Question

PKU is a rare autosomal recessive allele. 1/10,000 newborns are affected. What proportion of the population is expected to be carriers?

Answer 1

Around 75% of the population is expected to be carriers of the PKU allele, which is a rare autosomal recessive allele.

Step-by-step explanation:

In the case of PKU, which is a rare autosomal recessive allele, carriers can be identified by looking at the Punnett square. If two heterozygous individuals with PKU mate, there is a 25% chance that their offspring will express the PKU phenotype. This means that 75% of their offspring will not have PKU but will be carriers of the allele. Pheylketonuria (PKU) is a rare autosomal recessive genetic disorder. The question we are addressing is: What proportion of the population is expected to be carriers of the PKU allele, given the frequency of afflicted newborns (1 in 10,000)? To solve this, we can apply the Hardy-Weinberg principle which states that the frequency of the genotype can be found using the equation
`p2 + 2pq + q2 = 1`, where q2 is the frequency of the homozygous recessive individuals (those with PKU), and 2pq is the frequency of the carriers (heterozygous individuals). Since the homozygous recessive frequency (q2) is
`1/10,000`, we can take the square root of that value to find q, resulting in 1/100. P is thus calculated as
`1 - q`, meaning p = 99/100. Substituting these values into the formula for 2pq gives us
`2 * (99/100) * (1/100)`, which simplifies to approximately 1/50 or 2% of the population expected to be carriers for the PKU allele.