth |29 |20 |- 9 |81 |4.05 |

|Yellow, rough |9 |7 |2 |4 |0.5714286 |

|Chi-Square=x2=E (d2/e)| | | | | |

|= | | | | | |

Experimental Procedure: Wing Length and Body Color in Drosophila

Possible Chi-Square Results for Wing Length and Body Color in Drosophila

|Phenotype |Observed |Expected |Difference | |Partial Chi-Square |

| |Number |Results (e) |(d) |d2 |(d2/e) |

|Long, gray |78 |61 |-17 |289 |4.7377049 |

|Long, ebony |29 |20 |- 9 |81 |4.05 |

|Vestigial, gray |26 |20 |6 |36 |1.8 |

|Vestigial, ebony |9 |7 |2 |4 |0.5714286 |

|Chi-Square=x2=E (d2/e) = | | | | | |

Experimental Procedure: Red/White Eye Color in Drosophila

Possible Chi-Square Results for RedlWhite Eye Color in Drosophila

|Phenotype |Observed |Expected |Difference |D-squared |Partial Chi-Square |

| |Number |Results (e) |(d) |d2 |(d2/e) |

|Male, red-eyed |16 |16.5 |-0.5 |0.25 |0.0151515 |

|Male, white-eyed |17 |16.5 |0.5 |0.25 |0.0151515 |

|Chi-Square=x2=!, (d2/e) = | | | | | |

Conclusions: Chi-square Analysis (LM page 119)

• Do your results support your original prediction? Student (2 values should indicate differences are insignificant and their hypotheses/predictions are supported.

• If not, how can you account for this? Using a low number of individuals can lead to significant variations in data. Using data from a larger class or larger population of fruit flies would provide better results.

LABORATORY REVIEW 9 (LM page 120)

1. If offspring exhibit a 3:1 phenotypic ratio, what are the genotypes of the parents?

Aa x Aa

2. In fruit flies, which of the characteristics that you studied was X-linked? red/white eye color

3. If offspring exhibit a 9:3:3:1 phenotypic ratio, what are the genotypes of the parent generation? AaBb x AaBb

4. What was the phenotype of the parents if you counted 90 long-winged flies to 30 short-winged flies as the offspring? long-winged

5. Briefly describe the life cycle of Drosophila. The adults reproduce by laying eggs. The eggs hatch into larvae that feed. The larvae form pupae, in which the tissues are reorganized into an adult.

6. When doing a genetic cross, why is it necessary to remove parent flies before the pupae have hatched? It would be impossible to distinguish parents from offspring.

7. What is the genotype of a white-eyed male fruit fly? XrY

8. What may have been the genotypes of the parents if you counted 20 green tobacco seedlings and 20 white tobacco seedlings in one agar plate. Cc x cc

9. Suppose you counted tobacco seedlings for a monohybrid cross in six agar plates, and your data were as follows: 125 green plants and 39 white plants. According to the chi-square test, are your deviations from the expected values due to chance? Yes, the variations are due to chance.

10. Suppose that students in the laboratory periods before yours removed some of the purple and yellow corn kernels on the ears of corn as they were performing the Experimental Procedure. What effect would this have on your results? This would negate your results because the data had been tampered with.

Laboratory 10 Human Genetics

(LM pages 121–133)

Eleventh Edition Changes

The introduction now discusses karyotype preparation before giving a brief overview of the chapter. Section 10.1, Chromosomal Inheritance, was revised to show that nondisjunction is a cause of both autosomal and sex chromosome anomalies (a preferred term instead of abnormalities). Interest will increase due to the inclusion of new photos and exercises in this section. Section 10.2 has minor changes and remains essentially as it was.

New / Revised Figures: 10.1 Human karyotype preparation; 10.2 Nondisjunction; 10.3 Normal human male karyotype; numerous unnumbered photos and exercises

MATERIALS AND PREPARATION

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

10.1 Chromosomal Inheritance (LM pages 122–125)

_____ No supplies are needed

10.2 Genetic Inheritance (LM pages 125–132)

_____ Carolina Student Vision Kit (Carolina 69-4530) uses Ishihara test. An inexpensive alternative is to use the single-page plate found in many introductory psychology texts.

EXERCISE QUESTIONS

10.1 Chromosomal Inheritance (LM pages 122--125)

Observation: Autosome Chromosome Anomalies (LM pages 122–123)

With the assistance of Figure 10.2 explain how trisomy 21 comes about. Due to nondisjunction during either meiosis I or meiosis II an egg carries two number 21 chromosomes. Following fertilization, the zygote and then the individual has three number 21 chromosomes.

Which chromosome should have three copies? number 13 (In Figure 10.3, a student boxes the two number 13 chromosomes and adds a leader which reads: Patau syndrome has three #13 chromosomes.)

Which chromosome should have three copies? number 18 (In Figure 10.3, a student boxes the two number 18 chromosomes and adds a leader which reads: Edwards syndrome has three #13 chromosomes.)

Observation: Sex Chromosome Anomalies (LM pages 123--125)

Fill in the following diagram to show that nondisjunction during meiosis can lead to Turner, poly-X, or Klinefelter syndrome.

[pic]

Fill in the following diagram to show that nondisunction during meiosis I in a male can lead to this syndrome.

[pic]

Conclusion: Sex Chromosome Anomalies (LM page 125)

Complete Table 10.1 and label Figure 10.4

Table 10.1 Numerical Sex Chromosome Abnormalities

Syndrome Number Comparison with Normal Number

Turner: XO All Normal Males: XY All Normal Females: XX

Poly-X: XXX

Klinefelter: XXY

Jacob: XYY

Figure 10.4 a. Poly X b. Turner c. Klinefelter d. Jacob

10.2 Genetic Inheritance (LM pages 125–132)

Autosomal Dominant and Recessive Traits (LM pages 125–128)

For example, if W = widow's peak and w = straight hairline, what would be homozygous dominant? WW What would be heterozygous? Ww and what would be homozygous recessive? ww Which one of these would have a straight hairline? ww

Experimental Procedure: Autosomal Traits (LM pages 126–128)

4. Are dominant phenotypes always the most common in a population? no Explain. The prevalent phenotypes depend on the genes of the previous generations

Table 10.2 Autosomal Human Traits

Answers may vary according to the class members. Students may not know whether they are homozygous dominant or heterozygous. If so, they can use A? for their genotype,

Genetics Problems Involving the Traits in Table 10.2 (LM pages 127–128)

1. Nancy and the members of her immediate family have attached earlobes. Her maternal grandfather has unattached earlobes. What is the genotype of her maternal grandfather? Ee Nancy’s mother has the genotype ee (results in the recessive phenotype), therefore her maternal grandfather, who has unattached earlobes, must be Ee. Nancy’s maternal grandmother is no longer living. What could have been the genotype of her maternal grandmother? Nancy’s mother must have also inherited an e from her own mother, who could have been Ee or ee. Compare the chances that Nancy's maternal grandparents could have had a child with attached earlobes. Ee X Ee, = 25% chance and Ee X ee = 50% chance.

2. Joe does not have a bent little finger, but his parents do. What is the expected phenotypic ratio among the parents’ children? Joe’s genotype is ll (results in the recessive phenotype); therefore, his parents who have bent little fingers must be Ll. The expected ratio among their children would be 3:1, or 75% with bent little finger and 25% with straight little finger.

3. Henry is adopted. He has hair on the back of his hand. Could both of his parents have had hair on the back of the hand? yes Could both of his parents have had no hair on the back of the hand? no Explain. The presence of hair on the back of the hand is a dominant characteristic; at least one parent had to have hair on the back of the hand for Henry to have it.

Genetics Problems Involving Genetic Disorders (LM page 127)

1. Cystic fibrosis is an autosomal recessive disorder. If both of Sally’s parents are heterozygous for cystic fibrosis, what are her chances of inheriting cystic fibrosis? 25%

2. Nancy has cystic fibrosis, but neither parent has cystic fibrosis. What is the genotype of all people involved. Parents: Aa; Nancy: aa

3. Huntington disease is an autosomal dominant disorder. If only one of Sam’s parents is heterozygous for Huntington disease, and the other is homozygous recessive, what are his chances of inheriting Huntington disease? 50%

4. In Henry’s family, only his father has Huntington disease. What are the genotypes of Henry, his mother, and his father? Henry: aa; mother: aa; father: Aa.

Sex Linkage (LM pages 128–130)

Experimental Procedure: X-Linked Traits (LM pages 129–130)

3. Are you color blind? Results will depend on the individual student. If so, what is your genotype? Female: XbXb; Male: XbY

4. If your father is color blind, what is your genotype? XBXb

If your mother is color blind, what is your genotype? XBXb

If you know of no one in your family who is color blind, what is your

probable genotype? XBXB

Genetics Problems Involving X-linked Genetic Disorders (LM pages 129–130)

1. If a father is color blind, but the mother does not have a color-blind allele, what are the chances his daughters will be color blind? No chance Be carriers? 100%

2. Mary Jo is a carrier for hemophilia, an X-linked recessive disorder. Her mother is perfectly normal. What is her father’s genotype? XhY

3. If a boy is color blind, from which parent did he inherit the defective allele? mother

4. Mary has a color-blind son, but Mary and both of Mary’s parents have normal vision. Give the genotype of all people involved. Son: XbY; Mary: XBXb; Mary’s father: XBY; Mary’s mother: XBXb

5. In which parent and at what meiotic division did sex chromosome nondisjunction occur? Color blindness was inherited from his mother, who is XBXb. If nondisjunction occurred during meiosis I, he would have inherited an XB and would not be color blind. Since this male is color blind, we know that nondisjunction occurred during meiosis II.

6. In which parent did nondisjunction occur, considering that the single X came from the father? Nondisjunction occurred in the mother. Is it possible to tell if nondisjunction occurred during meiosis I or meiosis II? no Explain. Hemophilia was not passed on by the mother, and nondisjunction in the mother during meiosis I or meiosis II produces an egg without an X.

Pedigrees (LM pages 130–132)

Pedigree Analyses (LM pages 130–132)

1. a. What is the inheritance pattern for this genetic disorder? The inheritance pattern is autosomal recessive. The two parents of generation I are unaffected, and yet, two daughters are affected. These observations rule out autosomal dominant and X-linked recessive.

b. What is the genotype of the following individuals?

Generation I, individual 1: Aa This individual has to be heterozygous because some of the children are affected.

Generation II, individual 1: aa This individual has to be homozygous recessive because he is affected.

Generation III, individual 8: Aa This has to be the case because the mother is homozygous recessive, and the individual has to inherit at least one of her recessive alleles.

2. a. What is the inheritance pattern for this genetic disorder? The inheritance pattern is X-linked recessive. This has to be the case because only the males are affected, and the trait skips generations. (Note: if a trait is Y-linked, only the males are affected, but the trait is in every generation.)

b. What is the genotype of the following individuals?

Generation I, individual 1: XAXa This female has to be a carrier because she has an affected son.

Generation II, individual 8: XAX? Unable to determine whether this female is a carrier or not because she had no children.

Generation III, individual 1: XAY This male is unaffected; therefore, he must have received a dominant allele.

Construction of a Pedigree (LM page 132)

2. Construct two blank pedigrees. Student shows the four generations X2. Henry's grandparents, Henry's parents, Henry and his wife, Henry's daughter, Polly

3. Try out a pattern of autosomal dominant inheritance in Pedigree 1 and a pattern of X-linked dominant inheritance.

[pic]

4. Which pattern is correct? autosomal dominant What is your key for this trait? a = normal eyelashes; A = double row of eyelashes

5. Use correct genotypes to show a cross between Henry and Isabella and calculate the expected phenotypic ratio among the offspring:

Cross: Aa X aa Phenotypic ratio: 1:1

6. What are the percentage chances of Henry and Isabella having a child with double eyelashes? 50%

LABORATORY REVIEW 10 (LM page 133)

1. Name one pair of chromosomes not homologous in a normal karyotype. XY

2. If nondisjunction occurs in a heterozygous woman, would meiosis I or meiosis II produce an egg in which both X chromosomes carry the same alleles? Meiosis II Explain. Because only then do eggs receive two identical X chromosomes.

3. Which one could produce a sperm with two X chromosomes: nondisjunction during meiosis I or nondisjunction during meiosis II? Meiosis II Explain. Because only then do sperm receive two X chromosomes.

4. If an individual exhibits the dominant trait, do you know the genotype? no Why or why not? The individual could be heterozygous (Aa) or homozygous (AA).

5. A son is color blind, but both parents are normal. Give the genotype of the mother (XBXb) and the father (XBY). Explain the pattern of inheritance. A color-blind son inherits the X-linked, color blind gene from his mother and the Y from his father.

6. What pattern of inheritance in a pedigree would allow you to decide that a trait is X-linked? A trait is X-linked if it passes from grandfather to grandson.

7. What pattern of inheritance in a pedigree would allow you to decide that a trait is autosomal recessive? A trait is autosomal recessive if parents who do not show the trait have an affected child.

8. What is the difference between a Punnett square and a pedigree? A Punnett square shows the expected results of a cross when the pattern of inheritance is known. A pedigree shows the pattern of inheritance of a family trait.

9. What is the probability that two individuals with an autosomal recessive trait will have a child with the same trait? 100%

10. What is the probability that a woman whose father was color blind will have a son who is color blind? 50%

Laboratory 11 DNA Biology and Technology

(LM pages 135–148)

Eleventh Edition Changes

Section 11.3 now uses the term "gene strand" for the strand that is not transcribed because the mRNA transcript has a sequence of bases that matches that strand.

New and Revised Figures: 11.1 Overview of DNA structure; 11.4 Messenger RNA (mRNA)

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements: Section11.4 Isolation of DNA requires onion filtrate preparation.

All Exercises

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand

Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other

clothing protect

11.1 DNA Structure and Replication (LM pages 136–138)

_____ model, DNA; model kit, DNA-RNA; or puzzle kit, DNA (Carolina

17-1050)

11.2 RNA Structure (LM pages 139–140)

_____ puzzle kit, DNA (Carolina 17-1050)

11.3 DNA and Protein Synthesis (LM page 140–143)

_____ model kit, DNA-RNA Protein Synthesis (Lab Aids Kit #72)

Kits and models. For Sections 11.1 to 11.3, DNA kits are available from Carolina Biological Supply and Lab Aids, from which students construct models. The kits vary in degree of sophistication and in price. Descriptions and price information for the Carolina products can be found in the “Genetics” section of the Carolina catalog. Alternatively, students can simply use the figures in the lab manual to gain an understanding of the concepts.

11.4 Isolation of DNA (LM page 144)

To prepare onion filtrate:

_____ onion, fresh

_____ salt (NaCl) (Carolina 88-8880 -8901)

_____ dishwashing detergent

_____ ice-water bath

_____ blender

_____ spoon

_____ cheesecloth, #6 coffee filter, or metal kitchen strainer

_____ beakers, 500 mL and 1,000 mL

_____ large funnel or colander

_____ water bath, hot (60°C) or hot tap water

For laboratory use:

_____ test tubes, large

_____ test-tube rack

_____ ice-water bath

_____ meat tenderizer solution (Adolph’s meat tenderizer or a product containing

papain)

_____ stirring rods, glass or Pasteur pipettes

_____ graduated cylinder, glass, 10 mL

_____ 95% ethanol (Carolina 86-1281), ice cold (5 mL per student group)

Onion filtrate (LM page 144). Cut one large onion into medium-large pieces, and place the pieces in a 500 mL beaker. Cover the onion with 100 mL of a NaCl–detergent solution. (To prepare the NaCl–detergent solution, add distilled water to 1.5 g. noniodized salt to bring to 50 mL. Add 10 mL dish detergent and bring up to 100 mL with distilled water. Refrigerate the solution; keep on ice when in the laboratory. A cheap, watery dish detergent works best; concentrated detergent must be diluted with water, otherwise it will foam when blended and the mixture will be thick.) Stir mixture and let stand 15 minutes in a hot water bath at 60°C. (Temperature is not critical, but do not have the water boiling. Your water bath can be a sink containing hot tap water.) Cool the mixture in an ice-water bath for 5 minutes, stirring frequently with a spoon. Pour the mixture into a blender, and blend for 1 minute on low speed, then 30 seconds on high speed. Place four thicknesses of cheesecloth (or a #6 coffee filter or a small mesh metal kitchen strainer) in a large funnel or colander over a large beaker. (Be careful, and try to prevent foam from getting into the filtrate.) Put the whole setup into the refrigerator, and let it filter overnight.

Alternative: Peel ripe banana and place in blender. Cover banana with 15% NaCl plus 2 drops Dawn dishwashing liquid. Blend for one minute. Strain mixture through several layers of cheesecloth. Add a pinch of meat tenderizer. Mix. Divide mixture into test tubes, each about half full. Allow to stand for 15–20 minutes. Layer ice-cold ethanol on top of mixture. Stir interface of mixture and ethanol with glass stirring rod—DNA will adhere to the rod and can be spooled up by rotating the rod. This preparation yields a large amount of DNA. (This procedure may be easier to use and produce more DNA than the onion filtrate procedure.) Alternative: Kits, such as DNA Extraction of Micrococcus luteus Classroom Kit (Carolina 17-1090), are available for the DNA isolation procedure, however the instructions given in the lab manual are a simpler way to proceed.

6% meat tenderizer solution (LM page 144). Add 50 mL distilled water to 3g Adolph’s meat tenderizer or a product containing papain. Refrigerate the solution; keep on ice when in the laboratory. Alternative: Dissolve one pill of enzymatic contact lens cleaner to 5 mL of distilled water.

11.5 Genetic Disorders (LM pages 145–147)

Note: If desired, students can gain an understanding of the gel electrophoresis process by using the description and figures in the lab manual, rather than performing the actual procedures.

Gel Electrophoresis (LM pages 146--47)

_____ Electrophoresis Protein Separation Kit (Carolina 68-9800)

_____ horizontal gel electrophoresis apparatus:

_____ power supply

_____ cables

_____ electrophoresis chamber with gel

Horizontal gel electrophoresis apparatus (LM page 146). Biological suppliers have various types of electrophoresis apparatuses for sale. Biostar Corporation (P.O. Box 5756, Lafayette, In 47903) has Quadracell units (QEC-100) and power supply (MAB-125), which allow four gels of four lanes each per unit.

Electrophoresis Protein Separation Kit alternative (LM page 146). If a kit is not obtained, the following supplies will be needed:

Electrophoresis buffer (optional). If you have purchased a kit, the electrophoresis buffer will be included. Otherwise, make up a sterile 5% stock TBE buffer as follows:

54 g of Tris base (Tris aminomethane buffer), 27.5 g of boric acid, 20 mL of 0.5 M EDTA (disodium ethylene diamine tetraacetate 2H2O) (pH 8.0). Note: The wells also can be loaded before adding the buffer. Then they will need to be sealed with agarose solution.

Agarose solution (optional). Agarose powder can be purchased from biological suppliers. It also comes as part of a molecular biology experiment package, along with instructions for making the gel slab.

Gel slabs. Gel slabs can be used immediately, or they can be covered with plastic and left overnight (or longer) in the refrigerator.

Micropipettes and micropipette tips. Either adjustable or fixed pipettes are recommended. When using adjustable pipettes, you need only one (5–50 mL) per setup, with one kind of tip. To pipette 100 mL, just use the 50 mL adjustment level twice. VWR Scientific, with offices in many major cities, is a good supplier of adjustable pipettes.

The tip can be cleaned by rinsing three times, but when working with bacteria, using a new/sterile tip each time is preferable. (Tips can be reused after rinsing and resterilization in their dispenser boxes.)

EXERCISE QUESTIONS

11.1 DNA Structure and Replication (LM pages 136–138)

DNA Structure (LM pages 136–137)

Observation: DNA Structure (LM page 137)

1. Label phosphate, base pair, and deoxyribose in your drawing and 1–3 in Figure 11.1a.

1. phosphate; 2. base pair; 3. deoxyribose

2. Complete Table 11.1 by writing in the colors of the bases.

Table 11.1 Base Colors

In Figure 11.1b In Your Kit

Cytosine Red

Thymine Blue

Adenine Gold

Guanine Green

3. What type of molecules make up the backbone (uprights of ladder) of DNA (Fig. 11.1b)? sugar and phosphate molecules 

4. Dashes are used to represent hydrogen bonds in Figure 11.1b because hydrogen bonds are weak.

5. Notice that the base A is always paired with the base T, and the base C is always paired with the base G.

6. In Figure 11.1b, what molecules make up the rungs of the ladder? hydrogen-bonded bases adenine, cytosine, guanine, and thymine

7. Why is DNA also called a double helix (Fig. 11.1b)? The two strands making up DNA’s ladder configuration twist around one another in the form of a helix.

DNA Replication (LM pages 137–138)

Observation: DNA Replication (LM page 138)

1. What bonds are broken in order to unzip the DNA strands? hydrogen bonds

3. Complete Table 11.2. Are your molecules identical in Figure 11.2c? yes

4. Because of complementary base pairing, each new double helix is composed of an old strand and a new strand. Write old or new in 1–10, Figure 11.2a, b, and c. 1. old; 2. old; 3. old; 4. new; 5. new; 6. old; 7. old; 8., new; 9. new; 10. old Why is DNA replication called semiconservative? Because each new double helix is composed of an old (parent) strand and a new (daughter) strand.

5. In your own words how does replication provide a means for passing DNA from cell to cell and organism to organism? By replicating (making a copy of itself) daughter cells and also sperm and eggs receive a copy of the DNA.

Table 11.2 DNA Replication

Old strand G G G T T C C A T T A A A T T C C A G A A A T C  A  T  A

New strand C C C A A G G T A A T T T A A G G T C T T T A

G  T  A  T

11.2 RNA Structure (LM Pages 139–140)

1. Describe the backbone of an RNA molecule. RNA, like DNA, has a sugar phosphate backbone.

2. Where are the bases located in an RNA molecule? to the side

3. Complete Table 11.3 to show the complementary DNA bases for the RNA bases

Table 11.3 DNA and RNA Bases

RNA Bases C U A G

DNA Bases G A T G

Observation: RNA Structure (LM pages 139–40)

1. Label the ribose (the sugar in RNA), the phosphate, and the base in your drawing and in 1–3, Figure 11.3. 1. phosphate; 2.base; 3. ribose

2. Complete Table 11.4 by writing in the colors of the bases for Figure 11.3 and for your kit.

Table 11.4 Base Colors

In Figure 11.3 In Your Kit

Cytosine Red

Uracil Purple-blue

Adenine Orange

Guanine Green

3. The base uracil substitutes for the base thymine in RNA. Complete Table 11.5 to show the several other ways RNA differs from DNA.

Table 11.5 DNA Structure Compared to RNA Structure

DNA RNA

Sugar Deoxyribose Ribose

Bases Adenine, guanine, thymine, cytosine Adenine, guanine, uracil, cytosine

Strands Double stranded with base pairing Single stranded

Helix Yes No

11.3 DNA and Protein Synthesis (LM page 140)

Explain the role DNA, mRNA, and tRNA have in protein synthesis. DNA: stores information (i.e., proper sequence of amino acids); mRNA: carries information to ribosomes; tRNA: brings amino acids to ribosomes

Transcription (LM page 141)

Label Figure 11.4. For number 1, note the name of the enzyme that carries out mRNA synthesis. 1. RNA polymerase; 2.For number 2, note the name of this polynucleotide molecule. mRNA

Observation: Transcription (LM page 141)

3. Complete Table 11.6 to show the sequence of bases in mRNA.

Table 11.6 Transcription

DNA T A C A C G A G C A A C T A A C A T

mRNA A U G U G C U C G U U G A U U G U A

4. Locate the end of the strand that will move to the ribosome in the cytoplasm.

Translation (LM pages 142–143)

Label Figure 11.5. 1. amino acid; 2. tRNA; 3. anticodon

Observation: Translation (LM pages 142–143)

2. Complete Table 11.7.Why are the codons and anticodons in groups of three? The genetic code is a triplet code.

Table 11.7 Translation

mRNA codons AUG CCC GAG GUU GAU UUG UCU

tRNA anticodons UAC GGG CUC CAA CUA AAC AGA

Amino acid met pro glu val asp leu ser

3. In Figure 11.7, label the ribosome, the mRNA, and the peptide. Also, indicate the A, P, and E sites. 1. peptide; 2. ribosome; 3. mRNA; right to lift: A, P, E sites

11.4 Isolation of DNA (LM page 144)

Experimental Procedure: Isolating DNA (LM page 144)

7. Answer the following questions:

a. When did homogenization occur? during the preparation of the filtrate What was the purpose of homogenization? to release DNA from cells

b. Which of the preceding steps represents deproteinization? step 4

c. Which of the preceding steps represents precipitation of DNA? step 5

11.5 Genetic Disorders (LM pages 145–147)

Genomic Sequence for Sickle-Cell Disease (LM pages 145–146)

1. In what three-DNA-base sequence does HbA differ from HbS?

HbA   CTC HbS   CAC

2. What are the codons for these three bases?

HbA   GAG HbS   GUG

3. What is the amino acid difference?

HbA   glu HbS   val

Gel Electrophoresis (LM pages 146–47)

Experimental Procedure: Gel Electrophoresis (LM page 147)

Analyzing the Electrophoresed Gel

2. In Figure 11.11, label the lane that contains only HbS, signifying that the individual is HbSHbS. Lane 2

3. Label the lane that contains only HbA, signifying that the individual is HbAHbA. Lane 1

4. Label the lane that contains both HbS and HbA, signifying that the individual is HbAHbS. Lane 3

Conclusion: Genomic Sequences for Sickle-Cell Disease (LM page 147)

• What genotype do they each have? HbAHbS and HbAHbA What are the chances that this couple will have a child with sickle-cell disease (HbA is dominant and HbS is recessive)? 1 in 4 chance = 25%

LABORATORY REVIEW 11 (LM page 148)

1. Explain why DNA is said to have a structure that resembles a ladder. In an untwisted helix, DNA's structure resembles a ladder: the ladder sides are sugar phosphate molecules; the ladder rungs are hydrogen-bonded bases.

2. How is complementary base pairing different when pairing DNA to DNA than when pairing DNA to mRNA? When pairing DNA to DNA, A pairs with T, and G pairs with C. When pairing DNA to RNA, A pairs with U, and G pairs with C.

3. Explain why the genetic code is called a triplet code. Every three bases equals one codon which specifies one amino acid.

4. What role does each of the following molecules play in protein synthesis?

a. DNA: template for mRNA

b. mRNA: contains codons

c. tRNA: transfers amino acids, has anticodons

d. Amino acids: building blocks for proteins

5. Which of the molecules listed in question 4 are involved in transcription? DNA —> mRNA

6. Which of the molecules listed in question 4 are involved in translation? mRNA, tRNA, amino acids

7. During the isolation of DNA, what role was played by these substances?

a. Detergent: causes plasma membrane to precipitate out of solution

b. Meat tenderizer: strips protein from DNA

c. Ethanol: precipitates DNA

8. What is the purpose of gel electrophoresis? Gel electrophoresis separates molecules on the basis of their charge and size.

9. Why does sickle cell-hemoglobin (HBS) migrate slower than normal hemoglobin (HbA) during gel electrophoresis? Sickle cell hemoglobin contains nonpolar valine instead of polar glutamine in one location.

10. Why are red blood cells sickle shaped in a person with sickle-cell disease? The amino acid difference causes a polypeptide chain in hemoglobin to pile up as firm rods that push the plasma membrane, deforming the normal red blood cell shape.

Laboratory 12 Evidences of Evolution

(LM pages 149–64)

Eleventh Edition Changes

This lab remains essentially unchanged from the previous edition.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

All Exercises (Optional)

_____ “Evidence for Evolution” video (Carolina 49-2200V)

12.1 Evidence From the Fossil Record (LM pages 151–153)

_____ cenozoic era fossil collection

_____ mesozoic era fossil collection

_____ paleozoic era fossil collection

12.2 Evidence From Comparative Anatomy (LM pages 154–158)

skeletons or mounted limbs of:

_____ frog

_____ lizard

_____ bird (e.g., pigeon)

_____ bat

_____ cat

_____ human, adult (see Carolina’s “Skeletons: Human” section)

_____ chimpanzee

Comparison of Vertebrate Embryos (LM page 159)

_____ slide, prepared: chick embryo, 72-hour

_____ slide, prepared: chick embryo, 96-hour

_____ slide, prepared: pig embryo (8–12 mm)

_____ microscope, stereomicroscope

_____ lens paper

12.3 Molecular Evidence (LM pages 160–162)

_____ Immunology and Evolution Experiment kit (Lab-Aids, Inc.)

_____ stirring rod, plastic; or toothpicks

This kit tests the similarities of animal antigens. Students are supplied with antibodies that supposedly react against human antigens. The relatedness of an animal to humans is judged by the degree to which another organisms antigens also react to the prepared antibodies.

EXERCISE QUESTIONS

12.1 Evidences from the Fossil Record (LM pages 151-53)

Geologic Timescale (LM pages 151–153)

Divisions of the Timescale (LM pages 151-152)

List the four eras in the timescale, starting with Precambrian time: Precambrian, Paleozoic, Mesozoic, Cenozoic

How to Read the Timescale (LM page 151)

1. Why do you read the timescale starting at the bottom? The earliest dates are at the bottom.

2. During the Mesozoic era and the Jurassic period, the first flowering plants appear. How many million years ago was this? 199.6--145.5

3. How do you know that the plants in this forest were not flowering trees as most of our trees are today? Flowering trees had not evolved yet. What type animal was diversifying at this time? Amphibians

4. During what period and epoch did primates appear? Tertiary, Paleocene During what period and epoch did hominids appear? Tertiary, Pliocene What period and epoch is the age of Homo sapiens? Quaternary, Pleistocene

Dating Within the Timescale (LM pages 151–152)

Why wouldn’t you expect to find human fossils and dinosaur fossils together in rocks dated similarly? Humans had not evolved yet.

Limitations of the Timescale (LM pages 152)

Which of the animals shown in Figure 12.1 suffered the most during the P-T (Permian-Triassic) extinction? Poriferans (sponges)

Which of the animals shown in Figure 12.1 became extinct during the K-T extinction? Ammonoids and dinosaurs

Fill in the eras on the lines provided in Figure 12.1. Paleozoic, Mesozoic, Cenozoic

Observation: Fossils (LM pages 152–153)

1. One possible reason the Cambrian might be rich in fossils is that organisms before this time did not have shells and bones.

2. These fossils date back to which era and period? Answers will vary according to the kit .Fill in the title of Table 12.2

Table 12. 2 Invertebrate Fossils from the Era Period

Answers will vary according to the kit.

3. Which of the fossils available to you are vertebrates? Answers will vary according to the kit. Fill in Table 12.3 according to sequence of the time from the latest (top) to the earliest (bottom).

Table 12.3 Vertebrate Fossils

Answers will vary according to type of fossils observed.

4. Which of the fossils available to you are plants? Answers will vary according to the kit. Fill in Table 12.4 according to the sequence of the time frames from the latest to the earliest.

Table 12.4 Plant Fossils

Answers will vary according to type of fossils observed.

12.2 Evidence From Comparative Anatomy (LM pages 154–159)

Comparison of Adult Vertebrate Forelimbs (LM pages 154–155)

Observation: Vertebrate Forelimbs (LM page 154)

3. Fill in Table 12.5 to indicate which bones in each specimen appear to most resemble the ancestral condition and which most differ from the ancestral condition.

4. Relate the change in bone structure to mode of locomotion in two examples.

Example 1: Bat. The radius is long, relative to the humerus. The phalanges are extremely long, and the bat’s skin is stretched out over the forelimb forming a wing for flying.

Example 2: Humans. Because humans walk upright, their forelimbs are no longer used to bear weight. Their long upper limbs have carpals, metacarpals, and phalanges that are modified for object manipulation. The shape and angle of articulation of the first digit (the thumb) are particularly noteworthy. This opposable thumb allows for maximum manipulation. These capabilities could not be so specialized if humans were quadrupeds.

Table 12.5 Comparison of Vertebrate Forelimbs*

Animal Bones That Resemble Bones That Differ From

Common Ancestor Common Ancestor

Frog h, m u, r, c, p

Lizard h, u, r, c, m p

Bird h, u, r c, m, p

Bat h u, r, c, m, p

Cat h, c, m, p u, r

Human u, r, c, m, p h

* This comparison is subjective, and student answers will vary.

Conclusion: Vertebrate Forelimbs (LM page 154)

• Vertebrates are descended from a common ancestor, but they are adapted to various ways of life and this causes the bones to vary from one to the other.

Comparison of Chimpanzee and Human Skeletons (LM pages 156–158)

Observation: Chimpanzee and Human Skeletons (LM pages 157–158)

Posture (LM page 157)

2. (Referring to comparison of spines): How does this contribute to an erect posture in humans? Allows the weight to be balanced above the pelvis.

6. In which animal is the big toe opposable? chimpanzee How does an opposable toe assist chimpanzees? Allows them to grasp tree limbs with feet. Which foot has an arch? humans How does an arch assist humans? Helps them walk erect. Record your observations in Table, 12.6

7. How does the difference in the position of the foramen magnum, a large opening in the base of the skull for the spinal cord, correlate with the posture and stance of the two organisms? In a human, the foramen magnum is placed almost in the bottom center of the skull; in the chimpanzee, the opening is well to the rear. Humans walk upright, and chimpanzees use all four limbs for walking.

Table 12.6 Comparison of Chimpanzee and Human Postures

Skeletal Part Chimpanzee Human

Head and torso Thrust forward over hips and legs Balanced over hips and legs

Spine Short and stiff Long and curved

Pelvis Long and narrow Broad and short

Femur Straight, no angle Angled between articulations

Knee joint Femur about the same size top and Femur larger at bottom

bottom. Tibia about the same size Tibia larger at top top and bottom

Foot

opposable toe Yes No

arch No Yes

Conclusion: Chimpanzee and Human Skeletons (LM page 157)

• Do your observations show that the skeletal differences between chimpanzees and humans can be related to posture? yes Explain. All changes noted in human skeleton assist in walking erect.

Facial Features (LM page 158)

1-4 Record your observations in Table 12.7.

Table 12.7 Facial Features of Chimpanzees and Humans

Feature Chimpanzee Human

1. Supraorbital ridge more thick not as thick

2. Slope of frontal bone slope no slope

3. Teeth large smaller

4. Chin projects does not project

Conclusion: Facial Features (LM page 158)

• Do your observations show that diet can be related to the facial features of chimpanzees and humans? yes Explain. Chimpanzees eat more plant material than humans, and humans eat more meat than chimpanzees. Plant material is hard to chew and requires larger teeth facial muscles.

Comparison of Vertebrate Embryos (LM page 159)

Observation: Chick and Pig Embryos (LM page 159)

2. List five similarities of the embryos.

a. General shape of presumptive head

b. Shape and orientation of limb buds

c. Presence and shape of the tail

d. Appearance of somites on dorsal surface

e. Placement of the eyes in relationship to the rest of the head

Conclusion: Vertebrate Embryo (LM page 159)

• Vertebrate embryos resemble one another because all vertebrates have a common ancestor that developed in a particular way.

12.3 Molecular Evidence (LM pages 160–162)

Protein Differences (LM page 160)

Are the number of amino acid changes shown in Figure 12.6 consistent with the history of life as revealed by the fossil record and shown in the geolotical timescale (Table 12.1). yes

Protein Similarities (LM pages 161–162)

Experimental Procedure: Protein Similarities (LM pages 161–162)

5. Describe what you see. A distinct cloudiness or precipitate forms.

7. At the end of 10 and 20 minutes, record the amount of precipitate in each of the six wells in Figure 12.8.

[pic]

Conclusions: Protein Similarities (LM page 162)

• The last row of Figure 12.8 tells you that the test serum in well 3 is from a human. How do your test results confirm this? Because it has the same amount of precipitate as

well 6.

• Aside from humans, the test sera (supposedly) came from a pig, a monkey, an orangutan, and a chimpanzee. Which is most closely related to humans—the pig or the chimpanzee? chimpanzee

• Based on how closely the amount of reaction (precipitate) matches that of a human, fill in the names of the animals in the last row in Figure 12.8. See the figure. Use this chemical reaction to provide evidence of evolution. Closely related animals share the same chemical reactions.

• Molecular evidence shows us that of the vertebrates studied, chimpanzees and humans are most closely related.

12.4 Summarizing the Evidences of Evolution (LM page 163)

Evidence from the Fossil Record (LM page 163)Fossils are the remains of past life.

1. Fossils can be arranged in a sequential manner because organisms evolved sequentially.

2. Younger fossils and not older fossils are more like living organisms

3. In short, the fossil record shows evolution has occurred.

Evidence from the Comparative Anatomy (LM page 163)

1. The similarity between the bones in all vertebrate forelimbs and a common ancestor shows that today’s vertebrates are all related because they have a common ancestor. However, the various vertebrates are adapted to various ways of life.

2. A contrast in skeletons shows that the posture differences are due to the erect posture of humans as opposed to chimpanzees. A contrast in skulls shows that the facial differences are due to a difference in diet.

3. The similarity in the appearance of vertebrate embryos also shows that today’s vertebrates are related and can trace their ancestry to a vertebrate that had gill pouches.

Molecular Evidence (LM page 163)

1. In this section you learned that due to common descent, organisms share the same proteins (e.g. cytochome c). However, distantly related organisms have more amino acid differences than closely related organisms.

2. Two closely related organisms share the same antigen-antibody reactions.

LABORATORY REVIEW 12 (LM page 164)

1. List three types of evidence that various types of organisms are related through common descent. fossil record, comparative anatomy, molecular evidence

2. Why would you expect a fossil buried millions of years ago to not look exactly like a modern-day organism? Evolution has occurred.

3. What accounts for many of the skeletal difference between chimpanzees and humans? Only humans naturally walk erect and chimpanzees largely eat plant material.

4. If a characteristic is found in bacteria, fungi, pine trees, snakes, and humans, when did it most likely evolve? This characteristic most likely evolved in bacteria. Why? Bacteria preceded all these organisms.

5. What are homologous structures, and what do they show about relatedness? Homologous structures have similar anatomy because they are derived from a common ancestor. Organisms are related when they have a common ancestor.

6. The development of reptiles, chicks, and humans shows similarities. What can we learn from this observation? They are related through a common ancestor.

7. What do DNA mutations have to do with amino acid changes in a protein? Mutations are changes in DNA nucleotide base sequences and this sequence determines the sequence of amino acids in a protein.

8. How did antigen-antibody reactions help determine the degree of relatedness between species in this laboratory? The relatedness of an animal to humans was judged by the degree to which they shared the same antigen-antibody reaction.

9. Using plus (+) symbols, show the amount of reaction you would expect when a pig, monkey, and chimpanzee are tested for the same human antibody-antigen reaction. pig +, monkey ++, chimpanzee +++

10. Define the following types of evidence for evolution:

fossil: Any past evidence of an organism that has been preserved in the Earth’s crust. Fossils show that organisms have changed (evolved) over time.

common descent: Descent from a common ancestor. The concept of evolution is based on common descent.

comparative anatomy: Comparison of an organisms structure or development with other organisms. Similarity in anatomy or development shows which organisms are related.

adaptation: An organism’s modification in structure, function, or behavior suitable to the environment. Differences in adaptations support evolution.

molecular: Molecules found in organisms such as DNA and ATP. Similarity of these molecules supports evolution.

Laboratory 13 Natural Selection

(LM pages 165–174)

Eleventh Edition Changes

This lab was rewritten to stress gene pool changes during the process of natural selection. The introduction reviews the process of natural selection with pictorial examples of adaptation. In Section 13.1 the lab walks students through the Hardy Weinberg law and an illustration shows the five ways the law states evolution can occur. In Section 13.2, examples of selective agents are given before students mimic the process of natural selection. Students also mimic the process of the genetic drift founder effect as before. The author has modified a virtual lab on the Biology website called Knocking Out Genes to create a virtual lab experience that uses the scientific process to study the role of mutations in Natural Selection Note also a virtual lab called Natural Selection that is described on page 166.

New or Revised Figures: 13.1 Adaptations; 13.2 Agents of evolutionary change; 13.3 Selective agents.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

13.1 Hardy-Weinberg Law (LM pages 166–168)

The Parent Population (LM pages 167–168)

_____ PTC taste paper (Carolina 17-4010)

_____ calculators (or have students bring their own)

PTC (LM page 167). Seven out of ten persons can taste phenylthiocarbamide (PTC). Concern that PTC paper is carcinogenic has arisen in recent years. The following statement is from the Flinn Scientific, Inc. catalog:

"PTC paper is paper soaked with phenylthiocarbamide (PTC). The LD50 for PTC is 3.4 mg/kg. Such a low LD50 suggests this is a very toxic substance. The solution to make the taste test paper contains approximately 500 mg of PTC per liter of water. Through some very crude arithmetic, we calculate that each strip of PTC paper would contain approximately 3/10 of a milligram of PTC. A student, using our crude arithmetic, would have to ingest 500 2-inch x 1/4 inch strips of PTC paper to reach the LD50 for a student weighing about 50 kilograms. You will have to judge if you wish to use this taste test paper."

EXERCISE QUESTIONS

13.1 Hardy-Weinberg Law (LM pages 166–168)

The Parent Population (LM page 167)

Observation: Gene Pool Frequencies of Parent Populations (LM pages 167--168.

Are you homozygous recessive? Depends on student.

Table 13.1 Parental (P) Generation

Genotypes Genotype Frequency

Homozygous recessive (q2) 0.25

Homozygous dominant (p2) 0.25

Heterozygous (2pq) 0.50

Calculations: Homozygous recessive (q2) is determined by observation (proportion of students who cannot taste PTC). However, if q2 is .25 then q = .5. Since q plus p = 1, p = .5. Now it is possible to calculate p2 and 2pq = .50

13.2 Natural Selection and Genetic Drift (LM pages 168--171)

Natural Selection (LM pages168--169)

Experimental Procedure: Natural Selection (LM pages 168--69)

2. Write down your genotype which is your initial parental (P) generation genotype. Note: According to Table 13.1, out of a class of 36 students, 9 students are tt and cannot taste PTC; therefore, 9 students should be assigned the genotype TT and 18 students should be assigned the genotype Tt.

4., 6., and 7. Answers will vary according to the couple.

8. Fill in Table 13.3. The second column contains the number of students with each genotype. The third column is simply the proportion (written as a decimal) of students with each genotype

Table 13.3 F5 Generation

Genotypes Number of Students Genotype Frequencies

Homozygous recessive (q2) 0 0.00

Homozygous dominant (p2) 10 0.38

Heterozygous (2pq) 16 0.62

Note: Actual results will depend on class data

9. Compare Table 13.3 with Table 13.1 Do your results show that evolution occurred? The genotype frequencies have changed and therefore evolution has occurred. Was evolution due to natural selection? Yes, because tasters of PTC did not survive; the selective agent PTC chose the people who could taste PTC to survive and reproduce. More and more of the population can taste PTC.

Genetic Drift (LM pages 169–170)

Founder Effect (LM page 170)

Experimental Procedure: Founder Effect (LM page 170)

1. How many persons are in the new population? Example: 18 Record your genotype here: Depends on student.

2. Complete Table 13.5. Students complete Table 13.5 as they did Table 13.2.

4. After five generations, fill in Table 13.6. Students complete Table 13.6 as they did Table 13.4.

5. Compare Table 13.6 to Table 13.5. Do the results suggest that the size of the population affects genetic equilibrium and that genetic drift has occurred? yes Explain. Genetic drift occurs because the “founders” represent only a fraction of the total genetic diversity of the original gene pool. Also, a small population is subject to genetic drift more than a large population.

Genetic Drift (LM pages 169--171)

Founder Effect (LM pages 170--171)

1. How many persons are in the new population? Should be about ½ of the original P population. Use an arbitrary way to divide the students such as counting off using only the numbers 1 and 2 or dividing the class by a line down the middle

2. As before (see Table 13.1) simply calculate the proportion of student with each genotype to arrive at the current genotype frequencies.

Table 13.4 Parental (P) Generation

Genotypes Number of Students Genotype Frequencies

Homozygous recessive (q2) 5 0.22

Homozygous dominant (p2) 5 0.22

Heterozygous (2pq) 8 0.54

Note: This is a hypothetical example.

3. Compare Table 13.4 to Table 13.1. Do you expect the founder effect to occur? Yes, because the founder population has different genotype frequencies from the original P population Table 13.1.

4. Now proceed as before except all members of the population can reproduce. The answers vary according to the couple.

Table 13.5 F5 Generation

Genotypes Number of Students Genotype Frequencies

Homozygous recessive (q2) 2 0.11

Homozygous dominant (p2) 6 0.33

Heterozygous (2pq) 10 0.56

Note: This is a hypothetical example.

5. Compare Table 13.5 to Table 13.4. Do the results suggest that the size of the population affects a Hardy-Weinberg equilibrium and that genetic drift has occurred? Yes, the gene frequencies have simply drifted from the original gene frequencies.

Virtual Lab: Role of Mutations in Natural Selection (LM page 171)

Hypothesize whether you expect or do not expect the mutant plant to perform better than the wild-type plant under stressful environmental conditions. Mutant plants will perform better than wild-type plants under certain stressful environmental conditions.

Experiment 1 Optimum Conditions (LM page 171)

Confirm that both the wild-type plants and the mutant plants look the same —that is normal. Yes, both the wild-type plants and the mutant plants look normal under optimum conditions.

Experiment 2 High Salinity (LM page 172)

Table 1 Appearance of Plants Under Various Environments

Environmental Conditions

Plant High Salinity Drought UV Exposure

Wild type curled-up leaves curled-up leaves slightly thinner

green light green lighter green

Mutant curled-up leaves curled-up leaves thicker leaves

green light green darker green

Conclusion: Do your results (support or not support) your original hypothesis? My results do not support the hypothesis that mutant plants will perform better than wild-type plants under certain stressful environmental conditions because the mutant plants did not look normal under any of the environmental conditions.

Questions

1. Criticize the experimental design if you had kept the wild-type plants under optimum conditions but had changed the environmental conditions for mutant plants? You would not be able to tell if the mutant plants survived better than the wild-type plants.

2. Did this virtual laboratory show that mutations lead to adaptations when the environment changes and mutant plants are exposed to a new selective agent? no

If not, what advice do you have for the researchers based on the scientific method? Suggest that they keep on experimenting, perhaps they will find some other extreme environment in which the mutant plants will do better than the wild-type plants.

Laboratory Summary (LM page 173)

1. How did we recognize microevolution had occurred? Change in genotype frequency signified evolution has occurred.

2. What was this selective agent? Ability to taste PTC

3. Then we discovered that genetic drift, such as with the bottleneck and founder effect, occurs when a few of the original population reproduce together.

4. Finally, we carried out an experiment to determine if mutations can lead to natural selection when organisms are placed in an extreme environment that contains a selective agent.

LABORATORY REVIEW 13 (LM pages 173--174)

1. List the conditions necessary for the Hardy-Weinberg equilibrium. The conditions necessary for Hardy-Weinberg equilibrium are: there must be no mutation, no gene flow, only random mating, no genetic drift, and no natural selection.

2. What is the evidence that evolution is occurring in any given population? Change in genetic frequencies.

3. Assume a Hardy-Weinberg equilibrium.

a. If p = 0.8, what are the gene pool frequencies of a population?

p = dominant allele = 0.8

q = recessive allele = 0.2

q2 = homozygous recessive = 0.04 = 4%

p2 = homozygous dominant = 0.64 = 64%

2pq = heterozygous = 0.32 = 32%

b. What methodology was used in this laboratory to bring about natural selection? Use of a selective agent: PTC.

c. What methodology was used in this laboratory to bring about genetic drift? An original population (the class) was divided into two smaller populations. Explain. “Founders” most likely have a gene pool in which gene frequencies differ from the original population, and also a small population is subject to genetic drift more than a large population.

4. Natural selection results in bacteria adapting (becoming resistant) to the presence of antibiotics.

a. To what environment were the bacteria exposed to? presence of antibiotic

b. How do you know that at least some of the bacteria were adapted to this environment? They are able to grow in the presence of antibiotics.

5. How does the process of genetic drift differ from natural selection? As a result of natural selection certain phenotypes are selected to reproduce. Genetic drift occurs when certain phenotypes reproduce by chance.

6. How does the result of genetic drift differ from natural selection? Genetic drift does not necessarily result in adaptation to the environment.

7. Assume a Hardy-Weinberg equilibrium. If 49% of the population had a recessive phenotype for a trait, what does p equal? .49 = q2 so q = .7 and

p = 1–q = .3

8. How is the bottleneck like the founder effect? Both occur when only some portion of the parental population reproduce with one another.

9. Even though it is the phenotype that is exposed to a particular environment, the genotype frequencies of a population can change over time. Explain. When a phenotype survives so does its genes; it is the genes that are passed on the next generation.

10. Natural selection does not cause mutations to occur, yet mutations do play a role in the natural selection process. Explain. Mutations are the ultimate source of gene changes that lead to phenotypes that may survive better in new environments.

Laboratory 14 Bacteria and Protists

(LM pages 175–192)

Eleventh Edition Changes

This laboratory is improved by the addition of a virtual lab experience that gives the classification of bacteria an evolutionary emphasis and asks students to complete an evolutionary tree based on available data.

New and Revised Figures. 14.6 Gloeocapsa; 14.13 Diatoms

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Living material. If desired, order cultures of bacteria for Gram Stain and Colony Morphology and Cyanobacteria. Order cultures for observation of these protists: Green Algae, Brown Algae, Diatoms, Dinoflagellates, and Plasmodial Slime Molds. Order in advance and specify date of delivery.

Incubation. 48 hours prior to use for Observation: Colony Morphology.

Fresh material. Observation: Pond Water requires that students examine pond water. See this guide Lab 2 for how to obtain and maintain.

1. Bacteria (LM pages 176–183)

All Exercises

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand

Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other

clothing protection

_____ microscopes, compound light, oil immersion will be used

_____ lens paper

_____ microscope slides and cover slips

Gram Stain (LM pages 176--177)

_____ Gram Stain Kit (Carolina ER–82-1050)

_____ Gram Stain slide set

_____ bacteria, live cultures of representative types (Carolina’s “Living

Organisms: Bacteria Cultures” catalog section)

_____ water

_____ wax pencil

_____ bacterial loop (metal)

_____ Bunsen burner

_____ blotting paper

Demonstration (LM page 177). Instructor should demonstrate proper aseptic technique and inoculation technique.

Colony Morphology (LM page 178)

_____ bacteria, live cultures of representative types (Carolina’s “Living Organisms: Bacteria Cultures” catalog section)

_____ nutrient agar plates, sterile (Carolina 82-1862)

_____ nutrient agar plates, inoculated (incubated)

_____ incubator

Biohazard waste container. Because of increased awareness of hazards connected with bodily fluids, a biohazard waste container for swab disposal should be used.

Agar plates (LM page 178). Ready-to-use nutrient agar plates (Carolina 82-1860 or 82-1861) for the culture of bacteria can be ordered. See the section of the Carolina Biological Supply catalog entitled “Microbiological Media/Prepared Media.” If you wish to prepare your own plates, purchase nutrient agar (Carolina 78-5301), prepare according to package directions, and pour into sterile petri dishes.

Bacterial cultures (LM page 178). A variety of bacteria are available for the live cultures. Choose representative types from Carolina Biological Supply catalog’s “Bacteria—Single Cultures” section. Inoculate the demonstration agar plates with the cultures approximately 48 hours prior to use. Incubate in a warm incubator. After plate surfaces are covered with the organism, store the plates in a refrigerator until use. All bacteria and inoculated plates should be disposed of according to the guide lines established by your institution.

Shape of Bacterial Cell (LM page 179)

_____ slides, prepared: bacteria (coccus, bacillus, spirillum) (Carolina 29-3964)

Prepared slides (LM page 179). Carolina Biological Supply Company has an immense variety of prepared slides available. For prepared slides of bacteria, select representatives of bacillus (rod-shaped), coccus (sphere-shaped), and spirillus (spiral-shaped) bacteria.

Cyanobacteria (LM pages 182–183) Carolina Biological Supply Company harepresentative cyanobacteria available as live cultures or prepared slides.

_____ Gloeocapsa, live culture or prepared slide

_____ Oscillatoria, live culture or prepared slide

_____ Anabaena, live culture or prepared slide

14.2 Protists (LM pages 183–191)

Photosynthetic Protists (LM pages 184–188) Carolina Biological Supply Company has representative photosynthetic protists available as live cultures or prepared slides.

_____ Spirogyra, live culture or prepared slide

_____ Volvox, live culture or prepared slide

_____ Fucus, preserved specimen

_____ Sargassum, preserved specimen

_____ Laminaria, preserved specimen

_____ red algae, preserved specimens

_____ diatoms, live culture or prepared slide

_____ dinoflagellates, live culture or prepared slide

Heterotropic Protists (LM pages 189--190) Carolina Biological Supply Company has representative heterotrophic protists available as live cultures or prepared slides. Slime Molds (LM page 191)

_____ agar plate inoculated with Physarum or Physarum

Culture Kit

_____ microscopes, stereomicroscope

_____ lens paper

EXERCISE QUESTIONS

14.1 Bacteria (LM pages 176–183)

Gram Stain (LM pages 176--177

Observation: Gram Staining ((LM pages 176--177)

3. Why do the Gram-positive cells in Figure 14.2c appear purple? The thick peptidoglycan layer of Gram-positive organisms retain the purple crystal violet-iodine dye. Why do the Gram-negative cells appear reddish-pink? The thin peptidoglycan layer of Gram-negative organisms does not retain the crystal violet-iodine stain upon decolorization. They only retain the safranin counterstain, which is pink.

Experimental Procedure: Gram Stain (LM page 177)

Table 14.1 Gram Staining

Table results will depend on the organisms being tested.

Conclusions: Gram Stain (LM page 177)

• What do you know about the cell wall of Gram-positive organisms? Gram-positive bacteria have thick peptidoglycan layers outside their plasma membrane and these thick layers retain the purple Gram stain.

• What do you know about the cell wall of Gram-negative organisms? Gram-negative bacteria have thin peptidoglycan layers between the plasma membrane and an outer membrane. The thin layer does not retain the purple Gram stain.

Colony Morphology (LM pages 178–179)

Observation: Colony Morphology (LM pages 178–179)

Table 14.2 Agar Plates

Table data will depend on the bacteria cultures viewed.

3. Describe your plate. The description will depend on the culture observed.

4. Describe your plate. The description will depend on the culture observed.

Shape of Bacterial Cell (LM page 179)

Observation: Shape of Bacterial Cell (LM page 179)

1. View the microscope slides of bacteria on display. What magnification is required to view bacteria? 1,000X

3. Do any of the slides on display show bacterial cells with endospores? This depends on the type of bacteria present. Of the bacteria the students are most likely to see, the genus Bacillus is most likely to have endospores. If present, the endospore appears as a swelling at one end of the rod. What is an endospore, and why does it have survival value? An endospore is a bacterium that has shrunk its cell, rounded up within the former plasma membrane, and secreted a new and thicker cell wall when in an unfavorable environment for growth.

Virtual Lab Classifying Bacteria Using Biotechnology (LM page 180)

Record your data in Table 1.

Table 1 Classifying Bacteria*

Morphological Data Molecular Data Name of Organism

Appearance Gram Stain G-C percentage

cocci chains + 30% (low) Streptococcus mutans

oval single -- 67% (high) Bordatella pertussis

cocci cubicle pkgs + 45% (low) Sarcina sp.

cocci paired -- 75% (high) Velonella sp

rod single -- 51% (high) E. coli

cocci clusters + 25% (low) Staphylococcus aureus

spirilli single -- 35% (low) Vibreo cholerae

cocci chains + 67% (high) unknown

* Not necessarily in the order they were viewed.

1. What staining procedure allows bacteria to be divided into two main groups? Gram stain What are the two main groups? Gram positive and Gram negative

2. Complete the evolutionary tree.

[pic]

Cyanobacteria (LM pages 182–183)

Observation: Cyanobacteria (LM page 182)

Gloeocapsa (LM page 182)

2. What is the estimated size of a single cell? 5 mm

Oscillatoria (LM page 182)

2. If you have a living culture, are oscillations visible? The answer will depend on the sample used.

Anabaena (LM page 183)

2. If you have a living culture, what is its color? blue-green

14.2 Protists (LM pages 183–191)

Photosynthetic Protists (LM pages 184–188)

Observation: Green Algae (LM pages 184–185)

Spirogyra (LM page 184)

How do you think Spirogyra got its name? The chloroplasts of Spirogyra are ribbon-shaped and appear in the form of a spiral in the individual cells of a filament.

Green Algae Diversity (LM page 185)

Table 14.3 Green Algae Diversity

Have students include the green algae pictured in the lab manual and any others available. Have students decide if any of the photos on page 183 are green algae.

Observation: Brown Algae (LM pages 186–187)

Table 14.4 Brown Algae

Have students include the brown algae pictured in the lab manual (Fig. 24.1) and any others available. Have students decide if any of the photos on page 183 are brown algae.

Observation: Red Algae (LM page 187)

Table 14.5 Red Algae

Have students include the green algae pictured in the lab manual and any others available.

Observation: Diatoms (LM page 188)

Describe what you see. Most diatoms are unicellular; their cell wall composed of silica is often likened to a pill box: the lower half fits inside the upper half.

Observation: Dinoflagellates (LM page 188)

Describe what you see. These flagellated protists exhibit a great diversity of form. The lab manual features Gonyaulax

Heterotrophic Protists (LM pages 189–190)

In general, how do sporozoans differ from the protozoans shown in Figure 14.16? They are nonmotile and form spores during their life cycle.

Observation: Heterotrophic Protists (LM page 190)

Table 14.6 Heterotrophic Protists

Genus Name Structures for Locomotion Observation

Amoeba pseudopods depends on culture/student

Paramecium cilia depends on culture/student

Trypansoma flagella depends on culture/student

Plasmodium none

Slime Molds (LM page 191)

Observation: Plasmodial Slime Molds (LM page 191)

2. Describe what you see. The description will depend on the culture observed. Upon cursory examination, slime molds look like molds. Plasmodial slime molds exist as a plasmodium, a multinucleated, fan-shaped, cytoplasmic mass enveloped by a slime sheath.

LABORATORY REVIEW 14 (LM page 192)

1. What are two major differences between prokaryotic and eukaryotic cells? Prokaryotes lack the nucleus of eukaryotes and they lack membranous organelles such as mitochondria.

2. What chemical associated with the cell wall gives Gram-positive bacteria their final color? peptidoglycan

3. a. In what ways are cyanobacteria like land plants? They photosynthesize like other algae and plants.

b. How are they different? They are prokaryotes.

4. How can cyanobacteria, in contrast to saprotrophic bacteria, live in an environment that lacks organic nutrients? They are photosynthetic and can utilize nitrogen from air.

5. List the two major types of nutrition and give an example of a protist with this type of nutrition.

Nutrition Examples

a. Heterotrophic Protozoan

b. Photosynthetic Algae

6. List the three different structures for locomotion found among protozoans, and name an organism that utilizes each structure.

Structure Organism Name

a. Pseudopods Amoeba

b. Cilia Paramecium, Stentor, Vorticella

c. Flagella Trypanosomes (Trypanosoma, Trichomonas)

7. Name a protist that has both chloroplasts and flagella. Euglena Which makes this protist algal like. chloroplasts Which makes it protozoan like? flagella

8. Name two characteristics of slime molds by stating their mode of nutrition and how they survive conditions unfavorable for growth. Slime molds are heterotrophic by ingestion. They survive unfavorable conditions by forming sporangia and producing spores.

9. Complete the following sentence: Plasmodial slime molds usually exist as a plasmodium, multinucleated mass of cytoplasm.

10. Could an ecosystem that contains only protists remain in continued existence? possibly Photosynthesis is no problem considering how many protists are able to photosynthesize. However, decomposition may lag behind since you would have to depend on water molds in aquatic habitats and slime molds in terrestrial ones. Whether they could keep up is not known.

Laboratory 15 Fungi

(LM pages 193–204)

Eleventh Edition Changes

Introduction was rewritten to incorporate the latest approach to fungal evolution.

New or Revised Figures: Yeast illustration on page 197 now includes a micrograph.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Fresh material. Observation: Yeast requires dry baker's yeast; Observation: Cup Fungi and Observation: Mushrooms requires whole edible mushrooms for examination.

Culture or growth required. Observation: Black Bread Mold (LM page 196): Rhizopus on bread and in petri dishes;

Living material. Petri dishes of Rhizopus; Observation: yeast (LM page 197) requires Saccharomyces (yeast) culture; and Observation: Conidiophores and Conidiospores

(LM page 199) requires Aspergillus culture. Incubation may be required for these.

Additional study aids. Further understanding of fungi may be gained by viewing “The Biology of Fungi,” BioMEDIA Assoc., #BO129

All Exercises

_____ stereomicroscope

_____ compound light microscope

_____ lens paper

_____ slides and cover slips

_____ dropper bottles of distilled water

_____ dropper bottles of methylene blue

Methylene blue stain (LM page 199). To prepare a 1.5% stock solution of methylene blue stain, use 1.5 g of methylene blue dye powder per 100 mL of 95% ethyl alcohol. Dilute one part stock solution with nine parts water for laboratory. Avoid contact with this chemical. If any should spill on the skin, wash the area with mild soap and water.

15.1 Zygospore Fungi (LM pages 194–196)

Black Bread Mold (LM pages 194--196)

_____ white bread, fresh (without preservatives), one loaf for approx. 200

students

_____ petri dishes

_____ Rhizopus, whole mount slide (including sexual stages) (Carolina 29-7770,

-7776)

Growth of Rhizopus on white bread in petri dish (LM page 194). Purchase white bread containing no preservatives, because Rhizopus is not likely to develop on bread containing preservatives. Place a small amount of preservative-free white bread in the petri dish. Add one drop only of distilled water. (Any more than one drop, and yeast growth will be promoted instead of Rhizopus growth.) Sprinkle a small amount of dust from the corners of the room on the bread. Rhizopus growth occurs within two to three days and is at its peak within a week. Have students observe Rhizopus with a stereomicroscope. Rhizopus live culture also can be purchased if desired (Carolina 15-6222).

15.2 Sac Fungi (LM pages 196–199)

Yeasts (LM page 199)

_____ Saccharomyces culture or prepare using bakers yeast.

_____ slide, prepared: Schizosaccharomyces, showing ascospores, whole mount

(Carolina 29-8016)

Yeast culture (LM page 197). Add one packet of dry baker’s yeast to 100 mL of 5% sucrose

(5 g sucrose in 100 mL of distilled water). Incubate overnight at 37°C.

Slides (LM pages 197--199). Slides of yeast and of Peziza ascocarps are also available from Ripon Microslides and Triarch, Inc.

Cup Fungi (LM pages 197–198)

_____ cup fungi, Peziza, preserved

_____ slide, prepared: Peziza apothecium, cross section (Carolina 29-7980)

_____ morel, Morchella, preserved, plastomount, or prepared slide (Carolina 29-7962)

Conidiospores (LM page 199)

_____ Aspergillus culture (Carolina 15-5935, -5946) in petri dishes

_____ tape, clear

_____ slide, prepared: Penicillium, whole mount

_____ slide, prepared: Aspergillus, whole mount

Prepared slides (LM page 199). As an alternative to individual slides of Penicillium and Aspergillus, you might want to obtain a Penicillium slide for each student and a demonstration slide of Aspergillus.

Penicillium prepared slide (LM page 199). Most prepared slides of Penicillium contain a stained mass of mycelium. The student should locate the periphery of the mass under low power and then switch to high power to observe conidiophores.

15.3 Club Fungi (LM pages 199–201)

_____ mushroom, edible, fresh

_____ slide, prepared: Coprinus mushroom, showing pileus and gills, cross

section (Carolina 29-8176)

15.4 Fungal Diversity (LM page 201)

A whole mount of general fungi types—Rhizopus, Peziza, bracket fungus, mushroom, and wheat rust is available from Wards (56W1200). Carolina Biological Supply has a wide variety of fungi both fresh and preserved specimen, such as

_____ Pilobolus (cap-thrower fungus) (fresh)

_____ Erysiphe (powdery mildew) (fresh, or prepared slide)

_____ Cyathus (bird’s nest fungus) (fresh)

_____ Lycoperdon (puffball) (fresh, or preserved)

_____ Fomes (bracket). Obtain fresh locally if possible.

_____ Ustilago (smut) (fresh, or prepared slide)

_____ Puccinia (rust) (fresh, or prepared slide)

_____ Xylaria (dead man’s fingers)

15.5 Fungi As Symbionts (LM pages 202–203)

Lichens (LM page 202)

Observation Lichens (LM page 203)

(Carolina Biological Supply has a wide variety of lichen specimens and prepared slides.)

_____numbered crustose, fruticose, foliose lichens for students to identify.

_____prepared slide showing photosynthetic cells and fungal hyphae.

Mycorrhizae (LM page 203)

Observation Mycorrhizae (LM page 203)

_____ Carolina Biological Supply has both prepared slides and live specimens

Mycorrhizae If facilities are available, grow a flat of conifer seedlings in advance. Inoculate with mycorrhizae. Seedling root systems can then be examined for the whitish mycorrhizae.

EXERCISE QUESTIONS

15.1 Zygospore Fungi (LM pages 194–196)

Are the nuclei in the mycelium of Rhizopus haploid or diploid? haploid Where does meiosis occur in the Rhizopus life cycle? zygospore undergoes meiosis Where are spores produced? in the sporangium

Black Bread Mold (LM pages 194–196)

3. What structure accounts for the phylum name zygomycota (zygospore fungi)? the zygospore that develops during sexual reproduction

Observation: Black Bread Mold (LM page 196)

1. Do you recognize black bread mold on the bread? Students should be able to see the mold. Describe the mold you see. Students’ answers will vary.

2. Identify the three types of hyphae and the sporangia (black dots). Refer to Figure 15.2 for identification.

3. List the structures you can identify. They may be able to identify the mycelium, rhizoids, stolons, sporangiophores, and/or the sporangium.

4. In the micrograph on the left, label structures seen during asexual reproduction.

1. sporangium; 2. sporangiophore; 3. rhizoid In the micrograph on the right, label structures seen during sexual reproduction. 1. gametangium;

2. zygospore; 3. stolon

15.2 Sac Fungi (LM pages 196–199)

What structure accounts for the phylum name Ascomycota (sac fungi)? the saclike ascus that occurs during sexual reproduction

Yeasts (LM page 197)

Observation: Yeast (LM page 197)

3. Label the vegetative cell, bud, ascus, and ascospore in the following diagram of Saccharomyces. 1. bud; 2. ascus; 3. ascospore

4. How many ascospores are in each ascus? eight ascospores

Cup Fungi (LM pages 198–99)

Observation: Cup Fungi (LM page 199)

Are the ascospores inside or outside the asci? inside The “pits” of a morel are lined with asci.

Conidiospores (LM page 199)

Observation: Conidiophores and Conidiospores (LM pages 198 –99)

1. What do you see? Students’ observations will depend upon the slide used.

4. Label the following diagram of Aspergillus. 1. conidiospores (= conidia);

2. conidiophore; 3. hyphae

15.3 Club Fungi (LM pages 199–201)

What structure accounts for the phylum name Basidiomycota (club fungi)? the club-shaped basidium that occurs during sexual reproduction

Observation: Mushrooms (LM pages 200–201)

3. Can you see individual hyphae in the gills? It will depend upon the slide.

4. Are the basidiospores inside or outside of the basidia? outside

5. Can you suggest a reason for some of the basidia having fewer than four basidiospores? Some of the basidiospores may have dispersed already.

6. What type of nuclear division(s) took place just before the basidiospores were produced? meiosis

7. What happens to the basidiospores after they are released? Windblown to a new location, where they give rise to a mycelium.

15.4 Fungal Diversity (LM page 201)

Notice that the formal name refers to the sexual reproductive structure. Why is each groups so named?

Zygomycota A thick-walled zygospore is formed.

Ascomycota A structure called an ascus that contains eight ascospores is formed.

Basidiomycota A club-shaped basidium that produces basidiospores is formed.

Table 15.2 Fungal Diversity*

Type Common Name Scientific Name Description

Zygomycota Cap-thrower fungus Pilobolus Sporangium flung toward light source

Basidiomycota Bracket Fomes Flat, shelflike basidiocarp

Smut Ustilago Dark, powdery spores

Rust Puccinia Rust-colored, powdery spores

Bird’s nest fungus Cyathus “Eggs” in a nest

Puffball Lycoperdon StemLess basidiocarp

Ascomycota Dead man’s fingers Xylaria Club-shaped; grows in clusters

Powdery mildew Erysiphe Closed asocarp

Truffles Tuber Prune-shaped structure; large spores; grows underground

Scarlet cap Sarcoscypha Red, cup-shaped structures produced in clusters

*Answers will depend on which representative specimens are available.

15.5 Fungi As Symbionts (LM pages 202–203)

Lichens (LM pages 202–203)

Does it seem as if the fungus is dependent on the photosynthetic cell? yes What does this suggest? The fungus needs the algae to grow. Would you describe the relationship as mutualistic or parasitic? mutualistic Explain. The fungus and the alga benefit from each other. It is difficult for each to grow alone.

Observation: Lichens (LM page 203)

1. Observe numbered samples of lichens, and identify each as crustose, fruticose, or foliose. Answers will depend on order of samples set out by instructor.

2. View a prepared slide of a lichen that shows the placement of the photosynthetic cells and fungal hyphae. Describe the placement. Photosynthetic cells are positioned between two layers of fungal hyphae.

Mycorrhizae (LM page 203)

Observation: Mycorrhizae (LM page 203)

What type of mycorrhizae are you observing? It depends upon the slide.

LABORATORY REVIEW 15 (LM page 204)

1. Both fungi and animals are heterotrophs although they are markedly different in how they acquire nutrients. Explain. Fungi are saprotrophs that send out digestive juices into the environment and absorb the resulting nutrients; animals take in whole food and digest it internally.

2. Give the name of the sexual spore produced within or on each of the following structures, and cite the group of fungi.

Type of Spore Group Name

a. Sporangium Zygospore Zygomycota

b. Ascus Ascospore Ascomycota

c. Basidium Basidiospore Basidiomycota

3. The timing of meiosis during sexual reproduction of a fungus ensures that the adult is haploid. Explain Following the union of gametes, meiosis occurs during the production of spores and the spores develop into haploid individuals.

4. Ascomycota produce what type of spore during asexual reproduction? conidiospores

5. Which type of fungi studied is not at all like any other fungus? yeasts Explain. They are unicellular and usually reproduce by budding.

6. A student is observing fruiting bodies. If the spores are projecting from a club-shaped structure, the fungus is a club fungus (basidiomycota). If the spores are produced within a saclike structures, the fungus is a sac fungi (ascomycota).

7. Of what benefit is it to a typical mushroom if the cap opens up when mature? Cap produces spores and in this way spores can be dispersed by wind.

8. Mycorrhizae are sometimes known as “fungus roots.” Explain. Mycorrhizae coat and sometimes penetrate a root.

9. Of what benefit are mycorrhizae to the plant? Mycorrhizae are probably more efficient at passing water and minerals to the plants.

10. Explain the composition of lichens and why this composition is usually considered a mutualistic relationship. Algal cells surrounded by fungal hyphae. Mutualistic because algal cell pass nutrients to fungi and fungi provide water and minerals to the algae.

Laboratory 16 Nonvascular Plants and Seedless Vascular Plants

(LM pages 205–220)

Eleventh Edition Changes

This lab remains the same as the previous edition.

New/Revised Figures 16.15 Cross section of a frond leaflet

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials are used in several exercises.

Special Requirements

Living material, if desired. Section 16.1 Evolution and Diversity of Land Plants: Chara;. Section 16.2 Nonvascular plants: Moss gametophyte, moss sporophyte, moss minimarsh, and Marchantia; Section 16.3 Lycopodium, Equisetum, and assorted ferns and fern minimarsh.

Preserved Specimens, if desired. For 16.2 Seedless Vascular Plants: Lycopodium and Psilotum

16.1 The Evolution and Diversity of Land Plants (LM pages 206–208)

_____ Chara, living

16.2 Nonvascular Plants (LM pages 209–212)

Observation: Moss Gametophyte (LM pages 210–211)

_____ moss gametophyte and sporophyte, living or moss life

cycle plastomount

_____ slide, prepared: moss antheridia and archegonia, whole mount

Observation: Moss Sporophyte (LM page 211)

_____ slide, prepared: moss (Mniumcheck spelling?) sporophyte, longitudinal section

Observation: Liverworts (LM page 212)

_____ liverwort Marchantia, living (Carolina has several slides available.)

3. Seedless Vascular Plants (LM pages 212–219)

Living sets containing Selaginella, Lycopodium, and Equisetum may be available

Observation: Lycophytes (LM page 213)

_____ club moss Lycopodium, living or preserved

_____ slide, prepared: club moss (Lycopodium Strobilus)

_____ spike moss, Selaginella, living

_____ slide, prepared: spike moss (Selaginella Strobilus)

Observation: Pteridophytes (LM page 214)

Whisk Fern (LM page 214)

_____ whisk fern Psilotum, preserved

Horsetails (LM page 215)

_____ horsetail Equisetum, with rhizomes, living

Fern Diversity (LM page 217)

_____ ferns, assorted living (see Carolina’s Plants: Living Organisms, Other

Ferns), or herbarium mounts

Observation: Fern Sporophyte (LM pages 217–218)

_____ fronds, with sori, living (Carolina 15-6902), plastomount

_____ slide, prepared; fern frond leaflet with sori, cross section

Observation: Fern Gametophyte (LM pages 218--219)

_____ fern life cycle plastomount

_____ fern minimarsh

_____ slide, prepared: fern prothallium-archegonia, whole mount

_____ slide, prepared: fern prothallium-antheridia, whole mount

EXERCISE QUESTIONS

16.1 The Evolution and Diversity of Land Plants (LM pages 206–208)

Circle the evolutionary events that led to adaptation of plants to a land existence. Students should circle embryo protection, apical growth, vascular tissue, megaphylls, seeds, flowers, and fruit

Algal Ancestor of Land Plants (LM page 206)

Observation: Chara (LM page 206)

How does it superficially resemble a land plant? Has a stem and branches.

Measure the length of one cell. Will be several cm.

What does it feel like? rough

Conclusions: Chara (LM page 206)

• What characteristics cause Chara to resemble land plants? appearance, protects zygote

• Why are Chara called stoneworts? covered with calcium carbonate deposits

Alternations of Generations (LM page 208)

1. When does meiosis occur in the plant life cycle? Meiosis occurs in the production

of spores. In Figure 16.3.

2. When does meiosis occur in the animal life cycle? Meiosis occurs in the production of gametes.

3. Which generation produces gametes in plants? gametophyte

4. Which generation (sporophyte or gametophyte) is better adapted to a land environment? sporophyte Explain. The sporophyte has vascular tissue, which transports materials to all parts of the plant and provides structural support.

16.2 Nonvascular Plants (LM pages 209-212)

1. Put a check mark beside the phrases that describe nonvascular plants. The phrases that describe nonvascular plants are all in group I.

2. Which two of the features you checked cause nonvascular plants to be low-lying? No vascular tissue to transport water and flagellated sperm that swim to egg.

3. Which listing of features (I or II) would you expect to find in a plant fully adapted to a land environment? group II Explain. All of these features enable the plant to be successful on land.

4. How are windblown spores an adaptation to reproduction on land? Windblown spores, which lack flagella, no longer require water for dispersal.

Observation: Moss Gametophyte (LM page 210)

Living or Plastomount (LM page 210)

Obtain a living moss gametophyte or a plastomount of this generation. Describe its appearance. It has a stemlike structure covered with whorls of leaflike structures.

Microscope Slide (LM page 210)

1. What is the chromosome number (choose 2n or n) of the sperm (see Figure 16.4)? n (haploid) Are the surrounding cells haploid or diploid? n (haploid)

2. What is the chromosome number of the egg? haploid Are the surrounding cells haploid or diploid? haploid Is the sporophyte haploid or diploid? diploid

Observation: Moss Sporophyte (LM page 211)

Living Sporophyte (LM page 211)

2. Bracket and label the gametophyte and sporophyte in Figure 16.7a. Place an n beside the gametophyte and a 2n beside the sporophyte. The gametophyte (n) is the leafy shoot portion; the sporophyte (2n) is the capsule and stalk.

Microscope Slide (LM page 211)

Identify the stalk and the sporangium, where spores are being produced. By what process are the spores being produced? meiosis When spores germinate, what generation begins to develop? haploid gametophyte Why is it proper to say that spores are dispersal agents? Wind can carry the spores some distance.

Observation: Liverworts (LM page 212)

What generation is this sample? gametophyte

16.3 Seedless Vascular Plants (LM pages 212–219)

1. Place a check mark beside the phrases that describe seedless vascular plants. The phrases that describe seedless vascular plants are all in Group I.

2. Which listing (I or II) would you expect to find in a plant fully adapted to a land environment? Group II Explain. The gametophyte lacks vascular tissue; therefore, dependence on the sporophyte that has vascular tissue is beneficial. Protecting sperm from drying out is adaptive to a land environment. Are seedless vascular plants fully adapted to living on land? No. They have flagellated sperm which swim to the egg in external moisture.

Observation: Lycophytes (LM page 213)

4. Label the ground pine in Figure 16.9. 1. strobili; 2. leaves (microphylls)

3. Aerial stem; 4. root

Observation: Pteridophytes (LM pages 214–215)

Whisk ferns (LM page 214)

1. What generation are you examining? sporophyte

2. Label the whisk fern in Figure 16.11b. 1. sporangia; 2. aerial stem;

3. rhizome

Horsetails (LM page 215)

4. Label the horsetail in Figure 16.12. 1. strobilus; 2. branches; 3. node; 4. leaves;

5. rhizome; 6. root

Ferns (LM pages 215–219)

Ferns are largely restricted to moist, shady habitats because sexual reproduction requires adequate moisture. Why? The flagellated sperm require moisture to get to the egg. How do ferns disperse offspring? Ferns disperse offspring with windblown spores.

Observe the ferns on display, and then complete Table 16.1. Types and descriptions of ferns will vary.

Observation: Fern Sporophyte (LM pages 217–218)

Living or Preserved Frond (LM page 217)

What is being produced in the sporangia? spores Given that this is the generation called the fern, what generation is dominant in ferns? sporophyte

Microscope Slide of Sorus (LM page 218)

2. Does this fern have an indusium? Answer will depend on the species observed.

Observation: Fern Gametophyte (LM pages 218–219)

Plastomount (LM page 218)

2. Can you find this structure in your fern minimarsh (if available)? Answers will vary. What is the function of this structure (the prothallus)? The prothallus produces gametes.

Microscope Slide (LM page 219)

2. What is being produced inside the antheridia? sperm When sperm produced by the antheridia swim to the archegonia in a film of water, what results? Fertilization occurs, resulting in a zygote. This structure develops into what generation? sporophyte that becomes independent of the gametophyte

Conclusions: Ferns (LM page 219)

• How are ferns dispersed from one area to another? by windblown spores

• Is either generation in the fern dependent for any length of time on the other generation? no Explain. The sporophyte initially grows from the gametophyte but soon is independent. The gametophyte germinates directly from windblown spores and is also independent.

LABORATORY REVIEW 16 (LM page 220)

1. Name and describe the life cycle of plants. Plants have a two-generation life cycle called alternation of generations. The gametophyte (n) produces gametes by mitosis, while the sporophyte (2n) produces spores by meiosis.

2. Contrast the life cycle of plants to that of animals (e.g., human beings).

Plants Animals

Type of life cycle alternation of generations diploid

Meiosis results in spores gametes

Mitosis occurs when zygote to individual zygote to individual

spore to individual

Compare mosses and ferns by completing the following table:

Dominant Vascular Flagellated Method of

Generation Tissue Sperm Offspring Dispersal

3. Moss Gametophyte No Yes Windblown Spore

4. Fern Sporophyte Yes Yes Windblown Spore

5. Why do some biologists call nonvascular plants the “amphibians of the plant kingdom”? Nonvascular plants must have an aquatic environment to sexually reproduce.

6. Are ferns better adapted to a land environment than mosses? yes Why or why not? Ferns have vascular tissue in the dominant sporophyte.

7. How do both mosses and ferns disperse offspring? Mosses and ferns disperse offspring by windblown spores.

8. How is the gametophyte generation in ferns similar to that of mosses? Both produce flagellated sperm, and neither has vascular tissue in the gametophyte. How is it different (aside from appearance)? The gametophyte generation is dominant in the moss, but is separate and short-lived in the fern.

9. Why is it beneficial to have the sporophyte generation dominant in vascular plants? The sporophyte generation has vascular tissues.

10. Compare and contrast moss and fern sporophytes. The moss sporophyte, a stalk and capsule, is dependent on the gametophyte, which is the dominant generation. The fern sporophyte, the leafy frond, is independent and the dominant generation. Fern sporophyte has vascular tissue and true roots, stems, and leaves.

Laboratory 17 Seed Plants

(LM pages 221–236)

Eleventh Edition Changes

This lab has no changes this edition.

New or revised figures: 17.10 Embryo sac in a lily ovule.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Incubation. Experimental Procedure: Pollen Grains. This procedure should be done first to allow time for germination of pollen grains.

Fresh material. For Observation: Pine Leaf and Cones, order pine needles and cones in advance, or obtain locally. For Observation: A Flower obtain fresh flowers locally, close to time of use. Some flower shops, especially wholesale shops, will give away nonsaleable flowers for educational purposes.

Demonstration: prepared microscope slide of mature embryo sac of Lilium.

17.2 Gymnosperms (LM pages 223--229)

_____ representative preserved gymnosperm specimens (Carolina 22-3300 or

22-3305 includes cycads, ginkgoes, gnetophytes, and conifers)

Pine Trees (LM pages 226–229)

_____ stereomicroscope

_____ lens paper

_____ Pine Life Cycle Set (optional)

_____ fresh clusters of pine leaves (needles)

_____ fresh pine cones (pollen cones and seed cones)

_____ fresh pine seeds and scalpels

_____ slide, prepared: mature pine male (staminate) cone, longitudinal section

(Carolina 30-1436)

_____ slide, prepared: pine seed (ovulate) cone, longitudinal section (Carolina

30-1478)

Pine cones and seeds (LM pages 228–229). Pine cones and seeds are available from Carolina Biological Supply in the Pine Life Cycle Set (listed above). This set includes cones from a variety of life cycle stages, as well as seeds, a seedling, and a twig with needles. These are also available separately. See the “Preserved Plant” section of the catalog. If possible, collect cones and needles from various local identifiable species for display.

17.3 Angiosperms (LM pages 230–234)

_____ flower model

_____ compound light microscope

_____ lens paper

_____ flowers, fresh assortment of monocot and eudicot

_____ forceps or tweezers

_____ dropper bottle of water

_____ slides and coverslips

_____ scalpels

_____ slide, prepared: pollen grains showing tube cell and generative cell

(Carolina 30-4270)

_____ slide, prepared: germinated pollen grains with pollen tubes

_____ pollen, fresh

_____ petri dishes, 100 x 15 mm, four

_____ filter paper circles, four (or paper towel circles)

_____ supports, 2–3 cm long, eight: wooden splints (Wards) or small- diameter

_____ glass rods sufficiently thick to elevate inoculated slide above moistened filter paper in bottom of petri plate

_____ dissecting needle

_____ slides, coated on one side only with nutrient agar, four

_____ methylene blue or toluidine blue

_____ 10% glucose

_____ 1.5% agar

_____ slide, prepared: Lilium female gametophyte (megagametophyte), mature

embryo sac

Fresh flowers (LM page 230). Obtain examples of monocot and eudicot flowers for viewing and dissection. Some examples of monocots are tulips, hyacinths, and lilies. Roses, geraniums, impatiens, and snapdragons are examples of eudicots.

Fresh pollen (LM page 233). Use a variety of species for best results. Excellent percentages of pollen germination have been experienced with Tradescantia ohiensis and Crotalaria spectabilis. Magnolia grandiflora yields good results, while Helianthus angustifolius (after 12–24 hours) and Callicarpa americana yield fair results. Albizia julibrissin, Lonicera japonica, Hymenocallis species, Campsis radicans, Oenothera laciniata, and Brassica species produce negative results.

Nutrient agar (LM page 233). Use 10% glucose in 1.5% agar. While the type and concentration of sugar may be varied, the best results for most species studied have been obtained by using 10% glucose. Varying degrees of germination success have been achieved using 5% concentrations of sucrose and fructose. For every1 liter of stock media, add 100 g of glucose and 15 g of agar. Stock media are stored in the refrigerator in 1 liter autoclavable bottles.

Coating of slides (LM page 233). Dip standard glass microscope slides into the hot nutrient agar. Immediately upon removal of the slide from the hot agar, wipe one surface clean, using a moistened Kimwipe. Allow the coated slides to cool to room temperature. Water content and degree of hydration of the agar appear to be important variables, and best results have been obtained using freshly coated slides. If students coat and inoculate their own slides at the beginning of a 3-hour laboratory period, they still will have enough time to observe germination. Unfortunately, a 2-hour laboratory period does not allow enough time for students to do both activities. In this case, the slides must be coated just prior to the lab.

Inoculation of slides (LM page 233). Select mature anthers that are actively shedding pollen. With some species, you must “smudge” pollen onto the nutrient agar (that is, make direct contact between a mature anther and the surface of the nutrient agar). However, the preferred technique is to sprinkle pollen over the surface of the agar by holding the anther with forceps and sharply tapping it with a dissecting needle. Do not let the agar coating dry out excessively.

Culturing male gametophytes (LM page 233). A moist culture chamber must be provided to promote germination and male gametophyte development. Culture chambers are kept in light and at room temperature. Petri dishes are prepared as follows for use as culture chambers: Place a filter paper circle (or paper towel circles several sheets thick) into the petri dish. Moisten the filter paper thoroughly with distilled water. Place two supports (wooden splints or small-diameter glass rods) parallel with one another on opposite sides of the moistened filter paper. The supports elevate the inverted microscope slide slightly so that its inoculated, agar-coated surface does not touch the moistened filter paper. Place a lid on the petri dish, and set it aside for at least 1 hour. Some species do not show much germination for several hours, and one, Helianthus angustifolius, shows no evidence of germination for 12–24 hours.

Observation of developing male gametophytes (LM page 233). Carefully remove the inverted microscope slide from the culture chamber. Make certain that it is repositioned with the agar-coated surface up before placing it onto the microscope stage. Examine the slide without a coverslip at low power. If time and interest allow, reinvert the slide, and place it back into the culture chamber. Observations may be made at regular intervals for up to 24 hours, if precautions are taken to maintain a moist culture chamber and to prevent dehydration of the agar.

Just before completing the exercise, students should examine their preparations with the compound light microscope at high-power magnification. Although not absolutely necessary, a coverslip can be placed on the agar to gain better resolution. Remind students that the greater thickness of these preparations limits working distance even more than usual and that special care should be taken to protect the high-power objective.

Staining (LM page 233). Toluidine blue and methylene blue stains can be used to enhance sperm cells and tube nuclei. These are applied at the end of the observation period. Stains are flooded over the agar surface and allowed to penetrate for 5–10 seconds; then the excess stain is washed with distilled water.

EXERCISE QUESTIONS

Introduction (LM page 221)

Which generation—gametophyte or sporophyte—is better adapted to a land environment because it contains vascular tissue? sporophyte

17.1 Life Cycle of Seed Plants (LM page 222)

1. In which life cycle, nonseed or seed, do you note pollen sacs (microsporangia) and ovules (megasporangia)? seed In which life cycle, nonseed or seed, do you note two types of spores, microspores, and megaspores? seed Label heterospores where appropriate in

Figure 17.2. a. In which life cycle do you note male gametophyte (in pollen grain) and female gametophyte (embryo sac in ovule)? seed

2. Label pollination where appropriate in Figure 17.2. b.

3. In which life cycle does a seed appear between the zygote and the sporophyte? seed What generation is present in a seed? sporophyte

17.2 Gymnosperms (LM pages 223–229)

Cycads (LM page 223)

Observation: Cycads (LM page 223)

After examining cycads and the following photograph, give three characteristics you could use to recognize a cycad.

1. large divided leaves

2. presence of cones

3. short, thick trunk

Ginkgoes (LM page 224)

Observation: The Maidenhair Tree (LM page 224)

After examining ginkgo leaves and the following photograph, give three characteristics you could use to recognize a male ginkgo tree.

1. can be quite tall

2. fan-shaped leaves

3. vivid yellow leaves (fall) or pendulous lower branches

Gnetophytes (LM pages 224–225)

Observation: Gnetophytes (LM page 225)

If available, examine gnetophyte specimens and then label the following photographs on the lines provided. photo on left: Ephedra; photo on right: Welwitschia mirabilis

Conifers (LM pages 225–229)

Pine Trees (LM pages 227–229)

1. Which part of the pine life cycle is the adult sporophyte? pine tree

2. Which part of the pine life cycle is the male gametophyte? mature pollen grain The female gametophyte generations? egg-bearing ovule.

3. Where does fertilization occur and what structure becomes a seed? in the female gametophyte and the ovule becomes the seed.

Observation: Pine Leaf (LM page 227)

How many leaves are in the cluster you are examining? What is the common name of your specimen? Answers depend on the species being observed.

Observation: Pine Cones (LM pages 228–229)

Preserved Cones (LM page 228)

1. Compare a pine pollen cone to a pine seed cone. Pollen cones are quite small, while seed cones are larger.

2. What do the pollen sacs produce? microspores, which become pollen grains

3c. See if any seeds are present. Where are they located? on scales What is the function of the seed coat? protection What is the wing for? to assist dispersal by wind

4. Can you find an embryo inside? most likely, yes

Microscope Slides (LM page 229)

1. Label a pollen sac in Figure 17.7a and a pollen grain in Figure 17.7b.

1. Pollen sac; 2. pollen grain How do these help in the dispersal of pine pollen? make the pollen grain buoyant

2. Label the ovule and the megaspore mother cell in Figure 17.8. b. ovule;

c. megaspore mother cell. Also, label the pollen grains that you can see just outside the ovule. a. pollen grain

17.3 Angiosperms (LM pages 230–234)

Observation: A Flower (LM pages 230–231)

2. What is the common name of your flower? Answer will depend on the flower being observed.

3. How many sepals and petals are there? Answer will depend on the flower being observed.

4. Are the stamens taller than the carpel? Most likely, the stamens will be taller.

5. What are you observing? pollen grains

6. What are these bodies? ovules

7. Is your flower a monocot or eudicot? Answer will depend on the flower being observed (see Table 17.1).

The Male Gametophyte (LM page 233)

In flowering plants, pollination is the transfer of the pollen grain from the anther to the stigma. By what means is pollination accomplished in the flowering plant life cycle? Most flowers are animal pollinated, some are by wind.

Observation: Pollen Grain Slide (LM page 233)

2. What portion of a germinated pollen grain tells you that it is the mature male gametophyte? presence of sperm.

Experimental Procedure: Pollen Grains (LM page 233)

Have any of the pollen grains germinated? If so, describe. Answers will depend on the experimental results, but the pollen grains may show a tube growing from the pollen.

The Female Gametophyte (LM page 234)

What signifies that the embryo sac is the female gametophyte? It produces an egg.

Observation: Embryo Sac Slide (LM page 234)

2. Due to the first fertilization, what happens to the egg? It becomes fertilized and is now a zygote, which develops into the embryo. Due to the second fertilization, what happens to the polar nuclei? Joins with a sperm and becomes the 3n endosperm, which is food for the embryo.

3. What are the three parts of a seed? embryo, stored food, and seed coat

17.4 Comparison of Gymnosperms and Angiosperms (LM page 235)

1. Beneath the photos, list ways to tell a gymnosperm from an angiosperm.

a. evergreen tree, needlelike leaves, production of cones; b. broad leaves, deciduous in the temperate zone, production of flowers

Table 17.2 Comparison of Gymnosperms and Angiosperms

Heterospores Pollen grains/ Cones Flower Fruit

Ovule

Gymnosperms yes yes yes no no

Angiosperms yes yes no yes yes

3. What structure in gymnosperms and angiosperms delivers sperm to the vicinity of the egg? pollen grains Does delivery require external water? no

4. What structure in gymnosperms and angiosperms becomes a seed? ovule

5. Is a sporophyte or gametophyte embryo in a seed? sporophyte

6. What innovation in angiosperms led to the production of seeds covered by fruit? flower

LABORATORY REVIEW 17 (LM page 236)

1. Name two differences between the life cycles of seedless vascular plants and seed plants. Nonseed plants have an independent gametophyte and flagellated sperm. Most seed plants have heterogametophytes which are dependent upon the sporophyte and the sperm are not flagellated.

2. Name the structures that precede the ones listed in the life cycle of a seed plant.

a. Megaspore Female gametophyte

b. Microspore Male gametophyte (pollen grain)

c. Seed Adult sporophyte

3. The pine tree, unlike a fern, is able to reproduce sexually in a dry environment. Explain. In ferns, eggs are fertilized by flagellated sperm that need water to reach the egg. In the pine tree, pollen grains, which produce nonflagellated sperm, are carried by the wind to the ovule, where the sperm reach the egg via the pollen

Compare conifers to flowering plants by completing the following table:

Dominant Vascular Tissue Dispersal of Fruit

Generation (Present or Absent) Offspring (Present or Absent)

4. Conifer Sporophyte Present Seed Absent

5. Flowering Sporophyte Present Seed Present

plant

6. a. Compare the location of the pollen sacs in the pine and the flowering plant. In the pine, the pollen sacs are on the underside of the pollen cone scales. In the flowering plant, the pollen sacs are in the anther.

b. Compare the location of the ovule in the pine and the flowering plant. In the pine, the ovule is the upper surface of seed cone scales. In the flowering plant, the ovule is within an ovary.

7. Suppose you wanted to show a friend the female gametophyte of the pine. What would you do? Use a microscope to view the interior of the ovule just before fertilization.

8. Name two innovations (aside from fruit) in the flowering plant life cycle not found in the pine tree life cycle, and explain why they may have contributed to the success of flowering plants. Flowering plants use pollinators for pollination, which is more specific than wind pollination. Animals also help dispense fruits.

9. What is the difference between pollination and fertilization? Pollination is the transfer of pollen from pollen cone to seed cone in gymnosperms, or from anther to stigma in flowering plants. Fertilization is the union of egg and sperm.

10. What is the difference between monocot and eudicot flowers? Monocots have flower parts in threes and multiples of three, while eudicots have flower parts in fours or fives and multiples of four or five.

Laboratory 18 Organization of Flowering Plants

(LM pages 237–253)

Eleventh Edition Changes

In Section 18.1, this edition, students examine the shoot system and the root system of a plant. The diversity sections of roots, stems and leaves are now more student friendly, and students are no long required to fill out extensive tables. The lab is much improved because the illustrations are now placed on the same or facing page to their reference. In several places, reinforcement questions/tables have been added.

New/Revised Figures: 18.1 Organization of a plant

New/Revised Tables: 18.3 Primary Growth Versus Secondary Growth

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Fresh material. Observation: A Living Plant requires a tomato or geranium plant or any bean plant. Experimental Procedure: Monocot Versus Eudicot requires any available plants, flowers, or seeds. Observation: Root Diversity requires various root types; in particular examples of taproots and fibrous root systems. Observation: stem diversity requires examples of stem modification. Observation: Anatomy of Winter Twig requires winter twigs. For Observation: Leaf Diversity, requires available fresh leaf types, or use a leaf types set (Carolina 23-8805), which shows twelve specimens on a single herbarium mount.

All Exercises

_____ compound light microscope

_____ lens paper

18.1 External Anatomy of a Flowering Plant (LM page 238)

_____ tomato or geranium plant, living (obtain locally if in season, or Carolina 15-7380, 15-7381) or any convenient bean specimens

18.2 Major Tissues of Roots, Stems, and Leaves (LM pages 239–241)

_____ prepared slide of eudicot (dicot) root tips; longitudinal section shows

apical meristems

_____ prepared slide of eudicot (dicot) shoot tip; longitudinal sections shows

apical meristems

_____ prepared slide of leaf, cross section (maple, Carolina 30-3814 or any convenient slide)

18.3 Root System (LM pages 242–245)

_____ model, root tip (Carolina 56-8743)

_____ prepared slide: longitudinal section of eudicot (dicot) root tip

_____ prepared slide: cross section of buttercup root (Ranunculus) eudicot (dicot) root (Carolina 30-2090)

_____ representative root types for display

Root Diversity (LM page 245). Representative fibrous root types are: grasses, ornamental pepper; representative taproots are: sunflower, carrot, radish, dandelion ; representative adventitious roots are: ivy, corn, philodendron, Spanish moss. If the suggested plants are not available, select alternative plants that represent similar root type diversity.

18.4 Stems (LM pages 246–250)

_____ prepared slide of herbaceous stems: herbaceous eudicot (dicot), e.g., sunflower, and monocot, e.g., corn, stem set (Carolina 30-264

_____ winter twig, collected or purchased (Carolina 23-8810)

_____ prepared slide of woody stem : Liriodendron cross section Carolina 30-3134

_____ representative stem types for display

Stem Diversity (LM page 248).In addition to strawberry, some vines have above ground stolons. Underground stolons are produced by lily of valley, asters and white potato. Tubers (potato) are produced at the end of stolons and in turn give rise to new stolons. Begonias are well known for having tubers. In addition to irises, bamboos, ginger and many grasses form rhizomes; lilies, tulips, and daffodils grow from bulbs; dahlia, crocus, gladiolus, and cyclamen produce corms. Other stems produce tendrils such as wisteria, common ivy, and morning glory , or thorns, e.g., roses or suckers.

Winter twigs (LM page 249). Purchase, or collect in winter, identify them, and save

them for future use.

18.5 Leaves (LM pages 251–252)

_____ leaf model, angiosperm leaf (Carolina 56-8801)

_____ prepared slide of leaf, cross section (maple, Carolina 30-3814 or any convenient slide)

_____ representative leaf types for display (fresh or herbarium mounts)

Leaf Diversity (LM page 252) The leaves of cactuses are spines; the tendrils of peas, cucumbers, and squash are modified leaves; the bud scales of winter twigs are modified leaves; and an onion is composed leaves modified for storage. Insectivorous plants (i.e. pitcher plant) have modified leaves for catching insects.

EXERCISE QUESTIONS

18.1 External Anatomy of a Flowering Plant (LM page 238)

Observation: A Living Plant (LM page 238)

Shoot System

What structures are in the shoot system? stems ,leaves, nodes, internodes, axillary buds, terminal bud, vascular tissue

The Leaves

What part of leaf is the most expansive part? blade

The Stem

2. Does the internode get larger or smaller toward the apex of the stem? smaller Toward the roots? larger

Based on the fact that a stem elongates as it grows, explain your observation. As new tissue is added at the terminal bud, the internodes become longer.

3. Where is the terminal bud of a stem? At the apex of the stem

Where are axillary buds? At nodes

Root Stem

DoeDoes this plant have a strong primary root or many roots of the same size? Depends on plant.

What structures are in the root system? branch root, root hairs, primary root, root tip, vascular tissue

Where is the root tip? At the apex of the root.

18.2 Major Tissues of Roots, Stems, and Leaves (LM pages 239–241)

Observation: Major Tissues of Roots, Stems, and Leaves (LM pages 239–241)

1. What is the function of meristematic tissue? To divide and produce more cells.

Mature Tissues

1. What is the function of epidermis in all three organs? Protection

2. Which type of cell carries on photosynthesis or stores the products of photosynthesis in a leaf? Parenchyma Which one lends strength to ground tissue in roots and stems? Sclerenchyma

3. Which type tissue transport sugars in a plant? phloem

Which type tissue transports water? xylem

Monocots Versus Eudicots (LM pages 240–241)

Experimental Procedure: Monocot Versus Eudicot (LM pages 240--241)

1. Is the plant in Figure 18.1 a monocot or eudicot? eudicot. The leaf pattern in Figure 18.3 appear to be parallel. Is this the leaf of a monocot or eudicot? monocot

3. Is the plant that produced the peanut a monocot or eudicot? eudicot. If a flower has three petals or six petals, or any multiple of three, is the plant that produced this flower a monocot or a eudicot? monocot

18.3 Root System (LM pages 242–245)

Anatomy of Eudicot and Monocot Roots (LM pages 243–244)

Observation: Cross-Section Anatomy of Eudicot and Monocot Roots (LM pages 243–244)

Eudicot Root (LM page 243)

3. The endodermis regulates what materials that enter a plant though the root? minerals in the water Use this illustration to trace the path of water and minerals from the root hairs to xylem. root hair of epidermis, cortex, endodermis, pericycle, xylem

Monocot Root (LM page 244)

3. Contrast the arrangement of vascular tissue (xylem and phloem) in the vascular cylinder of monocot roots with that in eudicot roots.

Eudicot Xylem has the appearance of wheel and phloem is between spokes of wheel

Monocot Ring of xylem (inside) and phloem (outside) surrounds pith.

Root Diversity (LM page 245)

Observation: Root Diversity (LM page 245)

Taproots and fibrous roots. Examine the taproots on display, and name one or two in which the taproot is enlarged for storage. Most likely, carrot, radish The dandelion on display has which of these types of roots? Taproot because there is one main tapered root.

Adventitious roots. Which plants on display have adventitious roots? Most likely, grasses, ivy, corn.

Other types of roots . What root modifications not noted here are on display in your laboratory. Figure 18.8d shows pneumatophores.

18.4 Stems (LM pages 246–250)

What function do botanists assign the stems in addition to support for branches and leaves? transport of water and nutrients Explain why a limb and a branch of a plant cannot live if severed from the rest of the body/plant. The limb or branch is no longer receiving water and nutrients

Anatomy Herbaceous Stems (LM pages 246–247

Observation: Anatomy of Eudicot and Monocot Herbaceous Stems (LM pages 246–247)

Eudicot Herbaceous Stem (LM page 246)

1. Label the epidermis in Figure 18.9a. a

3. Label the vascular bundle in Figure 18.9a. b Which tissue (xylem or phloem) is closer to the surface? phloem

4. Both cortex and pith are composed of what type of tissue? ground tissue

Monocot Herbaceous Stem (LM page 247)

2. The vascular bundles in a monocot herbaceous stem are said to be scattered. Explain. In a monocot stem, vascular bundles are scattered because they do not occur in a regular ring formation, as they do in eudicot stems.

Comparison (LM page 247)

1. Compare the arrangement of ground tissue in eudicot and monocot stems. Ground tissue in eudicot stems is placed centrally. In monocot stems it fills the spaces between the scattered vascular bundles.

2. Compare the arrangement of vascular bundles in stems of eudicots and monocots. In eudicots the bundles occur in a ring and in monocot the bundles are scattered.

Stem Diversity (LM pages 247--248)

Observation: Stem Diversity (LM page 248)

Stolons. Label an adventitious root and a new shoot in Figure 18.11a 1. and 2. adventious root 3. new shoot

List any other plants on display that spread and produce new shoots by sending out stolons. Possibilities are various vines. The white potato, lily of the valley, and asters produce underground stolons

Rhizomes. List any other plants on display that have rhizomes. Possibilities include bamboo, ginger and many grasses.

Tubers. List any other plants on display that have tubers. Begonia tubers is a possibility

Corms List any other plants on display that have corms. Possibilities are dahlia, crocus, gladiolus, and cyclamen

Anatomy of Woody Stems (LM pages 249–250)

Table 18.3 Primary Growth Versus Secondary Growth

Primary Growth Secondary Growth

Active meristem apical meristem vascular cambium

Result increase in length increase in girth

Observation: Anatomy of a Winter Twig (LM page ) stet page 249

4. Note the vascular bundle scars. Complete this sentence: Vascular bundle scars appear where the vascular tissue previously extended into leaf petioles or a branch that dropped off.

5. The region between nodes is called an internode.

Observation: Anatomy of Woody Stem (LM page 250 )

3. How old is the stem you are observing? Count the number of annual rings. Are all the rings the same width? Answer depends on the stem examined, but not likely.

18.5 Leaves (LM pages 251–252)

Anatomy of Leaves (LM page 251)

Observation: Anatomy of Leaves (LM page 251)

3. If this is a model of a monocot, all the leaf veins will be circular and parallel. If this is a model of a eudicot, some leaf veins will be circular and some will be oval. Why? The net pattern of a eudicot leaf means that all the veins will be at various angles to the cut plane, and will show distortion ranging from circular to elliptical.

4. Label the layers of mesophyll in Figure 18.14. 2. palisade mesophyll; 3. spongy mesophyll Collectively, the mesophyll represents which of the three types of tissue found in all parts of a plant? ground

5. Label the two layers of epidermis in Figure 18.14. 1. upper epidermis 4. lower epidermis

Leaf Diversity (LM page 252)

Observation: Leaf Diversity (LM page 252)

Figure 18.15a. 1. simple; 2. palmately compound; 3. pinnately compound

Figure 18.15b. 4. alternate; 5. whorled; 6. opposite

LABORATORY REVIEW 18 (LM pages 253)

1. How would you distinguish between a monocot plant and a eudicot plant based on leaf vein pattern? The leaves of a monocot plant have parallel veins, while the leaves of a eudicot plant have a net vein pattern.

2. How is meristem different from all other types of plant tissue? Meristem is embryonic tissue. It continually produces new cells. This accounts for plants’ ability to grow their entire lives.

3. Show that a root has epidermal, ground, and vascular tissue by using these terms to label this illustration. a. vascular; b. ground; c. epidermal

4. In which zone of a eudicot root would you expect to find mature vascular tissue? Why? Vascular tissue is found in the zone of maturation because cells become specialized in this zone.

5. In a eudicot root, what structural feature allows the endodermis to regulate the entrance of water and materials into the vascular cylinder, where xylem and phloem are located? The Casparian strip, a layer of waxy material, prevents the passage of water and materials between root cells so that they must pass through the endodermis.

6. What type of root does a carrot have? A taproot modified for the storage of organic food. The main root is many times larger than the branch roots.

7. How would you microscopically distinguish a eudicot stem from a monocot stem? In a eudicot stem, the vascular bundles are arranged in a ring; in a monocot stem, they are scattered.

8. Compare how primary and secondary growth of a stem arises. Primary growth arises from the apical meristem in the terminal bud and adds to the length of a plant. Secondary growth arises from vascular cambium and adds to the girth of a plant.

9. Contrast how you could determine one year’s growth by looking at a winter twig with how you determine one year’s growth in a cross section of a tree stem. In a woody twig, the growth between terminal bud scars represents one year’s growth. In a cross section of a tree, one annual ring indicates one year’s growth.

10. Contrast the manner in which water reaches the inside of a leaf with the manner in which carbon dioxide reaches the inside of a leaf. Water enters xylem at the roots and then passes up the stem to a leaf vein. Carbon dioxide enters by way of stomata into the cells of spongy mesophyll.

Laboratory 19 Water Absorption and Transport in Plants

(LM pages 255–264)

Eleventh Edition Changes

In Section 19.2, The Water Column, no longer requires students to make freehand sections. In Section 19.3, students now follow the scientific method by first formulating hypotheses ; performing a transpiration experiment and coming to a conclusion. I have modified a transpiration virtual lab on the Biology website to also use the scientific method. The virtual lab is an alternative to doing the experiment hands-on.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Growth required (LM page 256). Radish or mustard seedlings for Observation: Root Hairs

Fresh material. Experimental Procedure: Absorption of Water by Osmosis requires fresh potatoes. Experimental Procedure: The Water Column requires fresh celery stalks. Experimental Procedure: Transpirational Pull at Leaves requires eudicot plants with leaves such as geranium plants. Observation: Number of Stomata, and Experimental Procedure: Open Versus Closed Stomata requires fresh leaves; obtain locally, close to time of use.

All Exercises

_____ dropper bottles of distilled water

_____ rulers, plastic millimeter

_____ microscopes, compound light

_____ microscope, stereomicroscope

_____ lens paper

_____ microscope slides and coverslips

19.1 Water Absorption by Root Hairs (LM pages 256–257)

Root Hairs

_____ seedlings (radish or mustard)

_____ petri dishes

_____ 0.1% neutral red solution (Carolina 87-6830)

Seedlings (LM page 256). Germinate seeds 2 to 5 days ahead of use in petri dishes lined with damp filter paper. Plant seeds such that you will have some of two, three, four, and five days’ growth for each section. Then select the ones that show the best root growth.

0.1% neutral red solution (LM page 256). This is sufficient for at least two student groups, depending on size of beaker. Dissolve 0.1 g of neutral red in 100 mL of distilled water.

Osmosis

_____ test tubes, 15 mL (5 per student group)

_____ test tube racks or beaker

_____ sucrose solutions (see note that follows)

_____ cork borer

_____ potatoes, fresh

_____ paper towels

_____ scalpel and forceps

Sucrose solutions (LM page 256). Prepare 50 mL of each concentration of sucrose per student group. To make sucrose solutions, equate the percent of sucrose to grams and add that amount to 100 mL. For example, to make a solution that is .05% sucrose, add .05 g to 100 mL of distilled water.

19.2 The Water Column in Stems (LM pages 257)

_____ red food coloring

_____ beakers or glass jars

_____ celery, fresh

_____ scissors

19.3 Transpirational Pull at Leaves (LM pages 259–262)

_____ rubber tubing, 4 cm length (to fit on glass tube and geranium stem)

_____ 1 mm glass pipettes

_____ large container of water

_____ geranium plant, enough for one cutting/transpirometer (local purchase or

Carolina 15-7380 or 15-7381)

_____ rubber bands or string

_____ ring stand with clamp

_____ wax pencils

_____ rulers, plastic millimeter

_____ heat lamp (light source)

_____ spray bottle

_____ plastic bag

_____ fan, small

This procedure requires a 45-minute repeat exercise. If both are set up simultaneously, 2.75 hours is the approximate amount of time required for the lab. If they are not done simultaneously, the lab could take 3 to 4 hours. As an alternative, this lab contains a virtual lab students can do regarding Plant Transpiration.

19.4 Stomata and Their Role in Transport of Water (LM pages 262–263)

_____ calculators

_____ leaves, eudicot (for epidermis strips)

_____ dropper bottles of distilled water

_____ 5% salt (NaCl) solution in dropper bottles

Eudicot leaves (LM pages 263)These can be from any plant of choice whose epidermis strips easily.

5% salt solution (LM page 263). Dissolve 5 g of sodium chloride (NaCl) in 100 mL of distilled water. This is enough for a class.

EXERCISE QUESTIONS

19.1 Water Absorption by Root Hairs (LM pages 256–257)

Observation: Root Hairs (LM page 256)

4. Does every epidermal cell have a root hair? not necessarily. How does the structure of a root hair aid absorption? provides increased surface area for absorption

Experimental Procedure: Absorption of Water by Osmosis (LM pages 256–257)

Table 19.1 Absorption of Water by Osmosis

Test Sucrose % Pretreatment Post-treatment Movement Tonicity of Original

Tube Weight Weight of Water Solution Compared

(into/out) to Potato

1 .05% Weight Weight Into Hypotonic

2 .07% can can Into Hypotonic

3 .09% vary vary Into or no Hypotonic or isotonic

net movement

4 .12% No net movement Isotonic

5 .14% Out of Hypertonic

Conclusions: Absorption of Water by Osmosis (LM page 257)

• Since potato cells (modified stems) are believed to behave much as root hairs do, this experiment indicates that only if groundwater is hypotonic to cytoplasm in root hairs will root hairs be able to absorb water.

• If, by chance, a plant’s roots are surrounded by a hypertonic solution, the roots will lose water.

19.2 The Water Column in Stems (LM pages 257–258)

Experimental Procedure: The Water Column in Stems (LM page 258)

Table 19.2 Celery Stalk Experiment

Stalk Speed of Dye (Minutes) Conclusion

Cut end placed in water Faster Water column was not broken.

prior to experiment

Cut end kept in air Slower Water column was broken.

prior to experiment

Conclusions: Water Column in Stems (LM page 258)

• Is a continuous water column helpful to the conduction of water in plants? yes Explain on the basis of your results. Speed of conduction was faster in the celery stalk whose cut end was not exposed to air.

19.3 Transpirational Pull at Leaves (LM pages 259–262)

Experimental Procedure: Transpiration (LM pages 259–261s)

Determining Transpiration Rate Under Standard Conditions (LM page 259)

Determining Transpiration Rates Under Varied Environmental Conditions (LM pages 260--261)

1. a. Hypothesize how an increase in temperature will affect the rate of transpiration? An increase in temperature will increase evaporation. Therefore, the transpiration rate should increase.

b. Hypothesize how humidity will affect the rate of transpiration? The transpiration rate should decrease with increasing humidity, since evaporation will decrease.

c. Hypothesize how wind will affect the rate of transpiration? The wind will cause an increase in evaporation. Therefore, the transpiration rate should increase.

2. Record your measurements in Table 19.3. Student results may vary individually; however, the general principles remain the same. The transpiration rate will increase with heat, decrease with humidity, and increase with wind. Example readings are provided in Tables 19.3.

Table 19.3 Effect of Temperature on Transpiration Rate

Standard Conditions Test Conditions

Time Reading Total Change Reading Total Change

(in mm) (in mm) (in mm) (in mm)

After 10 minutes 3.5 3.5 5.0 5.0

After 20 minutes 4.5 8.0 5.5 10.5

After 30 minutes 4.0 12.0 6.0 16.5

After 40 minutes 3.0 15.0 4.5 21.0

Table 19.3 Effect of Humidity on Transpiration Rate

Standard Conditions Test Conditions

Time Reading Total Change Reading Total Change

(in mL) (in mL) (in mL) (in mL)

After 10 minutes 4.5 4.5 2.0 2.0

After 20 minutes 4.5 9.0 2.0 4.0

After 30 minutes 3.0 12.0 1.5 5.5

After 40 minutes 2.0 14.0 1.5 7.0

Table 19.3 Effect of Wind on Transpiration Rate

Standard Conditions Test Conditions

Time Reading Total Change Reading Total Change

(in mL) (in mL) (in mL) (in mL)

After 10 minutes 4.0 4.0 6.5 6.5

After 20 minutes 4.5 8.5 6.5 13.0

After 30 minutes 4.5 13.0 6.0 19.0

After 40 minutes 4.0 17.0 6.0 25.0

[pic]

3. Plot the results of your two experiments on the graph provided. The transpiration rate will be the total change in mL between readings. These results are fairly typical

onclusion: Transpiration (LM page 261)

1. Did the results support or not support your hypothesis? Most likely support the hypothesis that the transpiration rate would increase due to an increase in temperature.

2. Did the results support or not support your hypothesis? Most likely support the hypothesis that the transpiration rate would decrease due to decreased evaporation in a humid environment.

3. Did the results support or not support your hypothesis? Most likely support the hypothesis that the transpiration rate would increase due to increased evaporation under windy conditions.

Virtual Lab Plant Transpiration (LM pages 261-262)

Hypothesis A fan will have the most effect on the rate of transpiration because evaporation will increase to a greater degree than the other environmental conditions.

Total Amount of Water (in mL) Transpired in One Hour

Normal Conditions With Fan With Heater With Lamp

Arrowhead 21.6 45.0 39.6 24.0

English Ivy 10.0 30.6 19.2 12.6

Geranium 7.2 28.2 33.8 14.4

Zebra Plant 25.2 45.6 36.6 20.2

Do the results support or fail to support your hypothesis? Usually, support although the heater produced more transpiration than the fan with the geranium Speculate on why the heater and the fan produced approximately the same results. Both of these environments bring about a drying effect that causes more evaporation. Speculate on why the rate of transpiration with the lamp approximated the rate under normal conditions.. The heat given off by the lamp was not enough to cause much of an increase in transpiration.

Journal Questions: as found online in the virtual lab. (LM page 262)

1. Describe the process of transpiration in vascular plants. (1) Water enters a plant at root hairs. (2) The water column is continuous from the roots to the leaves. (3) Water evaporates, pulling the water column from the roots to the leaves.

2. Describe any experimental controls used in the investigation. Plants were planted in same-type pot, exposed to same soil, and nutrient conditions, and data was collected for normal conditions in addition to experimental conditions.

3. What environmental factors that you tested increased the rate of transpiration? Was the rate of transpiration increased for all plants tested? The heater, fan, and lamp usually increased the rate of transpiration, but some to a greater degree than others.

4. Did any of the environmental factors (heat, light, or wind) increase the transpiration rate more than the others? Why? Not consistently.

5. Which species of plants that you tested had the highest transpiration rates? Under normal conditions, the zebra plant had the highest rate. Why do you think different species of plants transpire at different rates? Plants transpire through the stomata and they differ as to the number of stomata they have.

6. Suppose you coated the leaves of a plant with petroleum jelly. How would the plants rate of transpiration be affected? It would decrease because the stomata would no longer allow water to escape.

7. Of what value to a plant is the ability to lose water through transpiration? Transpiration allows water to be transported about a plant otherwise they would dehydrate and could not photosynthesize.

19.4 Stomata and Their Role in Transport of Water (LM pages 262–263)

Observation: Number of Stomata (LM page 262)

1. Calculate the area of the high-power microscopic field. If the diameter of your high power is 0.355 mm (r = 0.178 mm), r2 = 0.032 mm2, x = 3.14, area = 0.10 mm2.

4. Count the number of stomata you see in the high-power field. Example: six stomata

5. Divide the number of stomata by the area of the field calculated in number 1. This will tell you the number of stomata in 1 square millimeter. Example: 6 (stomata) divided by 0.10 mm2 = 60 stomata/mm2

6. If the underside surface area of your leaf were 400 square millimeters, how many stomata would be present on its surface? Example: 60 x 400 = 24,000 stomata

Experimental Procedure: Open Versus Closed Stomata (LM page 263)

4. Which slide contains open stomata, and which slide contains closed stomata? Stomata will be open on the slide with the distilled water. Stomata will be closed on the slide with the salt treatment.

5. Explain your results. Stomata will open when water is abundant (that is, the distilled water treatment). When water is in short supply (that is, the 5% salt solution treatment), stomata will close.

Summary (LM page 263)

Table 19.4 Water Transport in Plants

Process Where Mechanism

Absorption of water Root hairs Osmosis

Formation of water Xylem from roots Cohesion of water molecules

column to leaves and adhesion of water to walls of xylem

Transpirational pull Leaves Evaporation of water from leaves

LABORATORY REVIEW 19 (LM page 264)

1. How is the epidermis of a root specialized to absorb water? root hairs

2. What tonicity promotes water absorption in a root? hypotonic

3. Name and briefly describe the two types of conducting cells in xylem. Tracheids are thin tapered cells that can transport water because they have pits that allow water to pass through. Vessel elements are larger, have perforation plates or lack end walls. They are arranged to form a continuous pipeline.

4. What characteristics of water cause it to fill the conducting cells of xylem from the roots to the leaves? Water’s cohesive and adhesive properties cause it to fill the conducting cells of xylem.

5. Why is it best to cut flower stems under water? The water column is not broken.

6. Name two reasons the water column is continuous in xylem. Water molecules cling together (are cohesive) and they adhere to the sides of the vessel elements or tracheids.

7. What is transpiration? Transpiration is the evaporation of water from leaves that pulls the water column up in xylem.

8. Name two environmental factors that can affect transpiration? Warm and dry conditions speed up transpiration. Humidity impedes transpiration

9. What is the function of stomata? Water evaporates through stomata, which carry on gas exchange; carbon dioxide enters and oxygen exits the leaf at the stomata.

10. How is transpiration prevented when a plant is water stressed? Guard cells expand and the stomata close.

Laboratory 20 Control of Plant Growth and Responses1

(LM pages 265–274)

1Lab preparation and growing instructions courtesy of Wisconsin Fast Plants( Program, University of Wisconsin—Madison, Wisconsin, Department of Plant Pathology.

Eleventh Edition Changes

This lab is essentially the same as last edition.

New/Revised Figures: 20.5b Phototropism; 20.8 Phytochrome; 20.9 Etiolation

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some material may be used in more than one exercise.

Special Requirements

Growth required. Experimental Procedure:Gravitropism requires Brassica rapa plants and seedlings; Experimental Procedure: Phototropism requires Brassica rapa seeds; Gibberellins and Stem Elongation requires Brassica rapa plants; Experimental Procedure: Bean Seedling Etiolation requires bean seedlings

Advance preparation (LM page 269). Phototropism chambers for Experimental Procedure: Phototropism

Other. Instructors may find it useful to purchase the Wisconsin Fast Plants™ Manual (Carolina 15-8950) in preparation for the experiment. This manual explains in detail equipment and setup needed. (Carolina 15-8950).

All Exercises

_____ four-chambered planting quads (Carolina 15-8960)

_____ potting soil, specially formulated for growing Brassica rapa (Carolina

15-8965, 15-8966)

_____ fertilizer pellets, specially sized (Carolina 15-8970, 15-8971)

_____ Brassica rapa seeds (normal) (Carolina 15-8810, 15-8811)

_____ quad wicks (Carolina 15-8978)

_____ pipette for watering

_____ watering system (Carolina 15-8974)

_____ plant light bank system (Carolina 15-8998)

Brassica rapa seedlings life cycle (LM pages 267–268). Approximate days after sowing: 24 hours—germination, radicle emerges; 24 to 28 hours—hypocotyl hook; 48 to 72 hours—cotyledons, root hairs; days 4 to 5—emerging true leaves, root development;

day 7—true leaves; day 9—flower buds; day 11—petiole, nodes, internodes; day 13—leaf blade, nodes; day 15—inflorescence, stem, leaf; day 18—axillary bud, withered petal, enlarging pod; days 28 to 35—seeds mature in pods.

Note: Always begin a planting cycle on a Monday or Tuesday. This allows three consecutive school days for watering from above. Some growing instructions may come with the seeds, but additional instructions can be obtained from Carolina Biological Supply in a three-ring notebook. If you have not used Wisconsin Fast Plants™ previously, you may find it beneficial to do a test run with four to eight quads to see how the plants respond under the conditions in your lab.

20.1 Gravitropism (LM pages 266–268)

_____ B. rapa quad-germinated seedlings (less than fourteen days old)

_____ B. rapa petri dish seedlings (germinated within the last two to three days)

_____ ring stand

_____ box to cover fourteen-day-old seedling

_____ protractor

B. rapa quad-germinated plants (LM pages 266–272). Allowing the appropriate amount of time for the stage of development required for the exercise, germinate seeds in specially designed quads, which contain four cells. Add a wick to each cell to draw water from the source into the soil. Add potting mix until each cell is about half full. Add three fertilizer pellets. Add more soil, and press to make a depression. Add two or three B. rapa seeds to each cell, and cover with potting mix. Carefully water, using a pipette, until the water soaks through the potting mix and drips from the wick. Place the quad on the watering tray under fluorescent lights. Stagger plantings as necessary.

The Wisconsin Fast Plants™ watering and light systems (containing six fluorescent lights) are recommended for best results. Plants under a regular, two-light fluorescent system may show slower and less uniform growth. The watering system ensures a constant supply of water, which would be impossible to provide through hand watering. After the seeds begin to germinate, you can manipulate the plants in many different ways to investigate plant growth.

B. rapa petri dish seedlings (LM pages 266–272). Place seeds on wet filter paper in the lid of a petri dish. Cover the lid with the bottom half of the petri dish. Stand the dish, tilted on its end, in a water reservoir, such as the bottom of a 2 liter soft-drink bottle. Place the dish and reservoir under fluorescent lights. Germination begins within 24 hours, and observations may be made for several days. Keep the filter paper moist by carefully adding water.

If you wish to make quantitative measurements of seed germination, tape a transparent grid sheet marked in measured increments to the outside of the petri dish lid, or place it in the petri dish lid before the filter paper. Place the wet filter paper in the lid as before, and plant the seeds at a particular position in relation to the grid. As the seeds germinate and grow, you can easily use the grid to measure their size.

Plant ahead of time, allowing the appropriate amount of time for the stage of development required for the exercise. Stagger plantings as necessary. Also, your students may find it helpful to have a control demonstration petri dish germination setup that has not been turned available for comparison.

20.2 Phototropism (LM pages 269–270)

_____ film canisters (phototropism chambers)

_____ hole punch

_____ B. rapa seeds, three per canister

_____ blotting paper (not filter paper)

_____ red, blue, green plastic film set (Carolina 15-8990)

_____ tape, clear cellophane

Phototropism chambers (LM page 269). B. rapa seeds may be germinated in empty 35-millimeter clear or black film canisters (black seems to work best). These usually can be acquired, free of charge, from film-developing businesses. The canister may be used as is or may be modified by punching holes in the side of black canisters to allow light inside. Using clear tape, cover holes in the canister with red, blue, and green filters. Hole size may be varied to alter the quality or quantity of light hitting the plants. Place small, appropriately sized squares of blotting paper in the lid, and place three seeds on the blotting paper. (Do not use filter paper because it dries out too quickly.) Moisten the paper well, invert the canister, and snap it into the lid.

20.3 Gibberellins and Stem Elongation (LM pages 270–272)

_____ B. rapa (normal) quad-germinated plants of increasing age (six to eighteen days’ growth)

_____ B. rapa (dwarf—Carolina 15-8832, 15-8833) quad-germinated plants of

increasing age, half treated with gibberellic acid (six to eighteen day’s

growth)

_____ gibberellic acid (Carolina 15-8987)

_____ spray bottle for gibberellic acid

_____ rulers, plastic millimeter

Gibberellin effect on dwarf B. rapa (LM page 272). Students will be measuring treated and untreated B. rapa plants at three ages and comparing them to normal plants of the same ages. Though the plants may be used repeatedly for measurement, they are delicate and may be easily damaged. Start enough seeds to allow for the probable damage of plants when used by more than one group. Plant twice as many dwarf plants so that half can be treated with the gibberellin and half be left untreated.

Stagger plantings at four-day intervals, beginning twenty, sixteen, twelve, and eight days before the class. Follow the general planting, thinning, and transplanting instructions. Start treating half of the plants with gibberellin as soon as the first true leaves begin to grow. Placing drops of gibberellin on the leaves will work, although spraying the entire plant will probably yield more dramatic results. The results shown here are by the drop method, so your students may see even better results. Plants should be treated at least every other day.

You may wish to have students measure overall height of the plants, as well as the internode distance. Internode distance in the dwarf untreated plants is negligible overall and in the treated plants is negligible for the eight- and fourteen-day readings. Height would be more uniformly compared.

20.4 Etiolation (LM pages 272–273)

_____ bean seedlings, dark- and light-germinated

Bean seedlings (LM page 273). Place bean seeds inside a petri dish between moist sheets of blotting paper for several days until germination occurs. Germinate one set under light conditions and one set under dark conditions. Keep the blotting paper moist. Remove the lid so that plants can grow, but keep the petri dish in moist, humid conditions.

EXERCISE QUESTIONS

20.1 Gravitropism (LM pages 266–268)

Experimental Procedure: Gravitropism (LM pages 266–268)

Do you predict that light will have any effect on gravitropism? Students will likely answer no. Why or why not? A likely response would be that gravity affects gravitropism while light affects phototropism. Knowing that gravitropism is dependent on the settling of plastids (that store starch) on the lower portion of the stem, hypothesize how light might affect gravitropism? Light promotes photosynthesis and starch formation. If time is allowed for this to happen, then light might affect gravitropism.

Response of Mature Stem (LM page 267)

3. Record the angle of bending for each plant in Table 20.1.

Table 20.1 Gravitropism of Stems

Mature Plant Angle of Bending/ Gravitropism Conclusion

Degrees of Stem

In the light Possibly greater Negative Light can affect the degree upward bending of upward bending.

In the dark Possibly less Negative Stems respond to gravity

upward bending when light is absent.

Conclusions: Response of Mature Stem (LM page 267)

• Do your data suggest that the presence of light/dark affects the degree of gravitropism in stems? possibly yes

• If your prediction was not supported, suggest a new model (scenario) that might explain your results. Data in the literature is consistent with the conclusion that light promotes gravitropism because of increased plastid size and sedimentation. See text, p. 482.

• Why is it adaptive for seedlings to exhibit negative gravitropism? It is adaptive for seedlings to bend upward to absorb sunlight so photosynthesis can occur.

Response of Seedlings (LM page 268)

4. Describe the position of the root and stem in Table 20.2

Table 20.2 Gravitropism in Seedlings

Time Position of Root Position of Stem Explanation

t0 Horizontal Horizontal Takes time for response to occur

t1 Roots bend down Stems bend up In response to gravity, cells elongate on the upper side of a root and the lower side of a stem. Therefore, roots bend in the same direction as gravity and stems bend opposite to gravity.

Conclusions: Response of Seedlings (LM page 268)

• Do roots exhibit negative or positive gravitropism? positive

• Do stems exhibit negative or positive gravitropism? negative

20.2 Phototropism (LM pages 269–270)

Experimental Procedure: Phototropism (LM pages 269–270)

3. Record the response of the seedlings in Table 20.3.

Table 20.3 Phototropism

Type of Canister Response of Seedlings Conclusion

Clear windows (control) Seedlings bend toward light. Stems exhibit positive phototropism

Colored window (red) None

Colored window (blue) Seedlings bend toward light. Blue light (component of

white light) brings about

photosynthesis

Colored window (green) None

Conclusions: Phototropism (LM page 270)

• Do your results support the model that blue light (but not red and green light) reception is involved in positive phototropism of stems? yes Explain how you came to this conclusion. See Table 20.3 results.

• Why is it adaptive for plants to have a way to increase the bending of stems n response to unidirectional light? Bending toward light aids photosynthesis.

20.3 Gibberellins and Stem Elongation (LM pages 270–272)

Experimental Procedure: Stem Elongation (LM pages 270–272)

Do you hypothesize that gibberellins would cause dwarf plants to grow taller? yes Explain. After application of gibberellins, plants have energy needed for growth because stored starch is broken down.

Normal Plants, Untreated Dwarf Plants, Treated Dwarf Plants (LM page 271)

 1–2. Record your data in Table 20.4. The results shown are based on seven plant averages. Students’ results will vary.

Table 20.4 Effect of Gibberellin*

Internode 1 Internode 2 Avg. Internode

Measurement Measurement Distance*

Normal Plants

Age 6 weeks Negligible

Age 8 weeks 13.3 mm

Age 14 weeks 37.0 mm

Age 16 weeks 42.0 mm

Untreated Dwarf Plants

Age 6 weeks Negligible

Age 8 weeks Negligible

Age 14 weeks Negligible*

Age 16 weeks Negligible*

Treated Dwarf Plants

Age 6 weeks Negligible

Age 8 weeks Negligible

Age 14 weeks 5.1 mm

Age 16 weeks 21.2 mm

*Total height has increased, due mostly to the flower head, not to internode length.

Conclusions: Stem Elongation (LM page 272)

• How do the three types of plants differ from each other? Change in internode distance is greatest for normal plants. Internode distance does not change for untreated dwarf plants. Internode distance does eventually increase in treated dwarf plants.

• Did your data support your hypothesis? yes

• If not, why not? Student provides explanation if necessary.

20.4 Etiolation (LM pages 272–273)

Experimental Procedure: Bean Seedling Etiolation (LM page 273)

2. Complete Table 20.5 by explaining why the etiolated plant differs from the normal plant in the characteristics listed.

Table 20.5 Comparison of Normal and Etiolated Plants

Normal Plant Etiolated Plant Explanation

Color of stem Green Pale yellow The synthesis of chlorophyll requires the

Color of leaf Green Pale yellow presence of light. This explains the lack of green color in the etiolated plants. The pale yellow color is due to the presence of other pigmented molecules.

Length of Short Very long Stem elongation has survival value for the

stem seedlings because it increases the chances of reaching light before the seedling’s stored energy supplies are used up.

Size of leaf Large, Small or none Energy has gone into stem length, not well-formed leaves.

Stiffness Very stiff Limp Energy has gone into stem length, not

of stem stem strength.

LABORATORY REVIEW 20 (LM page 274)

1. Contrast the response of a stem and root to gravity. Stems grow up in the opposite direction of gravity, and roots grow down in the same direction as gravity.

2. Does a stem or root show negative gravitropism? stem

3. Explain why the stem of a plant placed in the dark and one placed in the light both respond to gravity. The plant stems are responding to gravity and not to light.

4. Contrast the response of a stem to gravity and to unidirectional light. Stems respond negatively to gravity and positively to unidirectional light.

5. Explain why these stem responses can both be due to the presence of auxin. In response to gravity, auxin increases on the lower side of a stem. In response to unidirectional light, auxin increases on the shady side of a stem. In both cases, auxin brings about elongation and bending of the stem.

6. Which color light is the best stimulus for phototropism seen in stems? Explain. In plants a pigment detects unidirectional light and responds best to a blue light.

7. What happens to a dwarf plant sprayed with gibberellin? Internodes elongate.

8. Offer an explanation for your response in question 6. Gibberellins bring about activity of a gene that codes for amylase. Amylase breaks down starch, providing energy for growth.

9. What is phytochrome? Phytochrome is a blue-green pigment that absorbs activated when it absorbs red light. The active form of phytochrome causes a plant to grow normally and become green.

10. What is etiolation, and how is it related to phytochrome? Etiolation is the abnormal elongation of stems caused by insufficient light. The presence of phytochrome, which occurs in the dark, signals that growth is needed for the plant to reach daylight.

Laboratory 21 Reproduction in Flowering Plants

(LM pages 275–290)

Eleventh Edition Changes

In this edition, the introduction to Section 21.3 Fruits was rewritten to better explain how to use the Key to Major type of Fruits. The experimental procedure which studies the effects of acid rain on seed germination was moved to an ecology lab.

New/Revised Figures: 21.4 Lily

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Fresh material. Observation: A Flower requires monocot and eudicot flowers; Observation: Keying Simple Fruits requires several different types of fruits; Observation: Eudicot and Monocot Seeds requires bean seeds and corn kernels; Experimental Procedure: Seed Germination requires seedlings at various stages of growth.

Overnight preparation time Presoak bean seeds and corn kernels for Observation: Eudicot and Monocot Seeds

21.1 Flowering Plant Life Cycle (LM pages 276–282)

_____ flowers, fresh, monocots and eudicots,

Obtain fresh examples of monocot and eudicot flowers (LM page 280). Tulips, hyacinths, and lilies are examples of monocots. Roses, geraniums, daisies, are examples of eudicots.

21.2 Development of a Eudicot Embryo (LMpage 283)

_____ slides, Shepherd’s Purse embryo, l.s. (Carolina 30-4870 to 30-4894)

21.3 Fruits (LM pages 284–286)

_____ apple, fresh

_____ string bean or pea pod, fresh

_____ sunflower seed

_____ simple fruits, various

21.4 Seeds (LM pages 287--289)

_____ bean seeds, presoaked, one per student, plus twenty per group

_____ corn kernel, presoaked, one per student

_____ scalpel or razor, single blade

Seed Germination (LM page 289)

_____ sunflower seeds

_____ water, tap

_____ flower pots

Sunflower seedlings (LM page 289). Three weeks before the lab plant 4 sunflower seeds in each of two containers. Periodically thereafter, continue to plant sunflower seeds in order to achieve seedlings of increasing height for the lab.

EXERCISE QUESTIONS

21.1 Flowering Plant Life Cycle (LM pages 276–282)

Observation: Flowering Plant Life Cycle (LM page 276)

Use Figure 21.1 as a guide to describe the life cycle of flowering plants.

1. The parts of the flower involved in reproduction are the stamen and the carpel.

2. The anther at the top of the stamen has pollen sacs, which contain numerous microspore mother cells that undergo meiosis to produce microspores.

3. The carpel contains an ovary that encloses the ovules. Within an ovule, a megaspore mother cell undergoes meiosis to produce four megaspores.

4. A microspore undergoes mitosis to produce a pollen grain, which has two cells, the tube cell and the generative cell.

5. One megaspore undergoes mitosis and develops into a(n) embryo sac, which contains two polar nuclei and the egg cell.

6. Pollination is the transfer of pollen from the pollen sacs of the anther to the stigma of the carpel. After a pollen grain lands on the stigma of a carpel, it develops a pollen tube that passes down the style and takes two sperm to the embryo sac.

7. During double fertilization, one sperm from the pollen tube fertilizes the egg within the embryo sac, and the other joins with two polar nuclei.

8. The fertilized egg becomes an embryo, and the joining of polar nuclei and sperm becomes the triploid (3n) endosperm inside a seed. In angiosperms, seeds are enclosed by fruits.

Summary (LM page 278)

Complete the following diagram that summarizes the life cycle of flowering plants.

a. meiosis; b. microspore; c. megaspore; d. mitosis; e. male gametophyte (pollen grain);

f. female gametophyte (embryo sac); g. double fertilization; h. seed; i. mitosis

Where do you find and what happens to the male gametophyte? The male gametophyte is the pollen grain that travels between the stamen and the carpel.

Where do you find and what happens to female gametophyte? The female gametophyte is inside the ovule in the ovary of a carpel. It becomes the embryo plus endosperm.

Structure of a Flower (LM pages 279–280)

In what way is the anther analogous to the testes in human males? Testes produce sperm. Pollen sacs produce microspores that develop into sperm-bearing pollen grains.

In what way is the ovary of a carpel analogous to the ovary in human females? Female ovary produces eggs. Plant ovary contains ovules, which hold an embryo sac that contains an egg.

Observation: The Structure of a Flower (LM page 280)

1. What is the scientific and common name of the plant that produced this flower? Answers will vary depending on type of flower used. Label your drawing using only the main terms on page 279. See Figure 21.3.

a Compare the structure of your flower to that of Figure 21.4 Answers will vary depending on type of flower used.

b. What is the color of your flower and does it have a smell? Answers will vary.

c. Is this flower a monocot or eudicot? Explain. Students should examine their flower for monocot or eudicot characteristics and note them here.

2. What color are the sepals? Answers will vary. Do they resemble the petals in any way? depends on flower observed If yes, how? could resemble petals in color and shape

3. Count the number of stamens and record the number here. depends on flower Is this the same as the petal number or a multiple of the petal number? depends on flower. Can you find two different nuclei? perhaps

4. Describe. resembles a vase in shape Does this ovary contain chambers, each with ovules? depends on flower.

5. A fruit develops from what part of the carpel? ovary wall The seeds came from what structures in the ovary? ovule

Plants and Their Pollinators (LM pages 281–282)

How could it happen that particular bees and particular flowers are suited to one another? Coevolution occurred. How is this mutualistic relationship advantageous to both the plant and the bee? Bee gets nectar and flower obtains pollination.

Observation: Plants and Their Pollinators (LM pages 281–282)

1. If the flower has a smell, which of these two pollinators might pollinate your flower? depends on flower Based on this information, which of these pollinators might pollinate your flower? depends on flower Which of these pollinators might prefer your flower? depends on flower

Table 21.1 Plants and Their Pollinators

See examples in Figure 21.5.

21.2 Development of a Eudicot Embryo (LM page 283)

Observation: Development of the Embryo (LM page 283)

List the stages you were able to identify. Students should identify the stages seen in

Figure 21.6.

21.3 Fruits (LM pages 284–286)

Observation: Fruits (LM pages 285–286)

2. What type of ovary does an apple have? compound How could an animal, such as a deer, help disperse the seeds of an apple? Eat seeds; defecate at a different locale.

3. How many seeds (beans or peas) are in the pod? Answers will vary. Would it help disperse the seeds of a pea plant if an animal were to eat the peas? no Why or why not? Because it would digest them. Is this plant a monocot or eudicot plant? eudicot How do you know? two cotyledons

4. The outercoat of a sunflower seed is actually the part of the fruit. How can examining the seed tell you that the sunflower plant is a eudicot? number of cotyledons Except for the apple, all the fruits you have examined are dry fruits. What does this mean? nonfleshy

5. Examine other available fruits and complete Table 21.2.

Table 21.2 Identification of Simple Fruits

Common Name Fleshy or Dry Eaten as a Type of fruit

Vegetable, (from key)

Fruit, Other

1. Apple Fleshy Fruit Pome

2. String bean Dry Vegetable Legume

3. Pea pod Dry Vegetable Legume

4. sunflower seed Dry Other Achene

21.4 Seeds (LM pages 287–89)

Observation: Eudicot and Monocot Seeds (LM pages 287–289)

Bean Seed (LM pages 287–288)

1. b. What is the function of these cotyledons? Cotyledons store organic food for the seedling until it has leaves to produce organic food.

2. As observed in Figure 21.8, which organ emerges first from the seed—the plumule or the radicle? The radicle emerges first. Of what advantage is this to the plant? The plant can immediately begin to transport water and inorganic nutrients from the soil.

3. What is the advantage of the hypocotyl pulling the plumule up out of the ground instead of pushing it up through the ground? In this way, the hypocotyl protects the plumule.

4. Do cotyledons stay beneath the ground? no

Corn Kernel (LM page 288)

3. As observed n Figure 21.9, does the cotyledon of a corn seed stay beneath the ground? yes

Seed Germination (LM page 289)

1. Order the seedlings in a series according to increase stages of growth. What criteria did you use to order the seedlings? Height and maturity.

2. Is this plant a monocot or eudicot? eudicot What criteria did you use to decide? Leaf venation and two cotyledons are present.

3. Can you see the cotyledons? depends on age of seedlings Explain why the cotyledons of a eudicot shrivel as the seedling grows. The cotyledons contain stored organic food, which the seedling uses. As the seedling uses the organic food, the cotyledons shrivel.

4. Use the space below to draw two contrasting stages of growth. See Figure 18.1 for one possible stage.

LABORATORY REVIEW 21 (LM page 290)

1. What parts of a flower might you be tempted to call the malepart? Why? The filament and anther of the stamen. The anther produces the microspores which become the male gametophyte generation which produces two sperm.

2. Tell what part of a carpel is involved in (1) germination, (2) growth, and

(3) function of a pollen tube. (1) A pollen tube germinates on the stigma,(2) grows through the style, and (3) enters the ovary. In the ovary one sperm fertilizes the egg and the other contributes to the formation of endosperm.

3. How can you tell a monocot flower from a eudicot flower? Monocots have flower parts in threes or multiples thereof. Eudicots have flower parts in fours or fives or multiples thereof.

4. What process results in pollinators that are specific to a particular type of flower? coevolution

5. Explain why a fruit ordinarily contains seeds. An ovary contains ovules. Ovules become the seeds enclosed by the fruit, which develops from the ovary wall.

6. A string bean, tomato, okra, and cucumber are fruits. Explain. A fruit develops from an ovary. Since string beans, tomatoes, okra, and cucumbers all develop from an ovary, they are fruits.

7. How can you tell a monocot seed from a eudicot seed? A monocot seed contains one cotyledon; a eudicot seed contains two cotyledons.

8. Name three general parts of a seed. The three general parts of a seed are the seed coat, the stored food, and the embryo.

9. Relate the plumule and radicle to parts of an adult plant. The plumule becomes the leaves, and the radicle becomes the root.

10. Name two growth pattern differences between monocot and eudicot seeds after germination. In monocots, a coleoptile protects the plumule as it progresses upward through the soil; while in eudicots, the hypocotyl moves upward and pulls the plumule up out of the soil. In eudicots, the cotyledons emerge from the ground, while in monocots, the cotyledon remains below ground.

Laboratory 22 Introduction to Invertebrates

(LM pages 291–310)

Eleventh Edition Changes

The laboratory was rewritten to better stress a hands on examination of living hydras and

planarians, vinegar eels, and rotifers. Also included in this edition is coverage of the flatworm and roundworm parasites. Diversity of sponges, cnidarians, and planarians is covered but students are no longer required to fill out tables.

New/Revised Figures: 22.4 Life cycle of Obelia; 22.6 Hydra feeding; 22.10 Planarian feeding; 22.11 Life cycle of the tapeworm Taenia; 22.12 Anatomy of Taenia; 22.13 Chinese liver fluke; 22.14 Human blood fluke; 22.16 Larva of the roundworm Trichinella embedded in a muscle; 22.17 Elephantiasis.

MATERIALS AND PREPARATIONS

Instruction are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Preserved Specimens. Preserved sponge, Hydra, planarian, Taenia, Ascaris specimens are required for this lab.

Living material. Hydra (LM page 297); small protozoans or crustaceans (LM page 298); planarians (LM page 301); vinegar eels (LM page 308), if desired.

Fresh material. Small piece of meat, such as hamburger or liver (LM page 301).

All Exercises

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand

Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other

clothing protection

_____ microscopes: compound light, stereomicroscope

_____ lens paper

_____ slides (regular and concave depression (Carolina 63-2200) and coverslips

_____ dissecting needle and probe

_____ optional: The Shape of Life. This DVD from PBS () has

information and footage of all the phyla in Figure 22.1.

22.2 Sponges (Phylum Porifera) (LM pages 293–295)

_____ preserved sponge specimens, various (see Carolina’s “Preserved

Organisms, Invertebrates” section)

_____ slide, prepared: Grantia, cross section

_____ slide, prepared: Grantia spicules

22.3 Cnidarians (Phylum Cnidaria) (LM pages 296–299)

_____ Hydra, Life Cycle Set, preserved (see Carolina’s “Preserved Organisms, Invertebrates” section)

_____ slide, prepared: Hydra, cross section

_____ slide, prepared: Hydra, longitudinal section

_____ Hydra, living

_____ small protozoans or crustaceans for feeding a hydra

_____ preserved cnidarian specimens, various

_____ hand lens

_____ 5% acetic acid or vinegar in dropper bottle

Small crustaceans (LM page 298). Sample crustaceans that can be used are: copepods or ostracods.

22.4 Flatworms (Phylum Platyhelminthes) (LM pages 299–305)

_____ planarians, living

_____ hamburger or liver pieces

_____ preserved flatworm specimens, various (see Carolina’s “Preserved

Organisms, Invertebrates” section)

_____ slide, prepared: planaria, whole mount

_____ slide, prepared: planaria, cross section

_____ tapeworm, Taenia pisiformis, preserved (Carolina 22-4355, -4340)

_____ slide, prepared: tapeworm, Taenia pisiformis (Carolina 30-6712, -6718)

_____ flukes, preserved, or slides of various for display (see list below)

_____ slide, prepared: blood fluke, Schistosoma japonicum, male and female

during copulation (Carolina 30-6472)

22.5 Roundworms (Phylum Nematoda) (LM pages 305–306)

_____ preserved roundworm specimens, various (see Carolina’s “Preserved

Organisms, Invertebrates” section)

_____ Ascaris lumbricoides, preserved (Carolina 22-4405, -4410) for dissection

_____ dissecting pans, tools, and trays (see Carolina’s “Dissecting Instruments

and Supplies” section)

_____ pins, insect (for pinning Ascaris) (Carolina 65-4302 to 65-4309)

_____ muscle, preserved, infected with Trichinella (Carolina 22-4440)

_____ slide, prepared: muscle infected with Trichinella (Carolina 30-7044)

_____ vinegar eels, living (Carolina 13-3266)

_____ vinegar, unpasteurized, or vinegar eel medium

_____ dropper bottles of methyl cellulose or Protoslo®

22.6 Rotifers (Phylum Rotifera) (LM pages 308-309)

_____ living rotifer specimen, Brachionus

EXERCISE QUESTIONS

22.1 Evolution of Animals (LM pages 292–293)

Which of the early phyla in the tree have radial symmetry? cnidarians and comb jellies

Which of the phyla in the tree have only two germ layers (i.e., ectoderm and endoderm? cnidarians and comb jellies.

Which pattern of development do the flatworms, rotifers, and roundworms (animals included in this laboratory) have? protostome

22.2 Sponges (Phylum Porifera) (LM pages 293–295)

Anatomy of Sponges (LM pages 293–295)

Observation: Anatomy of Sponges (LM pages 293–295)

Preserved Sponge (LM pages 293–294)

1. Label the arrows in the left-hand drawing of Figure 22.2 to indicate the flow of water.

1. water in; 2. water out

2. Does this particular sponge have pore-lined canals? Answer will depend on specimen examined.

3. Does this sponge have spicules? Answer will depend on specimen examined.

Prepared Slides (LM page 294)

1. a. Explain the expression sessile filter feeder. A sessile filter feeder stays in one place and filters its food from the water.

b. Do you see any spicules? yes Do they project from the wall of a sponge? yes

c. Name two possible advantages of spicules to a sponge. stiffens body wall and protection against predators

2. Examine a prepared slide of sponge spicules. What do you see? Spicules are typically three-rayed, and each ray has a needlelike point.

Conclusions: Anatomy of Sponges (LM page 295)

• The anatomy and behavior of a sponge aid its survival and its ability to reproduce. How does a sponge:

a. Protect itself from predators? spicules

b. Acquire and digest food? Filter food particles, collar cells have digestive vacuoles.

c. Reproduce asexually and sexually? fragmentation; formation of gametes

22.3 Cnidarians (Phylum Cnidaria) (LM pages 296–299)

How can a life cycle that involves two forms, called polymorphism, be of benefit to an animal, especially if one stage is sessile (stationary)? Medusae disperse species.

How can radial symmetry be a benefit to an animal? An animal can reach out in all directions for food.

Anatomy of Cnidarians (LM pages 296–299)

Observation: Cnidarians (LM pages 297–298)

Preserved Hydra

1. Do you see any evidence of buds that are developing directly into small hydras? depends on specimen

Prepared Slide of Hydra

2. Do you find any cells? Answers will vary. Describe them. Flagellated columnar cells line the gastrovascular cavity; cuboidal cells make up the cell wall.

Living Hydra

1. What is the current behavior of your hydra? Answers will vary. What is the reaction of your hydra? Answers will vary.

4. Did your hydra discard any nematocysts? depends on hydra Describe. Student is to describe. Of what benefit is it to Hydra to have cnidocysts? prey capture and protection from predators

Conclusions: Anatomy of Cnidarians (LM page 298)

• The anatomy and behavior of a hydra aid its survival and its ability to reproduce. How does a hydra:

a. Acquire and digest food? Tentacles place food in gastrovascular cavity; digestive juices produce nutrient molecules. Digestion is completed in food vacuoles of nutritive muscular cells.

b. Protect itself from predators? Cnidocysts discharge nematocysts.

c. Reproduce asexually and sexually? budding; production of eggs and sperms

22.4 Flatworms (Phylum Platyhelminthes) (LM pages 299–305)

Why is it advantageous for an animal to be hermaphroditic? Any two animals that meet can reproduce.

Planarians (LM pages 299--302)

Complete the labels in both Figure 22.8b and d. 1. Excretory canal 2. brain 3. nerve cord

Why would you expect an animal that lives in freshwater to have a well-developed excretory system? Rid itself of excess water.

Observation: Planarians (LM page 300)

Preserved Specimen

1. What is the advantage of a gastrovascular cavity that ramifies through the body? distributes nutrients to all parts of body

Prepared Slide

2. Does a planarian have a body cavity? no Explain. The interior of the animal is packed with mesoderm.

Living Specimen (LM pages 301–302)

1. Describe the behavior of the animal. Answers will vary.

2. Does the animal move in a definite direction? probably, yes Why is it advantageous for a predator such as a planarian to have bilateral symmetry and a definite head region? Can move toward prey or away from a predator.

3. What three types of cells must be present for flatworms to be able to respond to stimuli and move about? muscular, nervous, and sensory cells

4. What part of the animal would be able to detect light? eyespots How would the animal respond to the light? moves away

5. Offer the worm some food, such as a small piece of liver, and describe its manner of eating. When a planarian is feeding, a muscular pharynx is extended through the ventrally placed mouth.

6. Describe the mode of locomotion. A planarian moves about in a characteristic slow, gliding fashion, with the head bending from side to side as though testing the environment. This gliding is both ciliary and muscular. If you prod the animal, it hurries away by marked muscular waves.

Conclusions: Planarians (LM page 302)

• Contrast a hydra with a planarian by stating in Table 22.1 any significant differences between them. See Table 22.1 below.

• As with cnidarians, each individual cell takes care of its own needs for these two life functions.

Table 22.1 Contrasts between a Hydra and a Planarian

Digestive System Excretory System Nervous Organization

Hydra Gastrovascular cavity ——— Nerve net

Planarian Gastrovascular cavity Flame cells Ladderlike

Tapeworms (LM pages 302–303)

1. How do humans get infected with a tapeworm? Humans become infected with tapeworms by eating raw or undercooked meat that is infected.

2. What is the function of a tapeworm’s hooks (if present) and suckers? These structures allow the tapeworm to attach itself to the wall of the host’s intestine.

3. Given the life cycle of the tapeworm, why might a tapeworm produce so many eggs? The large number of eggs is needed to disperse the species because the likelihood of the eggs being ingested by a new host is remote.

Fluke Diversity and Anatomy (LM page 304)

Observation: Fluke Anatomy (LM page 304)

1. Can you see the reduced gastrovascular cavity, the reproductive organs noted in

Figure 22.13. Most likely yes.

2. Label the male and female in Figure 22.14. Top leader = male, lower leader= female

Planarians Versus Tapeworms and Flukes (LM page 305)

Why would endoparasites such as flukes and tapeworms have a dispersal problem? They live inside a host and require a means to pass though the environment to acquire new hosts.

22.5 Roundworms (Phylum Nematoda) (LM pages 305–309)

1. What is the advantage of this [complete digestive tract]? The same opening is not used for both intake of food and elimination of waste.

2. For example how would a complete digestive tract lead to a greater number of specialized organs such as a small intestine that assists digestion and a large intestine that assists elimination? Because there are two openings, each part of the tract can become specialized for a particular digestive function. How would a spacious body cavity (instead of packed mesoderm) promote a greater number of diverse internal organs? Now there is room for these organs to evolve.

Observation: Roundworms (LM pages 306)

Ascaris: Dissection of Preserved Specimen (LM page 306)

5. Does this [digestive tract inside outer wall] explain the phrase tube-within-a-tube body plan? Yes, the outer tube is the body wall and the inner tube is the digestive tract.

Trichinella (LM page 307)

Observation: Trichinella (LM page 307)

2. How can trichinosis be prevented in humans? Trichinosis can be prevented by thoroughly cooking all pork and pork products.

3. How can pig farmers help to stamp out trichinosis so that humans are not threatened by the disease? Pig farmers should not feed their pigs fresh pork, thereby preventing passing on any infection.

Conclusion: Anatomy of Roundworms (LM page 309)

• From your knowledge of adaptive radiation, explain why there might be so many different types of nematodes. They have adapted to many different types of environments.

22.6 Rotifers (Phylum Rotifera) (LM page 308-309)

Observation: Rotifers (LM page 309)

2. Describe the rotifer’s behavior. The corona gives the appearance of a moving wheel.

6. What other protostome studied today has both eyespots and flame cells? planarians

7. With the help of Figure 22.19, try to identify some of the rotifer’s internal organs. List those you observe here. See Figure 22.19.

LABORATORY REVIEW 22 (LM page 310)

1. Distinguish invertebrate from vertebrate. Invertebrates lack a vertebral column whereas vertebrates do have a vertebral column.

2. A hydra has radial symmetry but a planarian has bilateral symmetry. Explain. In a hydra any longitudinal cut results in two equal halves but since a planarian has two definite right and left halves only one longitudinal cut results in two equal halves.

3. The animals studied today have some sort of defense. Contrast the defense of a sponge with that of a hydra. Sponges have spicules for defense while a hydra has stinging cells that discharge nematocysts.

4. If a cnidarian has both a polyp stage and a medusa stage, what is the life-cycle function of the medusa stage? The medusa stage disperses the species.

5. Why would you expect an animal with bilateral symmetry to be more active than one with radial symmetry? Bilateral symmetry is consistent with moving ahead in a forward direction.

6. Name two contrasting anatomical features (other than symmetry) that distinguish a planarian from a hydra. A hydra has only two germ layers (ectoderm and endoderm) whereas a planarian also has mesoderm. A hydra has a nerve net and a planarian has a ladder nervous system consisting of a brain and nerve cords.

7. Name a group of animals that is usually hermaphroditic and a group that is usually dioecious. The flatworms are hermaphroditic and the roundworms are dioecious.

8. How does the process of acquiring food in cnidaria differ from the process in sponges? Cnidaria capture whole prey and digest it in a gastrovascular cavity. Sponges are filter feeders--water that enters the osculum exits the pores and, in the meantime, collar cells take up particles from the water.

9. Why would you expect a roundworm to have a nervous system? Roundworms have a muscular body wall, therefore you would expect them to a brain and nerves to control their muscles.

10. Name two types of parasitic roundworm and type of roundworm that is free-living. Ascaris, Trichinella, and filarial worms are parasitic. Vinegar eels are free-living.

Laboratory 23 Invertebrate Coelomates

(LM pages 311–332)

Eleventh Edition Changes

The introduction was revised to better explain the title of this lab and the discussion is accompanied by a supporting illustration. Two virtual labs on the Biology website are called out: Earthworm Dissection (LM page 321) is particularly helpful and directly applicable to the lab; Arthropod Classification (LM page 323) asks students to recognize and use five classes of arthropods.

New/revised Figures: 23.1 Molluscan diversity; 23.3a, Anatomy of a bivalve art; 23.7 External anatomy of an earthworm; 23.14 Grasshopper metamorphosis; 23.16 Anatomy of a sea star

New/revised Tables: 23.1 Comparison of Clam to Squid; 23.2 Comparison of Clam to Earthworm; 23.3 Comparison of Crayfish to Grasshopper

Notes

To shorten the lab, have half the class dissect the clam and earthworm, and the other half dissect the crayfish and grasshopper. After completing the dissections, each half of the class would then identify the structures from their dissections for the other half.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Preserved Specimens. Preserved clams, earthworms, crayfish, grasshopper and sea stars are recommended for student dissections. A pre-dissected squid can be on display. See Carolina’s “Preserved Organisms, Invertebrates” section.

All Exercises

_____ dissecting pans, pins, tools, and trays (see Carolina’s “Dissecting

Instruments and Supplies” section)

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand

Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other

clothing protection

_____ microscopes, compound light and stereomicroscope

_____ lens paper

_____ dissecting tools

_____ optional: The Shape of Life. This DVD from PBS () has

information and footage of all the phyla in this laboratory.

23.1 Molluscs (Phylum Mollusca) (LM pages 312–317)

_____ video (optional): The Anatomy of the Freshwater Mussel (Carolina

49-2365V)

_____ mollusc collection (Carolina 22-4800)

_____ clam, Venus, preserved for dissection

_____ squid, preserved for predissection (optional)

23.2 Annelids (Phylum Annelida) (LM pages 317–322)

_____ video (optional): The Anatomy of the Earthworm (Carolina 49-2372V)

_____ annelid collection (Carolina 22-5000)

_____ earthworm, preserved for dissection

_____ slide, prepared: earthworm cross section

23.3 Arthropods (Phylum Arthropoda) (LM pages 323–29)

_____ video (optional): The Anatomy of the Crayfish (Carolina 49-2403V)

_____ arthropod collection (Carolina 22-5200)

_____ crayfish, preserved for dissection

_____ watch glasses and/or petri dishes to hold dissected organs

_____ grasshopper, Romalea, preserved

_____ Life cycle display showing incomplete (grasshopper) and complete

(tobacco hornworm) metamorphosis (Nasco 1999 catalog, p. 56, or

Carolina metamorphosis set 26-2362), and complete metamorphosis

plastomounts of one or more insects, such as:

mealworm

cabbage butterfly

mosquito

housefly

honeybee

23.4 Echinoderms (Phylum Echinodermata) (LM pages 329–331)

_____ video (optional): The Anatomy of the Starfish

_____ echinoderm collection

_____ sea star, preserved for dissection

EXERCISE QUESTIONS

23.1 Molluscs (Phylum Mollusca) (LM pages 312–317)

On the lines provided in Figure 23.1, write cephalization or no cephalization as appropriate for this mollusc. a. no cephalization; b. no cephalization; c. cephalization; d. cephalization

Anatomy of the Clam (LM pages 312–315)

Observation: Anatomy of the Clam (LM pages 313–315)

External Anatomy

4. What is the function of a heavy shell? protection

Internal Anatomy

1. What is a mantle? The mantle is a covering that partly covers the visceral mass and secretes the shell.

3. What is the advantage of powerful adductor muscles? They keep the two shells closed for protection.

6. Explain the term mantle cavity. The mantle cavity is the space between the folds of the mantle.

7. Explain how water enters and exits the mantle cavity. Water enters through the incurrent siphon, circulates through the mantle cavity, and then exits through the excurrent siphon.

8. Does the clam have a respiratory organ? yes If so, what type of respiratory organ? It has gills.

9. Why is the clam called a filter feeder? Only small particles of food can enter the siphon—large particles remain outside the body of a clam.

15. A clam has an open circulatory system. Explain. The blood is not always enclosed in vessels.

Conclusion: Comparison of Clam to Squid (LM pages 316–317)

• Compare clam anatomy to squid anatomy by completing Table 23.1

Table 23.1 Comparison of Clam to Squid

Clam Squid

Feeding mode Filter feeder Active predator

Skeleton Heavy shell for protection No external skeleton

Circulation Open Closed

Cephalization None Marked

Locomotion Hatchet-shaped foot Jet propulsion

• Explain how both clams and squids are adapted to their way of life. Clams are inactive filter feeders. They have no cephalization and are protected by a shell. Squids are active predators. They have tentacles and jaws to seize and tear apart prey. They lack a shell but have cephalization with a well-developed brain and eyes. An ability to move by jet propulsion helps squids escape from predators.

23.2 Annelids (Phylum Annelida) (LM pages 317–322)

In which group would you expect the animals to be predators based on the type of head region? b. clam worm

Anatomy of the Earthworm (LM pages 319–322)

Observation: Anatomy of the Earthworm (LM pages 319–322)

External Anatomy (LM page 319)

1. Has cephalization occurred? no Explain. There is no well-defined head with sense organs.

3. Do you feel the setae? Yes, tiny bristles can be felt. Setae, along with circular and longitudinal muscles, enable the worm to locomote. Explain the action. The worm moves by alternating contractions of the circular and longitudinal muscles. Setae are extended during longitudinal contraction and retracted during circular contraction.

Internal Anatomy (LM pages 320–322)

4. Does the digestive system show specialization of parts? yes Explain. It consists of a pharynx, esophagus, crop, gizzard, and intestine.

5. Does the earthworm have an open or closed circulatory system? closed Explain. Blood flows within a closed system of vessels.

7. Does the excretory system show that the earthworm is segmented? yes Explain. The nephridia and excretory pores are present in each repeating unit—that is, each segment. Repeating units in which the same body part is present indicates segmentation is present.

9. Is the earthworm hermaphroditic? yes Explain. It has both ovaries and testes. Cross-fertilization still occurs because sperm are passed to another individual.

10. Does the earthworm have a respiratory system? no How does it exchange gases? It exchanges gases through its moist body wall.

11. Why would you expect an earthworm to lack an exoskeleton? An exoskeleton would impede gas exchange.

Prepared Slide (LM page 322)

3. Does the typhlosole help in nutrient absorption? yes Explain. It increases the surface area of the intestine, which means that there is a larger area across which nutrients can be absorbed.

Conclusion: Comparison of Clam to Earthworm

• Complete Table 23.2 to compare the anatomy of a clam to that of an earthworm.

Table 23.2 Comparison of Clam to Earthworm

Clam Earthworm

Habitat In marine and freshwater In loose soil

sediments

Feeding mode Filter feeder Scavenger

Skeleton Bivalve shell (exoskeleton) Hydrostatic

Segmentation Lacking Present

Circulation Open, with internal sinuses Closed with two large vessels and blue blood and red blood

Respiratory organ Gills Body wall

Locomotion Hatchet foot Setae and muscular contraction

Reproductive Sexes separate, marine forms Hermaphroditic; clitellum organs have trochophore larva supplies mucus

23.3 Arthropods (Phylum Arthropoda) (LM pages 323–329)

For each animal in Figure 23.10, circle all obvious appendages. Student should circle any visible legs and antennae and tentacles and tails.

Anatomy of a Crayfish (LM pages 324–325)

Observation: Anatomy of a Crayfish (LM pages 324–325)

External Anatomy (LM pages 324–325)

1. Has specialization of segments occurred? yes Explain. Segments have fused, forming specific regions, and there are specialized appendages on various segments.

3. Do crayfish demonstrate cephalization? yes Explain. There is a distinct head region, although it may be fused with the thorax.

7. Has specialization of appendages occurred? yes Explain. The appendages have specialized functions, such as walking, swimming, reproduction, and feeding.

Internal Anatomy (LM page 325)

1. How many rows of gills are there in your specimen? two The outer row of gills is attached to the base of certain appendages. Which ones? the walking legs

11. Do you see any grinding stones ingested by the crayfish? It depends on the specimen, but most likely, yes. If possible, identify what your specimen had been eating. Answer depends on the specimen.

Anatomy of a Grasshopper (LM page 326–327)

Observation: Anatomy of a Grasshopper (LM pages 326–327)

External Anatomy

2. Which pair of legs is used for jumping? hind legs How many segments does each leg have? five

3. Is locomotion in the grasshopper adapted to land? yes Explain. The grasshopper has jointed legs for walking and jumping on land and wings for flying in air.

Internal Anatomy (LM page 327)

What is the function of Malpighian tubules? Excretion

Conclusion: Comparison of Crayfish to Grasshopper (LM page 327)

• Compare the adaptations of a crayfish to those of a grasshopper by completing Table 23.3. Put a star beside each item that indicates an adaptation to life in the water (crayfish) and to life on land (grasshopper). Check with your instructor to see if you identified the maximum number of adaptations. nine

Table 23.3 Comparison of Crayfish to Grasshopper

Crayfish Grasshopper

Locomotion *Legs and swimmerets *Jumping legs, wings

Respiration *Gills *Tracheae, *spiracles

Sense organs Antennae, compound eyes Antennae, compound eyes,

*tympanum

Nervous system Cephalization Cephalization

External Reproductive

features

Male *Modified swimmerets *Claspers and penis

Female _____ *Ovipositor

Grasshopper Metamorphosis (LM pages 327–328)

Observation: Grasshopper Metamorphosis (LM page 328)

Examine any specific life cycle displays or plastomounts that illustrate complete and incomplete metamorphosis and add these examples in Table 23.4.

Table 23.4 Insect Metamorphosis

Specimen Complete or Incomplete Metamorphosis

Grasshopper Incomplete

Moth Complete

Butterfly Complete

Beetle Complete

Other examples are possible

Conclusions: Insect Metamorphosis (LM page 328--329)

• In insects with incomplete metamorphosis, do the nymphs or the adult have better developed wings? adult What is the benefit of wings to an insect? The ability to seek out and find food sources in other locations; distribution of the species.

• What stage is missing when an insect has incomplete metamorphosis? pupa What happens at this stage? metamorphosis

• What form, the larvae or the adult, disperses offspring in flying insects? the adult How is this a benefit? A species that is widely dispersed is more likely to survive any environmental onslaughts.

• In insects that undergo complete metamorphosis, the larvae and adults utilize different food sources and habitats. Why might this be a benefit? This increases chances of survival of both the larvae and the adult because there is less competition for food and habitat.

23.4 Echinoderms (Phylum Echinodermata) (LM pages 329–331)

When appropriate in Figure 23.16, write “obvious radial symmetry” or “radial symmetry not obvious” beside each photo. a. radial symmetry not obvious; b. obvious radial symmetry; c. obvious radial symmetry; d. obvious radial symmetry; e. obvious radial symmetry; f. radial symmetry not obvious

Anatomy of a Sea Star (LM pages 330–331)

Observation: Anatomy of a Sea Star (LM pages 330–331)

External Anatomy

2. What type of symmetry does an adult sea star have? radial

3. Why is this side of the sea star called the oral side? This is the side where the mouth is located.

4. How many rows of feet are there? several rows on each ray

6. Why is this side of the sea star called the aboral side? The mouth is located on the opposite side.

Internal Anatomy (LM page 331)

6. To which stomach do the digestive glands attach? They attach to the pyloric stomach.

8. What is the function of gonads? Gonads produce the gametes.

10. What is the function of the water vascular system? locomotion

LABORATORY REVIEW 23 (LM page 332)

1. List two advantages of having a coelom. The coelom provides room for the internal organs; it makes a hydrostatic skeleton possible, and coelomic fluid is used in circulation and excretion.

2. What is a protostome, deuterostome? (Explain what these terms mean.) Protostomes are animals in which the first embryonic opening becomes the mouth; deuterostomes are animals in which the first embryonic opening becomes the anus and the second opening becomes the mouth.

3. What are the general characteristics of molluscs? Molluscs have bodies composed of a foot, a visceral mass, and a mantle. Most molluscs also have a shell and a radula.

4. Compare the locomotion of a clam to the locomotion of a squid. How is each suitable to the way the animal acquires food? The clam burrows in mud with its wedge-shaped foot, and is a continuous filter feeder. The squid has jet propulsion and moves quickly toward its prey.

5. Give evidence that earthworms are segmented by stating an organ that occurs in most every segment. paired nephridia

6. Externally, arthropods have specialized parts compared to annelids. Explain. Arthropod body segments are specialized into various body parts, with the appendages of each segment specialized for specific functions.

7. Compare respiratory organs in the crayfish and the grasshopper. How are these suitable to the habitat of each? The crayfish lives in water and has gills suitable to an aquatic habitat. The grasshopper lives on land and uses air tubes (tracheae) for respiration.

8. All animals studied in this laboratory have a coelom and the tube-within-a-tube body plan. Why would you expect all these animals to have specialized parts of the digestive tract? There is room for specialized parts in a coelom, and food passes one way, which lends itself to specialization.

9. For each of the following characteristics, name an animal with the characteristic, and state the characteristic’s advantages:

a. Closed circulatory system: Earthworm, squid: transport is faster with a closed system.

b. Respiratory organ: Clam, squid, and crayfish have gills; insects have tracheae. Gills are an adaptation to living in water; tracheae are an adaptation to living on land.

c. Jointed appendages: Crayfish and grasshopper: jointed appendages provide flexible movement on a solid surface.

d. Exoskeleton: Clam, crayfish: exoskeletons give protection and facilitate muscle attachment for specialized movement.

10. In general, describe the water vascular system of echinoderms. The water vascular system consists of various canals that takes water to the ampulla, which contracts and forces water into the tube feet, organs of locomotion.

Laboratory 24 The Vertebrates

(LM pages 333–352)

Eleventh Edition Changes

Section 24.1 in this edition gives an overview of chordate evolution which is supported by an illustration. Section 24.2 is entitled Invertebrate chordates making it clearer that the tunicates and lancelets are not vertebrates. A frog dissection virtual lab is available on the Biology website and is an alternative to having students actually perform this dissection. The rat has replaced the pig as the mammalian example in this lab.

New/Revised Figures: 24.1 Phylogenetic tree of the chordates; 24.15 Rat anatomy

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Preserved Specimens. Preserved lancelet, frog, perch, pigeon, and rat are required for this lab. Students may dissect the frog themselves or make use of virtual lab on the Biology website. Pre-dissected perch, pigeon, rat, and also the frog if needed, are put on display.

All Exercises

_____ dissecting pans, pins, tools, and trays (see Carolina “Dissecting

Instruments and Supplies” section)

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand

Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other

clothing protection

_____ optional: The Shape of Life. This DVD from PBS () has

information and footage of all the phyla in this laboratory.

24.2 Invertebrate Chordates (Phylum Chordata) (LM pages 335–336)

_____ lancelet (amphioxus), preserved

_____ slide, prepared: lancelet (amphioxus), cross section

24.3 Vertebrates (Subphylum Vertebrata) (LM pages 336–344)

_____ video (optional): The Anatomy of the Shark

_____ video (optional): The Anatomy of the Frog

_____ frog, preserved for dissection

24.3 Comparative Vertebrate Anatomy (LM pages 345–351)

_____ video (optional): The Anatomy of the Perch

_____ perch, preserved

_____ frog, preserved

_____ pigeon, preserved

_____ fetal pig, preserved

_____ model, vertebrate hearts set

EXERCISE QUESTIONS

24.1 Evolution of Chordates (LM page 334)

Explain the term invertebrate chordates. These chordates retain the notochord and do not have a vertebral column.

What three features called out in Figure 24.1 evolved among fishes? jaws, bony skeleton, and lungs

The limbs of tetrapods are jointed appendages. Do all animals develop in a water environment? Yes. Explain. The reptiles including birds and mammals develop within the aminotic membrane which contains aminotic fluid.

24.2 Invertebrate Chordates (Phylum Chordata) (LM pages 335-336)

Invertebrate Chordates (LM pages 335–336)

Observation: Lancelet Anatomy (LM pages 335–336)

4. Has cephalization occurred? no Explain. There is no head or specialized sensory organs, such as eyes, located on the head.

24.3 Vertebrates (Subphylum Vertebrata) (LM pages 336–344)

Anatomy of the Frog (LM pages 337–344)

Observation: External Anatomy of the Frog (LM page 338)

3. What is the function of a tympanum? The tympanum detects sound waves and allows hearing.

4. What is the function of the nares? breathing

5. Which pair of limbs is longest? hind limbs How does a frog locomote on land? Hopping: the hind limbs provide the thrust and the forelimbs withstand the impact of the body when it hits the ground. What is a frog’s means of locomotion in the water? Swimming: the hind limbs provide the thrust for movement.

Observation: Internal Anatomy of the Frog (LM pages 338–344)

Mouth (LM page 338)

5. What is the function of a glottis? Air passes through the glottis into and out of the trachea and lungs.

Respiratory System and Liver (LM page 339)

2. Sequence the organs in the respiratory tract to trace the path of air from the external nares to the lungs. Air goes from the external nares to the internal nares to the glottis to the trachea to the bronchi to lungs.

Circulatory System (LM page 339)

3. Which vessel lies above (ventral to) the other? The posterior vena cava is ventral to the dorsal aorta and the aorta is dorsal to the posterior vena cava.

Digestive Tract (LM page 339-340)

2. Sequence the organs in the digestive tract to trace the path of food from the mouth to the cloaca. Food goes from the mouth to the esophagus to the stomach to the small intestine to the large intestine to the cloaca.

Urogenital System (LM pages 341-344)

7. Explain the term urogenital system. The urogenital system consists of the urinary system and the reproductive system.

8. The cloaca receives material from (1) intestine, (2) bladder, and (3) gonads.

9. Beside each organ listed on the right (of the figureon page 341), tell how the comparable frog organ differs from that of a human. kidney: frog kidney is longer and slimmer than human kidney; mesonephric ducts appear to be similar to the ureters; urinary bladder: in the frog, the bladder is attached to cloaca; the frog has no urethra.

24.3 Comparative Vertebrate Anatomy (LM pages 345–351)

Observation: External Anatomy of Vertebrates (LM pages 345–346)

1. Answer the following questions and record your observations in Table 24.1.

Table 24.1 Comparison of External Features

Perch Frog Pigeon Rat

a. Skin Scaly Smooth Feathers Hairy

b. Segmentation Internally Internally Internally Internally

only only only only

c. Symmetry Bilateral Bilateral Bilateral Bilateral

d. Regions Head, body Head, body Head, body Bilateral

tail tail

e. Neck None None Present Present

f. Post-anal tail Present Internally Present Peesent

only

g. Nares Present Present Present Present

h. Cloaca Absent Present Present Absent

i. Eyelids None Present (one) Present (three) Present (two)

j. Appendages Fins (5) Forelimbs & Wings & Forelimbs &

hindlimbs hindlimbs hindlimbs

2. Which of these (that birds are reptiles) can you substantiate by external examination? a and b

3. What is this an advantage (of an impenetrable covering) in each case? The nearly impenetrable covering provides protection for all. A perch lives in fresh water, and an impenetrable skin prevents water from entering the body. A pigeon and pig live on land, and an impenetrable skin prevents water loss.

4. A frog uses its skin for breathing. The skin is thin and moist and this allows gas exchange to occur.

Observation: Internal Anatomy of Vertebrates (LM pages 347–351)

3. Which of these animals has a diaphragm dividing the body cavity into thorax and abdomen? rat

Circulatory Systems (LM pages 347–350)

Complete Table 24.2.

Table 24.2 Comparative Circulatory Systems

Animal Number of Heart Chambers Pulmonary System (Yes or No)

Frog Three Yes

Perch Two No

Pigeon Four Yes

Rat Four Yes

1. Do fish have a blood vessel that returns blood from the gills to the heart? no Would you expect blood pressure to be high or low after blood has moved through the gills? low

2. What animals studied have a pulmonary vessels that take blood from the respiratory organ and back to the heart ? frog, pigeon, and rat What is the advantage of a pulmonary circuit? Blood pressure can be maintained.

3 Which of these animals has a four-chambered heart? pigeon and pig What is the advantage of having separate ventricles? Separate ventricles keep O2-rich blood from mixing with O2-poor blood.

4. Is the anatomy of birds and mammals conducive to maintaining a warm internal temperature An efficient cardiovascular system supplies muscles with oxygen which give off heat when they contract and the efficient cardiovascular system delivers the heat to the body keeping the body warm.

Respiratory Systems (LM pages 350–351)

Compare the respiratory systems of the frog, perch, pigeon, and pig, and complete Table 24.3 by checking the anatomical features that appear in each animal.

Table 24.3 Respiratory Systems

Glottis Larynx Trachea Lungs Rib Cage* Diaphragm Air Sacs

Perch No No No No No No No

Frog Yes Yes Yes Yes No No No

Pigeon Yes No Yes Yes Yes No Yes

Rat Yes Yes Yes Yes Yes Yes No

*A rib cage consists of ribs plus a sternum. Some ribs are connected to the sternum, which lies at midline in the anterior portion of the rib cage.

1. Among the animals studied, only a perch breathes by gills. Can the particular respiratory organ be related to the environment of the animals? Yes Aquatic animals have gills and terrestrial animals have lungs.

2. Knowing that gills are attached to the pharynx (throat in humans), explain why fish have no trachea. Water enters the pharynx through the mouth and directly passes over the gills, oxygen moves into the capillaries.

3. Can this difference be related to the fact that frogs breathe by positive pressure, while birds and mammals breathe by negative pressure? yes Explain During inspiration in birds and mammals, the volume of the chest (thoracic) cavity increases because the rib cage expands. Because the air pressure decreases, relative to the pressure outside, air is drawn in. In the frog, the external nares close, the floor of the mouth cavity is raised, and air is forced into the lungs by positive pressure.

4. A diaphragm is present only in mammals? Of what benefit is this feature to mammals? Of benefit is this feature to the expansion of lungs in mammals?

The diaphragm makes breathing by negative pressure more efficient because its downward motion helps to increase the chest cavity and lower the pressure in the lungs.

Digestive Systems and Urogenital Systems (LM page 351)

1. Which (liver or pancreas) is the larger, more prominent organ in the pigeon and rat? liver

2. Which vertebrates studied have a cloaca? frog, pigeon

3. Urogenital systems. All the animals have gonads and kidney. Which of these organs is involved in urine production? kidneys . Which of these organs is involved with reproduction. gonads What is the chief biological benefit of the penis in terrestrial animals. Prevent sperm from drying out when male passes sperm to the female.

LABORATORY REVIEW 24 (LM page 352)

1. What are the four characteristics of all chordates?

a. notochord (at least during development)

b. dorsal tubular nerve cord

c. pharyngeal gill pouches

d. post-anal tail

2. Give an example of a

Fish Perch

Amphibian Frog

Reptile (i.e., bird) Pigeon

Mammal Rat

3. Why isn’t a lancelet a vertebrate? The lancelet retains the notochord as an adult and has no vertebrae.

4. Which groups of vertebrates are fully adapted to life on land? reptiles, including birds, and mammals

5. Complete each of the following sentences. In a frog,

a. the glottis allows air to enter the trachea.

b. the esophagus allows food to enter the stomach.

c. the cloaca receives material from the intestine, bladder, and gonads.

d. sperm reach the cloaca by way of the urinary ducts.

6. What is the major difference between the heart of a frog and that of a rat? A frog has one ventricle and a total of three chambers, while a pig has two ventricles, for a total of four chambers.

7. A pulmonary circuit is seen in vertebrate animals adapted to life on land. Explain. A pulmonary circuit takes blood to and from the lungs. Lungs are an adaptation to life on land.

8. What is the major difference between the respiratory system of a perch and that of a frog, a pigeon, and a pig? A perch being aquatic, breathes with gills, while the other animals, being terrestrial, breathe with lungs.

9. Name a major difference between bird and mammalian lungs. A bird has air sacs which allow a one-way flow of air through the lungs and a mammalian lung does not have air sacs and the air goes in and out the same way.

10. When frogs mate, the male presses his cloaca against the female's cloaca. Why does this pass sperm to the female? The reproductive system of frogs connects to the cloaca.

Laboratory 25 Animal Organization

(LM pages 353–368)

Eleventh Edition Changes

No significant changes were made to this laboratory.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

All Exercises

_____ microscopes, compound light

_____ lens paper

25.1 Tissue Level of Organization (LM pages 356–365)

_____ models (if available in your laboratory) or diagrams: epithelial tissue:

simple squamous, simple cuboidal, simple columnar, pseudostratified ciliated columnar

_____ slide, prepared: human simple squamous epithelium

_____ slide, prepared: human skin

_____ slide, prepared: human simple cuboidal epithelium

_____ slide, prepared: human simple columnar

_____ slide, prepared: human pseudostratified ciliated columnar epithelium

_____ slide, prepared: connective tissue sets

_____ slide, prepared: human adipose tissue

_____ model, compact bone or diagram

_____ slide, prepared: compact bone

_____ slide, prepared: human hyaline cartilage

_____ slide, prepared: human blood

_____ model, skeletal, cardiac, and smooth muscle (Wards 81W0650), or

diagram

_____ slide, prepared: human skeletal muscle (Carolina 31-3316, -3328, or

31-3460, which has all three muscle types on one slide)

_____ slide, prepared: human cardiac muscle

which has all three muscle types on one slide)

_____ slide, prepared: human smooth muscle (Carolina 31-3376 or 31-3460,

which has all three muscle types on one slide)

_____ model, neuron or diagram

_____ slide, prepared: neuron

25.2 Organ Level of Organization (LM pages 366–367)

_____ slide, prepared: intestinal wall, cross section

_____ model, human skin or diagram

_____ slide, prepared: human skin

EXERCISE QUESTIONS

25.1 Tissue Level of Organization (LM pages 356–365)

Observation: Simple and Stratified Squamous Epithelium (LM page 356)

Simple Squamous Epithelium (LM page 356)

1. What does squamous mean? flat

2. What shapes are the cells? The cells are thin, flat, and many-sided.

Stratified Squamous Epithelium (LM page 356)

2. Approximately how many layers of cells make up this portion of skin? 40–45 layers

3. Which layers of cells best represent squamous epithelium? outermost layer

Observation: Simple Cuboidal Epithelium (LM page 357)

2. Are these cells ciliated? no

Summary of Epithelial Tissue (LM page 358)

Table 25.1 Epithelial Tissue

Type Structure Function Location

Simple squamous Flat, pancake-shaped Filtration, diffusion, Walls of capillaries,

osmosis lining of blood

vessels, air sacs of

lungs, lining of

internal cavities

Stratified squamous Innermost layers are Protection, repel Skin, linings of

cuboidal or water mouth, throat, anal

columnar; outermost canal, vagina

layers are flattened

Simple cuboidal Cube-shaped Secretion, absorption Surface of ovaries,

linings of ducts and

glands, lining of

kidney tubules

Simple columnar Columnlike—tall, Protection, secretion, Lining of uterus,

cylindrical nucleus absorption tubes of digestive

at base tract

Pseudostratified Looks layered but is Protection, secretion, Linings of respiratory

ciliated columnar not; ciliated movement of mucus passages

and sex cells

Connective Tissue (LM pages 359–362)

Observation: Connective Tissue (LM page 359)

1. What is the function of loose fibrous connective tissue? binds organs together

2. What two kinds of structures in the body contain dense fibrous connective tissue? tendons and ligaments

Observation: Adipose Tissue (LM page 360)

1. Why is the nucleus pushed to one side? The large, fat-filled vacuole, which occupies the center of the cell, pushes the nucleus to one side of the cell.

2. State a location for adipose tissue in the body. Adipose tissue is found in the subcutaneous layer beneath the skin. What are two functions of adipose tissue at this location? insulation and protection

Observation: Compact Bone (LM page 360)

2. What is the function of the central canal and canaliculi? Blood vessels in the central canal bring nourishment which is taken up by cell processes in canaliculi.

Observation: Hyaline Cartilage (LM pages 361)

2. Which of these types of connective tissue is more organized? compact bone Why? Cells are organized in concentric rings in compact bone, whereas cells in hyaline cartilage are in lacunae, which are scattered throughout a matrix.

3. Which of these two types of connective tissue lends more support to body parts? compact bone

Summary of Connective Tissue (LM page 362)

1. Complete Table 25.2 to summarize your study of connective tissue.

Table 25.2 Connective Tissue

Type Structure Function Location

Loose fibrous Fibers are widely Binds organs Between the muscles;

connective separated. together beneath the skin; beneath most epithelial layers

Dense fibrous Fibers are closely Binds organs Tendons, ligaments

connective packed. together, binds

muscle to bones,

binds bone to bone

Adipose Large cell with Insulation, fat Beneath the skin; around the fat-filled vacuole; storage, cushioning, kidney and heart; in the nucleus pushed and protection breast

to one side

Compact Concentric circles Support, protection Bones of skeleton

bone

Hyaline Cells in lacunae Support, protection Nose, ends of bones,

cartilage rings in walls of respiratory

passages; between ribs and sternum

Blood Red and white cells RBCs carry oxygen Blood vessels

floating in plasma and hemoglobin for

respiration; WBCs

fight infection

2. Working with others in a group, decide how the structure of each connective tissue suits its function.

Loose fibrous connective tissue Widely separated fibers allows for ease of movement.

Dense fibrous connective tissue Dense, closely packed fibers give strength to the tendons and ligaments.

Adipose tissue Large cells together make for insulation as well as good fat storage.

Compact bone The matrix gives strength to the bone.

Hyaline cartilage The matrix gives strength and resilience to cartilage.

Blood Blood fluidity allows it to flow through vessels.

Muscular Tissue (LM pages 362–364)

Observation: Cardiac Muscle (LM page 363)

2. What is the function of cardiac muscle? Cardiac muscle is found in the heart and is responsible for contraction of the heart, and thus, pumping of blood.

Observation: Smooth Muscle (LM page 364)

1. What does spindle-shaped mean? Fiber is thick in the middle and thin at the ends.

Summary of Muscular Tissue (LM page 364)

1. Complete Table 25.3 to summarize your study of muscular tissue.

Table 25.3 Muscular Tissue

Type Striations (yes/no) Branching (yes/no) Conscious Control (yes/no)

Skeletal Yes No Yes

Smooth No No No

Cardiac Yes Yes No

2. How does it benefit an animal that skeletal muscle is voluntary, while cardiac and smooth muscle are involuntary? Breathing and the beating of the heart occur all the time, even when we are sleeping; skeletal muscle contraction allows us to move from place to place, as when we are seeking food.

Nervous Tissue (LM page 365)

Observation: Nervous Tissue (LM page 365)

2. Identify the dendrites, cell body, and axon in Figure 25.3 and label the micrograph.

1. dendrite; 2. nucleus; 3. cell body; 4. axon

3. Explain the appearance and function of the parts of a motor neuron:

a. Dendrites short processes that take signals to the cell body

b. Cell body portion of the neuron that contains the nucleus, and therefore performs the usual functions of a cell

c. Axon long process that conducts nerve impulses away from the cell body

25.2 Organ Level of Organization (LM pages 366–367)

Observation: Skin (LM page 367)

2. List the structures you can identify on your slide. Answer will vary

LABORATORY REVIEW 25 (LM pages 367-368)

1. Name four major types of tissues, and state a general function for each. Epithelial tissue: protects internal organs; absorbs and secretes substances. Muscular tissue: voluntary and involuntary movement. Nervous tissue: conducts nerve impulses. Connective tissue: binds structures together, supports and protects organs, stores fat, transports substances, and protects against disease.

2. Describe the shapes of three types of epithelial tissue, and state a function for each. Simple squamous: flat, pancake-shaped; exchange. Simple cuboidal: cube-shaped; absorption. Simple columnar: columnlike; protection, absorption, secretion. Pseudostratified ciliated columnar: looks layered but is not; protection.

3. Describe the appearance of three types of muscular tissue, and state a function for each. Skeletal: striated; voluntary movement. Smooth: spindle-shaped; involuntary movement of visceral organs. Cardiac: striated with branching fibers; involuntary contraction of heart.

4. What is meant by the expression involuntary muscle? Involuntary muscle movement does not require conscious effort and is not under voluntary control.

5. List the three major parts of a neuron, and state a function for each part in a motor neuron. The cell body contains cytoplasm and the nucleus of the neuron. Dendrites conduct signals to the cell body. The axon conducts impulses away from the cell body.

6. Why are certain types of connective tissue called support tissues? Tissues such as cartilage and bone support the other tissues of an organ.

7. Describe how you would recognize a slide of compact bone. The cells are in concentric rings, separated by matrix.

8. The outer region of skin consists of stratified squamous epithelium. Define these terms:

a. Stratified layered

b. Squamous flattened

c. Epithelium a continuous layer of tissue

9. Identify the tissues a.–d. below as epithelial tissue, muscular tissue, nervous tissue, or connective tissue. On the same line, name the structure(s) shown inside the circles.

 a. smooth muscular tissue (muscle); b. connective tissue (blood cells); c. nervous tissue (neuron); d. epithelial tissue (ciliated columnar epithelium)

10. What is the shape of the cell in 9a? spindle shape

Laboratory 26 Basic Mammalian Anatomy I

(LM pages 369–382)

Eleventh Edition Changes

The introduction was revised to introduce the concept that organs work together in the body. Section 26.6 has been reorganized to facilitate student comprehension of organ functions.

Notes

Dissecting tools. As an alternative to a complete set of dissecting tools, two sizes of sharp scissors, forceps, and blunt probes can be used. Scalpels can be kept at the instructor’s bench to be used at the instructor’s discretion.

Safety. Fetal pigs are sometimes preserved by using formalin. Safety goggles, latex gloves, and lab coats or other clothing protection are recommended.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Notes

Preserved Specimens. Preserved fetal pigs are required for this lab. Note: In this laboratory, do not remove any organs because Laboratory 27, Basic Mammalian Anatomy II, contains exercises for the dissecting the organs of the urinary system, the male and female reproductive systems, the respiratory and digestive systems, and the cardiovascular system.

Alternate Procedure To reduce expenses and facilitate the lab, students can observe pre-dissected pigs. In Section 26.1 have students examine a non-dissected pig and use the other sections of the lab to identify the parts of the body and learn the functions of organs in predissected pigs. Save both non-dissected and pre-dissected pigs for the next laboratory classes.

26.1–26.6 All Dissection Exercises (LM pages 370–380)

_____ video (optional): The Anatomy of the Fetal Pig (Carolina 493075V)

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand

Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other

clothing protection

_____ fetal pigs, preserved, for dissection (Carolina 22-8420 to -8422)

_____ dissecting pans, pins, tools, and trays (see Carolina’s Dissecting

Instruments and Supplies section)

_____ pencils

_____ labels, for identifying a student's pig

_____ string, heavy, for tying pigs into dissection pans and for tying plastic bags

_____ plastic storage bags or containers for storing pigs

Fetal pigs (LM pages 370–380). Fetal pigs for dissection are available from many supply houses. Large, double-injected specimens are recommended. You may wish to soak the pigs in water overnight to decrease the smell and concentration of preservatives to which students and instructors are exposed. Before placing pigs into the plastic storage bags, have students pick them up by strings tied around the pigs’ hind legs. Then, if desired, dip pigs in a “pig dip” bucket containing preservative. Alternatively, spray pigs with preservative from a spray bottle before storing.

26.7 Human Anatomy (LM page 381)

_____ model, human torso or

_____ Dimensional Man paper model (ISBN 0-671-70342-0, Fireside Books,

Simon and Schuster 1992.)

Human torso model (LM page 381). Human torso models are available from a number of supply houses. The Carolina Biological Supply Company has a variety of torso models that vary widely in price. See Carolina’s “Models” section. The Dimensional Man paper model is an inexpensive alternative to the torso model.

EXERCISE QUESTIONS

26.1 External Anatomy (LM pages 370–371)

Observation: External Anatomy (LM pages 370–371)

Body Regions and Limbs (LM page 370)

5. Where is the heel of the pig? raised up, off the ground

Umbilical Cord (LM page 370)

3. What is the function of the umbilical cord? It contains the umbilical blood vessels that take blood to the placenta, where fetal blood gives up waste and receives oxygen and nutrients.

Nipples and Hair (LM page 370)

1. How many nipples does your pig have? Both male and female pigs have 16 nipples. When is it advantageous for a pig to have so many nipples? A nursing mother can suckle many offspring.

2. Can you find hair on your pig? yes Where? on the eyelashes and on the chin

Anus and External Genitals (LM page 371)

1. The anus is an opening for what system in the body? digestive system

4. What sex is your pig? Figure 26.1 will aid in distinguishing the sex of your pig.

26.2 Oral Cavity and Pharynx (LM pages 372–373)

Observation: Oral Cavity and Pharynx (LM pages 372–373)

Pharynx (LM page 373)

6. Explain why it is correct to say that the air and food passages cross in the pharynx. Air must pass from the back to the front of the pharynx to enter the trachea, and food must pass from the front to the back of the pharynx to enter the esophagus.

26.3 Thoracic and Abdominal Incisions (LM pages 374–375)

Preparation of Pig for Dissection (LM pages 374-375)

Thoracic Incisions (LM page 374)

3. List the organs you find in the thoracic cavity. The heart and lungs are readily apparent.

Abdominal Incisions (LM page 374)

10. Anatomically, the diaphragm separates what two cavities? the thoracic and abdominal cavities

11. List the organs you find in the abdominal cavity. The liver and intestines are readily apparent.

26.5 Thoracic Cavity (LM pages 376–377)

Observation: Thoracic Cavity (LM pages 376–377)

Heart and Lungs (LM page 376)

3. Sequence the organs of the respiratory tract to trace the path of air from the nasal passages to the lungs. nasal passages, pharynx, glottis, larynx, trachea, bronchi, lungs  

26.6 Abdominal Cavity (LM pages 378–380)

Observation: Abdominal Cavity (LM pages 378–380)

Liver (LM page 378)

2. Name several functions of the liver. destroying red blood cells, producing bile, storing glycogen, maintaining blood glucose levels, producing blood proteins

Stomach and Spleen (LM page 378)

4. The stomach is a part of what system? the digestive system What is its function? stores food, secretes gastric juice, contains an enzyme for protein digestion

5. The spleen is a part of what system? the lymphatic system What is its function? purifies blood and disposes of worn-out red blood cells

Small Intestine (LM page 378)

3. The small intestine is a part of what system? the digestive system What is its function? food digestion and absorption of the products of digestion

Gallbladder and Pancreas (LM page 380)

3. What is the function of the gallbladder? stores and releases bile

4. What is the function of the pancreas? As an exocrine gland it secretes pancreatic juice; as an endocrine gland it secretes insulin and glucagon.

Large Intestine (LM page 380)

4. The large intestine is a part of what system? the digestive system

5. What is the function of the large intestine? absorbs water, prepares feces for defecation

6. Trace the path of food from the mouth to the anus. mouth, pharynx, esophagus, stomach, small intestine, large intestine (colon and rectum), anus

26.7 Human Anatomy (LM page 381)

Observation: Human Torso (LM page 381)

2. In your studies so far, have you seen any major differences between pig internal anatomy and human internal anatomy? no

LABORATORY REVIEW 26 (LM page 382)

1. What two features indicate that a pig is a mammal? mammary glands and hair

2. Put the following organs in logical order: lungs, nasal passages, nasopharynx, trachea, bronchi, glottis. A logical order would be: nasal passages, nasopharynx, glottis, trachea, bronchi, lungs.

3. What difficulty would probably arise if a person were born without an epiglottis? When the individual swallows, food would enter the trachea.

4. What two cavities studied in this laboratory hold the internal organs? the thoracic and abdominal cavities

5. Name two principal organs in the thoracic cavity, and give a function for each. The heart pumps blood, while the lungs exchange gases.

6. What difficulty would arise if a person were born without a thymus gland? The person’s immunity would be reduced, and his or her susceptibility to infections would rise.

7. Name the largest organ in the abdominal cavity and list several functions. Liver. The liver removes poisonous substances from the blood, detoxifies them, and removes vitamins from the blood, which stores them. The liver makes plasma proteins and regulates the cholesterol and glucose level of the blood. The liver also produces bile, which aids digestion.

8. A large portion of the abdominal cavity is taken up with digestive organs. What are they? The digestive organs are the stomach, the small intestine, and the large intestine. The large intestine includes the cecum, colon, rectum, and anal canal.

9. Why is it proper to associate the gallbladder with the liver? The gallbladder is located on the underside of the liver and stores bile made in the liver.

10. Where would you find the pancreas? The pancreas is dorsal to the stomach in the upper left of the abdominal cavity.

Laboratory 27 Basic Mammalian Anatomy II

(LM pages 383–398)

Eleventh Edition Changes

This lab has been reorganized and now begins with the cardiovascular system with emphasis on a study of the heart model. There is some identification of blood vessels in the fetal pig. Dissection of individual organs follows and the lab now ends with the female and male reproductive systems.

Notes

Alternative procedure. To cut down on expenses and facilitate the lab, students can examine pre-dissected pigs instead of doing the dissection themselves. In that case, they can follow along in the lab manual to identify parts of the body and learn the functions of organs.

Dissecting tools. As an alternative to a complete set of dissecting tools, two sizes of sharp scissors, forceps, and blunt probes can be used. Scalpels can be kept at the instructor’s bench to be used at the instructor’s discretion.

Safety. Fetal pigs are sometimes preserved by using formalin. Safety goggles, latex gloves, and lab coats or other clothing protection are recommended.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Preserved Specimens. Preserved sheep heart (optional) and fetal pigs are required for this lab. Note: Laboratory 26, Basic Mammalian Anatomy I, contains exercises for the examination of fetal pig external anatomy and dissections of the oral cavity and pharynx, the neck region, the thoracic cavity, and the abdominal cavity.

27.1–27.4 All Dissection Exercises (LM pages 384–397)

_____ video (optional): The Anatomy of the Fetal Pig (Carolina 49–3075V)

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand

Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other

clothing protection

_____ fetal pigs, preserved, for dissection

_____ dissecting pans, pins, tools, and trays (see Carolina’s “Dissecting

Instruments and Supplies” sections)

_____ labels, for labeling individual pigs

_____ string, heavy, for tying pigs into dissection pans and for tying bags

_____ plastic bags or containers for storing pigs

Fetal pigs. Fetal pigs for dissection are available from many supply houses. Large, double-injected specimens are recommended.

Preparing fetal pigs. You may wish to soak the pigs in water overnight to decrease the smell and concentration of preservatives to which students and instructors are exposed.

Storing fetal pigs. Before placing pigs into the plastic storage bags, have students pick them up by strings tied around the pigs’ hind legs. You may wish to dip pigs in a “pig dip” bucket containing preservative. Alternatively, spray pigs with preservative from a spray bottle before storing.

27.1 Cardiovascular System (LM pages 384–387)

_____ model, heart or preserved sheep heart

27.2Respiratory, Digestive, and Urinary Systems (LM pages 388--389)

stet bold

Respiratory and Digestive Systems (LM pages 388--389)

_____ beaker, small, of water

_____ meter stick or measuring tape

EXERCISE QUESTIONS

27.1 Cardiovascular System (LM pages 384–387)

Observation: Heart (LM pages 384–386)

Heart Model (LM pages 384--386)

2. Which ventricle is more muscular? left Why is this appropriate? The left ventricle pumps blood throughout the body, while the right ventricle pumps blood to the lungs.

3.d. Why are the pulmonary veins colored red in Figure 27.2? They carry oxygen-rich blood. In the pulmonary circuit, the arteries are blue and the veins are red.

Tracing the Path of Blood Through the Heart (LM page 385)

To demonstrate the O2-poor blood is kept separate from O2-rich blood, trace the path of blood from the right side of the heart to the aorta by filling in the blanks with names of blood vessels and valves.

From Venae Cavae From Lungs

Right atrium Pulmonary veins

Right atrioventricular (tricuspid) valve Left atrium

Right ventricle Left atrioventricular (bicuspid or mitral) valve

Pulmonary semilunar valve Left ventricle

Pulmonary trunk Aortic semilunar valve

Pulmonary artery

To Lungs To Aorta

Blood Vessels (LM 386)

Pulmonary Circuit (LM page 386)

1. Sequence the blood vessels in the pulmonary circuit to trace the path blood from the right ventricle to the left atrium of the heart:

Right ventricle of the heart

pulmonary trunk

pulmonary artery

Lungs

pulmonary veins

Left atrium of the heart

2. Which of these blood vessels contains O2-rich blood? pulmonary veins

Observation: Systemic Circuit (LM pages 386–387)

1. With the help of Figure 27.3, complete Table 27.1.

Table 27.5 Other Blood Vessels in the Systemic Circuit

Body Part Artery Vein

Head Carotid Jugular

Front legs in pig Subclavian Subclavian

Kidney Renal Renal

Hind legs Iliac Iliac

2. With the help of Figure 27.3 and Table 27.1, trace the path of blood in the systemic circuit from the heart to the kidneys and from the kidneys to the heart.

Left ventricle

aorta

renal artery

Kidneys

renal vein

posterior vena cava

Right atrium

27.2 Respiratory, Digestive, and Urinary Systems (LM pages 388–391)

Respiratory System (LM page 388)

Name two locations from which air can enter the glottis? nostrils and mouth

Digestive System (LM page 389)

Sequence the organs large intestine, mouth, small intestine, esophagus, stomach, and anus, to trace the path of food from its entrance into the body to the elimination of the remains of digestion. Mouth, esophagus stomach, small intestine, large intestine, and anus.

Observation: Organs of the Digestive System (LM page 389)

2. Does it appear smooth or rough? rough

7. Measure and record in meters the length of the intestinal tract. about 8 meters

8. Considering the function of the small intestine, why would such a great length be beneficial to the body? The increased length allows more surface area for absorption of nutrients.

Observation: Urinary System in Pigs (LM page 389)

5. Sequence the organs in the urinary system to trace the path of urine. Urine travels from a kidney to a ureter, to the urinary bladder, to the urethra.

27.3 Male Reproductive System (LM pages 391–394)

Observation: Male Reproductive System in Pigs (LM pages 392–394)

Penis, Urethra, and Accessory Glands (LM pages 392--393)

6. Sequence the organs in the male reproductive system to trace the path of sperm from the organ of production to the penis. Sperm travel from the testis to the epididymis, to the vas deferens, to the urethra within the penis.

Comparison of Male Fetal Pig and Human Male (LM page 394)

Complete Table 27.3, which compares the location of the penis in these two mammals.

Table 27.3 Location of Penis in Male Fetal Pig and Human Male

Fetal Pig Human

Penis Underneath the ventral skin surface, Hangs in front

posterior to the umbilical cord of scrotum

27.4 Female Reproductive System (LM pages 395–397)

Vagina (LM page 395)

4. The vagina plays a critical role in reproduction even though development of the offspring occurs in the uterus. Explain. The vagina is the organ of copulation in pigs (sexual intercourse in humans) and it is necessary for receiving sperm from the male. The reception of sperm leads to formation of the zygote and a new life form.

Comparison of Female Fetal Pig to Human Female (LM page 397)

Complete Table 27.5, which compares the appearance of the oviducts and the uterus, as well as the presence or absence of a urogenital sinus in these two mammals.

Table 27.5 Comparison of Female Fetal Pig to Human Female

Fetal Pig Human

Oviducts Each leads to horn Each leads to upper

of the uterus portion of uterus

Uterus Two horns Fused, also called simplex

Urogenital sinus Present None

LABORATORY REVIEW 27 (LM page 398)

1. What are the four chambers of the mammalian heart? The four chambers are the right and left atria, and the right and left ventricles

2. Contrast the pumping function of the right and left sides of the heart. The right side of the heart pumps blood to the lungs; the left side pumps blood to the body.

3. Sequence the blood vessels in the systemic system, to trace the path of blood from the left ventricle to the kidneys and back to the right atrium. left ventricle, aorta, renal artery, capillaries in kidney, renal vein, posterior vena cava, right atrium

4. Sequence the organs of the respiratory system from the glottis to the lungs. glottis, trachea, bronchi, lungs

5. What's the difference between the ureters and the urethra in the urinary system? Both are tubular but the ureters) carry urine from the kidneys to the urinary bladder. The urethra carries urine from the urinary bladder to the urogenital opening.

6. Sequence the following organs: stomach, large intestine, small intestine, pharynx, mouth, esophagus, anus. A logical order would be: mouth, pharynx, esophagus, stomach, small intestine, large intestine, anus.

7. Sequence the path of sperm from the testes to the urogenital opening. Testis epididymis, vas deferens, urethra, urogenital opening

8. What organs enter the urogenital sinus in female pigs?

9. Which organ in males produces sperm, and which organ in females produces eggs? In the male, the testes produce sperm; in the female, the ovaries produce eggs

10. How and when do sperm acquire access to an egg in mammals? During copulation (in animals) intercourse (in humans) the male ejaculates sperm into the vagina of a female.

.

Laboratory 28 Chemical Aspects of Digestion

(LM pages 399–408)

Eleventh Edition Changes

With the addition of Figure 28.1, which gives an overview of the organs of digestion, this laboratory now provides a better correlation between the experiments and the digestive tract

of humans. In this edition also, the last section provides a summary for the lab.

New/Revised Figures: 28.1 Organs of the digestive tract and accessory organs

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Equipment. An incubator is required for Sections 28.1 and 28.2. Start these exercises at the beginning of the lab. In Section 28.1 protein digestion requires a 1 1/2 -hour incubation period; in Section 28.2 fat digestion requires incubation and monitoring at 20-minute intervals. To save time, and ensure digestion you may wish to perform the experiment prior to the lab and demonstrate the results of the experiments. The fat digestion tubes can be incubated overnight to ensure digestion occurring.

Enzyme solutions. For the best results, make each section’s enzyme solutions fresh, with no granules, just prior to the lab. However, if solutions must be made ahead of time, make them up as close as possible to the time of use and refrigerate. If incubation is then called for, warm the solution to room temperature in a water bath before incubating. To make a 1% enzyme solution, dissolve 1 g of the enzyme powder in 100 mL of distilled water.

All Exercises

_____ safety goggles (See Carolina’s Safety: Face Protection Section)

_____ latex gloves and/or non-latex gloves (See Carolina’s Safety: Hand

Protection Section)

_____ lab coats (See Carolina’s Safety: Body Protection Section) or other

clothing protection

_____ wax pencils

_____ test tubes and racks

_____ water, distilled

_____ thermometer, celsius (0–110°C)

_____ rulers, plastic millimeter

Test tubes. The exercises in this laboratory require students to add solutions to test tubes. Disposable 12 x 75 test tubes are less expensive than reusable test tubes and students achieve better results than when they wash and re-use test tubes. Disposable test tubes can be marked with sharpies—the marks do not come off in a hot water bath as do wax pencil marks. Measuring: As an expedient, students are asked to mark off the tubes at various centimeter levels with a ruler and then to fill to these marks. You may prefer to have students use a dropper and count the number of drops as per your instructions, or you may prefer to have students use a standard method of measuring volume, such as with a graduated cylinder or a pipette. Most experiments use the standard size test tube. A few experiments require the large size test tube. Mini test tubes can be substituted for most laboratory exercises as long as the total volume in a given tube does not exceed 9 cm. This will reduce the volume of reagents used by approximately one-third.

Test tube sizes/volumes are as follows:

Mini 13 x 100 mm 1 cm3 = 1.0 mL

Standard 16 x 150 mm 1 cm3 = 1.5 mL

Medium large 20 x 150 mm 1 cm3 = 2.4 mL

Large 25 x 150 mm 1 cm3 = 4.0 mL

28.1 Protein Digestion by Pepsin (LM pages 401–402)

_____ 1% albumin solution

_____ 1–2% pepsin solution

_____ 0.2% hydrochloric acid (HCl)

_____ incubator, 37°C

_____ biuret reagent

Order solutions or prepare your own:

Albumin solution (LM page 401). Prepare 10 mL per student group. Mix in a pH 7 buffer solution (Carolina 84-9380, -9683) as per directions on the vial, and dissolve 1 g per 100 mL of water. Allow time for precipitation to occur, and then decant. Swirl the stock prior to distribution to students. Also, check pH with indicator paper, and adjust to pH 7 with dilute acid or base.

1–2% pepsin solution (LM page 401). Prepare 20 mL per student group as close to time of lab as possible. To make a 1% solution, dissolve 1 g of pepsin in 100 mL of distilled water. Refrigerate between labs but then warm student tubes to room temperature in a warm, not hot, water bath.

0.2% hydrochloric acid (HCl) (LM page 401). Add 0.57 mL of concentrated HCl to

100 mL of distilled water.

Biuret reagent (LM page 401). 30 mL per student group should be sufficient (using standard test tubes for all procedures). If you buy prepared biuret, use only ten to fifteen drops; otherwise, the solution will be too dark, or dilute to a 10% solution (10 mL biuret with 90 mL distilled water). To prepare your own biuret reagent, maintain separate stock solutions of 3% copper sulfate—3 g of copper sulfate (cupric sulfate, Carolina 85-6550) per 100 mL of distilled water and 10% potassium hydroxide or sodium hydroxide—100 g of potassium hydroxide (Carolina 88-3488) or sodium hydroxide pellets (Carolina 88-9470) per 1,000 mL of distilled water). Adding five drops of copper sulfate solution and ten drops of potassium hydroxide solution to each experimental tube produces more consistent results. Biuret reagent should be prepared fresh for each lab.

28.2 Fat Digestion by Pancreatic Lipase (LM pages 403–404)

_____ vegetable oil, preferably canola, olive, or sunflower

_____ phenol red solution

_____ pancreatin (pancreatic lipase)

_____ bile salts (Wards Biology 38W2179). One gram is enough for a class.

_____ incubator, 37°C

Students should not vigorously shake tubes (LM page 404). Oil will float above phenol red solution and pancreatic lipase solution, and a color change may be observed in the transition zone; however, it will not be uniform unless bile salts are added.

Order solutions or prepare your own:

Phenol red solution (LM page 404). Prepare 20 mL per student group. Use a 0.04% solution. Dissolve 0.04 g of phenol red in 100 mL of distilled water.

Pancreatic lipase solution (LM page 404) (1% pancreatin in 0.1% Na2CO3). Prepare 30 mL per student group. As close to time of lab as possible, add 1 g pancreatin to every 100 mL of 0.1% Na2CO3 (0.1 g Na2CO3 per 100 mL of distilled water). Refrigerate between labs but then warm student tubes to room temperature in a warm, not hot, water bath.

28.3 Starch Digestion by Pancreatic Amylase (LM pages 405–406)

_____ pancreatic-amylase solution

_____ starch suspension

_____ boiling water bath

_____ test-tube holder

_____ hot plate

_____ beaker

_____ beaker tongs

_____ iodine-potassium-iodide (IKI) solution, premade

_____ Benedict’s reagent powder or Benedict’s reagent

solution

Order solutions or prepare your own:

Pancreatic-amylase solution (LM page 405). Prepare 20 mL per student group as close to time of lab as possible. Dissolve 1 g pancreatic-amylase in 100 mL distilled water. Refrigerate between labs but then warm student tubes to room temperature in a warm, not hot, water bath.

Starch suspension (LM page 405). Prepare 20 mL per student group. A fresh supply of this solution must be carefully prepared every day. To make a 1% starch suspension, dissolve 1 g of starch in a small amount of cold water to form a paste. Add this to 100 mL of boiling distilled water, and mix a few minutes. Cool. Add a pinch of sodium chloride (NaCl).

Iodine (IKI) solution (LM page 406). Prepare one dropper bottle per student group. For ease of comparison, the same amount should be used each time. Pre-made iodine-potassium-iodide solution can be purchased, or the ingredients can be purchased separately as potassium iodide (KI) (Carolina 88-3790, -3792) and iodine (I) (Carolina 86-8970, -8972). These dry ingredients have a long shelf life and can be mixed as needed, according to the instructions in Laboratory 2.

EXERCISE QUESTIONS

28.1 Protein Digestion by Pepsin (LM pages 401–402)

The stomach has a very low pH. Does this indicate that pepsin works effectively in an acidic or basic environment? acidic

Test for Protein Digestion (LM pages 401)

Experimental Procedure: Protein Digestion (LM pages 401–402)

Table 28.1 Protein Digestion by Pepsin

Tube Contents Temperature* Results Explanation

of Test

1 Albumin 37°C Pinkish-purple Digestion; enzyme and

Pepsin correct pH

HCl

2 Albumin 22°C Light purple Some digestion; temperature

Pepsin to pale is low HCl pinkish-purple

3 Albumin 37°C Purple No digestion; incorrect pH

Pepsin

Water

4 Albumin 37°C Purple No digestion (no enzyme);

Water control

* If solutions are cold, warm the tubes to room temperature in a warn, not hot, water bath before incubating and this will shorten the time of incubation.

Conclusions: Protein Digestion (LM page 402)

• Explain your results in Table 28.1 by giving an explanation why digestion did or did not occur. See Table 28.1. Now show here that Tube 1 met all the requirements for digestion.

Pepsin is the correct enzyme.

Albumin is the correct substrate.

37º C is the optimum temperature.

HCl provides the optimum pH

1 1/2 hours provides time for the reaction to occur.

• Which tube was the negative control? tube 4 Explain. Tube 4 contained no enzyme (pepsin).

• If this control tube had given a positive result for protein digestion, what could you conclude about this experiment? The experiment is invalid.

28.2 Fat Digestion by Pancreatic Lipase (LM pages 403–404)

With regard to the second step, would the pH of the solution be lower before or after the reaction? The pH would be lower after the reaction.

Test for Fat Digestion (LM pages 403)

Experimental Procedure: Fat Digestion (LM page 404)

Table 28.2 Fat Digestion by Pancreatic Lipase

Tube Contents Time* Color Change Explanation

Initial Final

1 Vegetable oil 60--90 min Red Yellow Digestion

Phenol red

Pancreatic lipase should be pancreatin per LM?

Bile salts

2 Vegetable oil Red Pink (red Limited digestion; no

Phenol red at border) emulsifier

Pancreatic lipase should be pancreatin per LM?

3 Vegetable oil Red Red No digestion (no

Phenol red enzyme or emulsifier);

Water control

*If solutions are cold, warm to room temperature in a warm, not hot, water bath before incubating; this will shorten the time needed for incubation.

Conclusions: Fat Digestion (LM page 404)

• Explain your results in Table 28.2 by giving a reason why digestion did or did not occur.

• What role did bile salts play in this experiment? Bile salts act as an emulsifier and break large drops of fat into very small droplets. This makes more fat molecules available for digestion.

• What role did phenol red play in this experiment? Phenol red was a pH indicator that turns yellow when digestion occurs.

• Which test tube in this experiment could be considered a negative control? tube 3

28.3 Starch Digestion by Pancreatic Amylase (LM pages 405–406)

1. If digestion does not occur, which will be present—starch or maltose? starch

2. If digestion does occur, which will be present—starch or maltose? maltose

3. What happens to enzymes when they are boiled? They denature. Will digestion occur? No.

Tests for Starch Digestion (LM pages 405)

Experimental Procedure: Starch Digestion (LM pages 405–406)

Why do you expect tube 1 to have a positive test for starch and tube 2 to have a negative test for starch? Not enough time was given for digestion take place.

Why do you expect tube 3 to have a negative test for starch and tube 4 to have a positive test for sugar? All requirements including enough time were provided for digestion to occur.

Why do you expect tube 5 to have a positive test for starch and tube 6 to have a negative test for sugar? The enzyme was denatured by boiling therefore digestion cannot occur.

Why do you expect tube 7 to have a positive test for starch and tube 8 to have a negative test for sugar? These tubes contain only starch and water.

Table 28.3 Starch Digestion by Amylase*

Tube Contents Time Type Results Explanation

of Test

1 Pancreatic-amylase 0 Iodine Black No digestion; not

Starch enough time

2 Pancreatic -amylase 0 Benedict’s No change No digestion; not

Starch enough time

3 Pancreatic -amylase 30 Iodine No change Digestion occurred

Starch

4 Pancreatic -amylase 30 Benedict’s Green-orange Digestion occurred

Starch

5 Pancreatic -amylase, 30 Iodine Black No digestion; enzyme

 boiled is denatured

Starch

6 Pancreatic -amylase, 30 Benedict’s No change No digestion; enzyme

boiled is denatured

Starch

7 Water 30 Iodine Black No digestion;

Starch negative control

8 Water 30 Benedict’s No change No digestion;

Starch negative control

*If solutions are cold, warm to room temperature in a warm, not hot, water bath, waiting the 30 minutes and/or testing directly.

Conclusions: Starch Digestion (LM page 406)

• This experiment demonstrated that, in order for an enzymatic reaction to occur, an active enzyme must be present, and time must pass to allow the reaction to occur.

• Which test tubes served as a negative control in this experiment? tubes 7 and 8 Explain. Tube 7 and tube 8 do not contain the enzyme pancreatic amylase.

28.4 Requirements for Digestion (LM page 407)

Explain in Table 28.2 how each of the requirements listed influences effective digestion.

Table 28.4 Requirements for Digestion

Requirement Explanation

Specific enzyme Each enzyme speeds only one type of reaction.

Warm temperature Chemical reactions occur at a faster rate at warm temperatures than at cold temperatures.

Time It takes time for the reaction to occur.

Specific pH Optimum pH maintains the shape of the enzyme so that the enzyme will combine with its substrate.

Fat emulsifier Fats are insoluble in water. The emulsifier breaks up fat so that fat droplets are exposed to the enzyme.

1. How is the amount of substrate reduced? Patient can eat very little at a time.

2. How is the amount of digestive enzymes reduced? Duodenum is shortened just

where pancreatic enzymes enter the duodenum. A smaller tract produces less enzymes,

such as peptidases, also.

3. How is the time reduced? Food passes more quickly through a short digestive tract.

4. What makes the pH of the small intestine higher than before? Pancreatic juices

which are basic no longer enter the tract as before.

5. How is fat emulsification reduced? Bile salts do not enter the duodenum as before.

6. How does surgery to reduce obesity sometimes result in malnutrition. The

requirements for digestion as outlined in Table 28.4 are not met.

LABORATORY REVIEW 28 (LM page 408)

1. Give a reason why amylase will not digest protein? Enzymes are specific. An enzyme that breaks down starch (i.e., amylase) cannot break down protein.

2. Enzymes perform better at room temperature than when they are boiled. Explain. Boiling destroys the shape of an enzyme.

3. Relate the expectation of more product per length of time to the fact that enzymes are used over and over. With time, each enzyme molecule can act more times; hence, more product.

4. Why do enzymes work better at their optimum pH? Optimum pH maintains the shape of the enzymes.

5. Why is an emulsifier needed for the lipase experiment but not for the pepsin and amylase experiments? Fat is insoluble in water, and the emulsifier makes it disperse in water.

6. Which of the following two combinations is most likely to result in digestion?

a. Pepsin, protein, water, body temperature

b. Pepsin, protein, hydrochloric acid (HCl), body temperature. Explain.

The second combination (pepsin, protein, hydrochloric acid [HCl], body temperature) is more likely to result in digestion because all requirements for digestion are present, including optimum pH for pepsin.

7. Which of the following two combinations is most likely to result in digestion?

a. Amylase, starch, water, body temperature, testing immediately

b. Amylase, starch, water, body temperature, waiting 30 minutes Explain.

The second combination (amylase, starch, water, body temperature, waiting 30 minutes) is more likely to result in digestion because time has been given for enzyme to act.

8. The test used for fat digestion was the presence of acids. Explain why this works. A fat consists of glycerol and fatty acids. Fatty acids released with digestion bring about acidic conditions, as detected by a pH indicator.

9. Given that, in this laboratory, you tested for the action of digestive enzymes on their substrates, what substance would be missing from a negative control sample? The enzyme is missing from a negative control sample.

10. Knowing that blood clotting is an enzymatic reaction, would you place a cut finger under warm or cold running water? warm water Explain.

Laboratory 29 Homeostasis

(LM pages 409–424)

Eleventh Edition Changes

The laboratory was rewritten to introduce more hands-on activities, including blood pressure and lung volume measurement and comparative urine analysis to diagnose particular illnesses.

New/Revised Figures: 29.3 The heartbeat sounds; 29.4 Measurement of blood pressure and pulse; 29.6 Healthy lungs tissue versus emphysema; 29.7 Inspiration and expiration; 29.8 Nine-liter student wet spirometer; 29.10 Nephron structure and blood supply.

New/Revised Tables: 29.1 Blood Pressure 29.2 Measurements of Lung Volumes

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

29.1 Heartbeat and Blood Flow (LM pages 410–412)

_____ stethoscope (Carolina 69-1634)

_____ automatic blood pressure monitor (Carolina 69-1120, -1124

Blood pressure monitors. Several models of automatic blood pressure monitors are available from Carolina Biological Supply Company.

29.2 Blood Flow and Systemic Capillary Exchange (LM pages 413–414)

_____ goldfish, live (Carolina 14-5280, -5282); or frog, live (Carolina’s “Living Organisms: Animals, Frogs” section)

_____ stereomicroscope

For fish tail viewing:

_____ cotton, absorbent

_____ petri dish

_____ glass slide

For frog foot viewing

_____ cheesecloth or paper toweling or absorbent cotton

_____ scotch tape

_____ corkboard

_____ pins, dissecting, and thread

Live goldfish. A variety of goldfish are available from the Carolina Biological Supply Company and from local aquarium shops. Wrap the body of the goldfish in wet absorbent cotton so that only the tail is exposed. Place it in a petri dish with a small amount of water. Cover the tail with a glass slide to hold in flat and expanded. Examine under low and high power of a stereomicroscope. Observe the swiftly moving blood in the arterioles and the more slowly moving blood in the venules (moving in the opposite direction). Look carefully for blood cells floating in the bloodstream. Set up similar demonstrations so that all students may observe circulation within a 15-minute period. Return fish to the aquarium.

Live frogs. A variety of frogs are available from the Carolina Biological Supply Company. Wrap a live frog in wet cheesecloth, paper toweling, or absorbent cotton. Lay the animal on a frog board made of cork, and secure in position with scotch tape. Punch a hole in one corner of the corkboard, and spread the web of one foot over this hole; use thread wound around the toes and secured with pins to keep the toes immobile. Keep the frog’s skin wet during the demonstration. Place the board on the stage of the stereomicroscope, and focus on the webbed foot. Observe capillaries, arterioles and venules, and the rate of blood flow. Notice the direction of the blood flow and the pulsations of the blood vessels. Estimate the diameter of the capillaries using the size of the red blood cells as a reference: 22 microns long, 15 microns wide, and 4 microns thick. Under high power, compare the movement of both red and white blood cells. Notice the appearance and disappearance of capillaries. To study the effect of temperature on circulation, place a small piece of ice on the web.

Alternative to live animals. View any video or DVD illustrating blood flow in the cardiovascular system.

29.3 Lung Structure and Human Respiratory Volumes (LM pages 414–417)

_____ slide, prepared: lung tissue

_____ microscopes, compound light

_____ lens paper

_____ spirometer, student wet, 9-L capacity

_____ mouthpieces, disposable

Spirometer. Student Wet Spirometer (Carolina 692661) is a classic physiology instrument used to determine pulmonary volumes and capacities. Durable and economically priced, this precision water bell spirometer has a 9-L capacity. It allows students to make easy and accurate measurements of vital capacity (VC), tidal volume (TV), inspiratory reserve volume (IRV), inspiratory capacity (IC), expiratory reserve volume (ERV), and expiratory capacity (EC). Includes instruction manual and 100 Disposable Mouthpieces

29.4 Kidneys (LM pages 418–423)

_____ model, kidney

_____ model, nephron

_____ Chemstrip—urine test strips (Carolina 69-5967)

_____ simulated urine sample

_____ small beakers for urine samples

_____ paper towel

Kidney model and Nephron model (LM page 418-419). Carolina Biological Supply has a large number of kidney and nephron models that vary widely in price. See the “Models” section of the Carolina catalog to select the most appropriate one for your needs.

Albumin solution (for urinespecimen, see below): Dissolve (while stirring) 1 g per 100 mL of distilled water. Decant if necessary. Check pH with indicator paper, and adjust to pH 5 with dilute acid or base.

Glucose solution (for urine specimen, see below): Add (while stirring) 40 g of dextrose (D-glucose) to 40 to 50 mL of heated distilled water. Increase volume to 100 mL.

Simulated urine specimens (LM page 422). Three specimens per student are needed:

(1) normal, (2) protein in urine, (3) sugar and ketones in urine.

Normal: It is easiest to prepare synthetic urine in 1,000 mL quantities. Using a low concentration hydrochloric acid solution (0.1 M suggested), adjust the pH of 1,000 mL distilled water to pH 5, using pH paper or a pH meter. Add phenol red solution to yield a slight “urine-yellow” color if desired

Protein in urine: Add the albumin solution, previously prepared, in 5 mL amounts to 1,000 mL "normal urine". Periodically test with a urine test strip and stop adding albumin when a positive test is obtained.

Sugar and ketones in urine: Add glucose solution, previously prepared, in 5 mL amounts and acetone in 4 mL amounts to 1,000mL "normal urine". Periodically test with a urine test strip, and stop adding glucose and acetone when positive tests are obtained. Check pH and add acid (0.1 M HCl) as needed to ensure a lower pH than 5.

EXERCISE QUESTIONS

29.1 Heartbeat and Blood Flow (LM pages 410–413)

Liver. The hepatic portal vein lies between the intestine and the liver.

Experimental Procedure: Heartbeat at Rest (LM page 412)

2. Which of the two sounds (lub or dub) is louder? Dub

Blood Pressure (LM pages 412--413)

Blood pressure is highest just after ventricular systole, and it is lowest during ventricular diastole. Why? During systole, a chamber contracts and during diastole a chamber rests.

We would expect a person to have lower blood pressure readings at rest than after exercise. Why? When a person is at rest, the heart does not pump as hard.

Experimental Procedure: Blood Pressure at Rest and After Exercise (LM pages 412--413)

Table 29.1 Blood Pressure

Data will depend on individual. Blood pressure of 120/80 is considered normal at rest.

Conclusions: Blood Pressure (LM page 413)

• Knowing that exercise increases the heart rate, offer an explanation for your results. Blood pressure increases when the heartbeat increases.

• Under what conditions in everyday life would you expect the heartbeat and the blood pressure to increase, even though you are not exercising? The heartbeat and blood pressure increase when you are under stress. When might this be an advantage? This is an advantage; stress is due to a physical danger because oxygen is getting to your muscles faster.

A disadvantage? This is a disadvantage when stress is not due to an emergency because high blood pressure can cause illness.

29.2 Blood Flow and Systemic Capillary Exchange (LM pages 413--414)

Systemic Capillary Exchange (LM page 414)

Conclusions: Systemic Capillary Exchange (LM page 414)

• What generates blood pressure? The heartbeat.

• Why are tissue cells always in need of glucose and oxygen? To carry on cellular respiration. Add glucose and oxygen at the of arrows in Figure 29.5

• Why are tissue cells always producing carbon dioxide? As a waste product of cellular respiration. Add carbon dioxide and wastes at the start of arrows in Figure 29.5. Do the same for metabolic wastes

Figure 29.5:

oxygen

 ( glucose

 ( carbon dioxide

 ( metabolic wastes

  (

29.3 Lung Structure and Human Respiratory Volumes (LM page 414-417)

Observation: Lung Structure (LM page 415)

4. In Figure 29.6a, add arrows to show an exchange between normal alveoli and pulmonary capillaries. One arrow should take O2 out of alveoli into blood and another arrow should take CO2 from blood into alveoli. [Note: This would be easily accomplished if students draw in a capillary at the bottom of the figure and then take arrows forward and backward between an alveolus and the capillary.]

Tidal Volume (TV) (LM page 416)

4. In your own words, what is tidal volume? The normal amount of air that moves into and out of lungs with each breath.

Expiratory Reserve Volume

4. In your own words, what is expiratory reserve volume? The amount of air you can force out of your lungs after a normal breath.

Vital Capacity

4. In your own words, what is vital capacity The amount of air that you can force out of your lungs after forcibly inhaling.

Inspiratory Reserve Volume

It will be necessary to calculate IRV. Explain The spirometer only measures expired air.

Table 29.2 Measurements of Lung Volumes*

Tidal Volume (TV) Expiratory Reserve Vital Capacity (VC) Inspiratory Reserve

Volume (ERV) Volume (IRV)

Average 600 mL 1,000 mL 5,200 mL 3,600 mL

*Average values for a male.

Conclusions: Human Respiratory Volumes (LM page 417)

How does your vital capacity compare to the typical? Answers will vary

How does good lung ventilation assist gas exchange? With good lung ventilation, greater gas exchange would occur.

29.4 Kidneys (LM pages 418–423)

Kidney Function (LM pages 420–422)

Glomerular Filtration (LM page 421)

2. In the list that follows, draw an arrow from left to right for the small substances that leave the glomerulus and become a part of the filtrate.

Glomerulus Glomerular Capsule (Filtrate)

Cells

Proteins

Glucose (

Amino acids (

Urea (

Water and Salts (

Complete the second column of Table 29.4 by using an X to show the substance is in the filtrate.

3. What substances are too large to leave the glomerulus and enter the glomerular capsule? cells and proteins

Tubular Reabsorption (LM page 421)

1. Use this information to state a way the kidneys help maintain homeostasis. Kidneys help maintain normal blood volume and pressure.

2. What would happen to cells if the body lost all its nutrients by way of the kidneys? Cells would die.

3. In the list that follows, draw an arrow from left to right for all those molecules passively reabsorbed into the blood. Use darker arrows for those actively reabsorbed:

Proximal Convoluted Tubule Peritubular Capillary

Water and Salts (

Glucose ( darker arrow

Amino acids ( darker arrow

Urea

4. Which of these substances is not reabsorbed and will become a part of urine? urea Urea is a nitrogenous waste. State here a way that kidneys contribute to homeostasis. Kidneys excrete nitrogenous wastes.

Tubular Secretion (LM page 422)

1. Complete the last column of Table 29.4. Check your entries against Figure 29.10.

2. Use this information to state a third way the kidneys contribute to homeostasis. The kidneys maintain the normal pH of the blood.

Table 29.4 Urine Constituents

Substance In Filtrate In Urine

Cells ___ ___

Protein (albumin) ___ ___

Glucose X ___

Amino Acids X ___

Urea X X

Water and Salts X X

NH4+ X

Kidney Function (LM page 422)

• The presence of urea in urine illustrates which of the kidney’s functions? Excretion. nitrogenous wastes The liver makes urea. What do the kidneys produce? urine

• The presence of NH4+ in the urine illustrates which of the kidney’s functions? Maintains pH of blood.

• Regulation of the blood’s water and salt content by the kidneys helps maintain the blood volume and pressure within normal limits.

Urinalysis: A Diagnostic Tool (LM pages 422–423)

Experimental Procedure: Urinalysis (LM pages 422–423)

Enter the test results in Figure 29.11. Use + and - symbols and a number for the pH.

Figure 29.11

Tests for: Normal Test 1 Test 2 Test 3

leukocytes = - - -

pH 5 5 5 3

protein - - + -

glucose - - - +

ketones - - - +

blood - - - -

Conclusions: Urinalysis (LM page 423)

• State below if the urinalysis is normal or indicates a urinary tract infection or the patient has diabetes mellitus (sugar in the urine).

Test strip 1 Normal

Test strip 2 Infection

Test strip 3 Diabetes Mellitus

• Explain why. The patient is metabolizing fat because glucose is not available. Why is the pH of urine lower than normal? Fats break down to acidic ketones.

• If urinalysis shows that proteins are excreted instead of retained in the blood, would capillary exchange in the tissues (see Fig. 29.5) be normal? yes, but the patient would lose water. Why? The loss of plasma proteins in the blood reduces the osmotic pressure of blood that draws water back into the capillaries.

LABORATORY REVIEW 29 (LM page 424)

1. In your own words what is homeostasis? Homeostasis is the dynamic equilibrium of the body’s internal environment.

2. Explain how the capillaries help to maintain homeostasis. Capillaries bring nutrients to cells and remove their wastes.

3. Relate a blood pressure of 120/80 to systole and diastole of the ventricles during a heartbeat. When would blood pressure be 120? During systole of ventricles. When would blood pressure be 80? During diastole of ventricles.

4. If a smoker has a low tidal volume, why might he or she feel tired and run down? Not enough air (oxygen) is entering the alveoli and therefore not enough oxygen is getting into the blood and to his or her cells.

5. How would you measure the effects of exercise on vital capacity? Determine vital capacity with a spirometer before exercise and after exercise.

6. What role do the kidneys have in maintaining blood pressure and volume? During urine formation the kidneys reabsorb water and salts and this maintains blood pressure and volume.

7. List the three steps in urine formation and define. Filtration (small molecule, i.e., nutrients and nitrogenous waste, enter glomerular capsule); reabsorption (nutrients water and salts are absorbed back into blood from rest of nephron); and tubular secretion (additional wastes enter proximal tubules).

8. With regard to urine formation, name a substance found in both the filtrate and the urine. water or salts or urea Explain. Water, salts, and urea are found in both since they are filtered, not reabsorbed completely, and then any remaining in the tubule are excreted.

9. With regard to urine formation, name a substance found in the filtrate and not in the urine. glucose or amino acids Explain why the substance is only in the filtrate and not in the urine. Glucose and amino acids are found in the filtrate but not in the urine because they are normally reabsorbed completely.

10. After a urinalysis test, what medical condition would be indicated by a positive test for glucose in the urine? diabetes mellitus.

Laboratory 30 Nervous System and Senses

(LM pages 425–443)

Eleventh Edition Changes

In Section 30.1, the sequence has been changed and the parts of the brain are now discussed from the cerebrum to the medulla oblongata. Student labeling of the human brain is now an included exercise. The spinal cord and spinal nerves are discussed separately allowing the use of the terms central nervous system and peripheral nervous system. Animal chemoreceptors was removed from the lab.

New/Revised Figures: 30.4 The human brain; 30.14 Evolution of the human ear

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

All Exercises

Models. Poster or good quality 35-mm color slides of models can be used in place of models.

30.1 Animal Nervous Systems (LM pages 426–432)

_____ sheep brain, preserved (Carolina 22-8710), or model, human brain (a

number of models are available; see the “Models” section in the Carolina catalog)

_____ models, vertebrate brains comparative set (Carolina POM8500)

_____ slide, prepared: spinal cord, cross section (Carolina 31-3726, 31-3738,

31-3744)

_____ microscopes, compound light

_____ lens paper

_____ index card, large enough to place behind a slide of the spinal cord

_____ meter stick

30.2 Animal Eyes (LM pages 432–437)

_____ model, human eye (a number of models are available; see the “Models”

section: the Carolina catalog)

_____ slide, prepared: insect compound eye, radial sec. (Carolina 30-7680)

_____ model, insect head showing compound eyes (Carolina 56-4565 (honey

bee), 56-4570 (housefly), 56-4580 (mosquito), 56-4585 (cockroach)

_____ pencil with eraser

_____ meter stick

30.3 Animal Ears (LM pages 438–440)

_____ grasshopper, preserved (display) (Carolina 22-5555)

_____ model, human ear (a number of models are available; see the “Models”

section in the Carolina catalog)

_____ tuning fork (various) or two spoons

30.4 The Senses of Human Skin (LM pages 441–442)

_____ model, human skin (Carolina 56-7665, 56-7671 to 56-7676)

_____ scissors, fine point; or hairpin

_____ beakers, 1,000 mL, three

_____ water: ice-cold, room-temperature, warm

_____ thermometer, centigrade

EXERCISE QUESTIONS

30.1 Animal Nervous Systems (LM pages 426–432)

Comparison of Vertebrate Brains (LM pages 428–429)

Observation: Comparison of Vertebrate Brains (LM pages 428–429)

3. Answer these questions (using Figure 30.3)

a. Based on the relative size of the optic lobe, in which animal is vision particularly important to survival? goose

b. Which animal would you predict to be the least agile? frog Why? Cerebellum is smallest.

c. Which two of the animals would be expected to exhibit more complex behaviors than the others? goose and horse Explain. Cerebrum is larger.

4. Which of the labeled structures appears to have undergone the greatest change in the horse brain compared to the frog brain? cerebrum Relate this to each animals’ lifestyle. Horses are social and live in groups; frogs do not.

5. Which animal has the proportionately largest cerebellum? goose How might this be explained? Flying in birds is a complex behavior that needs much coordination of muscles.

6. Which area of the human brain would you expect to be particularly large or small? Cerebrum is particularly large and optic lobe is quite small.

Figure 30.4 The human brain

a. Brain stem b.midbrain c. pons d. medulla oblongata e. cerebrum f. thalamus g. Diencephalon h. hypothalamus i. cerebellum

The Spinal Cord (LM page 430)

Observation: Spinal (Cord LM page 430)

Considering that the spinal cord runs the length of the body proper, why is it expected that it would have white matter. A neuron cell body at the base of the cord would have to have a long axon to communicate with the brain. Long axons are covered with a myelin sheath which is white and vice versa.

Spinal Nerves (LM page 430)

Observation: Spinal Nerves (LM page 430)

Describe the pathway of information, starting with the pain receptor in your foot, that would allow you to both feel and respond to this unwelcome stimulus. Sensory receptors in skin generate nerve impulses that move along sensory axon toward the spinal cord. Sensory neurons that enter the cord dorsally pass signals on to many interneurons. Some of these interneurons synapse with motor neurons. Nerve impulses travel along motor axons to muscle fibers, which bring about a response to the stimulus.

How would the brain become aware of your pain? Some interneurons send messages to the brain.

Spinal Reflexes (LM page 431)

Before we proceed, distinguish between the central nervous system and the peripheral nervous system. The central nervous system is the brain and spinal cord and the peripheral nervous system are all the nerves in the body. Do the two systems work together? Yes How so? They communicate by way of the spinal cord.

Experimental Procedure: Spinal Reflexes (LM page 432)

Ankle (Achilles) Reflex (LM page 432)

3. Which way does the foot move? Does it extend (move away from the knee) or flex (move toward the knee)? The foot extends.

Knee-Jerk (Patellar) Reflex (LM page 432)

3. In this relaxed state, does the leg flex (move toward the buttocks) or extend (move away from the buttocks)? The leg extends. Of what benefit are reflexes which involve only the spinal cord and not the brain? Allows quick action.

30.2 Animal Eyes (LM pages 432–437)

Anatomy of the Human Eye (LM pages 434–437)

Observation: Human Eye (LM pages 434–437)

2. Trace the path of light from outside the eye to the retina. Light goes from the cornea through the aqueous humor and pupil, through the lens to the vitreous humor, to the photoreceptors in the retina.

3. Specifically, what are the receptors for sight, and where are they located in the eye? The receptors are the rod cells and cone cells, which are located in the retina.

4. What structure takes nerve impulses to the brain from the rod cells and cone cells? the optic nerve

Physiology of the Human Eye (LM pages 435–437)

Experimental Procedure: Accommodation of the Eye (LM page 436)

5–7. Measure the distance (in centimeters) between the pencil and your eye. Distances will vary with the individual.

8. How “old” is the eye you tested? Answers will vary with the individual.

Experimental Procedure: Blind Spot of the Eye (LM page 436-437)

Left Eye/Right Eye (LM page 437)

6. Measure the distance (with your partner’s help) from your eye to the paper when the circle (for left eye) and cross (for right eye) first disappeared. Distances will vary with the individual. Generally, the blind spot is within 30 cm (1 foot) of the eye.

Experimental Procedure: An Optical Illusion (LM page 437)

3. What do you see? The circles appear to rotate in opposite directions.

30.3 Animal Ears (LM pages 438–440)

Physiology of the Human Ear (LM page 440)

4–5. Record the subject’s perceptions. Is there an apparent difference in hearing between your two ears? Perceptions will vary with the individual.

30.4 The Senses of Human Skin (LM pages 441–442)

Observation: Human Skin (LM page 441)

With the help of Figure 30.17 and a model of human skin, locate the following areas or structures, and describe the location of each:

1. Subcutaneous layer Beneath dermis

2. Adipose tissue Within subcutaneous layer

3. Dermis Beneath epidermis

4. Epidermis Uppermost part of skin

5. Hair follicle and hair Begin in dermis and continue through epidermis to outside

6. Oil gland Opens into hair follicle in dermis

7. Sweat gland Begins in dermis and continues through epidermis to outside

8. Sensory receptors Largely found in dermis

Sense of Touch (LM page 442)

Experimental Procedure: Sense of Touch (LM page 442)

5. a–d. Record the shortest distance between the hairpin or scissor points for a two-point discrimination. Distances will vary with the individual.

6. Which of these areas apparently contains the greatest density of touch receptors? the fingers Why is this useful? This sensitivity enables humans to manipulate sophisticated tools.

Sense of Heat and Cold (LM page 442)

Experimental Procedure: Sense of Heat and Cold (LM page 442)

4. Record the sensation in the right and left hands.

a. Right hand feels cold

b. Left hand feels warm

5. Summarize and explain your results. The sense receptors are reacting to the difference in temperature.

LABORATORY REVIEW 30 (LM page 443)

1. Describe the cerebrum of the human brain, and state a function. The cerebrum of the human brain is the most superior and highly convoluted region that covers much of the rest of the brain. It functions in critical thinking and memory. The cerebrum controls the rest of the brain.

2. The brain stem includes the medulla oblongata, the pons, and the midbrain. Explain the expression brain stem as an anatomical term. The brain stem is the most inferior portion of the brain and is attached to and somewhat resembles the spinal cord.

3. Describe the location of the gray/white matter of the spinal cord, and give a function for each. Gray matter has a butterfly shape and is located in the center of the cord. It contains interneurons, which take messages from sensory neurons to motor neurons. White matter lies outside gray matter and contains long fibers, which take messages up and down the cord.

4. State, in order from receptor to effector, the neurons associated with a spinal reflex. The neurons associated with a spinal reflex are, in order: sensory neuron, interneuron, and motor neuron.

5. Trace the path of light in the human eye—from the exterior to the retina and then from retinal nerve impulses to the brain. The path of light is: cornea through the pupil to aqueous humor, to lens, to vitreous humor, to photoreceptors in the retina. Nerve impulses travel from the retina in the optic nerve to the brain.

6. Contrast the eye of an arthropod with the eye of a squid and human. An arthropod has a compound eye (many individual units of sight). In squids and humans, light is focused on a single layer of the eye that contains photoreceptors.

7. If you move an illustration that contains a dark circle and a dark cross toward an eye, one or the other may disappear. Give an explanation for this. The eye has a blind spot where the optic nerve enters the eye, and since there are no photoreceptors here, no sight is possible.

8. Trace the path of sound waves in the human ear—from the tympanic membrane to the receptors for hearing. The path of sound waves is: tympanic membrane to maleus, to incus, to stapes, to fluid within cochlea. Pressure waves in the fluid stimulate hair cells (receptors for hearing) in spiral organ.

9. Compare the manner in which a grasshopper “hears” to the way a human hears. In the grasshopper, vibration of the tympanum results in nervous stimulation. In humans, the receptors for hearing are far removed from the tympanic membrane.

10. Name four structures located in the dermis of the skin. Structures located in the dermis include: hair follicles, sensory receptors, sweat glands, and oil glands.

Laboratory 31 Musculoskeletal System

(LM pages 445–460)

Eleventh Edition Changes

In Section 31.1 the instructions for Comparison of Vertebrate Skeletons has been improved and gives more guidance to students enabling them to complete Tables 31.1 and 31.2.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Fresh material (LM pages 447–448). Chicken femur bones for Experimental Procedure: Compact Bone. Obtain locally, close to time of use. Three treatments are needed: normal chicken femurs; chicken femurs that have been soaked in acetic acid; and chicken femurs that have been baked.

Preserved Frog (LM page 455). Experimental Procedure: Types of Movement. Using rotation can reduce the number of required frogs.

Perishable material (LM page 459). ATP muscle kit for Experimental Procedure: Contraction of Muscle Fibers. Order in advance and specify date of delivery.

All Exercises

_____ microscopes, compound light

_____ lens paper

31.1 Animal Skeletons (LM pages 446–451)

_____ slide, prepared: compact bone

_____ slide, prepared: spongy bone

_____ slide, prepared: hyaline cartilage

_____ bones, chicken femur (enough for acetic acid treatment, baking, and

untreated)

_____ 10% acetic acid (Carolina 84-1313 comes in 10% strength)

_____ oven, high temperature

_____ adult human skeleton, or model

_____ frog skeleton

_____ bird skeleton

_____ cat skeleton

Chicken bones (LM page 447). Ask students to bring in clean, uncooked chicken femurs one to two weeks before this lab. For decalcified chicken bones, soak one-third of the femurs in 10% acetic acid for four to five days. For deproteinized chicken bones, bake one-third of the femurs in a high-temperature oven for several days. Retain one-third of the chicken femurs as untreated.

Adult human skeleton (LM page 448). A number of articulated human skeletons are listed in the Carolina Biological Supply Company catalog, along with other similar publications. The skeletons come in plastic and natural bone, with and without cabinets and rod supports.

31.2 Vertebrate Muscles (LM pages 452–459)

_____ slide, prepared: smooth muscle

_____ slide, prepared: cardiac muscle

_____ slide, prepared: skeletal muscle

Types of Movement (LM page 454)

_____ frogs, preserved, for muscle dissection demonstration

Preserved frogs (LM page 455). A variety of bullfrog preparations are available. They differ according to the method of preservation and degree of injection. No color injection is necessary for this lab. (See the “Preserved Animals” section in the Carolina Biological Supply catalog.)

Contraction of Muscle Fibers (LM page 459)

_____ rulers, plastic millimeter

_____ slides

_____ ATP muscle kit (Carolina 20-3525)

_____ lens paper

ATP muscle kit (LM page 459). The kit contains glycerinated muscle, ATP solution, KCl and MgCl2 solution, and ATP plus KCl and MgCl2. These solutions are perishable and should be ordered for date of use. (Some instructors report that glycerinated muscle fibers may be stored in the freezer for several months with good results, and that solutions stored in the refrigerator also yield positive results after being stored for several months.)

EXERCISE QUESTIONS

31.1 Animal Skeletons (LM pages 446–451)

Tissues of the Human Skeleton (LM pages 446–448)

Experimental Procedure: Compact Bone (LM pages 447–448)

2. How does the shape of the treated femur compare with that of the untreated femur? The shape of the treated femur is similar to that of the untreated.

3. Is the treated bone more flexible or less flexible than the untreated bone? The treated bone is more flexible.

4. Is the treated bone able to withstand more direct downward compression or less direct downward compression than the untreated bone? The treated bone can withstand less direct downward compression.

5. What is the effect of this treatment on bone strength? The treatment reduces the bone’s tensile strength. The bone becomes brittle and will fracture easily.

Comparison of Vertebrate Skeletons (LM pages 449–451)

Observation: Vertebrate Skeletons (LM pages 450–451)

Forelimb Comparisons (LM page 450)

1. A bird uses it's forelimb for flying.

2. Compare the forelimb bones in the animals listed in Table 31.1 to those of a human. At a compare humerus; at b compare the radius and ulna, and at c compare the carpels, metacarpals, and phalanges as a whole.

Give examples to show that the forelimbs of these animals are modified to suit their means of locomotion. A frog uses its forelimb for landing and it has sturdy forelimb bones and long metacarpals and phalanges that spread out to dissipate the shock.

Table 31.1 Comparison of Forelimbs

Animal Forelimb Forelimb Bones Compared to Function Human Forelimb

Frog Landing a &b. More sturdy; c. longer metacarpals and phalanges provide support.

Bird Flying a. & b. More slender; c. Fused carpels- metacarpals and phalanges reduced in number for attachment of feathers.

Cat Running a. Shorter b. Elongated; c. carpels- metacarpals raised so that weight rests on phalanges.

HindLimbs to Forelimbs Comparison (LM page 455)

3. Compare the structure of the hind limb bones to those in the forelimb in each animal. Explain any difference in Table 31.2 on the basis of the function of the hind limb.

Table 31.2 Comparison of Hindlimb to Forelimb

Animal Differences and Explanation

Frog The bones of the hindlimb are longer—an adaptation for jumping.

Bird The ankle is raised and long, wide spread phalanges assist balance.

Cat The forelimb and hindlimb serve the same function, i.e. running; and they are similarly adapted.

Human The hindlimbs contain sturdier bones—they support the weight of the body.

31.2 Vertebrate Muscles (LM pages 452–461)

Types of Movement (LM pages 454–458)

Experimental Procedure: Types of Movement (LM pages 455–458)

2. What is the action caused by the muscle under these conditions? The action is flexion of the shank.

3. Now what is the action caused by the gastrocnemius? The action is extension of the foot.

4. Is this muscle a shank extensor or flexor? The muscle is an extensor.

5. Where does the biceps femoris insert? The biceps femoris inserts on the bone of the lower leg.

6. What action does the biceps femoris cause? The biceps femoris causes flexion of the shank.

Human Muscles (LM page 456)

Locate the following antagonistic pairs (muscles that act in opposition to each other) in Figure 31.12. In each case, state their contrary actions by inserting one of these functions—flexes, extends, adducts, or abducts—in the following:

1. The biceps brachii flexes the forearm.

The triceps brachii extends the forearm.

2. The sartorius abducts the thigh.

The adductor longus adducts the thigh.

3. The quadriceps femoris group extends the leg.

The biceps femoris flexes the leg.

Muscular Contractions (LM page 456)

Experimental Procedure: Types of Contraction (LM page 456)

Isometric Contraction (LM page 456)

2. Is the biceps brachii or the triceps brachii located on the ventral surface of the arm? biceps brachii

3. What change did you notice in the firmness of this muscle as it contracted? It became firmer.

4. Did your hand or forearm move as you pushed up against the table? no

5. Given your answer to question 4, did this muscle’s fibers shorten as you pushed up against the tabletop? no

Isotonic Contraction (LM page 456)

2. What makes this contraction isotonic rather than isometric? If a muscle contraction produces movement, it is an isotonic contraction.

Contraction of Muscle Fibers (LM pages 458–459)

The current model for muscle contraction states that as muscle contraction occurs, actin filaments slide past myosin filaments. This causes sarcomeres to shorten.

Experimental Procedure: Contraction of Muscle Fibers (LM page 459)

1–6. Record your results in Table 31.3

Table 31.3 Glycerinated Muscle Contraction

Length

Slide 1 Slide 2

Glycerol alone no change no change

K+/Mg2+ salt solution alone no change —-------

ATP alone —-------- no change

Both salt solution and ATP shortens shortens

7. To demonstrate that you understand the requirements for contraction, state the function of each of the substances listed in Table 31.4.

Table 31.4 Summary of Muscle Fiber Contraction

Substance Function

Myosin filament Thick filament: breaks down ATP

Actin filament Thin filament: slides during muscle contraction

K+/Mg2+ Cofactors needed for myosin to function

salt solution as an enzyme

ATP Supplies the energy needed for muscle contraction

LABORATORY REVIEW 31 (LM page 460)

1. An earthworm has no hard parts; why is it said to have a skeleton? An earthworm has a hydroskeleton in which the pressure of muscle contraction is applied to fully filled body cavity.

2. Name and describe three features that would allow you to identify a slide of compact bone. 1. presence of osteons (concentric units); 2. a central canal in the center of each osteon;

3. lacunae with canaliculi

3. What bones make up the pectoral girdle of humans? clavicles and scapulae

4. What bones protect the thoracic cavity? thoracic vertebrae, ribs, and sternum

5. Diversity of function explains why the bones of a bird’s forelimb differ from its

hindlimb. Explain. The forelimbs of birds are used to fly, and the hindlimbs are used for standing and walking.

6. Name the features that would allow you to identify the following muscular tissues:

a. Smooth muscle spindle-shaped cells with a single nucleus; nonstriated

b. Cardiac muscle striated cells that are branched

c. Skeletal muscle striated, multinuclear cells in long, parallel fibers.

7. In frogs, the triceps femoris and the biceps femoris are an antagonistic pair. The triceps femoris extends the hindlimb. What does the biceps femoris do? flexes the hind limb.

8. Sequence the components of a skeletal muscle fiber from the largest to the smallest component. myofibril, sarcomere, myosin filament, actin filament

9. When you observe glycerinated muscle shorten, what is happening to actin filaments microscopically? Actin filaments are moving past myosin filaments.

10. When you observe glycerinated muscle shorten, what happens to sarcomeres? They shorten.

Laboratory 32 Animal Development

(LM pages 461–479)

Eleventh Edition Changes

In this edition, Table 32.3. Comparison of Embryonic Features of a Developing Sea Star, Frog, and Chick was relocated and placed in the Laboratory Review.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Preserved Specimens (LM page 463). Preserved frog embryos are required for Observation: Preserved Frog Embryos.

Fresh material (LM page 468). Unfertilized and fertilized chicken eggs for 32.3 Chick Development.

Equipment (LM page 470). Incubator for fertilized chicken eggs

32.1 Early Embryonic Stages (LM pages 462–465)

____ slide, prepared: sea star (starfish) unfertilized egg, whole mount

_____ slides, prepared: sea star early cleavage; two-, four-, eight, and sixteen-cell stages, whole mount

_____ slide, prepared: sea star late cleavage; sixty-four-cell stage, whole mount

_____ slide, prepared: sea star blastula, whole mount

_____ slide, prepared: sea star gastrula (early, mid, late), whole mount

_____ microscopes, compound light

_____ lens paper

_____ frog embryos, preserved, or frog embryo models at various stages of development

_____ microscopes, stereomicroscope

32.3 Chick Development (LM pages 466–76)

_____ finger bowls (culture dishes), 4.5-inch (Carolina 74-1004)

_____ unfertilized chicken eggs

_____ fertilized chicken eggs

_____ incubator (available from many suppliers or locally; see the “Laboratory Equipment and Supplies” section in the Carolina catalog)

_____ Ringer solution, chicken, warmed (Carolina 88-6509) or prepare with NaCl, KCl, CaCl2, and NaHCO3 using recipe below

Fertilized chicken eggs (LM page 470). Fertilized chicken eggs should be purchased locally, if possible, or from a supplier, and placed in an incubator. Turn once a day. To keep track of daily egg turning for many sections, it is helpful to mark each egg with the section and duration of incubation on one side (e.g., 48 hours 001), and with an “X” on the opposite side before placing it into the incubator. Also, it helps to remind students that the egg must not be rotated as it is removed from the incubator in the lab, as the embryo will be found “on top.” With careful planning, successive labs can be run so that each class can see the various stages of living chick embryos using only one forty-two-egg incubator.

If you want to do this as a demonstration, carefully crack the eggs into individual finger bowls with warmed chicken Ringer solution. Then put the bowls under a dissecting microscope, and keep the illuminator on, which adds some heat. Chick embryos will often stay alive for a half a day or longer. Students can then see 24-, 48-, 60-, 72-, and 96-hour chick development.

Chicken Ringer solution (LM page 470). This solution is perishable; therefore, prepare within a few days of use. To prepare chick ringers, dissolve the following salts in 1 liter of distilled water:

NaCl, 9 g CaCl2 (anhydrous), 0.24 g

KCl, 0.4 g NaHCO3, 0.2 g

32.4 Human Development (LM pages 476–478)

_____ models of human development (see “Models” section in Carolina catalog)

Note: This laboratory is greatly enriched by showing a DVD depicting human development, Carolina 49-3556 (DVD).

EXERCISE QUESTIONS

32.1 Early Embryonic Stages (LM pages 462–465)

Sea Star Development (LM page 462–463)

Explain why you would not expect a sea star’s egg to be heavily laden with yolk. The embryo is free-living and develops quickly to a stage where it gets nourishment on its own.

Observation: Sea Star Embryos (LM page 462–463)

2. Cleavage. How does the size of the two- to eight-cell stage of a sea star compare to the size of the unfertilized egg? same size Does growth occur during cleavage? no

3. Morula Explain why the morula and zygote are about the same size. As cleavage is occurring, the cells get smaller. This is cell division without growth.

4. Blastula Label the blastocoel in Figure 32.1f. at the only leader

5. Early gastrula Why are the ectoderm and endoderm called germ layers? They give rise to various structures in the adult. Recall that the sea star is a deuterostome. In the sea star, does the blastopore become the mouth or the anus? anus

6. Late gastrula Recall that a sea star has a true coelom, do you expect that mesoderm will line the coelom of a sea star? yes Explain. A coelom is lined by mesoderm.

Frog Development (LM pages 463–465)

Observation: Preserved Frog Embryos (LM pages 463–465)

1. Fertilized egg Did the sea star unfertilized egg have an animal pole and a vegetal pole? no Explain. The sea star has a small amount of yolk equally distributed throughout the egg.

2. Morula Which pole, animal or vegetal, contains the largest cells? vegetal Explain. These cells contain yolk.

3. Blastula Compare the frog blastula to that of the sea star. In the sea star, the blastula is a hollow ball of cells. The frog blastula is not completely hollow because the blastocoel is at the animal pole.

5. Late gastrula Compare formation of the mesoderm in the frog to that in the sea star. In the sea star, the mesoderm begins as outpocketings from the primitive gut. In the frog, cells from the dorsal lip of the blastopore migrate between the ectoderm and endoderm to form the mesoderm.

6. Neurula Draw a series of sketches that show how the neural tube develops. See Figure 32.3c What has to happen to the ectoderm in order for a neural tube to form? Folds fused to produce an outer layer and the tube.

32.2 Germ Layers (LM page 466)

In Table 32.1, list the three germ layers and the major organs that develop from each in the frog.

Table 32.1 Germ Layer Organization

Germ Layers Organs/Systems Associated with Germ Layer

1. Ectoderm Skin, nervous system

2. Mesoderm Cardiovascular, muscular, and skeletal systems; connective tissue

3. Endoderm Digestive and respiratory systems

Induction (LM page 466)

1. A chick is a chordate. What is the function of the notochord in a chick? becomes vertebral column

2. For each of these scenarios, tell whether a neural tube will develop:

a. Notochord is removed. Does ectoderm in this location become a neural tube? no

b. Ectoderm above notochord is replaced with belly ectoderm. Does belly ectoderm become a neural tube? yes

c. Develop a hypothesis to explain these suggested outcomes. Notochord induces formation of neural tube in ectoderm.

32.3 Chick Development (LM pages 466–475)

Extraembryonic Membranes (LM pages 466–467)

Complete Table 32.2

Table 32.2 Functions of Extraembryonic Membranes in Chick and Human

Membrane Chick Human

Amnion Protects against drying Protects against drying

Allantois Collects waste Contributes to umbilical

blood vessels

Chorion Gas exchange Part of placenta

Yolk sac Contains nutrients First site of blood cell formation

Observation: Raw Chick Egg (LM page 468)

2. What could be its function [ egg white]. The primary function of the egg white is to cushion the embryo and keep it from drying out.

3. Why does a chick have need of a large amount of yolk? food has to be sufficient for full development because a chick has no larva form to assume the job of acquiring food.

4. Why would you expect a chick’s egg to be porous (permeable) instead of impervious (impermeable)? To allow oxygen to enter through the shell to the embryo.

Observation: Twenty-Four-Hour Chick Embryo (LM pages 468–470)

3. What two organ systems develop first in a chick? nervous and muscular Would you expect a human to also have a stage that looks like Figure 32.6? yes Why or why not? Birds are reptiles and humans are closely related to reptiles.

32.4 Human Development (LM pages 476–478)

Stages of Human Development (LM pages 476–478)

1. Review frog development on pages 463–465. What are the stages you would expect to see during early embryonic development in humans? zygote, morula, blastula, gastrula, neurula

2. Review chick development and decide what two organs will most likely be first to make their appearance during human development. brain, heart per Figure. 32.7.

3. Figure 32.11 shows in particular the development of the extraembryonic membranes in humans. Circle the labels for the membranes in Figure 32.11. chorion, amnion, yolk sac, allantois

LABORATORY REVIEW 32 (LM page 479)

Complete this table by placing an X in the appropriate square if the feature pertains to the organism.

Feature Sea Star Frog Chick

1.a. Has the most yolk X

b. Blastula has a circular cavity X

2.a. Germ layers are present X X X

b. Primitive streak is present X

3.a. Notochord is present X X

b. Waste is deposited in water X X

4. Describe how a sea star embryo becomes a morula, blastula, and gastrula. Cleavage results in a morula, a solid ball of cells. Movement of cells results in a blastula, a hollow ball of cells. Invagination of cells results in a gastrula, which has two germ layers.

5. What factor causes a frog’s morula, blastula, and gastrula to appear differently from that of a sea star? The yolk-laden cells cannot move, so the blastocoel is not as large, and the mesoderm develops by invagination between ectoderm and endoderm rather than outpocketing from the archenteron.

6. Describe how induction may control development. One tissue induces another, and so forth, until all organs are formed.

Name the four extraembryonic membranes, and state the function of each in birds and mammals.

Extraembryonic Function in Birds Function in Mammals

Membrane

7.a. Amnion Protects against drying Protects against drying

b. Chorion Gas exchange Part of placenta

8.a. Yolk sac Contains nutrients First site of blood cell

formation

b. Allantois Collects waste Contributes to umbilical

blood vessels

9. What three features are quite noticeable in a forty-eight-hour chick embryo? brain, heart, somites

10. List the two stages of human development, and state a reason for dividing human development into these two stages. During embryonic development, internal organs form. By the end of embryonic development, the embryo has a human appearance and is now referred to as a fetus. During fetal development, organs become more refined, and the fetus grows larger. Human development is divided into the two stages on the basis of when a human appearance becomes apparent.

Part VIII Ecology

The two labs in this part are preceded by a description of three virtual labs available on the Biology website. Ecology does not lend itself to traditional laboratory settings and I recommend these virtual labs as a way to have students experience the vitality of ecosystems without having to travel to the site. Laboratory 33 does assume that you are able to transport your students to a location of your choice.

Laboratory 33 Sampling Ecosystems

(LM pages 481–490)

Eleventh Edition Changes

This was lab 34 in the tenth edition. It has no changes this edition.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

All Exercises

At the site:

_____ meter sticks

_____ collection jars or bottles, empty, with labels

_____ collection jars or bottles, with 70% alcohol

_____ sample bags, small and large, plastic and paper

_____ pencils, wax

_____ collecting nets

_____ petri dishes, sterile, with nutrient agar

_____ forceps

_____ buckets

_____ thermometers, centigrade (0--110°C)

_____ light and humidity meters

_____ pH meter, soil, portable

_____ pH paper (see “Chemistry-pH Test Papers” section in Carolina catalog)

_____ recording materials

In the lab:

_____ microscopes, stereomicroscope

_____ microscopes, compound light

_____ lens paper

_____ slides and coverslips

_____ medicine droppers

_____ balance scale accurate to 0.01 g

Identification keys. When you return to the lab, identification keys may be required. One identification key is: Borror and White, A Field Guide to Insects (Boston: Houghton Mifflin). The series How to Know the Insects (and other titles) (Dubuque, Iowa: Wm. C. Brown Publishers) covers nearly everything. See also Laboratory 2 of this Resource Guide for a list of pictorial guides to freshwater organisms.

33.1 Terrestrial Ecosystems (LM pages 482–485)

A terrestrial ecosystem is studied at the site (page 482) with a follow-up in the lab (page 484). For Berlese funnel analysis: Soil and/or leaf litter is placed in a funnel and a heat source such as a light lamp heats the soil and/or litter. The animals escape through a filter into a flask that contains alcohol.

33.2 Aquatic Ecosystems (LM pages 486–489)

An aquatic ecosystem is studied at the site (page 486 ) with a follow-up in the lab (page need page number)

This equipment in addition to the above pertains to the aquatic site:

_____ plankton filter nets (Carolina 65-2158 to -2167)

_____ dissolved oxygen and temperature meter

_____ Lugol’s solution

_____ large sieve to place over bucket of water

Ecology Field Kit (LM page 487). This field kit (Carolina 65-2000) contains a general collection of aquatic sampling equipment, some of which is noted individually in the previous list.

Dissolved oxygen measurement (LM page 487). The dissolved oxygen meter and BOD (biological oxygen demand) probe (YSI 5905), which also gives temperature, can be purchased from YSI, Inc. YSI has a full line of environmental sampling equipment.

Sieves (LM page 487). Use sieves to sort invertebrates from sediment samples taken with a dredge. Sieves can be purchased or made with a wood frame (about 30 by 30 cm) and graded sizes of hardware cloth.

EXERCISE QUESTIONS

Answers to all questions in the laboratory will depend upon the data collected and student observations.

LABORATORY REVIEW 33 (LM page 490)

1. What is an ecosystem? An ecosystem is a biological community together with the associated abiotic environment.

2. Define and give several examples of abiotic components of an ecosystem. Abiotic (nonliving) components include water, soil, and sunlight.

3. Define and give several examples of biotic components of an ecosystem. Biotic (living) components include plants, animals, bacteria, and fungi.

4. In a forest ecosystem, what type of organisms are the predominant producers? trees

5. In an aquatic ecosystem, what type of organisms are the predominant producers? algae

6. What is the role of producers in an ecosystem? They produce organic food for themselves, and indirectly for all populations in an ecosystem, from inorganic material.

7. Define a consumer, and give examples of consumers in terrestrial and aquatic ecosystems. A consumer is an organism that must obtain organic food. Examples include rabbits (terrestrial ecosystem); zooplankton (aquatic ecosystem).

8. In an ecological pyramid. describe the final consumers in terms of their diet. Final consumers are usually heterotrophic organisms that need to take in preformed organic material.

9. In an ecological pyramid, where would you place parasites of humans such as a tapeworm? There is no recognized location for such parasites.

10. Regardless of the ecosystem, what type of organisms would be the predominant decomposers? bacteria and fungi

Laboratory 34 Effects of Pollution on Ecosystems

(LM pages 491–498)

Eleventh Edition Changes

This lab was 35 in the previous edition. The first section was revised to better pertain to ecosystems in general.

MATERIALS AND PREPARATIONS

Instructions are grouped by exercise. Some materials may be used in more than one exercise.

Special Requirements

Living material. Hay infusion cultures, sunflower seeds, and also a supply of Gammarus, Chlorella (common pond algae), and Daphnia are needed.

Equipment (LM page 496). Spectrophotometers.

All Exercises

_____ droppers

_____ slides and coverslips

_____ lens paper

_____ microscopes, compound light

_____ microscopes, stereomicroscope

_____ litmus paper or pH meter

34.1 Studying the Effects of Pollutants (LM pages 492--495)

Experimental Procedure: Study of Hay Infusion Cultures (LM pages 492–93)

_____ hay or grass clippings

_____ peptone solution (Carolina 82-1436)

_____ large beaker, or wide-mouthed glass jar, with screw-top lid

_____ 1% sulfuric acid (H2SO4) (Carolina 89-3300, -3301)

Hay infusion cultures (LM page 492). Add tap water and a pinch of peptone (a nutrient) to hay (or grass clippings) in an open beaker. Do not aerate. In about 24 hours or less, a variety of bacteria, including motile bacteria, will develop. In a few days, protists will develop, and can be seen feeding on the bacteria. Alternately, add a small amount of pond water including some mud from the bottom of the pond. The motility of microorganisms can be observed using a depression slide.

A more expensive alternative is Carolina’s hay infusion culture (culture media set, Carolina 13-1206). This set includes all materials for producing one gallon of hay infusion, sufficient for a class of 30 students. The infusion can be used to mass culture many species of protozoans and other invertebrates. The culture media set includes timothy hay, a plastic culture aquarium, spring water, and instructions. Use a 1-gallon, wide-mouth jar covered loosely with screw-cap, and maintain the culture in a well-lighted place, but not in direct sunlight.

To prepare the various types of cultures needed, follow these directions:

Control culture: Prepare the culture according to the instructions provided. Do not cover.

Oxygen-deprived culture: Prepare the same as the control culture, but use minimal oxygen—use saran wrap to tightly cover the culture, if in a beaker. Use a screw-top lid if the culture is in a wide-mouthed glass jar. Do not aerate.

Acidic culture: Adjust a portion of the control culture to pH 4 using 1% sulfuric acid.

Enriched culture: Add additional peptone to the culture.

Experimental Procedure: Effect of Acid Rain on Seed Germination (LM Page 493)

_____ five containers of sunflower seedlings

_____ five beakers containing tap water (0% vinegar), 1% vinegar, 5% vinegar, 20% vinegar, 100% vinegar for student testing

Sunflower seedlings (LM page 493). Two weeks before the lab, label and plant twenty sunflower seeds in each of five containers. Water each container as follows: container 1 (control): tap water; container 2: 1% vinegar solution; container 3: 5% vinegar solution; container 4: 20% vinegar solution; container 5: 100% vinegar. Students determine the pH of these solutions in the lab.

Experimental Procedure: Gammarus (LM Page 494)

_____ Gammarus culture (Carolina 14-2355, amphipods, or Nasco Science

LM00205, fairy shrimp)

_____ spring water for 2 beakers per group

_____ beakers to hold 4 Gammarus each

Gammarus (LM page 494). Large quantities of spring water (50 mL per group) will be needed for the Gammarus culture. Expect some Gammarus in each laboratory to die. Therefore, purchase half again as many as you determine you will need. Other species may also be used.

Acidic spring water (LM page 495). Optional. Add 1 mL of sulfuric acid to 100 mL of spring water. To save on the number of Gammarus needed and to prevent the exposure of many Gammarus to acid conditions, this could be on demonstration.

34.2 Studying the Effects of Cultural Eutrophication (LM pages 495--497)

Observation: Daphnia Feeding (LM page 496)

_____ petroleum jelly

_____ petri dishes, small

_____ Daphnia culture (Carolina 14-2314)

_____ indigo carmine (Carolina 86-8550)

_____ probe or dissecting needle (to turn Daphnia over)

Observation: Daphnia Feeding on Chlorella (LM page 496)

_____ spectrophotometers

_____ kimwipes

_____ wash bottles of distilled water, and ethanol

_____ two cuvettes for spectrophotometer

_____ Pasteur pipette

_____ Chlorella (common pond algae) culture (10 cc) (Carolina 15-2069)

_____ Daphnia culture (Carolina 14-2314)

_____ test-tube racks

0.1% carmine solution (LM page 496). Dissolve 0.1 g of indigo carmine in 100 mL of distilled water.

Spectrophotometer and tubes (cuvettes) (LM page 496). The most common instrument for teaching use is the Bausch and Lomb Spectronic 20. For this laboratory, the wave-length indicator should be set at 635 nm (or 0.635 mm). The instrument must be calibrated for zero and infinite absorbance. Calibrate for infinite absorbance by using the left-hand knob to line up the needle with the left-most marker (an infinity symbol). Then insert a clean cuvette of distilled water and adjust for 100% transmittance using the right-hand knob. The Chlorella culture should be just visibly green.

The Daphnia should be starved in clean, aged water (stale tap water that has been allowed to stand for a few days to allow chlorine to escape) for an hour or so prior to the lab. This clears their gut, since feces production would foul the experimental vessel and reduce the apparent feeding rate.

Preparing cuvettes (LM page 496). Next to each spectrophotometer, place a kimwipe-lined beaker containing two optically matched cuvettes, a Pasteur pipette, a wash bottle of distilled water, and a wash bottle of ethanol. Have students clean cuvettes with the ethanol thoroughly after use, and place cuvettes upside down in the beaker to drain.

EXERCISE QUESTIONS

34.1 Studying the Effects of Pollutants (LM pages 492--495)

Experimental Procedure: Effect of Pollutants on a Hay Infusion Culture (LM pages 492–493)

1. See Table 34.1 for answers.

2-4 Record your estimation [of species composition and species diversity] in the second and third column of Table 34.1.

Table 34.1 Hay Infusion Culture

Type of Culture Species Composition Species Diversity Explanation

Control High High Normal conditions

Oxygen-deprived Medium to Low Medium to Low Low cell respiration

Acidic Medium to Low Medium to Low Enzymes altered

Enriched Medium to Low Medium to Low Low oxygen

Conclusions (LM page 493)

All bullets. What could be a reason for the adverse effects of the test cultures? Enter explanations in Table 34.1. See Table above.

Experimental Procedure: Effect of Acid Rain on Seed Germination (LM pages 493--494)

In other words, it is hypothesized as acidity increases, the more likely seeds will not germinate.

1. Record the pH of each solution in Table 34.2

2. Count the number of germinated sunflower seeds in each container, and complete

Table 34.2.

Table 34.2 Effect of Increasing Acidity on Germination of Sunflower Seeds.

Concentration of Vinegar pH Seed and Percent Germination

0% -- 100% Decreasing pH Decreasing seed and percent germination

Conclusions: Effect of Increasing Acidity on Germination of Sunflower Seeds (LM page 494).

• Explain why acid rain is expected to inhibit metabolism, and therefore, seedling development. A low pH changes the shape of enzymes and decreases their activity

• Do the data support your hypothesis. Most likely, yes.

FL? Study of Gammarus (LM pages 494--495)

Experimental Procedure: Gammarus (LM pages 494–495)

Control Culture (LM page 494)

2. Measure the pH. Close to neutral pH.

3. Observe the behavior of Gammarus for 10 to 15 minutes, and then answer the following questions.

a. Where do the Gammarus spend their time in the container? near the bottom

b. How do they spend their time? They swim around in an apparently random fashion. Occasionally, they lie still on the bottom of the container.

c. What percentage of their time is spent moving? 90 to 95%

d. Do they use all their legs in swimming? No, they move by flexing the appendages on the side of their body, one side at a time.

e. Which legs are used in jumping and climbing? the thoracic appendages

f. Do Gammarus avoid each other? yes

g. What do Gammarus do when they “bump” into each other? They pass each other quickly, going off in a new direction.

4. Create a food chain: algae — protozoans —Gammarus —fish —humans.

a. Predict what would happen to this food web if the water was

oxygen deprived. Populations would be smaller and it's possible it would die out before humans benefit from it.

acidic. Populations would be smaller and it's possible it would die out before humans benefit.

enriched with inoranic nutrients, Short term. Populations are expected to increase in size. Long term: When algae die off bacteria of decomposition use up all the oxygen and the fish populations die to the detrement of humans

Acidic Culture (LM page 495)

If so directed by your instructor, put a Gammarus in a beaker of spring water adjusted to pH4 or examine a sample on display. How does Gammarus react to acid conditions? At pH 4, the Gammarus swim more actively than previously. At the beginning, they spend all of their time at the surface. Later, they swim to the bottom. Almost no time is spent lying still on the bottom

Conclusion: Studying the Effects of Pollutants (LM page 495)

• Give an example to show that the hay infusion study pertains to real ecosystems? Ecosystems are sensitive to temperature and pH variations.

• What are the potential consequences of acid rain on crops that reproduce by seeds? Yield will be less than expected On the food chains of the ocean? Animals will have less to eat.

• How does the addition of nutrients affect species composition and species diversity of an ecosystem diversity over time? Initially, the addition of nutrients causes increase in producers and consumers, but over time, it causes a decrease when the algae die off.

34.2 Studying the Effects of Cultural Eutrophication (LM pages 495--497)

Observation: Daphnia Feeding (LM page 496)

5. Does the carmine travel completely through the gut in 30 minutes? yes

Experimental Procedure: Daphnia Feeding on Chlorella (LM page 496)

4. Record your reading in the first column of Table 34.3.

7. Record your data in the second column of Table 34.3, and explain your results in the third column.

Table 34.3 Spectrophotometer Data of Daphnia Feeding on Chlorella

Absorbance Absorbance Explanation

Before Feeding After Feeding

Exact measurements will vary. Absorbance is lower after allowing Daphnia to feed, because the Daphnia have eaten the algae.

Experimental Procedure: Case Study in Cultural Eutrophication (LM page 497)

1. Using this information, complete Table 34.4.

Table 34.4 Daphnia Filtering

Number of Daphnia/Liter Percent of Lake Filtered

10 24%

50 120%

2. Using this information, complete Table 34.5.

Table 34.5 Cultural Eutrophication

Number of Condominiums Phosphorus Added Increase in Algal Population

10 1 kg 30%

20 2 kg 60%

30 3 kg 90%

40 4 kg 120%

50 5 kg 150%

Conclusion: Cultural Eutrophication (LM page 497)

• How many condominiums would you allow the developer to build? no more than forty

• What other possible impact could condominium construction have on the condition of the lake? cultural eutrophication due to fertilizer runoff

LABORATORY REVIEW 34 (LM page 498)

1. What type of population would you expect to be the largest in most ecosystems? Algae or plants Explain. They produce the food for the other organisms in an ecosystem.

2. What causes acid rain? When fossil fuels are burned, acids enter the atmosphere and then fall to the Earth’s surface in rain or snow.

3. Why is acid deposition harmful to organisms? It causes organisms to take in, for example, water that has a low pH. A low body pH is harmful to enzymes.

4. Name the type of pollution that results when water from rivers and ponds is used for cooling power plants, and explain why it has detrimental effects. Thermal pollution removes oxygen from the water because warm water does not hold as much oxygen as cold water.

5. Give an example to show that the pollutants studied in this laboratory can have an on the human. Any one of the pollutants can be used as an example. Stet AFFECT

6. When excess nutrients enter an aquatic ecosystem, long-term effects can result. Why? Algae overgrow, and when they die, bacteria use up all the oxygen to decompose them. Organisms that need oxygen die.

7. Describe how the cultural eutrophication supports the hypothesis that a balance of population sizes in ecosystems is beneficial? Only when algae populations remain at their normal level can consumers maintain them at a level which will not result in pollution.

8. When pollutants enter an ecosystem, they have far-ranging effects. Use acid rain and a food chain to support this statement. Acid rain can potentially reduce the size of all populations in any food chain, even humans.

9. Contrast species composition with species diversity of ecosystems. Species composition is the number of different species present and species diversity is the number of each type of species.

10. Suppose among sunflower seeds, a particular variety can germinate despite acidic conditions. What do you predict about the survival of that sunflower variety compared to the rest of the population? More and more of this variety will survive with each passing generation. What do we call a change in a population's phenotype composition due to the presence of an environmental agent? Evolution

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