Question

a) In Belgian terriers, the hernia allele (h) is a recessive allele with a frequency of 0.12 when the population is in Hardy-Weinberg equilibrium. What is the probability that your new Belgian puppy will develop a hernia?
b) What is the probability that the puppy will be a carrier? [3]

Answer 1

a) Here, the frequency of h (recessive allele), q = 0.12
If the population is in Hardy-Weinberg equilibrium: $p^2+2\mathrm{pq}+q^2=1$
The probability that new Belgian puppy will develop a hernia (Homozygous recessive) = $q^2$ = 0.12 X 0.12 = 0.0144 = 1.44% [1.5]

b) The frequency of H (dominant allele), p = 1-q = 1-0.12 = 0.88
The probability of carrier (heterozygous) will be 2pq = 2 X 0.88 X 0.12 = 0.2112 = 21.12% [1.5]