# Hardy Weinberg equilibrium [closed]

Hardy Weinberg would predict that the frequency of normal females, to carrier females, to affected females for colorblindness is equal to p$^2$ + 2pq + q$^2$=1. You were told that q=freq of the recessive allele. If p+q =1 (frequency of the dominant allele + frequency of the recessive allele=1 or 100%) and q=.08, what is the frequency of the normal allele in the population? What is the frequency of carrier females in the population (ie heterozygous females)?

I get that the frequency of the normal allele is p = 1 - .08 so 0.92. However the frequency of carrier females is giving me an unrealistic number. Using p$^2$ + 2pq + q$^2$ =1, the value for pq comes out to be a negative value. Is it safe to just multiply p and q since that is what we're essentially solving for using the full HW equation? This would make the value (.08*.92) = 0.0736. Would that be the way to approach the problem even though it asks for females only?

• Welcome to Biology.SE. Homework questions are off-topic on Biology unless you have shown your attempt at an answer. For more information see our homework policy. Feb 12, 2017 at 22:03
• Btw, although the question is quite clear, the phrasing is a bit inaccurate and misleading. Is it the question as written for your homework? Feb 12, 2017 at 22:33
• Yes, that is the exact wording of the problem! Feb 12, 2017 at 23:05
• check your math, the value for pq comes out to be a negative value is incorrect Feb 12, 2017 at 23:23
• Thank you, got the value to be .0736. Would this be correct as to being the frequency of carrier females? Feb 12, 2017 at 23:31