Answer:
0.4550
Explanation:
In this example, R allele is dominant, so individuals RR and Rr can roll their tongues. If in a population of 1000 gorillas, there are 575 gorillas who can roll their tongues, they will be RR and Rr.
In this case, there is Hardy Weinberg Equilibrium, so the following will be true:
[tex]p^{2}+2pq+q^{2}=1[/tex]
Where:
[tex]p^{2}=[/tex] frequency of RR
[tex]2pq=[/tex] frequency of Rr
[tex]q^{2}=[/tex] frequency of rr
The question is what is the frequency of heterozygotes, or, what is the value of 2pq.
We know that RR+Rr is 575 individuals in a population of 1000, or 0.575.
In other words:
[tex]p^{2}+2pq=0.575[/tex]
So, it is possible to find [tex]q^{2}[/tex]:
[tex]q^{2}= 1-2pq-p^{2} \\q^{2}=1-(0.575)\\q^{2}=0.425[/tex]
Now, there are two alleles in the population, so the following will be true:
[tex]p+q=1[/tex]
It is possible to find q (the frequency of allele r) and p (the frequency of allele p):
[tex]q=\sqrt{q^{2} }\\ q=\sqrt{0.425} \\q=0.652[/tex]
Therefore:
[tex]p=1-q\\p=1-0.652\\p=0.348[/tex]
Now, the frequency of heterozygotes or 2pq is:
[tex]2pq=2*0.652*0.348\\2pq=0.45[/tex]
