Question

In gorillas, the ability to roll the tongue is under the control of 1 gene. The R allele, which confers tongue-rolling ability, exhibits complete dominance over the r allele. In a population of 1000 gorillas in Uganda, there are 575 gorillas who can roll their tongues. Assuming Hardy Weinberg Equilibrium, what is the frequency of heterozygotes in this population?0.4250.1210.5750.5000.4550.3480.652

Answer 1

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:

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

Where:

$p^{2}=$ frequency of RR

$2pq=$ frequency of Rr

$q^{2}=$ 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:

$p^{2}+2pq=0.575$

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

$q^{2}= 1-2pq-p^{2} \\q^{2}=1-(0.575)\\q^{2}=0.425$

Now, there are two alleles in the population, so the following will be true:

$p+q=1$

It is possible to find q  (the frequency of allele r) and p (the frequency of allele p):

$q=\sqrt{q^{2} }\\ q=\sqrt{0.425} \\q=0.652$

Therefore:

$p=1-q\\p=1-0.652\\p=0.348$

Now, the frequency of heterozygotes or 2pq is:

$2pq=2*0.652*0.348\\2pq=0.45$