Deck 10: How Cells Divide, Sexual Reproduction and Meiosis

Question 1

What events normally contribute to genetic variation?

A) Crossing over during meiosis II.
B) Fusion of a random egg and sperm to generate a zygote.
C) Independent assortment of homologous chromosomes during mitosis.
D) Cytokinesis during mitosis.

Answer 1

Fusion of a random egg and sperm to generate a zygote.

Question 2

If it were possible to prevent crossing over, what would be the result of inhibiting this event?

A) Sister chromatids would not have cohesion during metaphase I.
B) Sister chromatids would not segregate properly during mitosis.
C) Homologous chromosomes would not align properly during metaphase I.
D) Homologous chromosomes would not segregate properly during mitosis.

Answer 2

Homologous chromosomes would not align properly during metaphase I.

Question 3

Some organisms have a life cycle that is NOT an alternation between haploid and diploid stages. What is an example of an organism that has a life cycle that is NOT an alternation between diploid and haploid chromosome numbers?

A) Dogs
B) The bacterium E. coli
C) Alfalfa plants
D) The mold N. crassa

Answer 3

The bacterium E. coli

Question 4

Why does sexual reproduction require both meiosis and syngamy?

A) The process of meiosis results in the production of gametes in which the number of chromosomes remains the same. 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.
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.

Question 5

If a germ-line cell from an owl contains 8 picograms of DNA during G₁ of interphase, how many picograms of DNA would be present in each cell during prophase I of meiosis? (Enter the number only, not the units.)

Question 6

If a somatic cell from a cat contains 40 picograms of DNA during G₂ of interphase, how many picograms of DNA would be present in each cell during metaphase II of meiosis? (Enter the number only, not the units.)

Question 7

If a germ-line cell from a salamander contains 10 picograms of DNA during G₁ of interphase, how many picograms of DNA would be present in each gamete produced by this species? (Enter the number only, not the units.)

Question 8

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

Question 9

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 an 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 pair of sister chromatids failed to separate during meiosis II.
C) 2 pairs of homologous chromosomes failed to separate during meiosis I.
D) 1 pair of homologous chromosomes failed to separate during meiosis I.

Question 10

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 an 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 with 20 chromosomes and 1 cell with 18.
B) 2 cells with 20 chromosomes and 2 cells with 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.

Question 11

A cell biologist examines a diploid cell from 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 from this species during anaphase II of meiosis?

Question 12

A cell biologist examines a diploid cell from a particular species of butterfly 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 from this species in metaphase I of meiosis?

Question 13

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 opposite poles during anaphase I.
B) Sister chromatids would migrate to 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.

Question 14

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) 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.
D) The eukaryotic genome is found on chromosomes and there are no chromosomes in bacterial cells.
E) Bacterial chromosomes are made up of RNA and eukaryotic chromosomes are made up of DNA.

Question 15

How is cytokinesis in plant cells different from cytokinesis in animal cells?

A) Cytokinesis in plant cells requires formation of a cell plate; cytokinesis in animal cells requires formation of a ring of actin.
B) Cytokinesis in plant cells occurs during G₂; cytokinesis in animal cells occurs in the M phase of the cell cycle.
C) Cytokinesis in plant cell results in one large cell with two nuclei; cytokinesis in animal cells results in two cells, each with one nucleus.
D) Cytokinesis in plant cells results in haploid cells; cytokinesis in animal cells results in diploid cells.

Question 16

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) Mother cells in the embryonic cell cycle spend the majority of their time in G_0.
B) Resulting daughter cells cannot form a mitotic spindle in the embryonic cell cycle.
C) Resulting daughter cells are smaller than the mother cell in the embryonic cell cycles.
D) Resulting daughter cells do not contain the same genetic information as the mother cell in the embryonic cell cycles.

Question 17

Consider the cell cycle of a human cell. During G2, what is the state of the homologous chromosomes?

A) The homologous chromosomes have not been replicated yet.
B) The homologous chromosomes are lined up on the equator of the cell.
C) The homologous chromosomes are now in the haploid or n condition.
D) The homologous chromosomes have all been copied through DNA replication and are beginning to condense.
E) The homologous chromosomes have been pulled to their respective poles by the spindle apparatus.

Question 18

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 undergo mitosis and its daughter cells would each have 18 chromosomes.
B) The G₁ cell would enter mitosis, but would likely arrest at the spindle checkpoint because the chromosomes have not been properly replicated.
C) The G₁ cell would undergo mitosis and its daughter cells would each have 36 chromosomes.
D) The G₁ cell would first go through S phase and then mitosis. Its daughter cells would have 36 chromosomes.

Question 19

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.