Phenylketonuria is an autosomal recessive disorder of man. If the frequency of affected newborn infants is about 1 in 14,000 assuming random matting, what is the frequency of heterozygotes?

Hint

• It is based on Hardy-Weinberg equation: p² + q² + 2pq = 1​

Key Points

• G.H. Hardy and Wilhelm Weinberg independently described the basic principle of population genetics that is known as the Hardy-Weinberg principle.
• It states that allele frequencies in a population remains constant from generation to generation.
• The gene pool remains constant and is referred to as genetic equilibrium or Hardy-Weinberg equilibrium.
• Gene pool - It is the total of all genes and alleles in a population.
• This is supported by a mathematical equation that can be used to calculate genotype frequencies in a population at equilibrium.

Important Points

Concept:

• Phenylketonuria is an autosomal recessive disorder, meaning the disease is expressed only in homozygous recessive condition.
• If we assume the dominant and recessive alleles as A and a, the genotypic conditions will be:
  • AA - Homozygous dominant
  • aa - Homozygous recessive
  • Aa - Heterozygous
• According to Hardy-Weinberg equation,

Sum of all allelic frequencies = 1

Therefore, p² + q² + 2pq = 1

or, (p + q)² = 1

where, p and q represent the frequencies of the alleles A and a respectively.

• Thus, the specific genotype frequencies are represented as:
  • p² - AA
  • q² - aa
  • 2pq - Aa

Calculation:

• It is given that the frequency of affected newborns is 1 in 14000.

Therefore, q² = 1/14000 = 0.0000714

or, q = 0.0084

Now, (p + q)2 = 1

or, p = 1 - q = 1 - 0.0084 = 0.9916

Therefore, 2pq = 0.0166 ≈ 0.017

• Thus, the frequency of heterozygotes will be 0.017.