37. A cell undergoing meiosis produces four daughter cells, two of which are aneuploids, while two are haploid. This can occur due to:
(1) Non-disjunction during first meiotic division only
(2) Non-disjunctin during second meiotic division only
(3) Non-disjunction during either first or second meiotic divisions
(4) Non-disjunction during both first and second meiotic divisions

Core concept

• Meiosis I normal: homologous chromosomes segregate properly → two secondary meiotic cells, each with $n$ chromosomes (still as sister chromatids).

• Meiosis II error (nondisjunction) in one of these cells: sister chromatids of one chromosome fail to separate.

  • That faulty cell produces:

    • one gamete with n + 1 (disomic for that chromosome)

    • one gamete with n − 1 (nullisomic for that chromosome)

  • The other secondary cell divides normally, making two normal haploid (n) gametes.

Result: two aneuploid gametes (n+1 and n−1) and two normal haploid gametes (n, n) — exactly what the question describes.

Option-wise explanation

1. Non-disjunction during first meiotic division only

   • If meiosis I fails, one daughter cell gets both homologs, the other gets none.

   • After meiosis II, all four gametes are aneuploid (two n+1 and two n−1); there are no normal haploid gametes, so this does not fit the scenario.

2. Non-disjunction during second meiotic division only – correct

   • Produces two normal haploid and two aneuploid gametes as explained above.

3. Non-disjunction during either first or second meiotic divisions

   • Too broad; meiosis I nondisjunction does not give the observed pattern, so “either” is incorrect.

4. Non-disjunction during both first and second meiotic divisions

   • Would create a more complex and severely abnormal set of gametes, not the clean 2 normal + 2 aneuploid pattern.

So, the described outcome—two aneuploid and two haploid gametes—arises from nondisjunction in meiosis II only (option 2).