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It's been a while since my last class in genetics, so I've forgot most of it. However, I found myself studying human reproduction and I've the following doubt concerning spermatogenesis:

the process begins with a spermatogonium, a diploid cell (i.e. it contains 23 chromosomes and the corresponding homologous), which, through mitosis, generates the primary spermatocytes. The latter undergoes meiosis generating the two secondary spermatocytes; these are haploid, one of them contains 23 chromosomes and the other the corresponding homologous.

Since a chromosome A and its homologous B contain the same genes but with (generally) different alleles, the two secondary spermatocites have different genomes. Now, each of them splits into two equal cells (spermatids), through mitosis. My textbook says that the four spermatids have all different genomes, but shouldn't there be only two genomes?

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    $\begingroup$ You're forgetting about homologous recombination during meiosis. $\endgroup$ – MattDMo Apr 8 '15 at 19:11
  • $\begingroup$ Look at this post. I explained what could be the result of homologous recombination in mitosis and this is similar to what would happen in meiosis. $\endgroup$ – cagliari2005 Apr 9 '15 at 5:48
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Spermatogonia contain 46 chromosomes (not 23) Spermatogonia increase their number by mitosis and one of Spermatogonia grow by increasing it's size to form primary spermatocyte. Primary spermatocyte undergo meiosis to produce spermatid. In 1st meiotic division secondary spermatocyte is formed following it in 2nd meiosis spermatids are formed. Both primary and secondary spermatocyte have 23 chromosomes.

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    $\begingroup$ Welcome to SE Biology. Your answer sounds, and may well be, correct, but as this is not my area, how am I to know? And that is the basic reason why we require answers supported by external authoritative sources (when it’s not just a question of logic). You just need to reference a Wikipedia article or book excerpt available on the internet, or quote from a printed edition. Have a look at the Help on answering questions. We look forward to your amended answer and future contributions. $\endgroup$ – David Aug 3 at 12:14

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