1. Meiosis and Sexual Life Cycles

2. Which of the following transmits genes from both parents to child, or from one generation of a family to another?
DNA
gametes
somatic cells
mitosis
nucleotides
Answer: B

3. Which of the following transmits genes from both parents to child, or from one generation of a family to another?
DNA
gametes
somatic cells
mitosis
nucleotides 3

4. Fertilization is to zygote as meiosis is to which of the following?
mitosis
diploid
chromosome
replication
gamete
Answer: E 4

5. Fertilization is to zygote as meiosis is to which of the following?
mitosis
diploid
chromosome
replication
gamete 5

6. Privet shrubs and humans each have a diploid number of 46 chromosomes per cell. Why are the two species so dissimilar?
Privet chromosomes undergo only mitosis.
Privet chromosomes are shaped differently.
Human chromosomes have genes grouped together differently.
The two species have appreciably different genes.
Privets do not have sex chromosomes.
Answer: D 6

7. Privet shrubs and humans each have a diploid number of 46 chromosomes per cell. Why are the two species so dissimilar?
Privet chromosomes undergo only mitosis.
Privet chromosomes are shaped differently.
Human chromosomes have genes grouped together differently.
The two species have appreciably different genes.
Privets do not have sex chromosomes. 7

8. Why is it more practical to prepare karyotypes by viewing somatic diploid cells rather than haploid gametes?
Somatic diploid cells do not contain organelles to interfere with karyotyping.
Both sets of chromosomes, which are present in somatic diploid cells, need to be examined.
DNA in haploid gametes will not stain.
The chromosomes are larger in a somatic diploid cell.
Haploid gametes do not have sex chromosomes.
Answer: B 8

9. Why is it more practical to prepare karyotypes by viewing somatic diploid cells rather than haploid gametes?
Somatic diploid cells do not contain organelles to interfere with karyotyping.
Both sets of chromosomes, which are present in somatic diploid cells, need to be examined.
DNA in haploid gametes will not stain.
The chromosomes are larger in a somatic diploid cell.
Haploid gametes do not have sex chromosomes. 9

10. Diploid cells may undergo either mitosis or meiosis
Diploid cells may undergo either mitosis or meiosis. Haploid cells may undergo mitosis (for certain species) but not meiosis because
the sister chromatids cannot separate.
the synaptonemal complex is too strong.
crossing over has occurred.
cohesins are no longer present.
homologous chromosomes cannot pair.
Answer: E 10

11. Diploid cells may undergo either mitosis or meiosis
Diploid cells may undergo either mitosis or meiosis. Haploid cells may undergo mitosis (for certain species) but not meiosis because
the sister chromatids cannot separate.
the synaptonemal complex is too strong.
crossing over has occurred.
cohesins are no longer present.
homologous chromosomes cannot pair.
Answer: E 11

12. How and at what stage do chromosomes undergo independent assortment?
meiosis I pairing of homologs
anaphase I separation of homologs
meiosis II separation of homologs
meiosis I metaphase alignment
meiosis I telophase separation
Answer: D 12

13. How and at what stage do chromosomes undergo independent assortment?
meiosis I pairing of homologs
anaphase I separation of homologs
meiosis II separation of homologs
meiosis I metaphase alignment
meiosis I telophase separation 13

14. What allows sister chromatids to separate in which phase of meiosis?
release of cohesin along sister chromatid arms in anaphase I
crossing over of chromatids in prophase I
release of cohesin at centromeres in anaphase I
release of cohesin at centromeres in anaphase II
crossing over of homologs in prophase I
Answer: D 14

15. What allows sister chromatids to separate in which phase of meiosis?
release of cohesin along sister chromatid arms in anaphase I
crossing over of chromatids in prophase I
release of cohesin at centromeres in anaphase I
release of cohesin at centromeres in anaphase II
crossing over of homologs in prophase I 15

16. Gametes produced from one meiotic event
are genetically identical to each other.
each have the same chromosome number.
are genetically identical to the cells produced from meiosis I.
are genetically identical to the parent cell.
each have the same mutations.
Answer: B 16

17. Gametes produced from one meiotic event
are genetically identical to each other.
each have the same chromosome number.
are genetically identical to the cells produced from meiosis I.
are genetically identical to the parent cell.
each have the same mutations. 17

18. Crossing over begins to occur during
anaphase I
anaphase II
prophase I
metaphase II
telophase II
Answer: C 18

19. Crossing over begins to occur during
anaphase I
anaphase II
prophase I
metaphase II
telophase II
Answer: C 19

20. In this cell, what phase is represented?
mitotic metaphase
meiosis I anaphase
meiosis I metaphase
meiosis II anaphase
meiosis II metaphase
Answer: C 20

21. In this cell, what phase is represented?
mitotic metaphase
meiosis I anaphase
meiosis I metaphase
meiosis II anaphase
meiosis II metaphase 21

22. Scientific Skills Exercise
When nutrients are low, cells of the budding yeast (Saccharomyces cerevisiae) exit the mitotic cell cycle and enter meiosis. In this exercise, you will track the DNA content of a population of yeast cells as they progress through meiosis.
Researchers grew a culture of yeast cells in a nutrient-rich medium and then transferred them to a nutrient-poor medium to induce meiosis. At different times after induction, the DNA content per cell was measured in a sample of the cells, and the average DNA content per cell was recorded in femtograms (fg; 1 femtogram  1 × 10–15 gram). Their data are shown in the table to the right.

23. Most of the yeast cells in the culture were in G1 of the cell cycle before being moved to the nutrient-poor medium to induce meiosis. How many femtograms of DNA are there in a yeast cell in G1? Estimate this value from the data in the table.
12 fg
24 fg
40 fg
48 fg
Answer: B

24. Most of the yeast cells in the culture were in G1 of the cell cycle before being moved to the nutrient-poor medium to induce meiosis. How many femtograms of DNA are there in a yeast cell in G1? Estimate this value from the data in the table.
12 fg
24 fg
40 fg
48 fg
Answer: B

25. How many femtograms of DNA are present in a cell in G2?
12 fg
24 fg
40 fg
48 fg
Answer: D

26. How many femtograms of DNA are present in a cell in G2?
12 fg
24 fg
40 fg
48 fg
Answer: D

27. How many femtograms of DNA are present in a cell at the end of meiosis I?
12 fg
24 fg
40 fg
48 fg
Answer: B

28. How many femtograms of DNA are present in a cell at the end of meiosis I?
12 fg
24 fg
40 fg
48 fg

29. How many femtograms of DNA are present in a cell at the end of meiosis II?
12 fg
24 fg
40 fg
48 fg
Answer: A

30. How many femtograms of DNA are present in a cell at the end of meiosis II?
12 fg
24 fg
40 fg
48 fg
Answer: A

31. A graph with labels indicating the different phases of the meiotic cell cycle (MI  meiosis I; MII  meiosis II) is shown to the right, based on the data from the table. Think carefully about the point on the graph where the line at the highest value begins to slope downward, indicated by the red arrow. What specific point of meiosis does this “corner” represent?
metaphase I
prophase II
cytokinesis
anaphase I
Answer: C

32. A graph with labels indicating the different phases of the meiotic cell cycle (MI  meiosis I; MII  meiosis II) is shown to the right, based on the data from the table. Think carefully about the point on the graph where the line at the highest value begins to slope downward, indicated by the red arrow. What specific point of meiosis does this “corner” represent?
metaphase I
prophase II
cytokinesis
anaphase I

33. Based on this data, how much DNA is present in a gamete of Saccharomyces cerevisiae?
12 fg
24 fg
48 fg
Answer: A

34. Based on this data, how much DNA is present in a gamete of Saccharomyces cerevisiae?
12 fg
24 fg
48 fg

35. Given the fact that 1 fg of DNA  9
Given the fact that 1 fg of DNA  9.78  105 base pairs (on average), you can convert the amount of DNA per cell to the length of DNA in numbers of base pairs. Millions of base pairs (Mb) is the standard unit for expressing genome size. Calculate the approximate number of base pairs of DNA in the haploid yeast genome.
0.08 Mb (8.0  104 base pairs)
1.2 Mb (1.2  106 base pairs)
12 Mb (12  106 base pairs)
23 Mb (23  106 base pairs)
Answer: C

36. Given the fact that 1 fg of DNA  9
Given the fact that 1 fg of DNA  9.78  105 base pairs (on average), you can convert the amount of DNA per cell to the length of DNA in numbers of base pairs. Millions of base pairs (Mb) is the standard unit for expressing genome size. Calculate the approximate number of base pairs of DNA in the haploid yeast genome.
0.08 Mb (8.0  104 base pairs)
1.2 Mb (1.2  106 base pairs)
12 Mb (12  106 base pairs)
23 Mb (23  106 base pairs)

37. 0.19 (1.9  10–1) base pair per minute
Given the fact that 1 fg of DNA  9.78  105 base pairs (on average), you can estimate the rate of DNA synthesis in Saccharomyces cerevisiae. Approximately how many base pairs per minute were synthesized during the S phase of these yeast cells?
0.19 (1.9  10–1) base pair per minute
100,000 (1.0  105) base pairs per minute
200,000 (2.0  105) base pairs per minute
11,000,000 (11.0  106) base pairs per minute
Answer: C
Because the S phase took place from approximately the 1-hour mark to the 3-hour mark, start by finding the difference between the amount of DNA at 3 hours and the amount at 1 hour: 47.0 fg – 24.0 fg = 23.0 fg. Now calculate the rate of fg synthesized per minute by dividing that amount by the number of minutes in two hours: 23.0 fg/ 120 minutes = fg/min. Finally, you need to convert from fg/min to base pairs/min: fg/min × (9.78 × 105 base pairs/fg) = 187,800, or approximately 200,000 base pairs/min.

38. 0.19 (1.9  10–1) base pair per minute
Given the fact that 1 fg of DNA  9.78  105 base pairs (on average), you can estimate the rate of DNA synthesis in Saccharomyces cerevisiae. Approximately how many base pairs per minute were synthesized during the S phase of these yeast cells?
0.19 (1.9  10–1) base pair per minute
100,000 (1.0  105) base pairs per minute
200,000 (2.0  105) base pairs per minute
11,000,000 (11.0  106) base pairs per minute
Answer: C
Because the S phase took place from approximately the 1-hour mark to the 3-hour mark, start by finding the difference between the amount of DNA at 3 hours and the amount at 1 hour: 47.0 fg – 24.0 fg = 23.0 fg. Now calculate the rate of fg synthesized per minute by dividing that amount by the number of minutes in two hours: 23.0 fg/ 120 minutes = fg/min. Finally, you need to convert from fg/min to base pairs/min: fg/min × (9.78 × 105 base pairs/fg) = 187,800, or approximately 200,000 base pairs/min.