Question

Hardy-Weinberg equilibrium:

· Hardy-Weinberg equation describes: The most possible distribution of the population genotype of the population, if the frequencies of the alleles are known.

· If allele frequencies are p and q, then the genotype frequencies are p2, q2, 2pq.

· A population is considered to be in Hardy-Weinberg equilibrium when: In a randomly mating population, allele frequency and genotype frequency remain constant.

· Thus, under Hardy-Weinberg equilibrium:

p+ q=1 and

p2+2pq + q2 = 1.

Problem 3.

Representation:

• Genotype for individuals with PKU = pp = recessive homozygous condition.
• Normal individuals, will have genotypes:
• PP = Homozygous dominant
• Pp = Heterozygous dominant (Carriers)
• Since 1 out of 10,000 babies in the population have PKU = pp
• Genotype frequency for pp = q2 = 1/ 10000 = 0.0001
• Allele frequency = q = 0.01
• As per Hardy-Weinberg equilibrium p+q = 1
• Thus, p = 1- 0.01 = 0.99
• Genotype frequency for PP = p2= (0.99)2= 0.9801
• Heterozygous genotype frequency (Pp) = 2 x p x q =2 x 0.01x 0.99 =0.0198

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