Confusion about Hardy-Weinberg Equilibrium!

Question

Doing a little HW equilibrium, and came across a something I just can't figure out.

 

There are 20 LL individuals, 24 Ll individuals, and 56 ll inds. in a pop of 100 total.

When I solve for p and q the "regular" way, I get 

p=((2x20)+24)/200 = 0.32 and therefore q=0.68 because p+q=1.

 

But you should be able to say that, from the given number of individuals, q^2 = 0.56 (because there are 56 ll individuals out of 100). If q^2 is 0.56, then q = 0.7483.... that's different than the other answer of 0.68.

 

What am I doing wrong??

Answer 1

Ah, so with the initial population we can’t just take the number of homozygous recessive individuals and square that and call it q^2... which you can do with a pop in HW equilibrium. Such mental gymnastics!

thanks for the answer, I appreciate it!

Answer 2

The reason is that q^2 when q=0.68 is actually 0.46. Even though it makes sense that 56 individuals out of 100 would be 0.56, this does not then account for the heterozygous individuals who have those recessive alleles as well. Therefore, your answers would be q^2=0.46, 2pq= 0.44, and p^2=0.10 which would all add up to 1. You must actually square the q and p values that you obtain or else you will not get the correct values.