Inheritance Test Preview

Multiple ChoiceIdentify the letter of the choice that best completes the statement or answers the question.

Figure 9-1

____ 1. The region labeled A in Figure 9-1 is called thea. centromere.b. centriole.c. sister chromatid.d. spindle.

____ 2. The structures labeled B in Figure 9-1 are calleda. centromeres.b. centrioles.c. sister chromatids.d. spindles.

____ 9. Unlike mitosis, meiosis results in the formation ofa. diploid cells.b. haploid cells.c. 2n daughter cells.d. body cells.

____ 10. Crossing-over rarely occurs in mitosis, unlike meiosis. Which of the following is the likely reason?a. Chromatids are not involved in mitosis.b. Tetrads rarely form during mitosis.c. A cell undergoing mitosis does not have homologous chromosomes.d. There is no prophase during mitosis.

____ 20. Unlike mitosis, meiosis usually results in the formation ofa. two genetically identical cells.b. four genetically different cells.c. four genetically identical cells.d. two genetically different cells.

____ 21. Gametes are produced by the process ofa. mitosis.b. meiosis.c. crossing-over.d. replication.

Figure 9-2

____ 22. What is shown in Figure 9-2?a. independent assortmentb. anaphase I of meiosisc. crossing overd. replication

____ 23. Gregor Mendel concluded that traits area. not inherited by offspring.b. inherited through the passing of factors from parents to offspring.c. determined by dominant factors only.d. determined by recessive factors only.

____ 24. The farther apart two genes are located on a chromosome, thea. less likely they are to be inherited together.b. more likely they are to be linked.c. less likely they are to assort independently.d. less likely they are to be separated by a crossover during meiosis.

____ 25. In the P generation, a tall plant was crossed with a short plant. No F1 plants were short. Short plants reappeared in the F2 generation becausea. some of the F2 plants produced gametes that carried the allele for shortness.b. the allele for shortness is dominant.c. the allele for shortness and the allele for tallness segregated when the F1 plants produced

gametes.d. they inherited an allele for shortness from one parent and an allele for tallness from the

other parent.

____ 26. The principles of probability can be used toa. predict the traits of the offspring produced by genetic crosses.b. determine the actual outcomes of genetic crosses.c. predict the traits of the parents used in genetic crosses.d. decide which organisms are best to use in genetic crosses.

Tt

T t

TT

T TT Tt

T TT Tt

T = tall

t = short

Figure 10-1

____ 27. In the Punnett square shown in Figure 10-1, which of the following is true about the offspring resulting from the cross?a. About half are expected to be short.b. All are expected to be short.c. All are expected to be of medium height.d. All are expected to be tall.

____ 28. What principle states that during gamete formation genes for different traits separate without influencing each other’s inheritance?a. principle of dominanceb. principle of independent assortmentc. principle of probabilitiesd. principle of segregation

____ 29. A cross of a black chicken (CBCB) with a white chicken (CWCW) produces all “blue” offspring (CBCW). This type of inheritance is known asa. intermediate inheritance.b. polygenic inheritance.c. codominance.d. multiple alleles.

____ 30. Variation in human skin color is a result ofa. intermediate inheritance.b. codominance.c. polygenic inheritance.d. multiple alleles.

____ 31. Which of the following assort independently?a. chromosomesb. two genes on the same chromosomec. multiple allelesd. codominant alleles

RrYy

RY Ry rY ry

RY RRYY RRYy RrYY RrYySeed ShapeR – roundr – wrinkled

RrYyRy RRYy RRyy RrYy Rryy

Seed ColorY – yellowy – greenrY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

Figure 10-2

____ 32. The Punnett square in Figure 10-2 shows that the gene for pea shape and the gene for pea colora. assort independently.b. are linked.c. have the same alleles.d. are always homozygous.

____ 33. When Gregor Mendel crossed a true-breeding tall plant with a true-breeding short plant, the F1 plants inheriteda. an allele for tallness from each parent.b. an allele for tallness from the tall parent and an allele for shortness from the short parent.c. an allele for shortness from each parent.d. an allele from only the tall parent.

____ 34. In the P generation, a true-breeding tall plant (dominant) was crossed with a true-breeding short plant. If alleles did not segregate during gamete formation,a. all of the F1 plants would be short.b. some of the F1 plants would be tall and some would be short.c. all of the F2 plants would be short.d. all of the F2 plants would be tall.

____ 35. A man and a woman who are both heterozygous for normal skin pigmentation (Aa) produce an albino offspring (aa). Which of Mendel’s principles explain(s) why the offspring is albino?a. dominance onlyb. independent assortment onlyc. dominance and segregationd. segregation only

____ 36. If the gene for seed color (yellow is dominant) and the gene for seed shape (round is dominant) in pea plants were linked,a. all of Mendel’s F1 plants would have produced wrinkled, green peas.b. Mendel’s F2 plants would have shown a different phenotype ratio for seed color and seed

shape.c. Mendel’s F1 plants would have shown a different phenotype ratio for seed color and seed

shape.d. all of Mendel’s P plants would have produced wrinkled, green peas.

____ 37. Offspring that result from crosses between true-breeding parents with different traitsa. are true-breeding.b. make up the F2 generation.c. make up the parental generation.d. are called hybrids.

____ 38. When Gregor Mendel crossed true-breeding purple-flowered plants with true-breeding white-flowered plants, all the offspring were purple becausea. the allele for purple-flowered plants is recessive.b. the allele for white-flowered plants is dominant.c. the allele for purple-flowered plants is dominant.d. they were true-breeding like their parents.

____ 39. If a pea plant has a recessive allele for green peas, it will producea. green peas if it also has a dominant allele for yellow peas.b. both green peas and yellow peas if it also has a dominant allele for yellow peas.c. green peas if it does not also have a dominant allele for yellow peas.d. yellow peas if it does not also have a dominant allele for green peas.

____ 40. Organisms that have two identical alleles for a particular trait are said to bea. hybrid.b. homozygous.c. heterozygous.d. dominant.

____ 41. A Punnett square shows all of the following EXCEPTa. all possible results of a genetic cross.b. the possible genotypes of the offspring.c. the alleles in the gametes of each parent.d. the actual results of a genetic cross.

____ 42. If two genes are on the same chromosome and rarely assort independently,a. crossing over never occurs between the genes.b. crossing over always occurs between the genes.c. the genes are probably located far apart from each other.d. the genes are probably located close to each other.

____ 43. Gregor Mendel removed the male parts (stamens) from the flowers of some plants in order toa. prevent hybrids from forming.b. prevent cross-pollination.c. prevent self-pollination.d. prevent blending.

____ 44. In the P generation, a true-breeding tall plant (dominant) is crossed with a true-breeding short plant (recessive). The probability that an F2 plant will be tall isa. 50%.b. 75%.c. 25%.d. 100%.

____ 45. What did Griffith observe when he injected a mixture of heat-killed, disease-causing bacteria and live harmless bacteria into mice?a. The disease-causing bacteria changed into harmless bacteria.b. The mice died.

c. The harmless bacteria died.d. The mice were unaffected.

____ 46. Avery’s experiments showed that bacteria are transformed bya. RNA.b. DNA.c. proteins.d. carbohydrates.

____ 47. Which of the following is a nucleotide found in DNA?a. ribose + phosphate group + thymineb. ribose + phosphate group + uracilc. deoxyribose + phosphate group + uracild. deoxyribose + phosphate group + cytosine

____ 48. DNA replication results in two DNA molecules,a. each with two new strands.b. one with two new strands and the other with two original strands.c. each with one new strand and one original strand.d. each with two original strands.

____ 49. During DNA replication, a DNA strand that has the bases CTAGGT produces a strand with the basesa. TCGAAC.b. GATCCA.c. AGCTTG.d. GAUCCA.

____ 50. What happens during the process of translation?a. Messenger RNA is made from DNA.b. The cell uses information from messenger RNA to produce proteins.c. Transfer RNA is made from messenger RNA.d. Copies of DNA molecules are made.

____ 51. How many codons are needed to specify three amino acids?a. 3b. 6c. 9d. 12

____ 52. What is produced during transcription?a. RNA moleculesb. DNA moleculesc. RNA polymerased. proteins

____ 53. During translation, the type of amino acid that is added to the growing polypeptide depends on thea. codon on the tRNA only.b. anticodon on the mRNA only.c. anticodon on the tRNA to which the amino acid is attached only.d. codon on the mRNA and the anticodon on the tRNA to which the amino acid is attached.

____ 54. Genes contain instructions for assemblinga. purines.b. nucleosomes.

c. proteins.d. pyrimidines.

____ 55. What would Hershey and Chase have concluded if both radioactive phosphorus and radioactive sulfur were found in the bacteria in their experiment?a. The virus’s protein coat was not injected into the bacteria.b. The virus’s DNA was not injected into the bacteria.c. Genes are made of protein.d. Both the virus’s protein coat and its DNA were injected into the bacteria.

____ 56. Which of the following are found in both DNA and RNA?a. ribose, phosphate groups, and adenineb. deoxyribose, phosphate groups, and guaninec. phosphate groups, guanine, and cytosined. phosphate groups, guanine, and thymine

____ 57. DNA is copied during a process calleda. replication.b. translation.c. transcription.d. transformation.

____ 58. Because of base pairing in DNA, the percentage ofa. adenine molecules in DNA is about equal to the percentage of guanine molecules.b. pyrimidines in DNA is about equal to the percentage of purines.c. purines in DNA is much greater than the percentage of pyrimidines.d. cytosine molecules in DNA is much greater than the percentage of guanine molecules.

____ 59. Which of the following statements is false?a. Some genes code for enzymes.b. The instructions for making some proteins are not specified by genes.c. An organism’s proteins determine its genes.d. An organism’s genes determine its inherited traits.

____ 60. Why is it possible for an amino acid to be specified by more than one kind of codon?a. Some codons have the same sequence of nucleotides.b. There are 64 different kinds of codons but only 20 amino acids.c. Some codons do not specify an amino acid.d. The codon AUG codes for the amino acid methionine and serves as the “start” codon for

protein synthesis.

____ 61. RNA contains the sugara. ribose.b. deoxyribose.c. glucose.d. lactose.

____ 62. Unlike DNA, RNA containsa. adenine.b. uracil.c. phosphate groups.d. thymine.

____ 63. Which of the following statements is true?

a. A exon is part of an intron.b. An initial RNA transcript is longer than the gene from which the molecule was

transcribed.c. Introns have complementary sequences in DNA.d. During RNA splicing, the exons are removed.

____ 64. Which of the following terms is LEAST closely related to the others?a. intronb. tRNAc. amino acidd. anticodon

____ 65. Which type(s) of RNA is(are) involved in protein synthesis?a. transfer RNA onlyb. messenger RNA onlyc. ribosomal RNA and transfer RNA onlyd. messenger RNA, ribosomal RNA, and transfer RNA

____ 66. Which type of RNA functions as a “blueprint” for protein synthesis?a. rRNAb. tRNAc. mRNAd. RNA polymerase

Modified True/FalseIndicate whether the sentence or statement is true or false. If false, change the identified word or phrase to make the sentence or statement true.

____ 67. Reproduction by simple cell division is called sexual reproduction, whereas the process in which two parents contribute genetic material is called asexual reproduction. ______________________________

Figure 9-1

____ 68. The structure shown in Figure 9-1 is a duplicated chromosome. _____________________________ 69. A cell splits into two daughter cells during cytokinesis. _____________________________ 70. Homologous chromosomes are the two sets of chromosomes found in a body cell—one set inherited from the

male parent and the other inherited from the female parent. _____________________________ 71. Mitosis results in two cells, whereas meiosis results in one cell. _____________________________ 73. A cell’s chromosomes are duplicated during interphase. _____________________________ 75. Cancer is a disease caused by the severe disruption of the mechanisms that normally control the cell cycle.

_____________________________ 76. If an organism has 16 chromosomes in each of its egg cells, the organism’s diploid number is 32.

_____________________________ 77. A trait is a variation of a particular character. _________________________

____ 78. An organism that is heterozygous for a particular character will generally show the recessive trait. _________________________

____ 79. If blue Andalusian chickens are mated, according to the idea of intermediate inheritance, 25% of the offspring are expected to be blue. _________________________

____ 80. Blood type in humans is determined by a single gene that has multiple alleles. _____________________________ 81. Gregor Mendel concluded that the true-breeding tall plants in the P generation passed the factor for tallness to

the F1 generation. _____________________________ 82. True-breeding plants that produced axial flowers were crossed with true-breeding plants that produced

terminal flowers. The resulting offspring produced terminal flowers because the allele for terminal flowers is recessive. _________________________

____ 83. When alleles segregate from each other, they join. _____________________________ 84. If the alleles for a trait did not segregate during gamete formation, offspring would always show the trait of at

least one of the parents. _____________________________ 85. A cross between an individual with an unknown genotype but dominant phenotype and a homozygous

recessive individual is called a testcross. _____________________________ 86. The probability that a gamete produced by a pea plant heterozygous for stem height (Tt) will contain the

recessive allele is 100%. _____________________________ 87. The heritable genetic information of an organism is stored in the molecule called ribonucleic acid (RNA).

_____________________________ 88. In eukaryotes, DNA replication proceeds in one direction from the origin of replication.

_____________________________ 89. The enzymes that make covalent bonds between the nucleotides of DNA strands are called DNA

polymerases. _____________________________ 90. Genes determine a person’s eye color by coding for nitrogenous bases that affect eye color.

_____________________________ 91. During DNA replication, only one strand of DNA serves as a template. _____________________________ 92. The replication of a DNA molecule results in four copies of the same gene. _____________________________ 93. If a nucleic acid contains uracil, it is DNA. _____________________________ 94. A codon consists of four nucleotides. _____________________________ 95. The anticodon AGA is complementary to the codon TCT. _____________________________ 96. The key enzyme involved in transcription is RNA polymerase. _________________________

CompletionComplete each sentence or statement.

104. In the disease _________________________, the mechanisms that normally control the cell cycle are disrupted.

105. An organism’s gametes have _________________________ the number of chromosomes found in the organism’s body cells.

106. Crossing over occurs during the stage of meiosis called ____________________.107. The principle of independent assortment states that ____________________ for different traits can segregate

independently during the formation of gametes.108. The alternative forms of a gene are called ____________________.109. Pea plants that are TT, ____________________, and tt have different genotypes.

Tt

T t

TT

T TT Tt

T TT Tt

T = tall

t = short

Figure 10-1

110. In the Punnett square shown in Figure 10-1, the possible genotypes of the offspring are ____________________.

111. Any gene that is located on a sex chromosome is called a ________________ gene.112. The plants that Gregor Mendel crossed to produce the F1 generation made up the ____________________

generation.113. If the allele for white flowers in pea plants were dominant, all the pea plants in Mendel’s F1 generation would

have had ____________________.114. If the alleles for traits in pea plants did not segregate during gamete formation, offspring that had the

recessive trait could be produced only by crossing two plants that had the ____________________ trait.115. When two heterozygous purple-flowered pea plants are crossed, the expected genotype ratio of the offspring

is _________________________.116. If pea plants that are homozygous for round, yellow seeds (RRYY) were crossed with pea plants that are

heterozygous for round, yellow seeds (RrYy), the expected phenotype(s) of the offspring would be _________________________.

117. The Watson and Crick model of DNA is a(an) _________________________, in which two strands are wound around each other.

118. The order of nitrogenous bases in DNA determines the order of _________________________ in proteins.119. Suppose that part of an amino acid sequence of a protein changed from tyrosine-proline-glycine-alanine to

tyrosine-histidine-glycine-alanine. This change was most likely caused by a _________________________.120. The triplet of bases on the end of a tRNA molecule is called a(an) _________________________.121. The process of joining exons together to form an mRNA molecule is called _________________________.122. In RNA, _________________________ and _________________________ are pyrimidines.123. A stop codon on an mRNA molecule does not code for any _________________________.124. During transcription, the _________________________ between base pairs are broken.125. A mutation will cause the cell to make an incomplete polypeptide if the mutation results in a(an)

_________________________.126. The _________________________ portion of a tRNA molecule determines the type of amino acid that bonds

with the tRNA.

Short Answer128. Distinguish between chromatids and chromatin.129. How are homologous chromosomes alike? How are they different?130. How many sets of chromosomes are in a diploid cell?131. How does crossing over contribute to genetic variation? Use the term genetic recombination in your answer.133. How do cancer cells differ from normal cells?

134. What might happen if the gametes of a species had the same number of chromosomes as the species’ body cells?

135. What happens to the number of chromosomes per cell during meiosis?136. Contrast the cells produced by mitosis with those produced by meiosis.137. Define genetics.138. What are monohybrid and dihybrid crosses?139. What is polygenic inheritance? Give two examples.140. What is genetic linkage?141. Why are most sex-linked disorders more common in males?142. What might have caused Gregor Mendel NOT to conclude that biological inheritance is determined by factors

that are passed from one generation to the next?143. How many recessive alleles for a trait must an organism inherit in order to have that trait?

RrYy

RY Ry rY ry

RY RRYY RRYy RrYY RrYySeed ShapeR – roundr – wrinkled

RrYyRy RRYy RRyy RrYy Rryy

Seed ColorY – yellowy – greenrY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

Figure 10-2144. What is the phenotype ratio of the offspring in the Punnett square shown in Figure 10-2?145. A pea plant heterozygous for height and seed color (TtYy) is crossed with a pea plant heterozygous for height

but homozygous recessive for seed color (Ttyy). If 80 offspring are produced, how many are expected to be tall and have yellow seeds?

146. The gene map of a fruit fly’s chromosome 2 shows the relative locations of the star eye, dumpy wing, and black body genes to be 1.3, 13.0, and 48.5, respectively. Between which two genes is crossing over likely to occur most frequently?

147. Explain how the bacteria in Griffith’s experiment were transformed.148. What is a bacteriophage?149. What causes translation to stop?150. What is the template mechanism of DNA replication?151. What is the “one gene–one polypeptide” hypothesis?152. What are the three main parts of an RNA nucleotide?153. If codons consisted of fewer than 3 bases, could they still code for 20 amino acids?154. What must happen to a DNA molecule before RNA polymerase can make RNA?155. If the percentage of guanine in the DNA of a certain species decreased by 5 percent over time, what would

you expect to have happened to the percentage of adenine in that DNA?156. What is a mutation?

Figure 9-5

163. Comparing and Contrasting Explain how cancer cells are different from normal cells. Then, relate these characteristics to the diagram in Figure 9-5 that shows cancer cells.

164. Predicting Look at the cancer cells shown in Figure 9-5. What can happen if these cells are left untreated? What is this process called?

165. Predicting How might the cancer cells shown in Figure 9-5 be prevented from doing more harm to the organism they are a part of?

Figure 9-6

166. Observing List the stages in Figure 9-6 in which the cells are 2n and those in which the cells are n.

167. Interpreting Graphics In Figure 9-6, what is the structure labeled X in stage B?168. Observing In Figure 9-6, during which stage might genetic recombination occur? Identify the stage.

Heterozygous male guinea pigs with black, rough hair (BbRr) are crossed with heterozygous female guinea pigs with black, rough hair (BbRr). The incomplete Punnett square in Figure 10-3 shows the expected results from the cross.

BbRr

BR Br bR br

BR BBRR BBRr BbRR BbRrHair ColorB – blackb – white

BbRrBr BBRr BBrr BbRr Bbrr

Hair TextureR – roughr – smoothbR BbRR BbRr

×bbRr

br BbRr Bbrr bbRr bbrr

Figure 10-3

169. Inferring In Figure 10-3, what is the genotype of an offspring that has white, smooth hair?170. Inferring Identify the genotype of the offspring that would be represented in the square labeled X in Figure

10-3.171. Inferring Identify the phenotype of the offspring represented in the square labeled X in Figure 10-3.172. Inferring In Figure 10-3, what are the different possible phenotypes of the offspring?173. Inferring In Figure 10-3, what are the possible genotypes of offspring that have black, rough hair?174. Calculating What fraction of the offspring in Figure 10-3 would be expected to have white, smooth hair?

Figure 10-4

175. Inferring In Figure 10-4, what is the genotype of the pink-flowered snapdragons?176. Inferring What do the letters R and I represent in Figure 10-4?177. Inferring In Figure 10-4, what is the genotype of the red-flowered snapdragons?178. Inferring Explain whether the alleles in Figure 10-4 show dominance, intermediate inheritance, or

codominance.179. Predicting According to Figure 10-4, if red-flowered snapdragons and ivory-flowered snapdragons are

crossed, what percentage of their offspring would be expected to be pink-flowered?180. Predicting According to Figure 10-4, if two pink-flowered snapdragons are crossed, what percentage of their

offspring would be expected to be pink-flowered?

Figure 11-1

181. Observing How are structures A and B in Figure 11-1 related?

182. Observing What is structure E in Figure 11-1? What does it code for?183. Predicting What would happen to structure F in Figure 11-1 if structure C were deleted?184. Observing Identify structure F in Figure 11-1. What does it specify?185. Inferring From which labeled structure in Figure 11-1 is structure D made? Identify that labeled structure.186. Predicting In Figure 11-1, what effect would the deletion of structure C have on the process that occurs

during step Y?

Figure 11-2

187. Observing Identify structure X in Figure 11-2.188. Observing What molecule’s structure is illustrated in Figure 11-2?189. Inferring What are the dashed lines shown in Figure 11-2? How do they relate to the rules for base pairing?190. Observing What is the twisting shape of the molecule in Figure 11-2 called?191. Observing What is a pyrimidine? Identify two pyrimidines in Figure 11-2.192. Observing What is a purine? Identify two purines in Figure 11-2.

Essay195. How can cancer be treated at the cellular level?196. Explain why the daughter cells produced by meiosis are genetically different from each other, whereas the

daughter cells produced by mitosis are not.197. Assume that prophase begins with eight chromatids in the nucleus of a cell. When telophase ends, how many

chromosomes will be present in each new nucleus? Explain your answer.198. Describe how cancer cells are different from other cells. Based on these differences, explain why cancer has

been such a difficult condition to cure.199. The stages of meiosis are classified into two divisions: meiosis I and meiosis II. Compare and contrast these

two divisions.200. Suppose the homologous chromosomes that make up a tetrad fail to separate during anaphase I of meiosis.

Predict the results of this event.201. How are the blending hypothesis and the particulate hypothesis of inheritance similar? How are they

different?202. You wish to determine whether a tall pea plant is homozygous or heterozygous for tallness. What cross

should you perform to arrive at your answer? Explain your choice of cross. What is this type of cross called?203. How can environmental conditions affect phenotype expression?204. How do sex-linked genes produce different inheritance patterns in males and females?205. A pea plant with yellow seeds was crossed with a plant with green seeds. The F1 generation produced plants

with yellow seeds. Explain why green seeds reappeared in the F2 generation.206. A cross between two organisms heterozygous for two different genes (AaBb) results in a 9 : 3 : 3 : 1

phenotype ratio among the offspring. Is the offspring’s genotype ratio the same? Explain your answer.207. Explain the difference between intermediate inheritance and codominance.208. What is the chromosome theory of inheritance? How does this theory explain Mendel’s results?

209. Describe the process in which an RNA transcript is converted into a final mRNA.210. Contrast the functions of the three main types of RNA.211. Why do some kinds of mutations cause greater changes in proteins than others?212. Describe some causes of mutations. Use the term mutagen in your answer.213. Describe the Hershey-Chase experiment. Why were the results important?214. Describe the structure of a DNA molecule.215. Describe the process of DNA replication in eukaryotes.216. How does transcription differ from DNA replication? Describe at least four differences.

Inheritance Test PreviewAnswer Section

MULTIPLE CHOICE

1. ANS: A DIF: Below average/Average OBJ: 9.2.1, 9.2kt2. ANS: C DIF: Below average/Average OBJ: 9.2.1, 9.2kt3. ANS: B DIF: Average/Above average OBJ: 9.3.14. ANS: A DIF: Below average/Average OBJ: 9.2.2, 9.2kt5. ANS: D DIF: Average/Above average OBJ: 9.2.26. ANS: C DIF: Average/Above average OBJ: 9.3.1, 9.3kt7. ANS: C DIF: Average/Above average OBJ: 9.3.28. ANS: B DIF: Below average/Average OBJ: 9.4.1, 9.4kt9. ANS: B DIF: Below average/Average OBJ: 9.5.3, 9.6.3

10. ANS: B DIF: Average/Above average OBJ: 9.6.311. ANS: D DIF: Average/Above average OBJ: 9.2.212. ANS: C DIF: Below average/Average OBJ: 9.2.213. ANS: D DIF: Below average/Average OBJ: 9.2.314. ANS: A DIF: Average/Above average OBJ: 9.2kt15. ANS: A DIF: Average/Above average OBJ: 9.2kt16. ANS: D DIF: Average/Above average OBJ: 9.2kt17. ANS: C DIF: Below average/Average OBJ: 9.3.118. ANS: A DIF: Below average/Average OBJ: 9.3kt19. ANS: C DIF: Average/Above average OBJ: 9.4kt20. ANS: B DIF: Average/Above average OBJ: 9.5.3, 9.6.321. ANS: B DIF: Below average/Average OBJ: 9.5kt22. ANS: C DIF: Below average/Average OBJ: 9.6kt23. ANS: B DIF: Average/Above average OBJ: 10.1.124. ANS: A DIF: Average/Above average OBJ: 10.4.225. ANS: C DIF: Average/Above average OBJ: 10.2.126. ANS: A DIF: Below average/Average OBJ: 10.2.227. ANS: D DIF: Average/Above average OBJ: 10.2.228. ANS: B DIF: Below average/Average OBJ: 10.2.429. ANS: C DIF: Below average/Average OBJ: 10.3.130. ANS: C DIF: Average/Above average OBJ: 10.3.331. ANS: A DIF: Below average/Average OBJ: 10.4.132. ANS: A DIF: Below average/Average OBJ: 10.2.433. ANS: B DIF: Average/Above average OBJ: 10.2.134. ANS: D DIF: Average/Above average OBJ: 10.2.135. ANS: C DIF: Average/Above average OBJ: 10.2.136. ANS: B DIF: Average/Above average OBJ: 10.4.237. ANS: D DIF: Average/Above average OBJ: 10.2kt38. ANS: C DIF: Below average/Average OBJ: 10.2kt39. ANS: C DIF: Average/Above average OBJ: 10.2kt40. ANS: B DIF: Below average/Average OBJ: 10.2kt41. ANS: D DIF: Average/Above average OBJ: 10.2kt42. ANS: D DIF: Below average/Average OBJ: 10.4.243. ANS: D DIF: Average/Above average OBJ: 10.1.244. ANS: B DIF: Average/Above average OBJ: 10.2.245. ANS: B DIF: Below average/Average OBJ: 11.1.146. ANS: B DIF: Below average/Average OBJ: 11.1.247. ANS: D DIF: Average/Above average OBJ: 11.2.148. ANS: C DIF: Below average/Average OBJ: 11.3.249. ANS: B DIF: Average/Above average OBJ: 11.3.150. ANS: B DIF: Average/Above average OBJ: 11.4.2, 11.4kt51. ANS: A DIF: Below average/Average OBJ: 11.4.352. ANS: A DIF: Below average/Average OBJ: 11.5.153. ANS: D DIF: Average/Above average OBJ: 11.5.354. ANS: C DIF: Below average/Average OBJ: 11.5.455. ANS: D DIF: Average/Above average OBJ: 11.1.356. ANS: C DIF: Average/Above average OBJ: 11.2.157. ANS: A DIF: Below average/Average OBJ: 11.3kt58. ANS: B DIF: Average/Above average OBJ: 11.2.259. ANS: B DIF: Average/Above average OBJ: 11.4.160. ANS: B DIF: Average/Above average OBJ: 11.4.361. ANS: A DIF: Below average/Average OBJ: 11.4kt

62. ANS: B DIF: Below average/Average OBJ: 11.4kt63. ANS: C DIF: Average/Above average OBJ: 11.5.264. ANS: A DIF: Average/Above average OBJ: 11.5.365. ANS: D DIF: Below average/Average OBJ: 11.5.466. ANS: C DIF: Below average/Average OBJ: 11.5kt

MODIFIED TRUE/FALSE67. ANS: F, asexual, sexual DIF: Average/Above average OBJ: 9.1.2, 9.1kt68. T DIF:Average/Above average OBJ: 9.2.1, 9.2kt69. ANS: T DIF:Below average/Average OBJ: 9.3.170. T DIF:Average/Above average OBJ: 9.5kt71. F, four cells DIF:Below average/Average OBJ: 9.6.372. F, G1 DIF:Average/Above average OBJ: 9.2.273. T DIF:Below average/Average OBJ: 9.2.274. F, prophase DIF:Average/Above average OBJ: 9.3.175. T DIF:Below average/Average OBJ: 9.4kt76. T DIF:Below average/Average OBJ: 9.5.277. ANS: T DIF:Below average/Average78. F, dominant DIF:Below average/Average OBJ: 10.2kt79. F, 50% DIF:Average/Above average OBJ: 10.3.180. T DIF:Below average/Average OBJ: 10.3.281. T DIF:Below average/Average82. F, dominant DIF:Average/Above average OBJ: 10.2kt83. F, separate DIF:Below average/Average OBJ: 10.2.184. T DIF:Below average/Average OBJ: 11.1.485. T DIF:Below average/Average OBJ: 10.2kt86. F, 50% DIF:Average/Above average OBJ: 10.2.287. F, deoxyribonucleic acid (DNA) DIF:Below average/Average OBJ: 11.2kt88. F, two directions DIF:Below average/Average OBJ: 11.3.289. T DIF:Average/Above average OBJ: 11.3kt90. F, proteins DIF:Below average/Average OBJ: 11.4.191. F, transcription DIF:Average/Above average OBJ: 11.5.192. F, two DIF:Below average/Average OBJ: 11.3.293. F, RNA DIF:Below average/Average OBJ: 11.4kt94. F, three DIF:Below average/Average OBJ: 11.4.295. F, UCU DIF:Average/Above average OBJ: 11.5.396. T DIF:Average/Above average OBJ: 11.5.4

COMPLETION97. spindle microtubule DIF:Average/Above average OBJ: 9.3.198. mitosis DIF:Below average/Average OBJ: 9.3kt99. plant DIF:Average/Above average OBJ: 9.3.2

100. cell plate DIF:Below average/Average OBJ: 9.3kt101. cytokinesis DIF:Below average/Average OBJ: 9.3kt102. 10 DIF:Average/Above average OBJ: 9.2.1103. interphase DIF:Below average/Average OBJ: 9.2.2104. cancer DIF:Below average/Average OBJ: 9.4kt105. half DIF:Average/Above average OBJ: 9.5.2106. prophase DIF:Average/Above average OBJ: 9.6.2107. genes DIF:Below average/Average OBJ: 10.2.3108. alleles DIF:Below average/Average OBJ: 10.2kt109. Tt DIF:Average/Above average OBJ: 10.2kt110. TT and Tt DIF:Below average/Average OBJ: 10.2kt111. sex-linked DIF:Average/Above average OBJ: 10.5kt112. P parental DIF:Below average/Average OBJ: 10.2.1113. white flowers DIF:Below average/Average OBJ: 10.2.1114. recessive DIF:Average/Above average OBJ: 10.2.1115. ANS: 1 PP : 2 Pp : 1 pp DIF:Average/Above average OBJ: 10.2.3116. round yellow seeds DIF:Average/Above average OBJ: 10.2.4117. double helix DIF:Below average/Average OBJ: 11.2.2118. amino acids DIF:Below average/Average OBJ: 11.4.2119. mutation, substitution DIF:Average/Above average OBJ: 11.6kt120. anticodon DIF:Below average/Average OBJ: 11.5kt121. RNA splicing DIF:Average/Above average OBJ: 11.5kt122. cytosine, uracil DIF:Average/Above average OBJ: 11.4.2123. amino acid DIF:Below average/Average OBJ: 11.4.3

124. hydrogen bonds DIF:Average/Above average OBJ: 11.5.1125. stop codon DIF:Average/Above average OBJ: 11.5.3126. anticodon DIF:Average/Above average OBJ: 11.5.3

SHORT ANSWER

127. When you are injured, your body produces cells that heal the wound. All of the growth of your body is also a result of new cells produced by cell reproduction.DIF: Below average/Average OBJ: 9.1.1

128. Chromatids are two identical DNA strands joined by a centromere, and chromatin is the material (DNA and proteins) that makes up chromosomes.DIF: Average/Above average OBJ: 9.2kt

129. Homologous chromosomes are similar in size and shape. Each one contains the same sequence of genes controlling the same inherited characteristics. However, the two genes may be slightly different versions.DIF: Below average/Average OBJ: 9.5.1

130. A diploid cell has two sets of chromosomes.DIF: Below average/Average OBJ: 9.5kt

131. Crossing over is the exchange of genetic material between homologous chromosomes. During crossing over, segments of two chromatids can be exchanged at one or more sites. Crossing over can thus produce a single chromosome that contains a new combination of genetic information from different parents, a result called genetic recombination.DIF: Average/Above average OBJ: 9.6.2, 9.6kt

132. During cytokinesis, a cell plate forms in the cytoplasm midway between each new nucleus. The cell plate gradually develops into a separating cell wall.DIF: Average/Above average OBJ: 9.3.2

133. Cancer cells have severe disruption of the mechanisms that normally control the cell cycle. As a result, cancer cells form tumors and can spread throughout the body.DIF: Below average/Average OBJ: 9.4kt

134. When the gametes fused during fertilization, the offspring would have more chromosomes in their body cells than their parents have. As a result, the species’ chromosome number would not be constant.DIF: Average/Above average OBJ: 9.5.2

135. The number of chromosomes is cut in half.DIF: Below average/Average OBJ: 9.5.3

136. Mitosis produces diploid body cells, whereas meiosis produces haploid gametes.DIF: Below average/Average OBJ: 9.6.3

137. Genetics is the study of heredity.DIF: Below average/Average OBJ: 10.1kt

138. A monohybrid cross is a pairing in which the parent plants differ in only one character. A dihybrid cross is a cross between organisms differing in two characters.DIF: Below average/Average OBJ: 10.2kt

139. Polygenic inheritance is a situation in which two or more genes affect a single character. Two examples in humans are height and skin color.DIF: Below average/Average OBJ: 10.3kt

140. Genetic linkage is the tendency for the alleles on one chromosome to be inherited together.DIF: Average/Above average OBJ: 10.4kt

141. Males only have one X chromosome, so they can have a sex-linked disorder even if they have only one copy of the allele for the disorder.DIF: Average/Above average OBJ: 10.5.2 STO: SC.F.2.4.1, SC.F.2.4.2

142. Answers may vary. If the F1 pea plants had traits of neither parent, Mendel might not have concluded that factors for traits are passed from one generation to the next.DIF: Average/Above average OBJ: 10.1.1 STO: SC.H.1.4, SC.H.2.4.1

143. An organism must inherit two recessive alleles for a trait in order to have that trait.DIF: Below average/Average OBJ: 10.2.1 STO: SC.F.2.4.1

144. The phenotype ratio is 9 round, yellow peas : 3 round, green peas : 3 wrinkled, yellow peas : 1 wrinkled, green pea.DIF: Below average/Average OBJ: 10.2.4 STO: SC.F.2.4.1

145. Thirty of the offspring are expected to be tall and have yellow seeds.DIF: Average/Above average OBJ: 10.2.4

146. Crossing over likely occurs most frequently between the star eye gene and the black body gene.DIF: Average/Above average OBJ: 10.4.2

147. The harmless living bacteria took in pneumonia-causing DNA (genes) from the heat-killed, pneumonia-causing bacteria, as a result of which the harmless bacteria changed into bacteria that cause pneumonia.DIF: Average/Above average OBJ: 11.1.1

148. A bacteriophage is a kind of virus that infects bacteria.DIF: Below average/Average OBJ: 11.1kt

149. A stop codon on the mRNA causes translation to stop.DIF: Below average/Average OBJ: 11.5.3

150. The base-pairing rules mean that if you know the sequence of bases on one strand of DNA, you can determine the sequence on the other strand. During DNA copying, the two strands of DNA separate and each strand acts as a template for producing a new, complementary strand.DIF: Average/Above average OBJ: 11.3.1

151. The “one gene–one polypeptide” hypothesis is the idea that each gene dictates the production of a single polypeptide, which may make up part of an enzyme or another kind of protein.DIF: Below average/Average OBJ: 11.4.1

152. A ribose molecule, a phosphate group, and a nitrogenous base are the three main parts of an RNA nucleotide.DIF: Average/Above average OBJ: 11.4.2

153. If codons consisted of fewer than 3 bases, the bases could not combine in enough ways to produce codons for specifying the 20 different amino acids.DIF: Average/Above average OBJ: 11.4.3

154. The DNA molecule must be separated into two strands.DIF: Below average/Average OBJ: 11.5.1

155. The percentage of adenine would have increased by about 5 percent.DIF: Average/Above average OBJ: 11.2.2

156. A mutation is any change in the nucleotide sequence of DNA.DIF: Below average/Average OBJ: 11.6kt

OTHER

157. FourDIF: Below average/Average OBJ: 9.2.1

158. It shows various stages of mitosis in an animal cell. We know this is an animal cell because of the presence of centrioles and the shape of the cells.DIF: Average/Above average OBJ: 9.3.1

159. X is the centriole; Y is a spindle fiber.DIF: Average/Above average OBJ: 9.3.1

160. D, A, C, BDIF: Average/Above average OBJ: 9.3.1

161. The next step would be cytokinesis. It would show two daughter cells forming.DIF: Below average/Average OBJ: 9.3.2

162. anaphaseDIF: Below average/Average OBJ: 9.3kt

163. Cancer cells do not respond to the signals that control the growth of most cells. As a result, cancer cells form masses (tumors). These signals include growth factors that stimulate cell division at a proper rate and signals that prevent excessive growth so that tissues do not disrupt each other. Diagram A shows cells that have divided until they have formed a tumor. These cells are dividing more quickly than normal cells do. They have started disrupting adjacent cells.DIF: Average/Above average OBJ: 9.4.1

164. They can break loose from the mass they are now a part of and spread throughout the body, disrupting normal activities and causing serious medical problems. This process is called metastasis.DIF: Below average/Average OBJ: 9.4kt

165. Students may suggest removing the cancerous cells in hopes of preventing their continued division, growth, and spread throughout the body, treating them with radiation or chemicals that will destroy the cells, etc.DIF: Average/Above average OBJ: 9.4.2

166. The cells in stages A, B, and C are 2n. The cells in stages D, E, F, and G are n.DIF: Average/Above average OBJ: 9.5.3

167. The structure is a tetrad.DIF: Below average/Average OBJ: 9.5kt

168. Genetic recombination might occur during stage A, which is prophase I.DIF: Below average/Average OBJ: 9.6kt

169. The offspring has genotype bbrr.DIF: Below average/Average OBJ: 10.2.3

170. The genotype of the offspring is bbRR.DIF: Below average/Average OBJ: 10.2.3

171. The phenotype of the offspring is white, rough hair.DIF: Below average/Average OBJ: 10.2.3

172. The possible phenotypes of the offspring are black, rough hair; black, smooth hair; white, rough hair; and white, smooth hair.DIF: Average/Above average OBJ: 10.2.3

173. Offspring with black, rough hair could have the genotypes BBRR, BBRr, BbRR, or BbRr.DIF: Average/Above average OBJ: 10.2.3

174. One sixteenth of the offspring would be expected to have white, smooth hair.DIF: Average/Above average OBJ: 10.2.3

175. The genotype of the pink-flowered snapdragons is RI.DIF: Below average/Average OBJ: 10.2kt

176. R represents the allele for red flowers. I represents the allele for ivory flowers.DIF: Below average/Average OBJ: 10.3.1

177. The genotype of the red-flowered snapdragons is RR.DIF: Below average/Average OBJ: 10.3.1

178. The alleles show intermediate inheritance, because a cross between red-flowered snapdragons and ivory-flowered snapdragons produces snapdragons with an in-between trait—pink flowers.DIF: Average/Above average OBJ: 10.3.1

179. One hundred percent of the offspring would be expected to be pink-flowered.DIF: Average/Above average OBJ: 10.3.1

180. Fifty percent of the offspring would be expected to be pink-flowered.DIF: Average/Above average OBJ: 10.3.1

181. Structures A and B are complementary strands of DNA.

DIF: Below average/Average OBJ: 11.2.2

182. Structure E is the start codon, which codes the amino acid methionine.DIF: Average/Above average OBJ: 11.4.3

183. The base sequence of the codon (structure F) would change from GCU to GUG.DIF: Average/Above average OBJ: 11.4.3

184. Structure F is a codon that specifies the amino acid alanine.DIF: Below average/Average OBJ: 11.4kt

185. Structure D is made from structure A, which is one of the strands of DNA.DIF: Below average/Average OBJ: 11.5.1

186. The deletion of structure C would alter the codons during translation. As a result, the sequence of the amino acids that make up the growing polypeptide would change.DIF: Average/Above average OBJ: 11.6.1

187. Structure X is a nucleotide (cytosine).DIF: Below average/Average OBJ: 11.2.2

188. Figure 11-2 shows a DNA molecule.DIF: Below average/Average OBJ: 11.2.2

189. The dashed lines are hydrogen bonds. Pairings between nucleotides are partly determined by the ability of nucleotides to form hydrogen bonds with each other.DIF: Average/Above average OBJ: 11.2.2

190. The shape is called a double helix.DIF: Below average/Average OBJ: 11.2kt

191. A pyrimidine is a single-ringed nitrogenous base. The structures labeled C and T are pyrimidines.DIF: Average/Above average OBJ: 11.2kt

192. A purine is a double-ringed nitrogenous base. The structures labeled A and G are pyrimidines.DIF: Average/Above average OBJ: 11.2kt

ESSAY

193. Interphase consists of the G1 phase, S phase, and the G2 phase. During the G1 phase, the cell grows; during the S phase, the genetic material duplicates; during the G2 phase, the cell prepares to divide. The M phase or mitotic phase includes mitosis and cytokinesis. Figures should approximate Figure 9-5 on page 183 of the student text.DIF: Average/Above average OBJ: 9.2.2

194. Prophase—the chromatin condenses into chromosomes and the nuclear envelope breaks down; Metaphase—the chromosomes line up across the middle of the cell and each chromosome is attached to a spindle fiber; Anaphase—sister chromatids separate into daughter chromosomes; Telophase—chromosomes move to opposite sides of the dividing cell, and two new nuclear envelopes form.DIF: Below average/Average OBJ: 9.3.1

195. In radiation therapy, parts of the body with cancerous tumors are exposed to radiation that disrupts cell division. Chemotherapy involves drugs that also disrupt cell division. Because cancer cells divide more often than most normal cells, radiation and chemotherapy can often destroy cancer cells with less damage to normal cells.DIF: Below average/Average OBJ: 9.4.2

196. During meiosis, the pairs of homologous chromosomes in the parent cell form tetrads and then separate. As a result, each daughter cell receives only one chromosome from each homologous pair, and the particular chromosomes that it receives are random. Thus, each daughter cell has a different combination of chromosomes. Also, crossing over may occur during meiosis and may result in new combinations of alleles on the chromosomes in the daughter cells. In contrast, during mitosis, homologous chromosomes usually do not form tetrads and separate, and therefore crossing-over usually does not occur.DIF: Average/Above average OBJ: 9.6.3

197. When telophase ends, each new nucleus will contain four chromosomes. The eight chromosomes form from the replication of four chromosomes. When mitosis is over, the chromatids have separated at their centromeres to form eight chromosomes, half of which move to each nucleus.DIF: Below average/Average OBJ: 9.2.1

198. Unlike normal cells, cancer cells divide uncontrollably, forming tumors and spreading throughout the body. A cure for cancer includes a way to both prevent cancer cells from dividing uncontrollably and to allow normal cells to continue dividing normally. Finding a

way to stop the cell cycle in some cells, but not interfere with the cell cycle in other cells, has made it difficult to cure cancer.DIF: Below average/Average OBJ: 9.4.1

199. Both meiosis I and meiosis II contain a prophase, a metaphase, and an anaphase. However, chromosomes replicate prior to meiosis I but not prior to meiosis II. Also, during meiosis I, tetrads form and align along the center of the cell. Then, the homologous chromosomes are separated and two haploid daughter cells form. During meiosis II, sister chromatids align along the center of the cell and are then separated. Four haploid daughter cells form.DIF: Average/Above average OBJ: 9.5.3

200. If the homologous chromosomes of a tetrad fail to separate, half the gametes formed will have an extra chromosome. The other half will lack a chromosome. When one of these gametes fuses with a normal gamete during fertilization, the offspring will have an abnormal number of chromosomes (not 2n) in its cells.DIF: Average/Above average OBJ: 9.5.3

201. Both hypotheses are attempts to explain how offspring inherit traits from both parents. According to the blending hypothesis, hereditary material from the parents blends, producing an intermediate trait. In contrast, the particulate hypothesis states that parents pass on separate and distinct factors that are responsible for inherited traits.DIF: Below average/Average OBJ: 10.1.1

202. The tall pea plant should be crossed with a short pea plant. If the tall pea plant is homozygous, all of the offspring will be tall. If the tall pea plant is heterozygous, it is likely that about half of the offspring will be tall and half will be short. This type of cross is a testcross.DIF: Average/Above average OBJ: 10.2kt

203. The product of a genotype is generally not a single, rigidly defined phenotype, but a range of possibilities influenced by the environment. For example, in humans, nutrition affects height, exercise affects build, and exposure to sunlight darkens the skin.DIF: Below average/Average OBJ: 10.3.4

204. Most sex-linked genes are found on the X chromosome. A female will have a trait controlled by a recessive allele only if she has the allele on both X chromosomes. A male only has one X chromosome, so he will have the recessive trait even if he has only one allele for it.DIF: Average/Above average OBJ: 10.5.1

205. When the heterozygous yellow-seed F1 plants produced gametes, their dominant allele for yellow seeds segregated from their recessive allele for green seeds. As a result, some of their gametes had the dominant allele, and others had the recessive allele. When the F1 plants self-pollinated, some male gametes with the recessive allele fused with female gametes with the recessive allele during fertilization. Some of the offspring that resulted had two alleles for green seeds and therefore had green seeds.DIF: Average/Above average OBJ: 10.2.1

206. The genotype ratio of the offspring is not the same as their phenotype ratio. The same phenotype can be produced by several different genotypes.DIF: Average/Above average OBJ: 10.2.3

207. In intermediate inheritance, one allele is not completely dominant over another. As a result, the heterozygous phenotype is intermediate between the two homozygous phenotypes. In codominance, both alleles are dominant. As a result, the heterozygous phenotype is a combination of each homozygous phenotype.DIF: Below average/Average OBJ: 10.3kt

208. ANS:The chromosome theory of inheritance states that genes are located on chromosomes and the behavior of chromosomes during meiosis and fertilization accounts for inheritance patterns. It is chromosomes that undergo segregation and independent assortment, which explains Mendel’s results.DIF: Below average/Average OBJ: 10.4.1, 10.4kt

209. After being transcribed from DNA, the RNA transcript is reduced in size by having noncoding regions, called introns, cut out from it. The remaining coding regions, called exons, are then spliced together.DIF: Below average/Average OBJ: 11.5.2, 11.5kt

210. Messenger RNA carries copies of instructions for assembling proteins from DNA to the ribosomes. Ribosomal RNA is a component of the ribosomes. Transfer RNA translates the codons of mRNA into the amino acids that make up proteins.DIF: Average/Above average OBJ: 11.5kt

211. Mutations include base substitutions, base insertions, and base deletions of single nucleotides in DNA. Insertions and deletions have a greater effect on proteins than do substitutions, because insertions and deletions affect every amino acid that is specified by the nucleotides that follow the point of mutation. In contrast, a substitution affects a single amino acid. A change in more than one amino acid is more likely to alter the ability of the protein to function normally than is a change in a single amino acid.

DIF: Below average/Average OBJ: 11.6.1

212. Mutations can be caused by chemical or physical agents, called mutagens. One physical mutagen is high-energy radiation, such as X-rays and ultraviolet light. One type of chemical mutagen consists of chemicals that are similar to normal DNA bases but cause incorrect base-pairing when incorporated into DNA.DIF: Below average/Average OBJ: 11.6.2, 11.6kt

213. Hershey and Chase grew bacteriophages in cultures containing radioactive isotopes of phosphorus and sulfur. The radioactive phosphorus became incorporated into the bacteriophage’s DNA, because DNA contains phosphorus. The radioactive sulfur became incorporated into the bacteriophage’s protein coat, because proteins contain sulfur. After the bacteriophages were allowed to infect bacteria, Hershey and Chase found that nearly all the radioactivity in the bacteria was from phosphorus. This indicated that the bacteriophage’s DNA was injected into the bacteria. The results were important because they showed that the bacteriophage’s genetic material was DNA, not protein.DIF: Average/Above average OBJ: 11.1.3

214. A DNA molecule has the shape of a double helix, or that of a twisted ladder. Each strand of the helix is a chain of nucleotides. The two strands are held together by hydrogen bonds between the nitrogenous bases of the nucleotides on opposite strands. The nitrogenous bases form hydrogen bonds with one another in pairs. Adenine forms hydrogen bonds with thymine, and guanine forms hydrogen bonds with cytosine.DIF: Below average/Average OBJ: 11.2.2

215. DNA replication begins with the two DNA strands separating and unwinding. Each strand serves as a template for the attachment of complementary bases, forming a new strand of DNA. As a result, two identical DNA molecules are formed, each with one original strand of DNA and one new strand of DNA. DNA replication begins at hundreds of points on the chromosomes (origins of replication) and proceeds in two directions away from these points.DIF: Average/Above average OBJ: 11.3.2

216. RNA polymerase is involved in transcription, whereas DNA polymerase is involved in DNA replication. During transcription, free nucleotides base pair with the nucleotides on only one strand of a DNA molecule, not both strands as in DNA replication. In transcription, the free nucleotides are RNA nucleotides, not DNA nucleotides. Transcription continues until a stop signal is reached on the DNA strand. DNA replication continues until the entire chromosome is replicated. At the end of transcription, one single-stranded RNA molecule is formed, not two double-stranded DNA molecules. The newly formed RNA molecule leaves the nucleus, whereas the newly formed DNA molecules stay in the nucleus.

DIF: Average/Above average OBJ: 11.4.2, 11.5.1