If q² equals 0.16 in the Hardy-Weinberg equation, what is the frequency of Rr?

To determine the frequency of the heterozygous genotype Rr in the context of the Hardy-Weinberg principle, we start with the provided value of q², which represents the frequency of the homozygous recessive genotype (rr). In this case, q² equals 0.16.

First, we calculate q, the frequency of the recessive allele (r), by taking the square root of q²:

[

q = \sqrt{0.16} = 0.4

]

Next, we can find the frequency of the dominant allele (R), which is represented as p. The relationship between p and q is defined by the equation p + q = 1. Solving for p gives us:

[

p = 1 - q = 1 - 0.4 = 0.6

]

Now, we need to find the frequency of the heterozygous genotype (Rr). The frequency of Rr is given by the formula 2pq:

[

Rr = 2pq = 2(0.6)(0.4)

]

[

= 2(0.24)

]

[

= 0.48