Wednesday, 30 September 2009

Hardy-Weinberg Question #4

This is the latest of the Hardy-Weinberg questions that I've received - see the rest of them here.

If the frequency of a homozygous dominant genotype is 0.49, what is the frequency of the homozygous recessive genotype?
The relevant equation is: p2 + 2pq + q2 = 1

Remember that the equation provided in the question isn't just a random algebraic expression, it means something. 'p2' is traditionally taken to refer the frequency of the homozygous dominant genotype, '2pq' stands for the frequency of the heterozygous genotype, and 'q2' is the homozygous recessive genotype's frequency.

Therefore, from the question:

p2 = 0.49
∴ p = 0.7

OK, now that we have the value for p, how do we get q's value? Well, for that we need the second of the two important Hardy-Weinberg equations. Recall that, since there are only two alleles assumed for this gene, all the 'p's and all the 'q's must collectively account for 100% of the total alleles. Hence:

p + q = 1
∴ 0.7 + q = 1 (substituting in the above value)
∴ q = 0.3

From here, it's really easy to get to the q2 [the homozygous recessive frequency]: (0.3)2 = 0.09.

Hope that helps!

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