34. Meiosis produces n, n+l and n-1 gametes. The probable reason is?
(1) Non-disjunction during metaphase I
(2) Non-disjunction during metaphase-II
(3) Non-disjunction during metaphase I and II
(4) Non-disjunction during Anaphase I and II
The correct answer is (2) Non-disjunction during metaphase-II (more precisely, the missegregation that becomes evident at anaphase II).
Concept:
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Normal meiosis: one diploid cell → four gametes, each with haploid number n.
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If meiosis I is normal but sister chromatids fail to separate in meiosis II for one chromosome (nondisjunction in meiosis II), the outcomes from that cell are:
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1 gamete with n + 1 chromosomes.
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1 gamete with n – 1 chromosomes.
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2 gametes with normal n chromosomes.
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That gives exactly the set of gametes mentioned: n, n+1 and n–1.
Option-wise explanation
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Non-disjunction during metaphase I
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Nondisjunction is actually expressed at anaphase I, but if it occurs in meiosis I, all four gametes are abnormal: two n+1 and two n–1, with no normal n gametes.
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So you would not see the mixture of n, n+1 and n–1 together.
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Non-disjunction during metaphase-II – correct
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Again, the actual separation error occurs at anaphase II, but this option corresponds to nondisjunction in meiosis II.
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This produces two normal n gametes plus one n+1 and one n–1 gamete, matching the question.
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Non-disjunction during metaphase I and II
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If errors happen in both divisions, the pattern of gametes will be more complex and not simply n, n+1, and n–1 in the classic 2:1:1 ratio; many gametes will be more severely abnormal.
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Non-disjunction during Anaphase I and II
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This suggests errors in both divisions as well; again, the outcome would not cleanly give the standard n, n+1 and n–1 gamete set.
Therefore, the probable reason meiosis produces n, n+1 and n–1 gametes is nondisjunction in meiosis II, represented by option (2).
