AP Biology Quiz: Meiosis

What this quiz covers

This quiz focuses on Meiosis, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Sister chromatids separate and move to opposite poles, producing diploid daughter cells.
2. Homologous chromosomes separate and move to opposite poles, reducing chromosome number in each resulting cell. (correct answer)
3. Chromosomes replicate again to ensure each pole receives a complete diploid set.
4. Centromeres divide so chromatids can assort independently into four nuclei immediately.
5. Tetrads dissociate without movement, and cytokinesis randomly partitions chromosomes into cells.

Explanation: This question assesses the skill of analyzing meiosis by predicting anaphase I events from spindle attachments. In a diploid cell, tetrads align at metaphase I with homologs attached to opposite poles and sister chromatids to the same pole, leading to homolog separation in anaphase I, which reduces the chromosome number in each resulting cell, as in choice B. This reduction division halves ploidy while keeping sisters together. The attachment details ensure proper segregation of homologs. A tempting distractor is choice A, which describes sister chromatid separation producing diploid cells, based on the misconception that anaphase I mimics mitosis rather than separating homologs. For meiosis questions, examine spindle connections to homologs versus chromatids to distinguish between division types.

Question 2

1. Two gametes receive three maternal chromosomes and two receive three paternal chromosomes, each haploid. (correct answer)
2. All four gametes receive three maternal chromosomes only, because meiosis I determines chromosome origin.
3. Each gamete receives a mixture of maternal and paternal chromosomes due to crossing over in metaphase I.
4. Two gametes are diploid with six chromosomes and two gametes are haploid with three chromosomes.
5. All four gametes are identical because sister chromatids separate only in meiosis I.

Explanation: This question assesses the skill of analyzing meiosis by examining the independent assortment of chromosomes during meiosis I and its impact on gamete composition. In this scenario, with 2n=6, all maternal homologs align toward one pole and paternal toward the other in meiosis I, resulting in one daughter cell receiving all three maternal chromosomes and the other all three paternal chromosomes after separation. During meiosis II, sister chromatids separate normally, so each of these daughter cells divides into two gametes, producing two gametes with three maternal chromosomes and two with three paternal chromosomes. Each gamete is haploid with n=3 chromosomes, as the diploid number is halved through the meiotic process. A tempting distractor is choice C, which incorrectly suggests that crossing over in metaphase I mixes maternal and paternal chromosomes in each gamete, misunderstanding that crossing over occurs in prophase I and does not affect the overall origin of whole chromosomes in this alignment. A transferable strategy for meiosis questions is to track chromosome origins and counts through each division, distinguishing between homolog separation in meiosis I and chromatid separation in meiosis II.

Question 3

1. Each gamete will have $2n=4$ chromosomes because sister chromatids separated in meiosis II.
2. Each gamete will have $n=2$ chromosomes because chromatids become individual chromosomes after separation. (correct answer)
3. Each gamete will have $n=4$ chromosomes because chromatids are counted as chromosomes only in meiosis I.
4. Each gamete will have $2n=8$ chromosomes because chromatids double the chromosome number in meiosis II.
5. Each gamete will have $n=1$ chromosome because homologs separate in meiosis II.

Explanation: This question assesses the skill of analyzing meiosis by determining chromosome number in gametes after meiosis II. In a species with 2n=4, homologous chromosomes separate in meiosis I, leaving each meiosis II cell with n=2 chromosomes composed of two chromatids each, and the observation of four chromatids total in anaphase II indicates the separation into two per pole. As sister chromatids separate in anaphase II, they become individual chromosomes, resulting in each gamete having n=2 chromosomes, as stated in choice B. This process halves the chromosome number overall while ensuring haploid gametes. A tempting distractor is choice C, which claims n=4 per gamete, arising from the misconception that chromatids are only counted as chromosomes in meiosis I and not after separation in meiosis II. For meiosis questions, visualize the chromatid movements in each division to accurately count chromosomes in the final gametes.

Question 4

1. Each daughter cell is haploid with $n=3$ chromosomes, each still composed of two sister chromatids. (correct answer)
2. Each daughter cell is diploid with $2n=6$ chromosomes, each composed of a single chromatid.
3. Each daughter cell is haploid with $n=3$ chromosomes, each composed of a single chromatid.
4. Each daughter cell is diploid with $2n=6$ chromosomes, each still composed of two sister chromatids.
5. Each daughter cell has $n=6$ chromosomes because homologs separate before sister chromatids.

Explanation: This question assesses the skill of analyzing meiosis by focusing on chromosome segregation and ploidy after meiosis I. In a diploid organism with 2n=6, anaphase I involves the separation of three homologous pairs to opposite poles, with sister chromatids remaining joined, resulting in each daughter cell receiving one chromosome from each pair. Since the chromosomes are still composed of two sister chromatids, the cells are haploid (n=3) but with duplicated chromosomes, matching choice A. This reflects the reduction division of meiosis I, where ploidy halves while chromatids stay intact until meiosis II. A tempting distractor is choice C, which incorrectly states each chromosome has a single chromatid, stemming from the misconception that sister chromatids separate in meiosis I rather than homologs. For meiosis questions, always track both the number of chromosomes and the chromatid composition separately to determine ploidy and cell state.

Question 5

1. Four gametes each with $n=4$, because meiosis II corrects nondisjunction from meiosis I.
2. Two gametes with $n=5$ and two gametes with $n=3$, due to homolog nondisjunction in meiosis I. (correct answer)
3. Two gametes with $n=4$ and two gametes with $n=6$, because sister chromatids fail to separate.
4. Four gametes each with $n=2$, because chromosome number is reduced twice across meiosis I and II.
5. All gametes with $n=4$, but each has duplicated chromosomes because chromatids remain attached.

Explanation: This question assesses the skill of analyzing meiosis by examining the consequences of nondisjunction during meiosis I on gamete chromosome numbers. With 2n=8, nondisjunction of one homologous pair in anaphase I means one daughter cell receives both homologs (5 chromosomes total) and the other receives none for that pair (3 chromosomes total). In meiosis II, sister chromatids separate normally, so the cell with 5 chromosomes produces two gametes each with n=5, and the cell with 3 produces two with n=3. This results in two gametes with n=5 and two with n=3, reflecting the aneuploidy from the meiosis I error. A tempting distractor is choice A, which wrongly claims meiosis II corrects the nondisjunction, misconstruing that meiosis II only separates chromatids and cannot redistribute missing or extra homologs from meiosis I. A transferable strategy for meiosis questions is to count chromosomes at each stage, noting how errors in one division propagate without correction in the subsequent division.

Question 6

1. Each daughter cell is diploid with four unreplicated chromosomes, because homologs separated.
2. Each daughter cell is haploid with two replicated chromosomes, because homologs separated but chromatids stayed joined. (correct answer)
3. Each daughter cell is haploid with two unreplicated chromosomes, because DNA replication is reversed in meiosis I.
4. Each daughter cell is diploid with four replicated chromosomes, because sister chromatids did not separate.
5. One daughter cell is haploid and the other is diploid, because homologs assort independently.

Explanation: This question assesses the skill of analyzing meiosis by examining chromosome composition after meiosis I in a cell with 2n=4. After DNA replication, each of the four chromosomes consists of two sister chromatids, and in anaphase I, homologous chromosomes separate to opposite poles while sister chromatids remain attached. Thus, each daughter cell receives two chromosomes (one from each homologous pair), each still composed of two joined sister chromatids. These daughter cells are haploid (n=2) but contain replicated chromosomes ready for meiosis II. A tempting distractor is choice D, which incorrectly states each cell is diploid with four replicated chromosomes, misunderstanding that homolog separation in meiosis I reduces the chromosome number to haploid despite chromatids remaining joined. A transferable strategy for meiosis questions is to distinguish between chromosome number (based on centromeres) and replication state, tracking what separates in each division.

Question 7

1. Exactly two gamete types occur, because meiosis always separates maternal from paternal chromosomes together.
2. Exactly four gamete types occur, because independent assortment can produce $2^2$ combinations of homologs. (correct answer)
3. Exactly eight gamete types occur, because each chromatid assort independently during meiosis I.
4. Only one gamete type occurs, because meiosis produces genetically identical haploid cells like mitosis.
5. Four gamete types occur only if homologs fail to pair during prophase I.

Explanation: This question assesses the skill of analyzing meiosis by examining independent assortment in a cell with 2n=4 and two homologous pairs. Without crossing over, the orientation of each pair at metaphase I is independent, allowing for 2^2 = 4 possible combinations of maternal and paternal chromosomes in gametes. Across many cells, all four combinations occur as meiosis I separates homologs and meiosis II separates chromatids, producing diverse but predictable gamete types. This reflects the principle that each pair assorts independently, generating genetic variety. A tempting distractor is choice C, which wrongly states eight gamete types due to independent chromatid assortment, misconstruing that chromatids do not assort independently in meiosis I but stay with their homolog. A transferable strategy for meiosis questions is to calculate gamete diversity using 2^n for n homologous pairs, excluding crossing over unless specified.

Question 8

1. Homologous chromosomes separate, reducing chromosome number from $2n$ to $n$.
2. Sister chromatids separate when centromeres divide, forming single-chromatid chromosomes. (correct answer)
3. Tetrads form as homologous chromosomes synapse and exchange segments.
4. Chromosomes replicate again, restoring two sister chromatids per chromosome.
5. Homologous chromosomes pair at the metaphase plate to ensure identical gametes.

Explanation: This question focuses on the specific events of anaphase II in meiosis. After meiosis I, each daughter cell has the haploid number of chromosomes (n=2), but each chromosome still consists of two sister chromatids joined at the centromere. During anaphase II, centromeres divide and sister chromatids separate, converting each double-chromatid chromosome into two single-chromatid chromosomes that move to opposite poles. This is the same mechanism as mitotic anaphase but occurs in haploid cells rather than diploid cells. Students choosing A confuse this with meiosis I events, incorrectly thinking homologs separate again in meiosis II. To distinguish meiosis I from meiosis II events, remember that homolog separation (meiosis I) reduces chromosome number while sister chromatid separation (meiosis II) changes chromosome structure from double to single chromatids.

Question 9

1. Gametes contain recombinant chromatids because synapsis alone causes segment exchange.
2. Gametes contain chromatids that match the original parental chromatids for each chromosome. (correct answer)
3. Gametes contain diploid sets because crossing over is required for homolog separation.
4. Gametes contain identical chromatids because meiosis II is equivalent to mitosis.
5. Gametes contain extra chromosomes because preventing crossing over causes nondisjunction.

Explanation: This question examines meiosis when crossing over is prevented but synapsis still occurs. Without crossing over, homologous chromosomes pair normally during prophase I but do not exchange DNA segments between non-sister chromatids. Each chromatid remains identical to its original parental form throughout meiosis. When homologs separate in meiosis I and sister chromatids separate in meiosis II, each gamete receives chromatids that exactly match the original parental chromosomes without any recombination. Students choosing A incorrectly believe synapsis alone causes DNA exchange, not recognizing that crossing over is a separate process requiring specific enzymatic machinery. The key insight is that crossing over and synapsis are distinct processes: chromosomes can pair without exchanging segments, producing gametes with purely parental chromosome combinations.

Question 10

1. Each gamete receives one homolog because spindle attachment is only required in meiosis II.
2. All gametes lack that chromosome because unattached chromosomes are excluded from nuclei.
3. Two gametes receive both homologs and two gametes receive neither homolog. (correct answer)
4. All gametes receive both homologs because homologs remain paired through meiosis II.
5. Gametes receive four copies because chromosomes replicate again between meiosis I and II.

Explanation: This question analyzes the consequences of spindle attachment failure for an entire homologous pair during meiosis I. Without spindle attachment, the affected homologous pair cannot separate and both homologs remain together, likely moving randomly to one pole or remaining in the middle during cytokinesis. If both homologs end up in one daughter cell, that cell will have both homologs while the other cell has neither. During meiosis II, the cell with both homologs produces two gametes each receiving both (as they separate as sister chromatids), while the cell lacking those chromosomes produces two gametes with neither. Students choosing B incorrectly assume unattached chromosomes are always lost rather than potentially being included in one daughter cell. When spindle attachment fails for entire homologous pairs, the result is typically all-or-nothing distribution creating reciprocal imbalances.

Question 11

1. All four gametes have $n=4$ chromosomes because meiosis I was normal.
2. Two gametes have $n=5$ and two gametes have $n=3$ chromosomes.
3. One gamete has $n=5$, one has $n=3$, and two have $n=4$ chromosomes. (correct answer)
4. One gamete has $n=6$ and three gametes have $n=2$ chromosomes.
5. All four gametes have $n=5$ chromosomes because chromatids duplicated before meiosis II.

Explanation: This problem tests understanding of nondisjunction occurring in meiosis II rather than meiosis I. The cell starts with 2n=8, and meiosis I proceeds normally, producing two daughter cells each with n=4 chromosomes (still as sister chromatid pairs). During meiosis II, nondisjunction affects one chromosome in one of the two cells: instead of sister chromatids separating, both chromatids of that chromosome go to the same pole. This creates one gamete with n=5 (receiving both sister chromatids as an extra chromosome) and one gamete with n=3 (missing that chromosome) from the affected cell, while the unaffected cell produces two normal gametes with n=4. Students choosing B incorrectly apply the nondisjunction to both cells or miscount the chromosome distribution. Remember that meiosis II nondisjunction affects only the two gametes from the specific cell where it occurs, leaving the other cell's gametes unaffected.

Question 12

1. Four cells with n = 3 because sister chromatids separate in meiosis II
2. Two cells with 2n = 6 because meiosis I fails but meiosis II occurs
3. Four cells with 2n = 6 because homologs never segregate into different cells (correct answer)
4. Four cells with n = 6 because chromosome number doubles after S phase
5. Two cells with n = 3 because meiosis stops before meiosis II begins

Explanation: This question tests understanding of how preventing homolog separation affects meiosis outcomes. In a cell with 2n = 6, if homologous chromosomes cannot separate during anaphase I, both homologs of each pair remain together, so after meiosis I, each "daughter cell" still has all 6 chromosomes (remaining diploid). During meiosis II, sister chromatids separate normally, but since homologs never separated, each of the four final cells still contains both members of each homologous pair, maintaining the diploid number of 2n = 6. Choice A incorrectly assumes that preventing homolog separation would somehow still result in haploid cells. To analyze meiosis modifications, trace what happens when specific steps are blocked: preventing homolog separation in meiosis I means cells never become haploid, regardless of normal sister chromatid separation in meiosis II.

Question 13

1. Each daughter cell is haploid with $n=3$ chromosomes, and each chromosome still consists of two sister chromatids. (correct answer)
2. Each daughter cell is diploid with $2n=6$ chromosomes, and each chromosome consists of one chromatid.
3. Each daughter cell is haploid with $n=3$ chromosomes, and each chromosome consists of one chromatid.
4. Each daughter cell is diploid with $2n=6$ chromosomes, and each chromosome still consists of two sister chromatids.
5. Each daughter cell has $n=6$ chromosomes because chromosome number doubles during meiosis I.

Explanation: This question tests your ability to analyze the outcomes of meiosis I, specifically tracking chromosome number and chromatid composition. During meiosis I, homologous chromosomes (not sister chromatids) separate, which reduces the chromosome number from diploid (2n=6) to haploid (n=3) in each daughter cell. Importantly, at the end of meiosis I, each chromosome still consists of two sister chromatids joined at the centromere because sister chromatids don't separate until meiosis II. Students often confuse this with option C, thinking that meiosis I produces single-chromatid chromosomes, but this misconception stems from not distinguishing between the two meiotic divisions. A key strategy for meiosis questions is to remember that meiosis I separates homologs (reducing ploidy) while meiosis II separates sister chromatids (maintaining ploidy but changing chromatid structure).

Question 14

1. Four gametes: two with $n$ chromosomes, one with $n+1$, and one with $n-1$. (correct answer)
2. Four gametes: all with $n+1$ chromosomes due to nondisjunction in meiosis II.
3. Four gametes: two with $n+1$ and two with $n-1$ due to nondisjunction in meiosis I.
4. Four gametes: one with $2n$, one with $0$, and two with $n$ due to failed cytokinesis.
5. Four gametes: all with $n$ chromosomes because nondisjunction changes only chromatid number, not chromosomes.

Explanation: This question examines nondisjunction during meiosis II and its impact on the final gamete set. Starting with normal meiosis I completion, two haploid cells enter meiosis II. When nondisjunction occurs in one cell during anaphase II (sister chromatids of one chromosome fail to separate), that cell produces two abnormal gametes: one with n+1 chromosomes (receiving both sister chromatids) and one with n-1 chromosomes (receiving neither). The other cell undergoes normal meiosis II, producing two gametes with n chromosomes. Students often select option C, incorrectly applying meiosis I nondisjunction patterns to meiosis II, but the key difference is that meiosis II nondisjunction affects only the gametes from one of the two cells entering meiosis II. The strategy is to track nondisjunction effects separately for each meiotic division and each cell involved.

Question 15

1. All four gametes have n = 2 with one copy of each chromosome
2. Two gametes have n = 3 and two gametes have n = 1 (correct answer)
3. All four gametes have n = 4 because homologs did not separate
4. Two gametes have n = 2 and two gametes have n = 0
5. All four gametes have n = 1 because meiosis reduces chromosome number

Explanation: This question requires analyzing the consequences of nondisjunction during meiosis I. The organism has 2n = 4 (two homologous pairs), and when both chromosome 1 homologs go to the same pole while chromosome 2 separates normally, one cell after meiosis I gets both chromosome 1s plus one chromosome 2 (total of 3), while the other cell gets zero chromosome 1s plus one chromosome 2 (total of 1). After normal meiosis II, the first cell produces two gametes with n = 3, and the second cell produces two gametes with n = 1. Choice A incorrectly assumes normal separation would occur despite the stated nondisjunction event. To solve nondisjunction problems, track each chromosome pair separately through both meiotic divisions, remembering that errors in meiosis I affect all resulting gametes from that cell.

Question 16

1. Four gametes, each haploid, all with the same chromosome number as normal gametes.
2. Two gametes with $n+1$ chromosomes and two gametes with $n-1$ chromosomes. (correct answer)
3. Two gametes with $2n$ chromosomes and two gametes with $0$ chromosomes.
4. Three gametes with $n$ chromosomes and one gamete with $2n$ chromosomes.
5. One gamete with $n+2$ chromosomes and three gametes with $n-2$ chromosomes.

Explanation: This question tests your understanding of nondisjunction during meiosis I and its consequences for gamete chromosome numbers. When one homologous pair fails to separate during anaphase I while others separate normally, one daughter cell receives both homologs of that pair while the other receives neither. After meiosis II (with normal sister chromatid separation), the cell that received both homologs produces two gametes with n+1 chromosomes, while the cell that received neither produces two gametes with n-1 chromosomes. Students might choose option C, thinking nondisjunction creates extreme imbalances like 2n and 0, but this misconception ignores that only one pair failed to separate while others separated normally. The strategy for nondisjunction problems is to track each homologous pair separately through both meiotic divisions.

Question 17

1. Sister chromatids separate randomly at metaphase I, producing different chromatid sets in each cell.
2. Homologous chromosomes replicate again during anaphase I, changing which chromosomes enter each cell.
3. Independent orientation of tetrads at metaphase I leads to different combinations of homologs at each pole. (correct answer)
4. Chromosomes condense differently in each cell, causing unequal distribution of homologous pairs.
5. Cytokinesis partitions the cytoplasm unevenly, which determines which homologs enter each daughter cell.

Explanation: This question assesses the skill of analyzing meiosis by examining how genetic variation arises from chromosome alignment. With 2n=8, tetrads form in prophase I and align independently at metaphase I, allowing each homolog to face either pole randomly, leading to different combinations of homologs in the daughter cells after anaphase I separation. This independent assortment ensures the two cells after meiosis I can differ in their homolog content, as described in choice C. Sister chromatids remain attached, emphasizing that variation comes from homolog orientation rather than chromatid separation at this stage. A tempting distractor is choice A, which wrongly attributes variation to random sister chromatid separation at metaphase I, based on the misconception that meiosis I involves chromatid rather than homolog segregation. For meiosis questions, remember to identify sources of genetic diversity like independent assortment and crossing over to explain variability in gametes.

Question 18

1. Meiosis II produces daughter cells with 16 chromosomes, while mitosis produces daughter cells with 8 chromosomes.
2. Meiosis II produces daughter cells with 8 chromosomes, while mitosis produces daughter cells with 16 chromosomes. (correct answer)
3. Meiosis II and mitosis both produce daughter cells with 8 chromosomes because both separate sister chromatids.
4. Meiosis II and mitosis both produce daughter cells with 16 chromosomes because DNA replication occurs first.
5. Meiosis II produces daughter cells with 4 chromosomes, while mitosis produces daughter cells with 8 chromosomes.

Explanation: This question assesses the skill of analyzing meiosis by comparing it to mitosis in terms of chromosome outcomes. In a species with 2n=16, meiosis I separates homologs into two cells, and meiosis II separates sister chromatids into four haploid cells with 8 chromosomes each, while mitosis separates sister chromatids in one division to produce two diploid cells with 16 chromosomes each, matching choice B. This highlights meiosis's role in reduction versus mitosis's maintenance of ploidy. Both processes involve chromatid separation, but meiosis's two divisions achieve the halving. A tempting distractor is choice C, which states both produce cells with 8 chromosomes, stemming from the misconception that mitosis also reduces ploidy like meiosis II, ignoring the overall context. For meiosis questions, compare it to mitosis by noting the number of divisions and segregation types to clarify ploidy changes.

Question 19

1. Each final cell contains 10 chromosomes because meiosis produces genetically identical diploid cells.
2. Each final cell contains 20 chromosomes because DNA replication doubled the chromosome number.
3. Each final cell contains 5 chromosomes because meiosis reduces the chromosome number by half. (correct answer)
4. Each final cell contains 15 chromosomes because homologs separate in meiosis I and chromatids add in meiosis II.
5. Each final cell contains 2 chromosomes because meiosis produces cells with one tetrad each.

Explanation: This question assesses the skill of analyzing meiosis by calculating chromosome numbers through both divisions. Starting with 2n=10, DNA replication produces chromosomes with two chromatids, and five tetrads align in metaphase I, followed by normal segregation that halves the number in meiosis I and separates chromatids in meiosis II. This results in each of the four final cells having 5 individual chromosomes, reflecting the haploid state (n=5), as in choice C. The process ensures reduction from diploid to haploid without further replication between divisions. A tempting distractor is choice B, which suggests 20 chromosomes per cell, based on the misconception that DNA replication's doubling persists without the reductive effects of the two meiotic divisions. For meiosis questions, count chromosomes at each stage, remembering that meiosis I reduces ploidy and meiosis II separates sisters like mitosis.

Question 20

1. Two gametes will be n+1 and two gametes will be n−1. (correct answer)
2. All four gametes will be haploid with n chromosomes.
3. Two gametes will be diploid and two gametes will be haploid.
4. All four gametes will be n+1 because chromatids fail to separate.
5. All four gametes will be n−1 because homologs pair incorrectly.

Explanation: This question tests your ability to analyze the outcomes of nondisjunction during meiosis I. When nondisjunction occurs in meiosis I for one homologous pair, both homologs move to the same pole instead of separating, while the other two pairs segregate normally. This creates two cells after meiosis I: one with 4 chromosomes (normal 3 plus the extra homolog) and one with 2 chromosomes (missing one homolog). After meiosis II, the cell with 4 chromosomes produces two gametes with n+1 (4 chromosomes each), while the cell with 2 chromosomes produces two gametes with n-1 (2 chromosomes each). Choice B is incorrect because it assumes normal segregation occurred, missing the nondisjunction event entirely. When analyzing meiosis problems, track chromosome numbers through each division by counting what moves to each pole.