Question

Albinism is an autosomal recessive condition characterized by the absence of melanin pigment from the skin, eye, and hair. Two carriers of albinism marry and plan to have four children. Assume a 1:1 sex ratio. What is the probability that all their children will be normal girls?

Answer 1

The probability that two carriers of the albinism gene will have four normal daughters is 81/4096 or about 1.98%, calculated by multiplying the probability of having an unaffected girl (3/8) to the power of four.

Step-by-step explanation:

Albinism is an autosomal recessive condition, meaning that it only appears in individuals who inherit two recessive alleles for the trait. In this scenario, both parents are carriers of the albinism gene (heterozygous). They have the genotype Aa, where 'A' represents the normal allele and 'a' represents the albinism allele.

For each child, the probability that they will inherit two normal alleles (AA) is 1/4, the probability for one normal and one albinism allele (Aa) is 1/2, and the probability for two albinism alleles (aa) is 1/4. Therefore, the probability of a child being unaffected by albinism (AA or Aa) is 3/4. Considering the equal sex ratio, the probability of a child being a girl is 1/2. Combining these probabilities, the chance of one child being an unaffected girl is (3/4) × (1/2) = 3/8.

To find the probability of all four children being unaffected girls, we raise the single child probability to the power of 4, because the events are independent. Thus, the probability is (3/8)4, which is 81/4096 or about 0.0198. This means there is approximately a 1.98% chance that all four children will be normal girls.