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I was taught in biology class that a cell dies if it doesn't have at least one copy of a chromosome "type". If this is the case, that means that:

  1. Zygotes without a copy of a specific chromosome (for example chromosome 21) die. But this would also mean:

  2. Gametes without a copy of a specific chromosome would also die.

If this is the case, how is it possible for monosomies to occur? Wouldn't there have to be at least one haploid cell with (n-1) chromosomes when fertilization occurs?

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    $\begingroup$ Your assumption #2 is wrong. $\endgroup$
    – user24284
    Commented Jun 3, 2017 at 14:51
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    $\begingroup$ @GerardoFurtado Ok; can you please elaborate on this and turn it into an answer? $\endgroup$
    – user32893
    Commented Jun 3, 2017 at 14:52

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Gametes cannot be compared to regular (id est, somatic) cells for two main reasons:

  1. Their lifetime and...
  2. Their gene expression.

In human beings, gametes have a very short lifetime. Supposing the nondisjunction happened at meiosis I (which is the worst case scenario), a "nulissomic female gamete"1 has to be alive for some days, from the end of the dictyotene to the fertilisation. The male gamete would have to stay alive for a little more time, some weeks, from the spermatogonium stage to the end of meiosis. Those are not big time spans.

But the most important reason is that gametes are basically a structure to "deliver" genes. They normally don't express the genes they have, at least not as a somatic cell would do. Sperm cells, for instance, have densely packed chromosomes and gene expression, if any, is little and poorly understood.

Thus, the problem in your reasoning lies in assumption #2:

Gametes without a copy of a specific chromosome would also die.

Gametes lacking a given chromosome can and do exist, and they form aneuploid zygotes.

Note 1: the name "gamete" here is a wrong choice, since a primary oocyte is not a gamete, as it is the name "nullisomic", which doesn't apply to haploid cells.

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  • $\begingroup$ Had I more points I would bounty this answer. Signed – a technical editor $\endgroup$
    – lauir
    Commented Jun 4, 2017 at 5:46

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