*1. In bacterial cell division, the cell divides into two nearly equal halves. This process is referred to as:

A. binary fission
B. mitosis
C. fusion
D. meiosis
E. cytokinesis

* A. binary fission

2. How does the organization of the bacterial genome differ from the organization of the eukaryotic genome?
A. The compaction of the eukaryotic genome involves structural maintenance of chromosome (SMC) proteins, and the compaction of the bacterial genome does not.
B. Most bacterial chromosomes are circular and the eukaryotic chromosomes contained in the nucleus are not.
C. Bacterial chromosomes are made up of RNA and eukaryotic chromosomes are made up of DNA.
D. The eukaryotic genome is found on chromosomes and there are no chromosomes in bacterial cells.
E. Eukaryotic chromosomes have to be tightly packed to fit into the nucleus, and bacterial chromosomes do not require tight packing to fit into the cell.

B. Most bacterial chromosomes are circular and the eukaryotic chromosomes contained in the nucleus are not.

3. The division of a bacterial cell occurs as the:
A. cell wall develops cracks around the equator of the cell.
B. chromosomes are pulled toward the ends of the cell.
C. actin and microtubules constrict the cytoplasm.
D. new membrane and cell wall materials begin to grow and form a septum.

D. new membrane and cell wall materials begin to grow and form a septum.

*5. The accommodation of the very long DNA strands that are part of a chromosome into the limited space of the nucleus is achieved by coiling the DNA around beads of histones into repeating subunits. These DNA-wrapped histones are called:
A. Solenoids
B. Nucleosomes
C. Chromatin loops
D. Rosettes

*B. Nucleosomes

6. The point of constriction on chromosomes that contains certain repeated DNA sequences that bind specific proteins is called:
A. The kinetochore
B. The centromere
C. The cohesin complex
D. The centriole
E. The centrosome

B. The centromere

7. Eukaryotic chromosomes are composed of a complex of 60% protein and 40% DNA. This complex is referred to as:
A. The histone complex
B. Chromatin
C. The kinetochore
D. Cohesin

B. Chromatin

8. The number of chromosomes characteristic of diploid eukaryotic cells, in general:
A. is 46.
B. can be predicted by the size of the organism.
C. changes as each organism grows and ages.
D. varies considerably from 2 to over 1000 in different species.

D. varies considerably from 2 to over 1000 in different species.

9. A person whose cells all contain a chromosome number of 2n+1 most likely has what type of condition?
A. Monosomy
B. Trisomy
C. Cancer
D. Obesity
E. None, this is the normal chromosome number for humans.

B. Trisomy

10. In later chapters, you will learn more about the regulation of gene expression. One way to regulate gene expression is to make changes to the histone proteins to alter how tightly the DNA is coiled and wrapped. The more tightly coiled and wrapped a particular region of DNA is, the less likely it is that the genes in that region will be expressed. Bearing this in mind, how tightly do you think regions of heterochromatin are compacted?
A. Very tightly
B. Moderately
C. Loosely
D. Not at all

A. Very tightly

11. You are assembling a model of a chromosome, but begin having some trouble when you get to the step of forming chromatin loops. If you are unable to resolve this problem, what step of chromosome structure would you be unable to achieve?

A. Histone/DNA complex
B. Nucleosome
C. Solenoid
D. Rosettes

D. Rosettes

13. The two copies of each type of chromosome found in normal somatic (body) cells in an organism, throughout the cell cycle, are called:
A. Sister chromatids
B. Homologous chromosomes
C. Daughter chromosomes
D. Kinetochores

B. Homologous chromosomes

15. These structures are held together by cohesin:
A. Nucleosomes
B. Sister chromatids
C. Homologous chromosomes
D. Solenoids

B. Sister chromatids

16. A cell biologist produces a karyotype of mouse somatic cells arrested in mitosis. She sees 40 chromosomes, which is completely normal for mice. Based on this information, what is the haploid number of chromosomes for mice?
A. 10
B. 20
C. 40
D. 80
E. It cannot be determined from the information provided

B. 20

*17. If there are 32 sister chromatids in a normal somatic cell, what is the haploid number for that cell?
A. 8
B. 16
C. 32
D. 64

*A. 8

*18. If there are 32 sister chromatids in a normal somatic cell, how many chromosomes are there?
A. 8
B. 16
C. 32
D. 64

*B. 16

19. A somatic cell from a garden pea normally contains 14 chromosomes. How many sister chromatids would that cell contain during G1 of the cell cycle?
A. 0
B. 7
C. 14
D. 28

A. 0

20. A somatic cell from a corn plant normally contains 20 chromosomes. How many sister chromatids would that cell contain during G2 of the cell cycle?
A. 0
B. 10
C. 20
D. 40

D. 40

21. What is the sequence of events in a typical eukaryotic cell cycle?
A. G1 to G2 to S to mitosis to cytokinesis
B. G1 to S to G2 to mitosis to cytokinesis
C. G1 to S to G2 to cytokinesis to mitosis
D. G1 to G2 to mitosis to S to cytokinesis
E. S to G1 to G2 to mitosis to cytokinesis

B. G1 to S to G2 to mitosis to cytokinesis

22. The portion of the cell cycle when the cell is growing and does not contain a replicated genome is referred to as:
A. G1
B. S
C. G2
D. Mitosis
E. Cytokinesis

A. G1

23. The stage of the cell cycle during which the cytoplasm divides to form two cells is called:
A. G1
B. S
C. G2
D. Mitosis
E. Cytokinesis

E. Cytokinesis

*24. This stage of the cell cycle is characterized by growth and it contains a checkpoint to verify that all of the DNA has been replicated prior to mitosis.
A. G1
B. S
C. G2
D. Mitosis
E. Cytokinesis

*C. G2

25. A duplicate copy of all of the hereditary information contained in the nucleus of eukaryotic cells is made during what stage of the cell cycle?
A. G1
B. S
C. G2
D. Mitosis
E. Cytokinesis

B. S

26. The physical distribution of cytoplasmic material into the two daughter cells in plant cells is referred to as:
A. The gap phase
B. Cytokinesis
C. Binary fission
D. Interphase

B. Cytokinesis

27. If a cell has 32 chromosomes prior to S and undergoes mitosis followed by cytokinesis, each new daughter cell will have how many chromosomes?
A. 64
B. 32
C. 16
D. 8

B. 32

28. Embryonic cell cycles allow the rapid division of cells in the early embryo. These mitotic cell cycles are much shorter in length than the mitotic cell cycles of cells in a mature organism. In the embryonic cell cycles, mitosis takes approximately the same amount of time as it does in the cell cycles of mature cells. What do you think is a result of the embryonic cycle?
A. Resulting daughter cells are smaller than the mother cell in the embryonic cell cycles.
B. Resulting daughter cells do not contain the same genetic information as the mother cell in the embryonic cell cycles.
C. Resulting daughter cells cannot form a mitotic spindle in the embryonic cell cycle.
D. Mother cells in the embryonic cell cycle spend the majority of their time in G0.

A. Resulting daughter cells are smaller than the mother cell in the embryonic cell cycles.

29. What is the portion of the cell cycle during which the chromosomes are invisible under the light microscope because they are not yet condensed?
A. Interphase
B. Prophase
C. Metaphase
D. Anaphase
E. Telophase

A. Interphase

30. Interphase is made up of what stages of the cell cycle?
A. G1 + G2 + S
B. S + cytokinesis
C. prophase + metaphase + anaphase + telophase
D. cytokinesis + mitosis
E. G0 + G1 + G2

A. G1 + G2 + S

31. During what stages of the cell cycle are sister chromatids bound together by cohesin?
A. G1, S, G2
B. S, G2
C. G1, S
D. S, G2, prophase, metaphase
E. S, G2, prophase, metaphase, anaphase, telophase

D. S, G2, prophase, metaphase

*32. Following S phase, a human cell would have how many pairs of sister chromatids and individual DNA molecules?
A. 23 pairs of sister chromatids and 46 individual DNA molecules
B. 23 pairs of sister chromatids and 92 individual DNA molecules
C. 46 pairs of sister chromatids and 46 individual DNA molecules
D. 46 pairs of sister chromatids and 92 individual DNA molecules
E. 46 pairs of sister chromatids and 184 individual DNA molecules

*D. 46 pairs of sister chromatids and 92 individual DNA molecules

33. If a chromosome contains a mutation such that it cannot bind to the kinetochore complex, what would be the consequence?
A. That chromosome would not be able to be replicated.
B. That chromosome would not be able to condense.
C. That chromosome would not be able to bind to the mitotic spindle.
D. That chromosome would not be able to interact with histone proteins.

C. That chromosome would not be able to bind to the mitotic spindle.

34. Consider the cell cycle of a human cell. During G2, what is the state of the homologous chromosomes?
A. The homologous chromosomes are lined up on the equator of the cell.
B. The homologous chromosomes have all been copied through DNA replication and are beginning to condense.
C. The homologous chromosomes have been pulled to their respective poles by the spindle apparatus.
D. The homologous chromosomes have not been replicated yet.
E. The homologous chromosomes are now in the haploid or n condition.

B. The homologous chromosomes have all been copied through DNA replication and are beginning to condense.

35. This is the stage of mitosis characterized by the alignment of the chromosomes in a ring along the inner circumference of the cell:
A. Interphase
B. Telophase
C. Prophase
D. Metaphase

D. Metaphase

36. The stage of mitosis characterized by the physical separation of sister chromatids is called:
A. Anaphase
B. Metaphase
C. Prometaphase
D. Telophase

A. Anaphase

*37. This stage of mitosis is characterized by the disassembly of spindle apparatus, the reestablishment of the nuclear membrane, and the decondensation of the chromosomes:
A. Prometaphase
B. Telophase
C. Anaphase
D. Metaphase

*B. Telophase

38. During this stage of mitosis, the nuclear envelope begins to break down and the spindle begins to form.
A. Anaphase
B. Metaphase
C. Prophase
D. Telophase
E. Prometaphase

C. Prophase

39. In prophase, ribosomal RNA synthesis stops when the chromosomes condense, and as a result:
A. the chromosomes lengthen.
B. the nuclear envelope reforms.
C. the nucleolus disappears.
D. the chromosomes line up at the equator of the cell.

C. the nucleolus disappears.

40. During this stage of mitosis, the chromosomes become attached to the spindle at their kinetochores.
A. Prophase
B. Prometaphase
C. Metaphase
D. Anaphase
E. Telophase

B. Prometaphase

41. What happens during Anaphase B?
A. Kinetochores are pulled toward the poles.
B. The spindle poles move apart.
C. The spindle apparatus disassembles.
D. The nuclear envelope reforms.
E. The APC/C gets activated for the first time.

B. The spindle poles move apart.

42. You are conducting a genetic screen using Caenorhabditis elegans embryos to isolate mutations affecting anaphase (A). Therefore, you need to look for embryos in which
A. the centromeres do not move toward the poles.
B. the poles do not move apart.
C. the spindle apparatus does not disassemble.
D. sister chromatids are mismatched and therefore fail to separate.

A. the centromeres do not move toward the poles.

43. What stage of mitosis is essentially the reverse of prophase?
A. Anaphase
B. Prometaphase
C. Metaphase
D. Telophase
E. Cytokinesis

D. Telophase

44. The drug Taxol, or Paclitaxel, is used to treat patients with a variety of cancers, including breast, lung and ovarian cancers. The drug works by stabilizing microtubules, and preventing their disassembly. The goal of the drug is to prevent dividing cells from being able to complete mitosis. As a result, cancerous cells can no longer divide. In a cell treated with Taxol, at what stage of mitosis will the cells arrest?
A. Prior to metaphase
B. Anaphase
C. Telophase
D. Cytokinesis

A. Prior to metaphase

45. Why is it so important that all of the chromosomes align on the metaphase plate during metaphase?
A. This is the only place in the cell where the cyclins and Cdks are located.
B. If they cannot, it suggests that they aren't properly attached to the spindle microtubules, and thus won't separate properly during anaphase.
C. This is the location where the chromosomes can become attached to the spindle microtubules.
D. This allows asters to form.
E. This allows sister chromatids to form.

B. If they cannot, it suggests that they aren't properly attached to the spindle microtubules, and thus won't separate properly during anaphase.

46. If a cell was capable of bypassing metaphase and going directly from prometaphase to anaphase, what is the most likely consequence of this?
A. The resulting daughter cells would not have a nuclear envelope.
B. The resulting daughter cells would have significantly different quantities of cytoplasmic materials.
C. The resulting daughter cells would have different numbers of chromosomes.
D. The resulting daughter cells would be completely normal.

C. The resulting daughter cells would have different numbers of chromosomes.

47. Animal cells typically achieve cytokinesis by:
A. binary fission.
B. forming a cell plate across the middle of the cell.
C. forming a cleavage furrow that pinches the cell into two.
D. chromosome condensation.
E. chromosome elongation.

C. forming a cleavage furrow that pinches the cell into two.

*48. Plant cells typically achieve cytokinesis by:
A. binary fission.
B. forming a cell plate across the middle of the cell.
C. forming a cleavage furrow that pinches the cell into two.
D. chromosome condensation.
E. chromosome elongation.

*B. forming a cell plate across the middle of the cell.

49. If a drug that inhibited transport from the trans face of the Golgi was applied to plant cells, which stage of the cell cycle would be directly affected?
A. G2
B. S
C. Metaphase
D. Anaphase
E. Cytokinesis

E. Cytokinesis

50. The progress of the eukaryotic cell cycle is regulated primarily by what proteins?
A. Cyclins
B. Histone proteins
C. Condensins
D. FtsZ

A. Cyclins

51. At what checkpoint(s) does the cell arrest in response to DNA damage?
A. G1/S
B. G2/M
C. spindle
D. G1/S and G2/M
E. G2/M and spindle

D. G1/S and G2/M

52. You are examining the effect of maturation-promoting factor (MPF) in sea urchin cells, which have a diploid number of 36. If you fuse a dividing sea urchin cell with a G1 arrested oocyte, what would be the outcome?
A. The G1 cell would enter mitosis, but would likely arrest at the spindle checkpoint because the chromosomes have not been properly replicated.
B. The G1 cell would undergo mitosis and its daughter cells would each have 36 chromosomes.
C. The G1 cell would undergo mitosis and its daughter cells would each have 18 chromosomes.
D. The G1 cell would first go through S phase and then mitosis. Its daughter cells would have 36 chromosomes.

A. The G1 cell would enter mitosis, but would likely arrest at the spindle checkpoint because the chromosomes have not been properly replicated.

53. What time point represents G2?

A. 1
B. 2
C. 3
D. 4

C. 3

54. You are studying cell cycle progression in yeast cells. If you could prevent cdc2 from associating with the mitotic cyclin, the cells would:
A. arrest in G1.
B. arrest in G2.
C. arrest in S.
D. arrest in prometaphase.
E. arrest in metaphase.

B. arrest in G2.

55. You are studying cell cycle progression in an early frog embryo. If you were to inject a protein synthesis inhibitor into a cell during S phase, where do you predict that the cells would arrest?
A. G1
B. G2
C. Metaphase
D. Telophase

B. G2

56. In G2, there are typically high levels of the mitotic cyclin. Why is cdc2 not active during G2 if the mitotic cyclin is present?
A. Cdc2 is also regulated by phosphorylation.
B. Cdc2 does not bind to the mitotic cyclin.
C. Cdc2 requires ubiquitination to be activated.
D. Cdc2 also has to bind to cohesin to be activated.

A. Cdc2 is also regulated by phosphorylation.

57. What is separase?
A. A protein that marks a protein called securin for destruction.
B. A protein that is part of the cohesin complex.
C. A protein that destroys cohesin through its protease activity.
D. A protein that targets the mitotic cyclin for degradation.

C. A protein that destroys cohesin through its protease activity.

58. What is one of the roles of the APC/C during anaphase?
A. To directly target the mitotic cyclins for destruction.
B. To directly target cohesin for destruction.
C. To directly target separase for destruction.
D. To directly target microtubules for destruction.

A. To directly target the mitotic cyclins for destruction.

59. What would you expect to happen if the anaphase-promoting complex/cyclosome (APC/C) failed to ubiquitinate securin?
A. The cohesin complex will be destroyed, and the cell will remain in metaphase.
B. The cohesin complex will persist, preventing the cell from entering anaphase.
C. Separase will be marked for degradation by securin, preventing the cell from entering anaphase.
D. Securin will remain intact and therefore will degrade cohesin, allowing the cell to enter anaphase.

B. The cohesin complex will persist, preventing the cell from entering anaphase.

60. If you were to think of the cell as a car, and mitosis as a process that drives that car to go, what would be a good analogy for a cell that has a mutation in both copies of a tumor-suppressor gene?
A. The gas pedal of a car gets stuck while pushed down.
B. The gas pedal of a car does not work at all.
C. The brake pedal of a car gets stuck while pushed down.
D. The brake pedal of a car does not work at all.

D. The brake pedal of a car does not work at all.

4. If a cell contained a mutation in the gene that encodes FtsZ, what process would be affected?
A. Septation
B. Cytokinesis
C. Prophase
D. DNA Synthesis
E. Cohesin cleavage

A. Septation

61. You are leading a team of researchers at a pharmaceutical company. Your goal is to design drugs that help fight cancer. Specifically, you want to focus on drugs that bind to and inactivate certain proteins, thereby halting cell cycle progression. One of your team members suggests targeting the retinoblastoma (Rb) protein and inhibiting this protein. Will this approach be successful? Why or why not?
A. This approach will not be successful. Rb is tumor-suppressor protein, and functions to inhibit the action of a number of cell cycle regulatory proteins. A drug designed to inactivate the Rb protein would essentially create the same situation as in as a cell that lacks both copies of the Rb gene. Lack of Rb activity would release the inhibition of cell cycle regulatory proteins, thereby promoting cell cycle progression, rather than halting it.
B. This approach will be successful. Rb is an oncogene, and functions to activate a number of cell cycle regulatory proteins. A drug designed to inactivate the Rb protein would halt the cell cycle in cells that contain an active Rb. As a result, cancer cells expressing a constitutively active Rb protein would be good targets for this type of therapeutic.
C. This approach will be successful. Rb is tumor-suppressor protein, and functions to inhibit the action of a number of cell cycle regulatory proteins. A drug designed to inactivate the Rb protein would activate cell cycle inhibition. Lack of Rb activity would therfore inhibit the cell cycle regulatory proteins.
D. This approach will not be successful. Rb is an oncogene, and functions to activate a number of cell cycle regulatory proteins. A drug designed to inactivate the Rb protein would actually activate cell cycle progression. As a result, this drug would likely make this situation worse for patients whose cancer cells contain mutant Rb.

A. This approach will not be successful. Rb is tumor-suppressor protein, and functions to inhibit the action of a number of cell cycle regulatory proteins. A drug designed to inactivate the Rb protein would essentially create the same situation as in as a cell that lacks both copies of the Rb gene. Lack of Rb activity would release the inhibition of cell cycle regulatory proteins, thereby promoting cell cycle progression, rather than halting it.

62. This protein or protein complex functions in the cell to stop cell division if the cell has experienced extensive DNA damage:
A. APC/C
B. p53
C. FtsZ
D. Condensin

B. p53

11. What oxidizing agent is used to temporarily store high energy electrons harvested from glucose molecules in a series of gradual steps in the cytoplasm?
A. FADH2
B. ADP
C. NAD+
D. Oxygen

C. NAD+

14. What molecule can oxidize NADH?
A. Acetaldehyde
B. Lactate
C. Ubiquinone
D. Glucose
E. Isocitrate

A. Acetaldehyde

**

23. Homologous chromosomes pair along their length during prophase I of meiosis. While two homologues are paired, genetic exchange may occur between them in a process called ________.
A. syngamy
B. synapsis
C. independent assortment
D. crossing over

** crossing over

**

27. All animal cells are diploid except
A. gametes.
B. muscle cells.
C. nerve cells.
D. germ-line cells.
E. somatic cells

** gametes

**

31. At the end of meiosis II, each of the four resulting cells contains
A. one full set of chromosomes, each with 2 molecules of DNA.
B. two full sets of chromosomes, each with 2 molecules of DNA.
C. one full set of chromosomes, each with 1 molecule of DNA.
D. two full sets of chromosomes, each with 1 molecule of DNA.

**

c. one full set of chromosomes, each with 1 molecules of DNA

**

All of the following increase genetic variation EXCEPT
A. crossing over.
B. random fertilization.
C. independent assortment.
D. mitosis.
E. mutation.

mitosis

**

You are studying meiosis in an organism where 2n=24. How many chromosomes will each nucleus have after meiosis II is complete?
A. 24
B. 12
C. 6
D. 48

12

Edouard van Beneden proposed that an egg and a sperm each containing half the complement of chromosomes found in somatic cells, fuse to produce in single cell called
a. zygote
b. karyotype
c. embryo
d. oocyte

zygote

______ is a process of nuclear division which reduces the number of chromosomes per cell from sets to 1 set
a. mitosis
b. meiosis
c. binary fission
d. syngamy

meiosis

____ cells contain one set of chromosomes
a. germ-line
b. somatic
c. diploid
d. haploid

haploid

In life cycles that alternate between haploid and diploid stages, fertilization doubles the number of chromosomes per cell while_____ reduces it in half
a. mitosis
b. meiosis
c. binary fission
d. syngamy

meiosis

Homologous chromosomes pair along their length during prophase I of meiosis. While two homologous are paired, genetic exchange may occur between them in a process called
a. syngamy
b. synapsis
c. independent assortment
d. crossing over

crossing over

Compare to asexual reproduction, the main advantage of sexual reproduction is that it
a. requires less energy
b. increases the genetic diversity of the offspring
c. can produce more complex offspring
d. can produce a greater number of offspring

b. increases the genetic diversity of the offspring.

Why does sexual reproduction require both meiosis and syngamy (fusion of gametes to form a new cell)
a. The process of meiosis results in the production of gametes fuse to form a new cell, and the number of chromosomes is restored to the full amount. Therefore, by coupling meiosis and syngamy, the organism ensures that the number of chromosomes will be maintained
b. the process of meiosis results in the production of gametes in which the number of chromosomes is reduced by half. During syngamy, two gametes fuse to form a new cell, and the number of chromosomes is restored to the full amount. Therefore, by coupling meiosis and syngamy, the organism ensures that the proper number of chromosomes will be maintained.
c. the process of meiosis results in the production of gametes in which the number of chromosomes is doubled. During syngamy, gametes are reduced by half , and the number of chromosomes is restored to the full amount. Therefore, by coupling meiosis and syngamy, the organism ensures that the proper number of chromosomes will be maintained.

b. the process of meiosis results in the production of gametes in which the number of chromosomes is reduced by half. During syngamy, two gametes fuse to form a new cell, and the number of chromosomes is restored to the full amount. Therefore, by coupling meiosis and syngamy, the organism ensures that the proper number of chromosomes will be maintained.

Which cells never divide by meiosis
a. haploid cells
b. diploid cells
c. somatic cells
d. germ line cells
e. zygotes

a. haploid
b. somatic
e. zygotes

germ line cells will eventually go through meiosis so it will divide

If a germ-line cell from an owl contains 8 picograms of DNA during G1 of interphase, how many pictograms pf DNA would be present in each cell during prophase I of meiosis

16

If a somatic cell from a cat contains 40 picograms of DNA during G2 of interphase, how many pictograms pf DNA would be present in each cell during metaphase II of meiosis

20

If a germ-line cell from a salamander contains 10 picograms of DNA during G1 of interphase, how many pictograms pf DNA would be present in each gamete produced by this species

5

How many tetrads are present in a single elephant cell (2n=56) during metaphase I of meiosis

28

A geneticist examines a somatic cell from a fly during metaphase of mitosis and determines that 16 chromatids are present. If a germ-line cell from this species divides by meiosis then at the end of meiosis I each cell will contain
a. 8 chromosomes with 8 DNA molecules
b. 8 chromosomes with 16 DNA molecules
c. 4 chromosomes with 4 DNA molecules
d. 4 chromosomes with 8 DNA molecules

d. 4 chromosomes with 8 DNA molecules

Nondisjunction is the failure of homologous chromosomes to separate during meiosis I, or the failure of sister chromatids to separate during meiosis II or mitosis. As a result, both homologous chromosomes or both sister chromatids migrate to the same pole of the cell. This produces daughter cells with a imbalance of chromosomes. A cell biologist examines the final products of meiosis in an earthworm (2n=36) and finds 2 cells with 20 chromosomes and 2 cells with 16 chromosomes . Most likely this was because
a. 2 pairs of sister chromatids failed to separate during meiosis II
b. 1 pairs of sister chromatids failed to separate during meiosis II
c. 2 pairs of homologous chromosomes failed to separate during meiosis I
d. 1 pairs of homologous chromosomes failed to separate during meiosis I

c. 2 pairs of homologous chromosomes failed to separate during meiosis I

Nondisjunction is the failure of homologous chromosomes to separate during meiosis I, or the failure of sister chromatids to separate during meiosis II or mitosis. As a result, both homologous chromosomes or both sister chromatids migrate to the same pole of the cell. This produces daughter cells with a imbalance of chromosomes. If 18 pairs of sister chromatids segregate normally during meiosis II in cats (n=19) but we have nondisjunction of 1 pair, then at the end of meiosis II we will have
a. 3 cells of 20 chromosomes and 1 cell with 18
b. 2 cells of 20 chromosomes and 2 cells of 18
c. 2 cells with 19 chromosomes, 1 with 20 and 1 with 18
d. 3 cells with 18 chromosomes and 1 cell with 20

c. 2 cells with 19 chromosomes, 1 with 20 and 1 with 18

A cell biologist examines a diploid cell form a particular species of during metaphase of mitosis and determines that 8 centromeres are present. Based on this finding, how many centromeres should be present in a single cell form this species during anaphase II of meiosis

8

A cell biologist examines a diploid cell form a particular species of during prometaphase of mitosis and determines that 10 centromeres are present. Based on this finding, how many chromatids should be present in a single cell form this species during anaphase II of meiosis

20

Meiosis results in a reassortment of maternal chromosomes (inherited form the mother) and paternal chromosomes (inherited form the father). if n=4 for a given species and ignoring the effects of crossing over what is the probability that a gamete will receive only paternal chromosomes
a. 1/2
b. 1/4
c.1/8
d.1/16

1/16

A cell in G2 before meiosis begins, compared with one of the four cells produced at the end of meiosis II has
a. twice as much DNA and twice as many chromosomes
b. four times as much DNA and twice as many chromosomes
c. twice as much DNA but the same number of chromosomes
d. four times as much DNA and four times as many chromosomes
e. twice as much DNA and half as many chromosomes

b. four times as much DNA and twice as many chromosomes

In meiosis, sister kinetochores are attached to the same pole of the cell during meiosis I and sister chromatid cohesion is released during anaphase II. What would be the likely result if sister kinetochores were attached to different poles of the cell during meiosis I and sister chromatid cohesion was released during anaphase I
a. sister chromatids would migrate to the opposite poles during anaphase I
b. sister chromatids would migrate to the opposite poles during anaphase II
c.sister chromatids would migrate to the same pole during anaphase I
d. sister chromatids would migrate to the same pole during anaphase II

a. sister chromatids would migrate to the opposite poles during anaphase I