Pattern baldness in humans iscontrolled by a single autosomal locus with two all
ID: 4420 • Letter: P
Question
Pattern baldness in humans iscontrolled by a single autosomal locus with two alleles(B and B'), but pattern baldness is a sex influenced phenotype:baldness is dominant in males (BB and BB' males become bald), butrecessive in females (only BB females become bald). If a populationis in Hardy Weinberg equilibrium for this locus and 4% of the womenin the population become bald,
a) what percentage of the men are bald?
b) what percentage of the nonbald women could give a gene forbaldness to their sons?
c) what percentage of all possible marriages are expected toproduce daughters who develop baldness? (assume that the marriagesare not influenced by the baldness of the individuals)
Explanation / Answer
Well first you need to determine the allele frequency in thepopulation. If 4% of the women become bald, we know that they are homozygousBB. Therefore we can determine the frequency of the B allele, q^2.0.04 = q^2, and q = 0.2. Since it's autosomal, we know that itapplies to all people, not just females/males. p is the frequencyof B', which is equal to 1-0.2 = 0.8. The percentage of men who are bald (have either BB or BB') would be2pq for BB' (2(0.8)(0.2), or 0.32. 32% of males are BB' BB is the homozygous for q, or q^2. (0.2)(0.2) = 0.04 or 4% ofmales are homozygous. Adding both of these groups together, 32% +4% = 36% of males are bald. b) Then you would solve for the women and their son's baldnessdepending on the genotype of the female. Any female with BB' canpass on either allele with the combination from the father and makethe son bald. c) marriages between two individuals who are bald definitely give ahigher percentage of daughter baldness. However since this is anautosomal, not sex-linked characteristic, the baldness ofsons or daughters is merely a difference between the sensitivity tothe allele they receive. Sons will go bald with either BB or BB',and daughters only with BB. hope that helps!