Question

The trait for 'male-pattern baldness' is a recessive trait encoded for by "b". Non-balding is encoded for by a dominant allele encoded for by the letter "B". A street survey conducted by Hair Club for Men found that out of 1000 men, 360 had male pattern baldness, the other 480 who did not and were heterozygous, and 160 who did not and were homozygous for a full head of hair. What are the phenotype ratios? Using this information find the allele frequencies for B and b. What are the expected genotype frequencies?

Answer 1

Allele frequencies for B and b.

"b"(q) allele frequency = 0.60

"B"(p) allele frequency = 0.40

Genotype frequencies;

BB = 0.16

Bb = 0.48

Bb = 0.36

Step-by-step explanation:

Given

Non-baldness (B) is dominant on baldness(b), so B is dominant over b.

Homozygous pattern baldness male (bb) = 360,

Heterozygous non- baldness (Bb)= 480,

Homozygous non-baldness (BB)= 160.

So, we can also denote then by genotypes only,

BB= 160;

Bb= 480;

bb= 360;

Total= 1000

Allele frequency q² (bb) = 360/1000=0.36

allele fequency for q( b)= √o.36=0.60.

Allele frequency for p²(BB) = 160/1000=0.16

allele frequency p(B)= √0.16 = 0.4

Expected genotype frequencies;

BB = 160/1000 = 0.16

Bb = 480/1000= 0.48

Bb = 360/1000= 0.36