38. If a gamete produced following non disjunction of a chromosome at second meiotic division was fertilized by a normal gamete, what is the expected frequency of trisomic progeny?
(1) 1/4               (2) 2/4
(3) 3/4               (4) 1

Logic of the situation

• The cell produces four daughter cells.

• Two are haploid (n) → completely normal chromosome number.

• Two are aneuploid → one has n+1, the other n−1 for that chromosome.

This pattern arises when:

1. Meiosis I is normal

   • Homologous chromosomes segregate correctly → two secondary cells, each with n chromosomes (as paired sister chromatids).

2. Nondisjunction occurs in Meiosis II in ONE of those cells

   • Sister chromatids for a particular chromosome fail to separate.

   • That faulty cell produces:

     • One gamete with n+1

     • One gamete with n−1

   • The other secondary cell divides normally → two haploid (n) gametes.

So the four products are: n+1, n−1, n, n (two aneuploids, two haploids).

Option-wise explanation

1. Non-disjunction during first meiotic division only

   • Error at meiosis I sends both homologues to the same pole.

   • After meiosis II, all four gametes are abnormal: two n+1 and two n−1, with no normal n cells.

   • Does not match the described mix of 2 normal + 2 aneuploid.

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

   • As shown above, yields exactly two aneuploid and two haploid daughter cells.

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

   • Too broad; meiosis I errors give 4 aneuploid products, which contradicts the scenario.

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

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

Therefore, the situation described can only be explained by nondisjunction in the second meiotic division (option 2).