Question

In humans, the sickle-cell trait is caused by a single defective allele, but sickle-cell disease only occurs in individuals that are homozygous for the sickle-cell allele. A man and woman each carry the trait, but do not have sickle-cell disease. What is the probability that their first two children will both have sickle-cell disease?

Answer 1

Answer: The probability that the two children have the disease is 1/16 or 6,25%

Explanation: If the disease only occurs in homozygous individuals and the parents have the trait but not the disease that means is a recesive allele and the parents are heterozygous for this trait.

According to proportions stablished by Mendel, when there is a cross between 2 heterozygous (Ss x Ss) indivdials the probability to obtain a homozygous recesive (ss) descendant is 25% or 1/4 as shown in punnet square below:

$\left[\begin{array}{ccc}&S&s\\S&SS&Ss\\s&Ss&ss\end{array}\right]$

Taking into account that the condition of the first kid does not imply the condition for the second, it should be analyzed as 2 independent events.

1. First kid has 1/4 of probability have the dissease as shown in punnet square

2. For a second birth, child has the 25% chance to have the disease

The probability that both events to happen is the product of each event probabilty:  

p = 1/4 * 1/4 = 1/16 or 6,25%