Question

WI W IT 0.9 0.9 b 1 0 .95 0.90 e 1 0. 990 .98 6.15a Cystic fibrosis (CF) is a fatal disease caused by a recessive mutation. Individuals homozygous for the CF allele die before reproducing. The frequency of the CF allele is about 0.02 in some populations. Assume these populations are at equilibrium, and assume this CF allele is maintained in the population by mutation-selection equilibrium and the heterozygotes have no reduction in viability. Estimate mutation rate to the CF allele. Does this seem reasonable given what you know about mutation rates? (hint: dominance model)

This is Population Genetics.

Answer 1

Cystic fibrosis gene protects against tuberculosis. The gene that causes cytic fibrosis may persist in human populations. Cystic fibrosis is an example of a recessive disease. It is caused by mutation of the cystic fibrosis transmembrane conductance regulator (CFTR) gene. Mutation rate is the frequency of a new mutation in a single gene or organism over time. Mutation rate are constant and are not limited to a single type of mutation. Dominance is a relationship between two allels of a gene and their associated phenotypes. Deleterious recessive alleles may persist in a population at low frequency with most copies carried in heterozygotes, at no cost to those individuals. Assuming dominance model with two alleles for cytic fibrosis locus, the recessive allele frequency for different population estimated by

q = [N22/N]1/2

where N22 = Number of recessive

Homozygous, N = Total individuals

The dominant allele frequency - P was estimated as

P = 1 - q (q - frequency of recessive allele )