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: p*

^{2}

*+ 2pq + q*

^{2}

*= 1*

**Answer:**

Remember that the equation provided in the question isn't just a random algebraic expression, it

*means*something. 'p^{2}' is traditionally taken to refer the frequency of the homozygous dominant genotype, '2pq' stands for the frequency of the heterozygous genotype, and 'q^{2}' is the homozygous recessive genotype's frequency.Therefore, from the question:

p

∴ p = 0.7

^{2}= 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)

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

∴ q = 0.3

From here, it's really easy to get to the q

^{2}[the homozygous recessive frequency]: (0.3)^{2}= 0.09.Hope that helps!

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