Harvey Weinberg Equilibrium?

Question

In a herd of wild mustangs, 10 were Aa (cream coloured), 29 AA (brown and black), and 1 aa (white). What are the genotypic frequencies? What is the allele frequency of a? 

Is this group in HWE?

Answer 1

Favorite Answer

Who knows? 

Okay.  You have 10 + 29 + 1 = 40 horses.  Yeah, we're not talking about the muscle car of the 1960s.

freq(AA) = 29/40 = 0.725 (answer)

freq(Aa) =10/40 = 0.25 (answer)

freq(aa) = 1/40 = 0.025 (answer)

Let's count the alleles.  We have 40 horses and they're all diploid so that means 80 alleles.

We have 29 + 29 + 10 = 68 A alleles

We have 10 + 1 + 1 = 12 a alleles

let q = freq(a) = 12/80 = 0.15 (answer)

let p = freq(A) = 68/80 = 0.85

We have have been told that this is a herd.  We have not been told that it is a population.  Even if it were a population, it's too small to be in Hardy-Weinberg equilibrium.  Accidents happen (genetic drift).  Mutations happen.  Immigration/emigration happen (gene flow).  Selection is going to happen too.  So hey.  Unfortunately, saying "heck no" is not going to get you full marks from a teacher who is looking for a dumb calculation.  So, let's do the dumb calculation.

p + q = 1 (obvious)

(p + q)*(p + q) = 1 (Hardy and Weinberg were pretty clever)

p^2 + 2pq + q^2 = 1 (did the math)

Okay.  Let's see if that applies to our horse numbers:

0.85^2 + 2*0.85*0.15 + 0.15^2 = 1 ?  Yes, yes it does.

These numbers support "the group IS in Hardy-Weinberg equilibrium"