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In a meiotic division if non-disjunction occurs at second division, what would be the expected gametes if ‘n’ represents the haploid number of chromosomes.
  • a)
    only (n + 1) gametes
  • b)
    only (n – 1) gametes
  • c)
    All, (n), (n + 1) and (n – 1) types of gametes
  • d)
    Both (n + 1) and (n – 1) types of gametes
Correct answer is option 'C'. Can you explain this answer?
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In a meiotic division if non-disjunction occurs at second division, wh...
Non-disjunction at second division produces both normal and aneuploid gametes.
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In a meiotic division if non-disjunction occurs at second division, wh...
Non-disjunction occurs at the second division of meiosis, the expected gametes would be:

1. Normal gametes: These gametes would have the normal number of chromosomes, with each chromosome having undergone proper separation during meiosis. These gametes would be genetically normal.

2. Non-disjunction gametes: These gametes would have an abnormal number of chromosomes. They would either have one extra chromosome or one less chromosome compared to the normal gametes. This can lead to aneuploidy, where the individual has an abnormal number of chromosomes in their cells.

For example, if non-disjunction occurs during the second division of meiosis in a human male, the expected gametes would be:

- Normal gametes: 22 autosomes + 1 X chromosome or 22 autosomes + 1 Y chromosome.
- Non-disjunction gametes: 23 autosomes + 2 X chromosomes or 23 autosomes + 0 sex chromosomes (no Y chromosome).

Similarly, in a human female, the expected gametes would be:

- Normal gametes: 22 autosomes + 1 X chromosome.
- Non-disjunction gametes: 22 autosomes + 0 sex chromosomes (no X chromosome) or 23 autosomes + 2 X chromosomes.

It is important to note that the specific outcome of non-disjunction can vary depending on the chromosomes involved and the individual's sex.
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In a meiotic division if non-disjunction occurs at second division, what would be the expected gametes if ‘n’ represents the haploid number of chromosomes.a)only (n + 1) gametesb)only (n – 1) gametesc)All, (n), (n + 1) and (n – 1) types of gametesd)Both (n + 1) and (n – 1) types of gametesCorrect answer is option 'C'. Can you explain this answer?
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