Chi-Square/Hardy Weinberg Question (Biology)

Question

The question is :

Imagine a population of 800 individuals, of which 14% have a recessive genetic condition.

a. What is the frequency of homozygous recessive individuals? (think q²)

I got 0.14, since for the organism to actually have the recessive condition, it would have to be homozygous.

b. How many individuals have the condition?

14% of 800 is 112

c. What is the frequency of the allele causing the recessive condition? (think q)

I got 0.37. If q²= 0.14, then q= 0.37

parts d. and e. are where I'm getting lost.

d. What is the frequency of the homozygous dominant individuals? (think p²)

e. How many individuals are heterozygous?

If the recessive allele has a frequency of 0.37, then the frequency of the dominant should e 0.63, correct? but how do I find out which of these individuals are homozygous and which are heterozygous?

Answer 1

So Margaret B.,

If q=0.374 (as you correctly calculated) then p = 1-q = ?

You can figure p^2 (dominant homozygous) and 2pq (heterozygous) from there?

Cheers, -- Mr. d. (you'll note from my last name initial that my family is quite recessive!)

P.S. that 2pq is easily seen from expanding the binomial product: (p+q)*(p+q) . We show the cross with segregated alleles in the Punnett square display, which is just a handy way of visualizing the math of probabilities.