Is it true that for a single meiotic event when considering only two genes, there are only two possible genetic outcomes among the four gametes produced, regardless of whether the two genes are found on the same homologous pair or on two different pairs?

To simplify my question I am disregarding recombination, but feel free to mention the importance and implications of recombination and other factors that I may be disregarding, in gamete formation.

To explain my thought process further, I will use an example. I think that if one daughter cell produced from meiosis I gets the alleles A and b for example, then the other daughter cell must get the alleles a and B, because each gene only has two alleles, one of which is given to each daughter cell in meiosis I. Thus, 50% of the gametes will be the same as one another, again this happens only theoretically, when considering only two genes and disregarding recombination. In this case, two gametes will have the alleles Ab and the other two will have the alleles aB, irrespective of whether the two genes are found on the same homologous pair or on two different pairs. Of course my example is only considering two alleles. Below is an image which illustrates my thinking:

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I take it each band represents an allele and the colour of the band represents the gene. I think the image is overcomplicating my question, but it does seems to be consistent with my above simplified description.

I think when people say there are four unique gametes resulting from meiosis e.g., Ab, AB, aB and ab, when considering two genes and assuming independent assortment, they need to be more specific and emphasise that there are four possible gametes per meiotic division, or maybe more depending on whether recombination is considered. Now, that I think of it there are probably more than four possible gametes if you add recombination to the mix.

Regardless of independent assortment, I take it that the result of the meiotic division will actually be two genetically unique gametes, if you disregard recombination and only consider two genes. However, the number of possible unique gametes will be greater.

In summary, I think we need to specify whether we are talking about the observed number of genetically unique gametes resulting from one meiotic event or the possible number of genetically unique gametes, which could be observed given more than one meiotic division, and obviously we need to specify how many genes we are considering in order to determine the genetic uniqueness of the gamete. Lastly, we need to state whether we are assuming recombination or not.

Again, I am assuming everything in I have written based on what I have learned so I may be wrong at any point, I am just a Biology student at University.

To add to my confusion, I assume there are actually two possibilities for the specific combination of alleles found on each chromosome, when two genes are found on the same homologous pair. But these possibilities only express themselves at the individual level, rather than at the 'meiotic event' level. For example, if two genes are found on the same homologous pair, one individual may have the alleles Ab on one homologous chromosome and aB on the other homologous chromosome, whereas the other individual may have the alleles AB on one homologous chromosome and Ab on the other homologous chromosome. However, there is only one possibility for the gametes produced from each meiotic division for each individual. So, in this case, considering only individuals with these two specific genes on the same homologous pair of chromosomes, the gametes produced will be different between individuals for the two genes, whereas the gametes produced for individuals that have these two genes on different homologous pairs can be different between independent meiotic events.

My last inquisition is on how one would calculate the probability of a specific allelic composition for a gamete, with the correct interpretation. If two genes are found on different homologous chromosomes and if we disregard recombination, I assume there is a 50% chance that 50% of the gametes will be Ab and 50% will be aB or 50% will be AB or ab? Is this the correct interpretation?

Apologies for the length of this question!

  • $\begingroup$ Yes, correct. But there are many cells to begin with, not just 1, meaning that all 4 end-combinations will be present. Importantly, crossing-over is not a rare optional event, but a necessity with control mechanisms. $\endgroup$
    – KaPy3141
    Commented Aug 17, 2023 at 10:49
  • $\begingroup$ I am surprised because to me most of what I said was 'logical' assumptions and I suspect more of the later. Thank you. $\endgroup$ Commented Aug 17, 2023 at 11:29

1 Answer 1


A "single meiotic event" would include both meiosis I and meiosis II. So, if you start with an AaBb mother cell prior to any phase of meiosis, then observe one of the resulting final daughter cells, you would have an equal probability (25%) of finding AB, Ab, aB, or ab.

The line of reasoning from OP is only relevant after meiosis I has occurred, and in the absence of recombination. At that point the two intermediary daughter cells have either the genotypes AB and ab (the left side of the figure), or Ab and aB (the right side). From that point the possibilities of the final daughter cells are indeed limited, so if one knew the outcome of meiosis I, they would know the set of genotypes held by the four final daughter cells.

One would also know—again, in the absence of recombination—that a single AaBb mother cell will produce four daughter cells that contain only two possible genotypes among them (but not which genotypes they will be). However, this would only be useful in very narrow circumstances because, first, recombination is integral to meiosis, and second, usually there's a population of mother cells to consider. (And third, it doesn't seem like very useful information in the first place.)

As other comments point out, recombination is not a rare event. When recombination is considered, it is indeed possible to get all four genotypes among the daughter cells

  • $\begingroup$ Thank you so much for clarifying Darlingtonia! While the information is certainly not practical, it is very useful to know and helps in understanding the theoretical basis of human knowledge on the subject, especially for a student like me. I came to these conclusions independently because teachers would always 'breeze' past this stuff. Now my understanding is more thorough and I can utilise this. Furthermore, now I understand the mechanisms of recombination, rather than just the implications. Thank you for providing an excellent answer! $\endgroup$ Commented Aug 19, 2023 at 1:58

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