1. A population of capybaras contains 104 individuals. Studying them you discover that they have two types of alleles for a certain gene. The frequency of the recessive allele (c) is 0.3 and we assume that this is an ideal population with simple dominance.
a) Calculate the frequency of the dominant allele, C.
There are only 2 alleles, C and c, where C = p and c
= q
Remember our formula: p + q = 1, which we can
rearrange to p = 1 - q
We know from the question that the frequency of q
(i.e., c) = 0.3, therefore, p = 1 - 0.3.
The frequency of p (i.e. C) = 0.7
b) Calculate the genotype frequencies in the next generation (F1)
Because it is an ideal population, we can use
Hardy-Weinberg equilibrium (p² + 2pq + q² = 1) to calculate genotype
frequencies
We know from our answer to question1a that p = 0.7
and q = 0.3, therefore . . .
Frequency of CC = p² = 0.49
Frequency of Cc = 2pq = 0.42
Frequency of cc = q² = 0.09
c) Calculate the phenotype frequencies of F1
C_
= p² + 2pq = 0.42 + 0.49 = 0.91
cc = q² = 0.09
d) Calculate the heterozygosity (H)
H = 2pq = 0.42
e) Calculate the number of heterozygous individuals in the wild population
The number of heterozygous
individuals is calculated by multiplying the heterozygosity (from question 1d)
times the number of individuals in the population (i.e., N)
In other words, the number
of heterozygous individuals = H * N