AP Biology Question of the Day

1. All four gametes have $n=5$, because nondisjunction in meiosis II changes only chromatid identity.
2. Two gametes have $n=6$ and two gametes have $n=4$, because both meiosis II divisions are affected.
3. One gamete has $n=6$, one has $n=4$, and two have $n=5$, due to nondisjunction in one meiosis II cell. (correct answer)
4. Two gametes have $n=5$ and two have $n=10$, because meiosis II restores diploidy in one cell.
5. All four gametes are aneuploid, because a meiosis II error affects both sister cells equally.

Explanation: This question assesses the skill of analyzing meiosis by examining the effects of nondisjunction in meiosis II on gamete chromosome counts in a 2n=10 cell. After normal meiosis I, each daughter cell has 5 replicated chromosomes; one proceeds normally to produce two gametes with n=5 each. In the other cell, nondisjunction of one chromosome's sister chromatids in anaphase II sends both to one pole, resulting in one gamete with 6 chromosomes and one with 4. Overall, this yields one gamete with n=6, one with n=4, and two with n=5, as the error affects only one meiosis II division. A tempting distractor is choice B, which wrongly suggests two with n=6 and two with n=4, misconstruing that the nondisjunction impacts both meiosis I products equally instead of just one. A transferable strategy for meiosis questions is to follow errors through specific cells and divisions, calculating chromosome distribution separately for each branch of the process.