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I am looking at the following question

If $m$ alleles may occur at a given locus, how many distinct diploid genotypes are possible at that locus?

The obvious answer is that there their is $m$ possible paternal alleles and $m$ possible maternal alleles so the solution is $m^2$. However this considers inheriting allele $a$ maternally and allele $A$ paternally as the same as inheriting $a$ paternally and $A$ maternally.

If they are considered distinct we have double counted all matchings apart from matchings where the same allele is inherited maternally and paternally so I think the answer to the question if $Aa$ is the same as $aA$ would be $\frac{m(m-1)}{2}+m$

My question is which if any of my interpretations is correct?

Thanks in advance

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  • $\begingroup$ 1 allele (a) means one possible genotype (aa), 2 (a,b) means four possible combinations, but three distinct genotypes (aa, ab, ba, bb... I say three distinct genotypes becuase ab and ba would not be distinct when looking at the offspring unless you can determine parent of origin)... that would be my interpretation at least. What is the source of the question? $\endgroup$ – rg255 Apr 8 '15 at 12:36
  • $\begingroup$ Okay @GriffinEvo, so if we have 3 alleles then we have 9 genotypes, 6 of which are distinct? If we had two loci one with 2 alleles and another with 3 alleles would the number of distinct genotypes be $3 \times 6 = 18$ (just to check I understand) $\endgroup$ – HBeel Apr 8 '15 at 13:14
  • $\begingroup$ (a,b,c) = (aa, ab, ac, bb, bc, cc) so according to my logic, yes, but I don't know whether the question is looking for my logic. $\endgroup$ – rg255 Apr 8 '15 at 13:21
  • $\begingroup$ My instinct is inline with your logic but we'll see what others have to say if you aren't 100%. Thanks $\endgroup$ – HBeel Apr 8 '15 at 13:23
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If you have genomic imprinting then $k^n=m^2$ (with k=m and n=2 because of diploidy) is correct as inheriting a from the father and b from the mother (i.e. ab) is not equivalent to inheriting b from the father and a from the mother (i.e. ba). For 3 alleles (a,b,c) you have 9 possible genotypes (aa,ab,ac,ba,bb,bc,ca,cb,cc).

Under no genomic imprinting, ab and ba are equivalent. This is equivalent of calculating the number of elements of a triangular matrix which is $\frac{(m+1)*m}{2}$. So for m=3, you have 6 possible genotypes (aa,ab,ac,bb,bc,cc), for m=4 you have 10 possibilities (aa,ab,ac,ad,bb,bc,bd,cc,cd,dd) etc...

I hope this helps. Your formula is actually equivalent.

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