Hardy-Weinberg Question #4

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

Question:

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

Answer:

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!