NAME

  • Doc File 105.00KByte



NAME _________________________ sp15

HARDY-WEINBERG THEOREM

p2 + 2pq + q2 = 1 AND p + q = 1

|p = frequency of the dominant allele in the population |p2 = percentage of homozygous dominant individuals |

|q = frequency of the recessive allele in the population |q2 = percentage of homozygous recessive individuals |

| |2pq = percentage of heterozygous individuals |

PRACTICE PROBLEMS

1. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Calculate the following:

A. The frequency of the "aa" genotype. = 36%

B. The frequency of the "a" allele. = 60%

C. The frequency of the "A" allele. =40%

D. The frequencies of the genotypes "AA" = 16% and "Aa." = 48%

E. The frequencies of the two possible phenotypes if "A" is completely dominant over "a."

Dominant = 64% recessive = 36%

|q2 |q |p |p2 |2pq |

|.36 |.6 |.4 |.16 |.48 |

2. A very large population of randomly-mating laboratory mice contains 25% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population.

|q2 |q |p |p2 |2pq |

|.25 |.5 |.5 |.25 |.50 |

Dominant allele = 50% Recessive allele = 50%

AA = 25% Aa= 50% aa= 25%

3. If q=0.3 and there are Hardy-Weinberg proportions in the population, what is the most common genotype and what is its frequency? What is the least frequent genotype and its frequency?

|q2 |q |p |p2 |2pq |

|.09 |.3 |.7 |.49 |.42 |

Most common genotype = AA (p2) = .49

Least common genotype = aa (q2) = .09

|q2 |q |p |p2 |2pq |

|.36 |.6 |.4 |.16 |.48 |

4. . In Drosophila (fruit fly), the allele for normal wing length is dominant over the allele for short wings. In a population of 1000 individuals, 360 show the recessive phenotype. How many individuals would you expect to be homozygous dominant for the trait. AA = .16 X 1000 = 160 individuals

5. The allele for a widow's peak (hairline) is dominant over the allele for a straight hairline. In a population of 500 individuals, 9% show the recessive phenotype. How many individuals would you expect to be homozyous dominant and heterozygous for the trait? AA = 49% Aa= 42%

|q2 |q |p |p2 |2pq |

|.09 |.3 |.7 |.49 |.42 |

6. The allele for a hitchhiker's thumb is dominant over a straight thumb. In a population of 1000 individuals, 510 show the dominant phenotype. How many individuals would you expect for each of the three possible genotypes for this trait. AA = 9% Aa= 42% aa= 49%

|q2 |q |p |p2 |2pq |

|1000-510= .49 |.7 |.3 |..09 |.42 |

7. Let’s say that brown fur coloring is dominant to gray fur coloring in mice. If you have 168 brown mice in a population of 200 mice then . . . .

a. What is the predicted frequency of heterozygotes? 48%

b. What is the predicted frequency of homozygous dominant? 36%

c. What is the predicted frequency of homozygous recessive? 16%

|q2 |q |p |p2 |2pq |

|200-168=.32/200 = .16 |..4 |.6 |.36. |.48. |

8. In humans, Rh-positive individuals have the Rh antigen on their red blood cells, while Rh-negative individuals do not. If the Rh-positive phenotype is produced by a dominant gene (A), and the Rh-negative phenotype is due to its recessive allele (a), what is the frequency of the Rh-positive allele if 84% of a population is Rh-positive. Rh+ allele = 60%

|q2 |q |p |p2 |2pq |

|100-84= .16 |.4 |.6 |.36 |.48 |

14. The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies.

A= 45% a= 55% AA= 20% Aa= 50% aa= 30%

|q2 |q |p |p2 |2pq |

|65/215=. 30 |.55 |.45 |.2 |.5 |

10. Sickle-cell anemia is an interesting genetic disease. Normal homozygous individuals (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

Aa= 42%

|q2 |q |p |p2 |2pq |

|.09 |.3 |.7 |.49 |.42 |

11. In a large randomly mating population with no forces acting to change gene frequencies, the frequency of homozygous recessive individuals for the character extra-long eyelashes is 75 per 1000. What percentage of the population carries this trait, BUT DISPLAYS the dominant phenotype (short eyelashes)? Would the frequency of the extra-long-eyelash allele increase, decrease, or stay the same if extra-long-lashed individuals preferentially mated with each other and no one else?

|q2 |q |p |p2 |2pq |

|75/1000 = 0.075 |0.274 |0.726 |0.527 |.765 |

Heterozygotes = .765 = 77% q (extra long eyelashes) would increase in the population

12. A rather large population of Doodle bugs 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:

A. The allele frequencies of each allele. A = 35% a= 65%

B. The expected genotype frequencies. AA= 12 Aa=46% aa=42%

C. The number of heterozygous individuals that you would predict to be in this population. .46 X 953=438

D. The expected phenotype frequencies. RED= 42% TAN = 58%

E. Conditions happen to be really good this year for breeding and next year there are 1,245 young Doodle bugs. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided? RED= .42 X 1245 = 523 .TAN + 58X1245 = 722

|q2 |q |p |p2 |2pq |

|396/386+557= .42 |..65 |.35. |.12. |..46 |

13.  After graduation, you and 19 of your closest friends (10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island?

20 people X 2 alleles = 40 total alleles 2/40 = .05 = a aa= 5%

|q2 |q |p |p2 |2pq |

|.0025 |.05 |.95 |.9 |.095 |

14. In a population of 2000 earthworms, there is a condition governed by a recessive allele where the worms do not have any setae (tiny hair-like bristles used for traction to move through the ground). 530 worms were found not having setae. What percentage of the population were heterozygous for the setae? What was the actual number of worms containing setae?

Heterozygotes = 50% Worms with setae = .499+.236 = .735

|q2 |q |p |p2 |2pq |

|530/2000 = .265 |.514 |.486 |.236 |.499 |

15. A Pangorian trait which results from simple Mendelian inheritance is antenna shape. Corkscrew antennae (A) are dominant over straight antennae (a). When the entire Pangorian population was screened (all 9,904 of them),3,565 had corkscrew, while the rest had straight antennae.

a) What is the frequency of each allele? A = 20% a= 80%

b) What percentage of the population has each of the genotypes? AA=4% Aa=32% aa= 64%

c) How many Pangorians are heterozygous for antennae shape? .32 X 9904 = 3169

d) The great ruler of Pangoria has determined that Pangorians born with straight antennae have a greater

tendency toward violent behavior than do those with corkscrew antennae. He also determined that

neutering stops the violent behavior. He decrees that all Pangorians born with straight antennae shall be

neutered shortly after birth. In general, what will happen to the allele frequencies in the population over

the next six generations? a will decrease A will increase

|q2 |q |p |p2 |2pq |

|9904-3565/9904=.64 |,8 |,2 |..04 |.32 |

16. The allele for a hitchhiker's thumb is recessive compared to straight thumbs, which are dominant . In a population of 1000 individuals, 512 show the dominant phenotype. How many individuals would you expect for each of the three possible genotypes for this trait?

aa = 68% Aa = 43.5% AA = 10%

|q2 |q |p |p2 |2pq |

|1000-532 = 468/1000 =.468 |.68 |.32 |.10 |.435 |

17. If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes?   Aa = 42% A = 30% a= 70%

|q2 |q |p |p2 |2pq |

|98/200 = .49 |,7 |,3 |.09 |.42 |

B. Your original population of 200 was hit by a tidal wave and 100 organisms were wiped out, leaving 36

homozygous recessive out of the 100 survivors. If we assume that all individuals were equally likely to be wiped out, how did the tidal wave affect the predicted frequencies of the alleles in the population?

A = 40% (increased) a= 60% (decreased)

|q2 |q |p |p2 |2pq |

|36/100=.36 |,6 |,4 |.16 |.48 |

C. Is this population in Hardy-Weinberg equilibrium? NO allele frequencies changed.

D. If the allele frequencies did change as the result of the tidal wave, we would say that microevolution

has occurred. What do we call the phenomenon that caused this evolution? Genetic drift (bottleneck)

18. The gene for albinism is known to be a recessive allele. In Michigan, 9 people in a sample of 10,000 were found to have albino phenotypes. The other 9,991 had skin pigmentation normal for their ethnic group.

|q2 |q |p |p2 |2pq |

|9/10,000 = .0009 |,03 |.97 |.94 |.06 |

 

a.  Assuming hardy-Weinberg equilibrium, what is the allele frequency for the dominant pigmentation allele in

this population? 97%

    b. How many out of the 10,000 people in the sample above were expected to be heterozygous for

pigmentation? .06 X 10,000 = 600

    c.   A similar survey was carried out in Wisconsin, but only 2,500 people were surveyed. If allele frequencies

are the same in Wisconsin as they are in Michigan, how many people would you expect to have the albino

phenotypes in this sample. 2500 X .0009 = 2

    d.    Following are the data that were actually collected in Wisconsin. Perform a chi-square test on these data.

Can you accept the hypothesis that the allele frequency is the same in Wisconsin as in Michigan?

Give degrees of freedom and p values. Use the chi-square table below to interpret your results.

In Hardy-Weinberg chi-square analysis, the number of degrees of freedom is equal to the number of genotypes minus the number of alleles. (Once a value for q is determined, there is no freedom for choosing the value for p.)

|Table 2. Chi-square probability table |

|  |

|AA |

|Aa |

|aa |

|Total |

| |

|Observed |

|2345 |

|150 |

|5 |

| 2500 |

| |

|Expected |

| .94 X 2500= 2350 |

| .06 X 2500 = 150 |

| 0.0009 X 2500=2 |

| 2500 |

| |

X2 = (o-e) 2 = (2345-2350)2 + (150 – 150)2 + (5-2)2 = 25 + O + 9 = 0.01 + 0 + 4.5 = 4.51

E 2350 150 2 2350 2

Df = 3-2=1 4.51 > 3.84 REJECT the Null hypothesis

PROBLEMS FROM:







................
................

Online Preview   Download