ࡱ> ?A>'` "bjbj 70"T$,f 5$j        #######$&h)Z#     #  ####   ## ####  pNEwT##$05$#a)p#a)#a)#$  #     ###   5$    f f f  f f f 4" Section 10-1 VOCABULARY REVIEW 1. A virulent strain of a bacterium is one that causes disease. 2. Transformation is the transfer of genetic material from one cell to another cell or from one organism to another organism. 3. A bacteriophage is a virus that infects bacteria. MULTIPLE CHOICE 1. a 2. c 3. d 4. a 5. b SHORT ANSWER 1. to show that live R cells are not virulent 2. to show that live S cells are virulent and can kill a mouse 3. to show that heat-killed S cells do not cause disease 4. to show that something in the heat-killed S culture was acting on the live R cells to kill the mouse; experiment 3 showed that it was not the killed S cells themselves that killed the mouse. 5. The slippery capsule prevents the cells of the defense system from capturing and destroying the bacterial cells. STRUCTURES AND FUNCTIONS 1. Experiment 2 2. Experiment 1 Section 10-2 VOCABULARY REVIEW 1. A purine is a nitrogenous base with two rings of carbon and nitrogen atoms. Examples may include adenine or guanine. 2. A pyrimidine is a nitrogenous base with one ring of carbon and nitrogen atoms. Examples may include cytosine or thymine. 3. A complementary base-pair is a pair of nitrogenous bases connected to each other by hydrogen bonds. Examples may include adenine-thymine and cytosine-guanine. 4. A nitrogenous base is a base in DNA containing nitrogen and carbon. MULTIPLE CHOICE 1. c 2. d 3. a 4. b 5. b SHORT ANSWER 1. The three parts are a deoxyribose sugar, a phosphate group, and a nitrogenous base. The phosphate group and the base are connected to different parts of the sugar. 2. Since guanine and cytosine are complementary, another 15% of the nucleotides must contain cytosine. The remaining 70% of the nucleotides (100%30%) must contain adenine and thymine in equal proportions (35% each), since they are complementary to each other. 3. Complementary base pairing is important because the hydrogen bonds between the bases hold the two strands of DNA together and because it serves as a way for DNA to replicate. 4. The X-ray diffraction photographs showed that the shape of the DNA molecule was a double helix. STRUCTURES AND FUNCTIONS a, deoxyribose; b, guanine; c, adenine; d, phosphate group Section 10-3 VOCABULARY REVIEW 1. A replication fork is a Y-shaped region that results when the two strands of DNA separate during replication. 2. A helicase is an enzyme that separates the strands of DNA during replication. 3. Semi-conservative replication produces a new DNA molecule with one original strand and one new strand. MULTIPLE CHOICE 1. b 2. a 3. b 4. d 5. c SHORT ANSWER 1. Replication occurs simultaneously at many origins along the DNA. 2. Producing exact copies ensures that when a cell divides, the offspring cells will receive the same genetic information as the parent cell. 3. Cancer can result when errors occur in the replication of DNA in genes that control how a cell divides. A mass of cancerous cells called a tumor can result. 4. The hydrogen bonds break easily, making it easier for the two strands in the molecule to separate during replication. The strong covalent bonds ensure that the sequence of nucleotides remains fixed in each strand. STRUCTURES AND FUNCTIONS Part a: helicase enzymes separate DNA strands; Part b: DNA polymerase enzymes add complementary nucleotides to each original strand of DNA and covalent bonds form between adjacent nucleotides; Part c: DNA polymerases finish replicating DNA and fall off, two DNA molecules identical to original DNA molecule have formed. Section 10-4 VOCABULARY REVIEW 1. A codon is a sequence of three mRNA nucleotides that codes for a specific amino acid or a start or stop signal. 2. Translation is the process of assembling polypeptides from information encoded in mRNA. 3. An anticodon is a sequence of three tRNA nucleotides that pairs with a specific codon. MULTIPLE CHOICE 1. a 2. d 3. b 4. b 5. b SHORT ANSWER 1. The anticodons are UAC, GUA, CGU, and UCA. (The last three nucleotides in the mRNA sequence are a stop codon, which has no anticodon.) The polypeptide will initially contain four amino acids. 2. The tRNA that pairs with the start codon on mRNA carries methionine. 3. RNA contains ribose; DNA contains deoxyribose. RNA usually contains uracil in place of thymine. RNA is single stranded; DNA is double stranded. 4. All of the codons from the deletion point to the end of the transcript would be shifted by one nucleotide, so the sequence of amino acids specified from that point on would be different. Translation would terminate prematurely if the shift resulted in a new stop codon before the end of the transcript. Back Print 9 Copyright by Holt, Rinehart and Winston. All rights reserved. Modern Biology Study Guide Answer Key STRUCTURES AND FUNCTIONS a, polypeptide or protein; b, peptide bond; c, amino acid; d, tRNA; e, anticodon; f, codon; g, mRNA or transcript; h, ribosome  "_b " % ' * , / 1 4 6 9 ; H K ] _ v y   < > $ = @ M P ] j |  p s   Y i l n )h1H6B*CJOJQJ]^JaJph#h1HB*CJOJQJ^JaJph)h1H5B*CJOJQJ\^JaJph)h1H5B*CJOJQJ\^JaJph)h1H5B*CJ OJQJ\^JaJ ph< V_ " ; H v % W  $ = M ] j | 7$8$H$gd1H" ( S p  D Y i (6g,;n 7$8$H$gd1Hn q s v x { } 69;>Pi47  %(ilp[^ #%(*-/2479خخ)h1H5B*CJ OJQJ\^JaJ ph)h1H5B*CJOJQJ\^JaJph#h1HB*CJOJQJ^JaJph)h1H5B*CJOJQJ\^JaJphD"Pi'4j%Zi1Y 7$8$H$gd1H*Zp!J}N[ 9Ft  7$8$H$gd1H9FI  QT!#c !îr]D2#h1HB*CJOJQJ^JaJph1h1Hh1HB*CJOJQJ^JaJmHphsH)h1H5B*CJOJQJ\^JaJph#h1HB*CJOJQJ^JaJph)h1H5B*CJOJQJ\^JaJph)h1H5B*CJOJQJ\^JaJph)h1H5B*CJOJQJ\^JaJph#h1HB*CJOJQJ^JaJph)h1H5B*CJOJQJ\^JaJph)h1H5B*CJOJQJ\^JaJph 8QF{!#c!" 7$8$H$gd1H!"hZ50P:p1H/ =!"#$% @@@ NormalCJ_HaJmH sH tH DA@D Default Paragraph FontRi@R  Table Normal4 l4a (k@(No List"0 z z z" 2 V_";Hv%W$=M]j|(SpDYi(6g,;n"Pi' 4 j  % Z i 1 Y * Z p ! J } N[ 9Ft 8QF{!#c!$!!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!j!!j!R!j!j!j!j!j! V_";Hv%W$=M]j|(SpDYi(6g,;n"Pi' 4 j  % Z i 1 Y * Z p ! J } N[ 9Ft 8QF{!#c!$000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000X0!$0j00$0n 9!"  ""lw9 A  MW/4EOv$V]!%&*+/0459:KMy{%(Wa(*SZDLbhlmqrvw{|(*gn,.nq"'ij' 2 j l         Z _ 1 3 Y `  * 0 Z _ ! & } NR#$()-.23788<FP{#M$:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::!$$Z1H@!!8!!"0@Unknown Gz Times New Roman5Symbol3& z ArialE StoneSans-BoldGCheltenham-BoldGCheltenham-BookSCheltenham-BookItalicM StoneSans-Semibold; StoneSans"qh:&:&G "G "!243HX)?1H2 Section 10-1sclarksclarkOh+'0t  0 < HT\dlSection 10-1sclarkNormalsclark1Microsoft Office Word@F#@6Ew@|^EwG՜.+,0 hp  FHSD" '  Section 10-1 Title  !"#$%&'()*+,-/012345789:;<=@Root Entry F`jXEwB1Tablea)WordDocument70SummaryInformation(.DocumentSummaryInformation86CompObjq  FMicrosoft Office Word Document MSWordDocWord.Document.89q