Respuesta :
C = p, c = q, cc = q2
q2 = 49/540 = .091, srq2 = q = sr.091
q = .30
p = 1-.30 = .70
So frequency of C = .70
frequency of c = .30
q2 = 49/540 = .091, srq2 = q = sr.091
q = .30
p = 1-.30 = .70
So frequency of C = .70
frequency of c = .30
The frequency of the recessive genotype (thin fur) = 49/540= 9%. So q2 = 0.09
Take the square root of 0.09 to get the frequency of this allele: q = 0.3
p + q = 1, so p = 0.7
Frequencies for both alleles for the population of cats are p = 0.7, q = 0.3, where p is the allele for thick fur and q is the allele for thin fur.