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:
[tex]p^{2}+2pq+q^{2}=1[/tex]
Where:
[tex]p^{2}=[/tex]frequency of AA
[tex]2pq=[/tex] frequency of Aa
[tex]q^{2}=[/tex]frequency of aa
It is known that 64% of population is red, in other words:
[tex]p^{2}+2pq=0,64[/tex]
Or AA+Aa=0,64
So, it is possible to find the value of [tex]q^{2}[/tex] (aa):
[tex]p^{2}+2pq+q^{2}=1[/tex]
[tex](0,64)+q^{2}=1[/tex]
[tex]q^{2}=1-0,64[/tex][tex]q^{2}=0,36[/tex]
Now, there are two alleles in this population, therefore their frequencies will be:
[tex]p+q=1[/tex]
So, from [tex]q^{2}[/tex] it is possible to find q and p:
[tex]q=\sqrt{q^{2} }[/tex]
[tex]q=\sqrt{0,36}[/tex]
[tex]q=0,6[/tex]
And:
[tex]p+q=1\\p=1-q\\p=1-0,6\\p=0,4[/tex]
