Question

In trout, Gene A affects body color and yellow color (allele a) is recessive to red. In a particular river system you find that 64% of the fish are red. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of the dominant A allele?

Answer 1

Answer:

The frequency of the dominant A allele is 0,4 or 40%.

Explanation:

In this population, we have:

- "A" red body color Dominant

- "a" yellow doy color Recessive

So, individuals AA and Aa will be red and trouts aa will be yellow.  In this case, 64% of the fish are red, so they are AA or Aa.

If the population is in Hardy-Weinberg equilibrium the following will be true:

$p^{2}+2pq+q^{2}=1$

Where:

$p^{2}=$frequency of AA

$2pq=$ frequency of Aa

$q^{2}=$frequency of aa

It is known that 64% of population is red, in other words:

$p^{2}+2pq=0,64$

Or AA+Aa=0,64

So, it is possible to find the value of $q^{2}$ (aa):

$p^{2}+2pq+q^{2}=1$

$(0,64)+q^{2}=1$

$q^{2}=1-0,64$$q^{2}=0,36$

Now, there are two alleles in this population, therefore their frequencies will be:

$p+q=1$

So, from $q^{2}$ it is possible to find q and p:

$q=\sqrt{q^{2} }$

$q=\sqrt{0,36}$

$q=0,6$

And:

$p+q=1\\p=1-q\\p=1-0,6\\p=0,4$