MITOSIS & MEOSIS

True/FalseIndicate whether the statement is true or false.

____ 1. Diffusion over large distances is slow and inefficient because it relies on random movement of molecules and ions.

____ 2. The cell cycle is divided into interphase and mitosis.

____ 3. During prophase in an onion root tip cell, centrioles migrate to the poles of the cell.

____ 4. In plant cells, cytokinesis begins with a furrow that pinches the cell.

____ 5. Stem cells are only of one type: embryonic.

____ 6. Embryonic stem cells are found in various tissues in the body and might be used to maintain and repair the same kind of tissue in which they are found.

____ 7. A gamete has one-half the number of chromosomes of a regular body cell.

____ 8. Homologous chromosomes are two chromosomes with identical DNA sequences.

____ 9. Recent research suggests that beneficial mutations accumulate faster when species undergo sexual reproduction rather than asexual reproduction.

____ 10. During meiosis, chromosome number is reduced through three rounds of cell division.

____ 11. Sexual reproduction would be more advantageous than asexual reproduction for organisms living in an environment that is diverse and undergoes frequent changes.

____ 12. Mendel’s work on garden pea plants resulted in the discovery that genetic traits of parents always blend together in subsequent generations.

____ 13. In humans, the ability to roll one’s tongue is a dominant trait. Therefore, a tongue roller can only have children who are also tongue rollers.

____ 14. The separation of genes during crossing over occurs more frequently between genes that are far apart on a chromosome than for genes that are close together.

____ 15. Polyploidy is more common in plants than animals.

____ 16. Polyploid plants such as coffee and strawberries are often less healthy and smaller than diploid plants of the same species.

____ 17. During meiosis I, homologus chromosome pairs are separated when the centromeres split apart.

____ 18. Meiosis occurs during both sexual reproduction and asexual reproduction.

____ 19. Gregor Mendel’s research supports the idea each organism carries a pair of alleles.

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

____ 20. By the end of prophase, each of the following has occurred except ____.a. tighter coiling of the chromosomesb. breaking down of the nuclear envelopec. disappearing of the nucleolusd. lining up of chromosomes in the cell

____ 21. The chromosomes shown in Figure 9-1 are in which state of mitosis?

Figure 9-1

a. prophase c. anaphaseb. metaphase d. telophase

Figure 9-2

____ 22. Which of the cells depicted in the line graph in Figure 9-2 are most likely cancerous?a. A c. Cb. B d. D

____ 23. If cancer is present, what is the likely explanation for what happened to the cells depicted in the curves labeled B and D in Figure 9-2?a. They thrived with the cancerous cells.b. They were harmed by radiation therapy.c. They died off due to natural causes.d. They died off because the cancerous cells deprived them of nutrients.

____ 24. Which of the following does not occur as a cell grows larger and larger in size?a. difficulty obtaining nutrientsb. difficulty eliminating wastesc. ratio of surface area to volume increasesd. diffusion across the cell membrane is impaired

Figure 9-3

____ 25. Which of the graphs in Figure 9-3 shows the correct changes in the amount of DNA in a cell as it moves through one cell cycle?a. A c. Cb. B d. D

____ 26. Why is the synthesis stage called this?a. because protein synthesis is taking placeb. because DNA synthesis is taking placec. because it combines several smaller stages into oned. because the chromosomes come together

____ 27. Which of these has occurred by the end of prophase?a. Sister chromatids are separated.b. The spindle is beginning to form.c. The cell membrane has begun to pinch inward.d. The nuclear membrane has disappeared.

Figure 9-4

____ 28. Figure 9-4 illustrates which stage of mitosis?a. anaphase c. prophaseb. metaphase d. telophase

____ 29. How is the alignment of chromosomes, shown in Figure 9-4, on the equatorial plate of the cell maintained?a. They are always located there, since that is where the nucleus was.b. Tension between opposite spindle fibers pulls them there.c. The pressure of the cytoplasm moves them there.d. The chromosomes are attracted to each other and meet there.

____ 30. Which of the following occurs in telophase?a. chromosomes condenseb. chromosomes line upc. chromosomes move to opposite polesd. chromosomes relax

____ 31. A cell has 12 chromosomes. How many chromosomes will each daughter cell have?a. 4 c. 12b. 6 d. 24

____ 32. A cell that is undergoing mitosis is examined with a light microscope. An observation that would allow for identification of the cell as an animal cell rather than a plant cell would be a. the centrioles migrating.b. the chromosome pairs separating from each other.c. the chromosomes twisting about each other.d. the nucleoli disappearing.

____ 33. The cell cycle is regulated bya. cyclins c. hormonesb. enzymes d. sugars

____ 34. Cancer cells can reproduce rapidly because theya. are smaller than normal cells. c. undergo mitosis fasterb. bypass interphase. d. spend less time in interphase

____ 35. What is cancer caused by?a. cell-membrane damage c. mutationb. metabolic poisoning d. immune-system damage

____ 36. Prokaryotes divide by binary fission, a form of asexual reproduction in which

a. the nucleus divides into two nuclei.b. the number of chromosomes in the cell is reduced.c. a cell divides into two cells with identical genetic information.d. spindle fibers attach to the poles of the cell.

____ 37. It is often said that normal cells change into cancerous cells frequently in our bodies. Which of the following explanations accounts for the relative rarity of cancer?a. The cancerous cells die on their own.b. The DNA repair system fixes the mutation that causes cancer.c. The cancer cells grow only very slowly at first.d. The cancerous cells are normally crowded out by normal cell growth.

____ 38. What is the role of cyclin-dependent kinases in the cell cycle?a. They stop the cycle if something has gone wrong.b. They catalyze the condensation of the chromosomes.c. They provide the energy for the actions of the spindle fibers.d. They initiate various stages of the cell cycle.

Figure 9-5

____ 39. The cell in Figure 9-5 is undergoing mitosis. Which stage of mitosis will follow this one?a. anaphase c. prophaseb. metaphase d. telophase

____ 40. Which checkpoint has this cell just passed?a. near the end of gap 1, monitoring DNA damageb. during the S stagec. during the gap 2 staged. during mitosis, monitoring spindle formation

____ 41. DNA replication occurs duringa. anaphase. c. metaphase.b. interphase. d. prophase.

____ 42. If you were studying the causes of cancer, which topic might interest you?a. cyclin-dependent kinases c. spindle-fiber structureb. centromere structure d. cell membranes

____ 43. Which of the phases of mitosis has the shortest duration?a. anaphase c. metaphaseb. cytokinesis d. prophase

____ 44. A cell that undergoes repeated mitosis without cytokinesis would havea. many daughter cells. c. many nuclei.b. fewer chromosomes. d. cancerous properties.

____ 45. Why is it important for the chromosomes to condense during mitosis?a. to facilitate DNA replicationb. to facilitate chromosome movementc. to facilitate cytokinesisd. to facilitate spindle formation

____ 46. The typical growth period of a cell occurs during which stage of the cell cycle?a. Gap 1 c. synthesisb. Gap 2 d. mitosis

____ 47. Some cancers have a genetic component to them, if a parent has a cancer the children are more likely than the average population to develop the cancer. Why might this be?a. Cancers require more than one mutation to occur.b. Cancers are inherited but remain dormant until a certain age.c. Parents and children are often exposed to similar environmental factors.d. Cancers are often recessive traits and require alleles from both parents.

____ 48. Colchicine is a chemical that when applied to a cell during mitosis can be used to “freeze” cells in metaphase by preventing the chromosomes from moving away from the metaphase plate. What part of the cell does colchicine most likely affect?a. chromosome structure c. nuclear membraneb. spindle fibers d. cell membrane

____ 49. A stem cell has potential medical uses because ita. undergoes mitosis.b. is not specialized in structure and function.c. is similar to a cancer cell, providing a study system.d. undergoes apoptosis.

____ 50. The numbers in Figure 10-1 represent the chromosome number found in each of the dog cells shown. The processes that are occurring at A and B are ____.

Figure 10-1a. mitosis and fertilization c. mitosis and pollinationb. meiosis and fertilization d. meiosis and pollination

____ 51. A white mouse whose parents are both white produces only brown offspring when mated with a brown mouse. The white mouse is most probably ____.a. homozygous recessive c. homozygous dominantb. heterozygous d. haploid

____ 52. In chickens, rose comb (R) is dominant to single comb (r). A homozygous rose-combed rooster is mated with a single-combed hen. All of the chicks in the F1 generation were kept together as a group for several years. They were allowed to mate only within their own group. What is the expected phenotype of the F2 chicks?a. 100% rose combb. 75% rose comb and 25% single combc. 100% single combd. 50% rose comb and 50% single comb

____ 53. In mink, brown fur color is dominant to silver-blue fur color. If a homozygous brown mink is mated with a silver-blue mink and 8 offspring are produced, how many would be expected to be silver-blue?a. 0 c. 6b. 3 d. 8

____ 54. The diagram in Figure 10-2 shows a diploid cell with two homologous pairs of chromosomes. Due to independent assortment, the possible allelic combinations that could be found in gametes produced by the meiotic division of this cell are ____.

Figure 10-2a. Bb, Dd, BB, and DD c. BbDd and BDbdb. BD, bD, Bd, and bd d. Bd and bD only

____ 55. Using Figure 10-3, which process would result in the formation of chromosome C from chromosomes A and B?

Figure 10-3a. asexual reproduction c. crossing overb. independent assortment d. segregation

Figure 10-5

____ 56. What is the genotype of generation 1 in Figure 10-5?a. II c. iib. Ii d. I

Figure 10-7

____ 57. What fraction of this cross will be recessive for both traits?a. 1/2 c. 1/8b. 1/4 d. 1/16

Figure 10-8

____ 58. In Figure 10-8, which set of chromatids illustrates the result of a single crossover of the homologous chromosomes?a. A c. Cb. B d. D

____ 59. In Figure 10-8, which set of chromatids will result if each chromatid crossed with a nonsister chromatid?a. A c. Cb. B d. D

____ 60. Which event during meiosis leads to a reduction in chromosome number from 2n to n?a. Pairs of homologous chromosomes line up at the equator.b. DNA undergoes replication.c. Homologous chromosomes travel to opposite sides of the cell.d. Sister chromatids are pulled apart at the centromere.

____ 61. Crossing over would most likely occur during which stage of the cell cycle?a. when DNA is being replicatedb. when homologous chromomosomes line up in pairsc. when centromeres are separatedd. when cytokinesis begins

____ 62. Which is the best description of the events that take place during anaphase II?a. The replicated chromosomes become visible.b. Homologous chromosomes line up along the equator.c. Sister chromatids are separated and pulled to opposite sides of the cell.d. Homologous pairs are separated and pulled to opposite sides of the cell.

____ 63. What is the role of the spindle fibers?a. to join homologous chromosomes togetherb. to store nucleotides prior to DNA synthesisc. to initiate the formation of the nuclear membraned. to move chromosomes in the cell

____ 64. The typical human body cell contains 46 chromosomes. How many chromosomes are found in a typical human sperm?a. 23 c. 46b. 45 d. 92

____ 65. Suppose an animal is heterozygous AaBb, and the traits are not linked. When meiosis occurs, what is the total number of possible combinations of gametes that can be made for these traits?a. 2 c. 6b. 4 d. 8

____ 66. Suppose an animal is heterozygous AaBbCc, and the traits are not linked. When meiosis occurs, what is the total number of possible combinations of gametes that can be made for these traits?a. 3 c. 8b. 6 d. 12

____ 67. During which phase of meiosis do homologous pairs of chromosomes line up next to one another along the equator?a. anaphase I c. prophase IIb. metaphase I d. metaphase II

____ 68. Which stage of meiosis is responsible for the law of independent assortment?a. metaphase I c. telophase Ib. prophase I d. metaphase II

____ 69. A true-breeding tall pea plant is crossed with a true-breeding short pea plant, and all the offspring are tall. What is the most likely genotype of the offspring assuming a single-gene trait?a. tt c. TTb. Tt d. TT or tt

____ 70. In mice, black is dominant to white color and color is determined by a single gene. Two black mice are crossed. They produce 2 black offspring and one white offspring. If the white offspring is crossed with one of its parents, what percent of the offspring are expected to be white?a. 0 c. 50b. 25 d. 75

____ 71. Mendel took the pollen from a tall pea plant and pollinated the flower of a short pea plant. When he did this, he removed the male parts of the flower on the short plant. Why was it important that he remove the male parts from the flower of the short plant?a. because he wanted to prevent self pollinationb. because the short plant was heterozygousc. because he wanted to prevent the development of seedsd. because the short plant was not a true-breeding plant

____ 72. Mendel crossed a true-breeding plant that produced green seeds with a true-breeding plant that produced yellow seeds to produce an F1 generation. The entire F1 generation produced yellow seeds. Then he crossed the F1 offspring with each other to produce the F2 generation. From the F2 generation, he counted 6022 yellow seeds.Which of these is the most likely estimate of the number of green seeds he collected from the F2 generation?a. 0 c. 6000b. 2000 d. 18000

____ 73. A heterozygous organism is best described as which of these?a. dominant c. hybrid

b. genotype d. true-breeding

____ 74. In which situation are the phenotypes of F2 offspring expected to follow the ratio of 9:3:3:1.a. a monohybrid cross for 2 unlinked traitsb. a monohybrid cross for 2 closely linked traitsc. a dihybrid cross for 2 unlinked traitsd. a dihybrid cross for 2 closely linked traits

____ 75. If two heterozygous individuals are crossed, what percent of their offspring are also expected to be heterozygous?a. 0 c. 75b. 50 d. 100

____ 76. Two rabbits that were homozygous for both coat and eye color were crossed. The following are the phenotypes of their F2 generation. Which most strongly suggest that the genes for eye color and coat color are linked?a. 22 brown coat brown eye, 7 brown coat blue eye, 9 white coat blue eye, 4 white coat blue

eyeb. 62 brown coat brown eye, 21 brown coat blue eye, 18 white coat blue eye, 7 white coat

blue eyec. 92 brown coat brown eye, 35 brown coat blue eye, 31 white coat blue eye, 11 white coat

blue eyed. 95 brown coat brown eye, 12 brown coat blue eye, 16 white coat blue eye, 30 white coat

blue eye

____ 77. Of the following species used in agriculture, which is most likely a polyploid?a. cow c. henb. goat d. wheat

____ 78. A geneticist crossed fruit flies to determine whether two traits are linked. The geneticist crossed a fly with blistery wings and spineless bristles (bbss) with a heterozygous fly that had normal wings and normal bristles (BbSs). Which results in the next generation would suggest these traits are linked?a. 35 normal wings, normal bristles, 28 normal wings, spineless bristles, 23 blistery wings,

normal bristles, 30 blistery wings, spineless bristlesb. 105 normal wings, normal bristles, 101 normal wings, spineless bristles, 111 blistery

wings, normal bristles, 115 blistery wings, spineless bristlesc. 198 normal wings, normal bristles, 200 normal wings, spineless bristles, 185 blistery

wings, normal bristles, 189 blistery wings, spineless bristlesd. 222 normal wings, normal bristles, 27 normal wings, spineless bristles, 22 blistery wings,

normal bristles, 228 blistery wings, spineless bristles

Figure 10-9

____ 79. Consider the cell labeled X in Figure 10-9 containing 4 chromosomes. Which of the four cells below it represents a healthy gamete that could be produced from this cell?a. A c. Cb. B d. D

Figure 10-10

____ 80. Which stage of meiosis is represented in Figure 10-10?a. anaphase I c. anaphase IIb. metaphase I d. metaphase II

CompletionComplete each statement.

81. Mitosis is the division of ____________________.

82. Cytokinesis is division of the _____________________.

83. If cells do not respond to the normal mechanisms that control cell division, ____________________ results.

84. ____________________ are unspecialized cells that can develop into specialized cells under the right conditions.

85. An individual with the genotype Aa is ____________________ for the trait.

86. Genes on separate chromosomes follow Mendel’s law of ____________________.

87. The possible combination of genes in any gamete due to independent assortment equals 2n, where n equals the number of ____________________.

88. The term ____________________ refers to the occurrence of one or more sets of extra chromosomes in an organism.

Short Answer

The large size of many fruits and flowers is the result of polyploidy, a condition in which the nuclei of an organism's cells contain extra sets of chromosomes. Polyploidy often occurs naturally, but it can also be artificially induced by plant breeders. How have breeders been able to mimic a naturally occurring phenomenon?Researchers have determined that the chemical colchicine suppresses cell division by preventing the formation of spindle fibers. Without these fibers, the sister chromatids cannot become properly oriented for separation into individual nuclei. In effect, mitosis is stopped after prophase. However, the cell may continue to make copies of its chromosomes. As a result, the nucleus of the cell contains multiple sets of chromosomes.Suppose a researcher wished to investigate how extra sets of chromosomes are produced. First, she treated two onion roots with a colchicine solution and left two roots untreated. After a period of several days, she placed thin slices from each root tip on separate slides, stained the specimens, and examined the slides under a microscope at high power.

89. How do you predict the slides of treated and untreated root tips will differ?

90. Describe what happens in each stage of interphase.

91. How does meiosis maintain a constant number of chromosomes in the body cells of organisms that reproduce sexually?

92. A researcher crossed a brown-eyed flightless fruit fly (bbff) with a heterozygous normal fruit fly (BbFf). Of the 235 offspring, 220 appeared either brown-eyed flightless or normal. What is the best explanation for this observation?

Problem

93. In guinea pigs, the allele for rough coat (R) is dominant to the allele for smooth coat (r), and the allele for black fur (B) is dominant to the allele for white fur (b). If two guinea pigs that are heterozygous for rough, black fur (RrBb) are mated, what are the possible phenotypes and what is the frequency of each? Show your work in a Punnett square, Figure 10-4.

Figure 10-4

Some biology students wanted to determine whether a pair of brown mice purchased at a pet store was homozygous dominant or heterozygous for fur color. They let the mice mate and examined the offspring. Six mice were born. All six had brown fur.Some of the students felt that this was enough evidence to prove that the mice were homozygous for brown fur color. Other students did not, so another experiment was planned.

94. Describe the next experiment the students could conduct to determine whether the parent mice are homozygous brown or heterozygous. Explain your answer.

95. In fruit flies, the allele for normal body (H) is dominant to the allele for hairy body (h), and the allele for red eye color (B) is dominant to the allele for brown (b). Use a Punnett square to determine the possible phenotypes and frequencies of the offspring from the cross Hhbb hhBb.

Essay

96. A squash plant that produces white fruit was crossed with a squash plant that produces yellow fruit to produce 25 offspring. Fourteen of the offspring produced white fruits, while 11 produced yellow fruit. Does this evidence support the idea that white fruit color is dominant to yellow? Defend your answer with specific information about the likely genotypes of the individual plants involved.

97. In humans, albinism is recessive to normal coloration. Assume that an albino woman and a man who is homozygous for normal coloration are planning to have children. What is the likelihood that their children will be albino? Defend your answer by including specific genotypes of the individuals involved.

Answer Section

TRUE/FALSE

1. ANS: TMovement of substances in a cell is more easily managed if the cell is small.

2. ANS: FThe cell cycle has three stages: interphase, mitosis, and cytokinesis.

3. ANS: FCentrioles are not present in plant cells. Only spindle fibers are present.

4. ANS: FPlants have a rigid cell wall that covers their plasma membrane. Therefore, they must form a new structure called a cell plate between the daughter cells.

5. ANS: FThere are two basic types of stem cells: embryonic and adult.

6. ANS: FEmbryonic stem cells are unspecialized cells. They have the capability of developing into a wide variety of specialized cells. Adult stem cells are found in various tissues of the body and used to repair the tissue in which they are found.

7. ANS: TDuring meiosis, the number of chromosomes is reduced to one-half the original number.

8. ANS: FHomologous chromosomes contain the same genes as one another, but the DNA sequences vary. One homolog comes from an individual’s father, while the other comes from the mother.

9. ANS: TSexual reproduction allows for more variety in offspring, allowing natural selection to play a stronger role in favoring those individuals with beneficial traits.

10. ANS: FOnly two rounds occur.

11. ANS: TSexual reproduction allows for greater variety in offspring. Thus, in a variable and diverse environment, there is a greater chance of success for organisms that reproduce sexually as compared to organisms that reproduce asexually.

12. ANS: FHe discovered that traits were discrete and could be “hidden” by other traits, but could reappear in another generation in their original form.

13. ANS: FThis statement would only be true if the tongue roller was homozygous (TT). If the tongue roller is heterozygous (Tt) and has children with someone else that carries the t allele, then it is possible for them to have a child who cannot roll his or her tongue.

14. ANS: TThe farther apart two loci are, the more likely it is that they will be separated by a crossover.

15. ANS: TPlants are less likely to suffer negative effects due to multiple chromosome sets.

16. ANS: FPolyploid plants are used often in agriculture because they often have increased vigor and size.

17. ANS: FHomologous chromosomes separate, but centromeres do not.

18. ANS: FMeiosis occurs in sexual reproduction only.

19. ANS: THe concluded the genes come in pairs, such as round versus smooth seed, or green versus yellow seed.

MULTIPLE CHOICE

20. ANS: DIn prophase, the cell’s chromatin tightens into chromosomes. Near the end of this stage, the nuclear envelope breaks down and disappears. Finally, the spindle starts to form.

FeedbackA See page 248 for help.B Good try.C That occurs during prophase.D That's correct!

21. ANS: CIn anaphase, the chromatids are pulled apart. The microtubules of the spindle apparatus begin to shorten, which pulls the centromere and breaks it. This causes the sister chromatids to separate.

FeedbackA Prophase is an earlier stage.B Try againC That's correct!D No, telophase is the final stage before cell division.

22. ANS: ACancer is the uncontrolled growth and division of cells.

FeedbackA That's correct!B No, the number of cells is going down.C Try again.D Did you consider all the factors?

23. ANS: DWhen unchecked, cancer cells can kill an organism by crowding out normal cells, resulting in loss of tissue function.

FeedbackA Look at the graph again.B Try again.C See page 254 for help.D That's correct!

24. ANS: CIf a cell continues to grow, the ratio of surface area to volume will decrease. This means the cell will have difficulty supplying nutrients and expelling enough waste products. Diffusion across the cell membrane will not be affected.

FeedbackA Try again.B Did you consider all the factors?C That's correct!D See page 245 for help.

25. ANS: BThe amount of DNA doubles during the S phase and is then halved after cytokinesis.

FeedbackA Does the DNA increase four times at any point?B Do you end up with half the amount of DNA you started with as shown?C That's right!D The amount of DNA does change with time during the cell cycle.

26. ANS: B“Synthesis” refers to the synthesis of DNA during this phase.

FeedbackA This is not what is being made.B That's correct!C It refers to something being made.D Review the material on page 247.

27. ANS: DBy the end of prophase the nuclear membrane has disintegrated.

FeedbackA This occurs during anaphase.B This occurs at the beginning of prophase.C This occurs in telophase.D That's right!

28. ANS: BDuring metaphase, the spindle apparatus aligns the sister chromatids in the center, or equator, of the cell.

FeedbackA Try again.B That's correct!C See page 250 for help.D Look at what the chromosomes are doing.

29. ANS: BEach chromosome is attached to opposing spindle fibers at the centromere. These spindle fibers can grow or shrink and pull the chromosomes to the equatorial plate by maintaining tension in opposite directions.

FeedbackA The nucleus is not always at the center of the cell.B That's right!C See page 250 for help.D See page 250 for help.

30. ANS: DIn telophase, the chromosomes arrive at the poles and begin to relax, or decondense.

FeedbackA That's prophase.B That's metaphase.C That's anaphase.D That's correct!

31. ANS: CDuring mitosis, the cell’s replicated genetic material separates and the cell prepares to divide into two cells.

FeedbackA Did you consider all the factors?B Too low. This would occur in meiosis.C That's correct!D That's too many.

32. ANS: AAll of the events can happen in both plant and animal cells except choice A. Plant cells do not have centrioles.

FeedbackA That's correct!B Try again.C See page 250 for help.D Both plant and animal cells have nucleoli.

33. ANS: AThe normal cell cycle is regulated by cyclin proteins.

FeedbackA That's correct!B Good try.C Try again.D See page 253 for help.

34. ANS: DCancer cells spend less time in interphase than do normal cells.

FeedbackA Try again.B See page 254 for further information.C Mitosis takes as long in cancer cells.D That's correct!

35. ANS: CCancer can have diverse causes, all of which result in mutation in a cell’s DNA.

FeedbackA This will just kill a cell.B This doesn't cause cancer.C That's right!D Have a look at page 254.

36. ANS: CWhen prokaryotic DNA is duplicated, both copies attach to the plasma membrane. As the plasma membrane grows, the attached DNA molecules are pulled apart. The cell completes fission, producing two new prokaryotic cells.

FeedbackA See page 252 for help.B Try again.C That's correct!D Did you consider all the factors?

37. ANS: BThe mutations that might lead to cancer occur relatively frequently but DNA repair systems correct them. Cancer often is a result of a mutation in the DNA repair system itself that prevents it from fixing other mutations.

FeedbackA If they are cancerous they won't die.B That's right!C Think about what causes cancer.D The opposite is what happens with cancer cells.

38. ANS: DThe cyclin-dependent kinases initiate various phases of the cell cycle such as the synthesis phase.

FeedbackA This is the role of the cycle checkpoints.B Have a look on page 253.C This is not their role.D That's right!

39. ANS: BIn this stage, the chromosomes are condensing and the spindle is beginning to form. Therefore, this stage is prophase. The one to follow it is metaphase.

FeedbackA That's a later stage.B That's correct!C This cell is currently in prophase.D Try again.

40. ANS: CThis cell has recently left the gap 2 stage and has therefore passed the checkpoint near the end of that stage.

FeedbackA See page 254 for more information.B Which stage is this cell in?C That's correct!D The spindle has not completely formed yet.

41. ANS: BDuring the second stage of interphase, called synthesis, a cell copies its DNA in preparation for cell division.

FeedbackA DNA replication has already occurred by this point.B That's correct!C See page 247 for help.D Try again.

42. ANS: AOne of the causes of cancer is mutations in the proteins that regulate the cell cycle, some of which are cyclin-dependent kinases. These enzymes would be a useful place to begin understanding the causes of cancer.

FeedbackA That's correct!B The centromere is involved in proper movement of the chromosomes, not cancer.C Are the spindle fibers involved in cancer?D See pages 253–254 for further help.

43. ANS: CMetaphase is one of the shortest stages of mitosis, but when completed successfully, it ensures that new cells have accurate copies of the chromosomes.

FeedbackA Good guess!B That's not part of mitosis.C That's correct!D That's the longest phase!

44. ANS: C

Mitosis is the division of the nucleus. Without cytokinesis, or cell division, mitosis alone will lead to a multinucleate cell. This occurs in many organisms, especially fungi.

FeedbackA Cytokinesis is what produces the daughter cells.B The chromosome numbers would not decrease.C That's right!D Cancer is a result of uncontrolled cell division; in this case division is not occurring.

45. ANS: BThe decondensed chromosomes are so long and diffuse that they could not be disentangled when they are dragged to the poles of the dividing cell.

FeedbackA You're not considering when condensation takes place.B That's right!C The chromosomes aren't involved in cytokinesis.D The chromosomes aren't involved in spindle formation.

46. ANS: AThe normal growth of the cell occurs in Gap 1 before the chromosomes are replicated.

FeedbackA That's correct!B This is after synthesis, does that make sense?C The cell is busy replicating DNA during this period.D Have a look at page 247.

47. ANS: ACancers are often caused by multiple mutations, inheriting one mutation from a parent means the individual has to develop fewer mutations to develop the cancer.

FeedbackA That's correct!B This is not how it works.C But this isn't genetic.D This isn't the case; think about how mutations contribute to cancer.

48. ANS: BColchicine inhibits the formation of the microtubules or spindle fibers. Without the spindle fibers the chromosomes cannot move to the poles.

FeedbackA The movement of chromosomes isn't directly affected by their structure.B That's right!C The nuclear membrane isn't present at this stage.D The nuclear membrane doesn't affect chromosome movement.

49. ANS: B

A stem cell’s usefulness lies in the fact that it is unspecialized and therefore has the potential to develop into cells of different kinds that might be used to repair damaged tissue.

FeedbackA Many cells undergo mitosis; this is not their usefulness.B That's right!C They are not especially similar to cancer cells.D This is not their usefulness.

50. ANS: BMeiosis involves the reduction in chromosome number prior to fertilization.

FeedbackA If this were mitosis, then step A would not lead to a reduction in chromosome number.B Correct.C Pollination occurs in plants.D Pollination occurs in plants.

51. ANS: AThe most likely scenario is that the white mouse displays the recessive trait. If this is the case, then the white mouse must be homozygous. If the white mouse were either homozygous dominant or heterozygous, then it would likely produce white offspring when mated with a brown mouse.

FeedbackA Correct.B If the white mouse is heterozygous, then this would mean white is dominant and its

brown mate is homozygous recessive. Such a cross would yield about half white and brown offspring.

C If the white mouse were homozygous dominant, then its offspring would have to be white.

D Sperm and eggs are haploid. Individual mice are not.

52. ANS: BThis is the classic situation in which the F1 generation (all heterozygous) are crossed to produce offspring in a 3 to 1 ratio of dominant to recessive.

FeedbackA Remember that the F1 individuals are heterogygous.B Well done.C Remember that the F1 individuals are heterogygous.D About 3/4 of the offspring will have rose comb.

53. ANS: A

The brown mink will always donate a dominant allele to its offspring, which means all the offspring will have the dominant phenotype. None of the offspring would be silver-blue, which is the recessive color.

FeedbackA Well done.B Remember, the brown mink is homozygous.C Remember, the brown mink is homozygous.D Brown is dominant to silver-blue.

54. ANS: BChromosomes separate so that one of each gamete contains one member of each pair. Thus, B and b would not be found together, nor would D and d. However B can be combined with either D or d, and similarly b can be combined with either D or d.

FeedbackA Homologous pairs would separate from one another.B Correct.C Each gamete would have one copy of each homologous pair.D Four combinations are possible.

55. ANS: CCrossing over leads to new combinations of alleles on a given chromomosome.

FeedbackA Asexual reproduction does not lead to new combinations of alleles.B Review independent assortment.C Correct.D Check page 272.

56. ANS: BThe first generation are all heterozygotes, or Ii.

FeedbackA The individuals in P1 are homozygous.B Well done.C Review pages 277–281.D The individuals have two alleles, not one.

57. ANS: DThis is a classic F1 cross for two independent traits. The ratio is 9:3:3:1. Of the sixteen possible combinations, only one or 1/16 has a genotype with completely recessive alleles.

FeedbackA Check page 282.B See page 282.C Only one in 16 are expected to have both recessive traits.D Well done.

58. ANS: A

A single crossover will cause B and b to change positions relative to A and a on just two chromatids. The other two chromatids will be unaffected, one would remain AB and the other aa. The chromatids that experience the crossover will contain Ab and aB.

FeedbackA This is correct.B This shows the effect of zero crossovers.C This shows the effect of two crossovers.D There should be 2 copies of B.

59. ANS: CThis would result in recombination on all the chromatids, producing two Ab chromatids and two aB chromatids.

FeedbackA This is the result of one crossover.B Try again.C Correct.D There must be two B alleles.

60. ANS: CReduction in chromosome number occurs when homologous pairs of chromosomes are separated during meiosis I.

FeedbackA The cell is still 2n at this point.B Check page 272.C Correct.D The cells are already n at this point.

61. ANS: BCrossing over refers to the exchange of DNA, which occurs when homologous chromosomes form pairs.

FeedbackA Check page 272.B Well done.C Crossing over occurs before this.D Try again.

62. ANS: CAnaphase is the phase in which chromosomes are separated. Homologous pairs are separated during anaphase I. Sister chromatids are separated during anaphase II.

FeedbackA This is prophase I.B This is metaphase I.C This is correct.D This is anaphase I.

63. ANS: DSpindle fibers attach to centromeres and move chromosomes.

FeedbackA See pages 272 and 274.B See pages 272 and 274.C Spindle fibers attach to chromosomes.D Correct.

64. ANS: AThe sperm contains half the number of chromosomes found in a body cell.

FeedbackA Correct.B Sperm contain a haploid number of chromosomes.C A body cell has twice the number of chromosomes as a sperm.D See pages 272–274.

65. ANS: BThere are four combinations: AB, Ab, aB, and ab.

FeedbackA Remember the traits are not linked.B Correct.C Check page 275.D Find all possible combinations of A with B and b and a with B and b.

66. ANS: CEach of the following 8 combinations are possible: ABC, ABc, AbC, aBC, Abc, aBc, abC, and abc.

FeedbackA Remember that the traits assort independently.B Consider all possible combinations of the three alleles.C Well done.D There are 2 x 2 x 2 possible combinations.

67. ANS: BHomologous pairs line up along the equator of the cell during metaphase I.

FeedbackA This is when homologous pairs move away from each other.B Correct.C See page 275.D Homologous pairs are separated before meiosis II.

68. ANS: A

The arrangement of alleles in the gametes is determined by the way the homologous pairs line up at the equator during metaphase I.

FeedbackA Well done.B Chromosomes are condensing, but have not yet lined up on the equator.C See page 275.D Homologous pairs were already separated before this stage.

69. ANS: BSince the parents are true breeding, they are most likely homozygous (TT and tt). This means the offspring are most likely heterozygous, Tt.

FeedbackA See page 280.B Correct.C But one parent was short.D If all offspring are tall, then they cannot be TT and tt.

70. ANS: CThe original parents must be heterozygous (Bb) since they produced both black and white offspring. The white offspring must be homozygous recessive (bb). If the white offspring is crossed with its parent then half the offspring will be white.

FeedbackA The black parent is heterozygous.B The parents of this cross are Bb and bb.C Correct.D Review pages 277–281.

71. ANS: AHe removed the male parts on the short plant because he wanted to avoid letting the plant self-pollinate. He wanted the pollen to come only from a tall plant.

FeedbackA CorrectB Both plants were true-breeding homozygous.C Mendel used the seeds.D Mendel used only true-breeding plants.

72. ANS: B

In the F2 generation, we expect a 3 to 1 ratio of yellow to green seeds. Thus, if there are approximately 6000 yellow seeds, then we expect about 2000 green seeds.

FeedbackA Check page 278.B Correct.C Try again.D There should be 3 times more yellow seeds than green seeds.

73. ANS: CA heterozygous (Aa) individual is produced by a cross of two different true-breeding (AA and aa) individuals.Thus the heterozgous individual is a hybrid.

FeedbackA This description refers to a trait or allele.B This description refers to the genetic makeup of an individual.C Correct.D This refers to a homozygous individual.

74. ANS: CThis ratio is expected in the F2 generation of a dihybrid cross when independent assortment takes place. This is the case when two traits are unlinked, or found on different chromosomes.

FeedbackA In this case, a 3 to 1 ratio is expected.B Try again.C Correct.D If the traits are linked the results will be skewed from this ratio.

75. ANS: BFour combinations are equally likely: AA, Aa, aA, and aa. Of these, half are heterozygous.

FeedbackA Check page 280.B Correct.C Not this many.D Some offspring will be either homozygous recessive or homozygous dominant.

76. ANS: DWith independent assortment, the ratio of phenotypes of the F2 generation from homozygous parents is expected to approximate 9:3:3:1. When linkage occurs, the traits do not assort independently and this ratio is not observed.

FeedbackA Try again.B Review pages 283–284.C The ratio of these traits approximates a 9:3:3:1 ratio.D Well done.

77. ANS: DPlants are often polyploids, while animals are not.

FeedbackA Animals are not usually polyploids.B Polyploid animals are not usually robust.C Check page 284.D Correct.

78. ANS: DLinkage is evident if two of the phenotypic combinations occur at a much greater rate than the other two.

FeedbackA The frequencies of all groups are about equal.B This result supports independent assortment.C Try again.D Well done.

79. ANS: AThe healthy gamete would have half the original number of chromatids. In this case, the number is A.

FeedbackA Correct.B The chromatids in the gamete would not be joined at the centromere.C There should be two chromatids in the gamete.D Try again.

80. ANS: DChromosomes line up at the equator of the cell during metaphase II.

FeedbackA The chromosomes are lined up along the equator.B The chromosomes are not in pairs.C Check page 275.D Correct.

COMPLETION

81. ANS: a nucleus

82. ANS: cytoplasm

83. ANS: cancer

84. ANS: Stem cells

85. ANS: heterozygous

86. ANS: independent assortment

87. ANS: chromosome pairs

88. ANS: polyploidy

SHORT ANSWER

89. ANS:Answers may vary. The slides of untreated root tips will show cells in interphase and in various stages of mitosis. The slides of treated root tips will show cells only in interphase and in prophase, and they may show some cells with multiple sets of chromosomes.

90. ANS:During G1, a cell is growing, carrying out normal functions, and preparing to replicate its DNA. During synthesis, the cell copies its DNA in preparation for mitosis. Finally, G2 is the stage in which the cell prepares for division of the nucleus.

91. ANS:Meiosis reduces the number of chromosomes to n or half in the sperm and egg. When fertilization occurs, the 2n number of chromosomes is restored.

92. ANS:The traits are linked closely on the same chromosome. As a result, there is little crossing over between them, and they tend to stay together. (In fact, these traits are both found on chromosome number 2).

PROBLEM

93. ANS:The phenotypic ratio is 9 rough, black fur: 3 rough, white fur: 3 smooth, black fur: 1 smooth, white fur. See Solution 10-1.

Solution 10-1

94. ANS:

There are a number of ways students could respond correctly to this problem.a) The parent mice could be permitted to mate several more times and produce large numbers of offspring.

The larger F1 population would increase the likelihood of the recessive phenotype being expressed.b) The students could allow several of the F1 mice to interbreed. If the parents are heterozygous, about 50%

of their offspring should be heterozygous and with the larger numbers, there would be a greater chance of the recessive phenotype showing in the F3 population.

c) The parent mice each could be mated with a homozygous recessive mouse. In that way, if a parent were heterozygous, there would be a 50% chance of the offspring showing the recessive trait. There are other possible correct responses. All of the responses should in some way indicate the need for more offspring because heredity operates according to the laws of probability.

Both mice could be homozygous brown, and that is why the recessive allele does not segregate out and appear in the offspring. BUT, only one mating and six offspring are not enough to prove this mathematically. One of the pair of mice could be heterozygous and the other homozygous brown and again, the recessive trait would not be seen in the offspring. Another possibility is that both mice are heterozygous. There would be only a 25% chance that the recessive alleles would segregate out and combine during fertilization. Six offspring may not be a large enough sample mathematically to reasonably expect the 25% chance of white mice to be expressed.

95. ANS:The phenotypic ratio is 1 normal body hair red eye: 1 normal body brown eye: 1 hairy body red eye: 1 hairy body brown eye. See Solution 10-2.

Solution 10-2

ESSAY

96. ANS:The evidence does not necessarily support the idea that white color is dominant to yellow. The results of this cross support the idea that one parent is a heterozygote (Aa) and the other is a homozygous recessive (aa). Similarly, each offspring has the same genotype of the parent that produces the same color fruit. It is not clear, however, which parent is the heterozygote. The ratio of offspring is nearly 1 to 1.

97. ANS:Aside from the small chance of mutation, the likelihood is zero. The mother has a genotype of aa and the father’s genotype is AA. Each child will have a heterozygous genotype, Aa. The child will carry the allele for albinism but will appear normal.(If the man were heterozygous Aa, then half the children would be expected to be albino.)