36. What kind of aneuploid gametes will be generated if meiotic non-disjunction occurs at first division? (‘n’ represents the haploid number of chromosomes)
(1) onlyn + 1 and n
(2) only n – 1  and n
(3) bothn + 1 and n – 1
(4) either n + 1 or n – 1

Concept: nondisjunction in first division

In meiosis I, homologous chromosomes should segregate to opposite poles.
If nondisjunction happens at this stage:

• Both homologs move to the same pole.

• After meiosis I:

  • One daughter cell has n + 1 chromosomes.

  • The other has n – 1 chromosomes.

• Meiosis II proceeds normally, so each of these cells produces two gametes with the same chromosome count.

Therefore the four gametes are:

• Two gametes with n + 1 (disomic for that chromosome).

• Two gametes with n – 1 (nullisomic for that chromosome).

All are aneuploid, and you see both n+1 and n–1.

Option-wise explanation

1. Only n + 1 and n

   • Would occur if some gametes were normal (n), which is not the case for nondisjunction in meiosis I.

2. Only n – 1 and n

   • Also incorrect; again implies some normal gametes, which do not arise when the error is in the first division.

3. Both n + 1 and n – 1 – correct

   • Matches the outcome described above: all four gametes abnormal, two of each aneuploid type.

4. Either n + 1 or n – 1

   • Suggests only one aneuploid type from a given meiosis, which is not correct for first‑division nondisjunction; both types appear.

So, when nondisjunction occurs in meiosis I, the aneuploid gametes produced are both n + 1 and n – 1 (option 3).