Expert: Dana Krempels, Ph.D. - 10/30/2008

Question
I cannot get this question for the life of me!!

In a population of crickets that is in Hardy‐Weinberg equilibrium, the allele for long leg bristles is completely dominant over the allele for short leg bristles. If there are 32 crickets with short leg bristles in this population and 168 crickets with long leg bristles:
(a) What is the most common genotype for the bristle‐length gene locus?
(b) What proportion of crickets possessing long leg bristles are expected to be heterozygous?

Answer
Dear David,

This is a pretty straightforward HW calculation.  I can't do your homework for you, but I can give you some hints that I hope will allow you to solve the problem.

The first thing you need to do is determine the frequency of the recessive allele.  To do this, determine the value of q-squared (sorry; can't do exponents in this format).  

If you have a population of 200 crickets, and 32 of them express the recessive condition, then each of those 32 crickets *must* have two alleles for the recessive condition.  Remember that the frequency of individuals in your population expressing the recessive condition is q-squared, so q-squared in your population is 32/200.  From this, you should be able to calculate q.  (If you can't, I'm not sure what to tell you.)

Once you have q, you can calculate p.  (remember:  p + q = 1.0)

And once you have both p and q, you just plug them into the HW equation to calculate the relative frequencies of homozygous dominants (p-squared), heterozyotes (2pq) and homozygous recessives (q-squared, which you already know) to expect in the population if it is in HW equilibrium.

The answer to (a) would be which of the three genotypes you calculated above has the highest frequency.

The answer to (b) would be the value of 2pq.

Go for it!  You should be able to solve this now.

Dana