Respuesta :
The Hardy-Weinburg equation is p + q = 1, or p^2 + 2pq + q^2 = 1, where p is dominant and q is recessive. If 10 out of 100 rabbits have white fur, 10% of the rabbits have white fur. Therefore, 90% of the rabbits have brown fur, which can be substituted into the first equation to become 0.9 + 0.1 = 1. Now that we know what p and q equal, we can solve the rest of the equation.
0.9^2 = 0.81
0.9 * 0.1 * 2 = 0.18
0.1^2 = 0.01
Therefore, the allele frequency of the recessive allele is 0.1
The allele frequency of the recessive allele is 0.1.
What do you mean by allelic frequency?
The number of times the allele of interest is observed in a population is divided by the total number of copies of all the alleles at that particular genetic locus in the population to get an allele frequency.
P + q = 1, or p2 + 2pq + q2 = 1, where p is dominant and q is recessive, is the Hardy-Weinburg equation. If 10 rabbits out of 100 have white fur, then 10% of the rabbit population has white fur. Due to the fact that 90% of the rabbits have brown fur, the first equation can be changed to read 0.9 + 0.1 = 1 by substituting this information. We can now answer the last problem because we know what p and q equal.
0.9^2 = 0.81
0.9 * 0.1 * 2 = 0.18
0.1^2 = 0.01
Consequently, the recessive allele's allele frequency is 0.1.
.Learn more about allelic frequency here:
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