Sickle cell anaemia is a recessive trait in humans.The gene that causes this disease is not located on the sex chromosomes.In a cross between a father who has sickle cell anaemia and a mother who is heterozygous for the gene,what is the probability that their first three children will have the normal phenotype?
A) 1/4
B) 1/2
C) none
D) 1/8
E) 1/16 will be albino

Question

Quizplus · Accepted Answer

In this cross, the father is homozygous recessive for sickle cell anaemia (ss) and the mother is heterozygous (Ss). The Punnett square for their cross is:

|   | s | s |
|---|---|---|
| S | Ss| Ss|
| s | ss| ss|

The offspring will have a 50% chance of inheriting the sickle cell allele (s) from their father and a 50% chance of inheriting the normal allele (S) from their mother. Therefore, the probability of each child having the normal phenotype (SS or Ss) is 1/2. The probability of the first child having the normal phenotype is 1/2. The probability of the second child having the normal phenotype is also 1/2. The probability of the third child having the normal phenotype is also 1/2. Therefore, the probability of all three children having the normal phenotype is (1/2) x (1/2) x (1/2) = 1/8.

Sickle Cell Anaemia Is a Recessive Trait in Humans