Question

Kritzik and colleagues (1998) discovered three alleles of the alpha2 gene that influence blood

platelet function and may be significant in thrombosis and/or bleeding in humans. Nucleotide

substitutions account for variants 1,2,and 3. Assume the following variant frequencies in a given

population from North America:

Variant 1= 0.5

Variant 2= 0.3

Variant 3= 0.2

Given these values, of 1000 individuals sampled, what would be the predicted population

distribution of genotypes under Hardy-Weinberg equilibrium assumption

Answer 1

-Let Variant 1 be p
Variant 2 be q
Variant 3 be r
According to Hardy-Weinberg equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1.
So p+q+r =1
p2+ q2+r2 +2pq+2qr+2pr =1
Given that frequency of p is 0.5
Given that frequency of q is 0.3
Given that frequency of r is 0.2
Freqeucny of varaint 1 = p2 = 0.5*0.5 = 0.25
Frequecny of vairnt 2 = 0.3*033 = 0.09
Freqeuncy of varint 3 = 0.2*0.2 = 0.04
Freqeucny of heterozygous variant 1 and 2 = 2pq = 2*0.5*0.3 = 0.3
Freqeucny of heterozygous variant 1 and 3 = 2pq = 2*0.5*0.2 = 0.2
Freqeucny of heterozygous variant 3 and 2 = 2pq = 2*0.2*0.3 = 0.12.
The total number of progeny = 1000.
Number of homozygous variant 1 = 0.25 *1000 =250
Number of homozygous variant 2 = 0.09 *1000 =90
Number of homozygous variant 3 = 0.25 *1000 =40
Number of heterozygous variant 1 and 2 = 0.3 * 1000 = 300
Number of heterozygous variant 1 and 3 = 0.2 * 1000 = 200
Number of heterozygous variant 3 and 2 =0.12*1000 = 120.

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