Question

12. A small portion of deer have a trait called leucism, which is white coat. The frequency of the leucitic phenotype is 0.00005 (or 1/20,000). Leucism is recessive to wild type color. Assuming that the deer population is in Hardy-Weinberg equilibrium fr A. Estimate the frequencies of the recessive leucistic allele (I) and the dominant wild type allele (L) B. Estimate the frequency of deer with heterozygous genotype C. For the calculations above, you assume that the population is in equilibrium for this gene. Of the characteristics of a population in equilibrium, state one that you think could not be valid for this trait and briefly explain why. Z-C

Please show work. Thank you.

Answer 1

Hardy-Weinberg equation is p² + 2pq + q² = 1

p= dominant allele frequency

q= recessive allele frequency

(In a phenotype there are 2 alleles)

Recessice, leucitic phenotype frequency (q²) = 0.00005 ; q = root of 0.00005 = 0.007

B) q is the allele freqency of homozygous recessive allele

p+q= 1

A) p= 1-0.007 = 0.993 , the frequency of homozygous dominant allele

Heterozygous genotype is 2pq, so its frequency is 2 x 0.993 x 0.007 = 0.0139

C) Five conditions are required in order for a population to remain at Hardy-Weinberg equilibrium:

• A large breeding population.
• Random mating.
• No change in allelic frequency due to mutation.
• No immigration or emigration.
• No natural selection.

Of these the dominant allele shows a large value, so may be a large breeding population is not the characteristic valid for this population.