Question

In a population of armadillos in Hardy-Weinberg equilibrium, the C allele/gene for long claws is completely dominant over the recessive c allele for short claws. If 16% of the population has short claws, what are the gene and genotypic frequencies at the claw type locus?

Possible equations to use:

p=P+1/2H; q=Q+1/2H; p+q=1; P+Q+H=1

P=p2; Q=q2 ; H=2pq

q=square root of Q

Answer 1

The frequency of the dominant allele (C) for long claws is 0.6 and the recessive allele (c) for short claws is 0.4. The genotypic frequencies are 0.36 for CC, 0.48 for Cc, and 0.16 for cc in the armadillo population.

Step-by-step explanation:

In a population showing Hardy-Weinberg equilibrium, where 16% have short claws (cc genotype), this means that q² (frequency of homozygous recessive genotype) is 0.16. The frequency of the recessive allele (q) is the square root of 0.16, which is 0.4. Thus, the frequency of the dominant allele (p), given p + q = 1, is 0.6. We can then calculate the genotype frequencies: homozygous dominant (CC) is p² (0.6² = 0.36), heterozygous (Cc) is 2pq (2 × 0.6 × 0.4 = 0.48), and homozygous recessive (cc) is q² (as given, 0.16).

The Hardy-Weinberg Principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of evolutionary influences. Here, the frequency of the recessive phenotype (short claws) provides us with q², the frequency of the homozygous recessive genotype, which is 0.16. To find q, we take the square root of 0.16, yielding 0.4. To find p, we subtract q from 1, giving us 0.6. The genotype frequencies can then be calculated: p² for homozygous dominant (CC), 2pq for heterozygous (Cc), and q² for homozygous recessive (cc). Putting the values into the equations provides the frequencies of each genotype, which will predict the distribution of phenotypes in the population.