Question

The allele for color blindness is inherited as a recessive allele. Using letter B (normalallele) and b (color blind allele)

Answer the questions below:

1.What genotype will be normal

2.What genotype willhave color blindness

3.Color blindness is found in about 8% of the population. Using the Hardy-Weinberg formula, determine the frequency of:

a.Homozygote dominant

b.Heterozygote

c.Homozygote recessive

Answer 1

The allele for color blindness is inherited as a recessive allele. Using letter B (normal allele) and b (color blind allele).

1.BB genotype will be normal as no 'b' is there and 'B' is a dominant allele (normal).

2. 'bb' genotype will have color blindness as the colorblindness allele is recessive.

3. Color blindness is found in about 8% of the population.

Using the Hardy-Weinberg formula, the frequency calculation of BB, Bb and bb is given below:

If there are only 2 alleles B and b with frequencies p and q then the frequencies of the 3 possible genotypes will be given by,

(p + q)² = p² + 2pq + q² where p + q = 1, therefore p² + 2pq + q² = ( p + q)² = 1.

p = frequency of dominant allele

q =frequency of recessive allele

Now, 8% population is colorblind means 'bb' genotype frequency or 'q²' is 8/100 i.e. 0.08

Frequency of 'b' allele or 'q' will be the square root of 0.08 i.e. 0.28

The frequency of the dominant allele B would be 1 - q or 1 - 0.28 = 0.72

Frequency of allele 'B' or 'p' is 0.72.

Frequency of 'Bb' genotype or '2pq' is 2 X 0.72 X 0.28 = 0.40

Here, p + q = 1 (0.72 + 0.28 = 1)

So,

the frequency of:

a.Homozygote dominant (BB) or q2 = 0.51.

b.Heterozygote (Bb) = 0.72 X .28 = 0.20

c.Homozygote recessive (bb) = 0.08