Question

2.3 Problem 3 The Hardy-Weinberg equation is useful for predicting the percent of a hu- man population that may be heterozygous carriers of recessive alleles for certain genetic diseases. Phenylketonuria (PKU) is a human metabolic dis- order that results in mental retardation if it is untreated in infancy. In the United States, one out of approximately 10.000 babies is born with the disor- der. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele? If you need help, look back at the Hardy-Weinberg equation. 3 Allelic Frequency vs. Genotypic Frequency 3.1 Allelic Frequency If you are told that the frequency of a recessive allele in a population is 10%. you are directly given q, since by definition q is the frequency of the recessive allele. This comprises all the copies of the recessive allele that are present in heterozygotes as well as all the copies of the allele in individuals that show the recessive phenotype. What is q for this population? If you need help, look back at the Hardy-Weinberg equation. 3.2 Genotypic Frequency Genotypic frequency is the frequency of a genotype (homozygous recessive, homozygous dominant, or heterozygous) in a population. If you don't know

Answer 1

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