Question

Beth and Tom eachhave a sibling with cystic fibrosis, but neither Beth nor Tom nor any of their parents have the disease. Calculate the probability that if this couple has a child, the child will have cystic fibrosis. What would be the probability of a test revealing that Tom is a carrier, but Beth is not? Explain your answers.

Short answer

The probability of the child of Beth and Tom acquiring cystic fibrosis is 1/9. Both parents have equal chances of being carriers. The disease can also be caused due to the mutation in the zygote of the non-cystic fibrosis parents.

Step-by-step solution

Description of cystic fibrosis

Cystic fibrosis is a disease condition that affects the lungs. This disease may result in breathing difficulty and lung infections. The accumulation of mucus results in the blockage of respiratory pathways. In particular, cystic fibrosis is a genetic condition.

Description of the recessive allele

An allele is different for a particular gene in which some alleles may get expressed, and some get masked. The allele that is getting masked is known as the recessive allele. It may show its expression once it is transferred to the offspring.

Inheritance of cystic fibrosis to the child

The recessive allele is the reason for causing the cystic fibrosis condition. Beth and Tom’s siblings, who have cystic fibrosis, have a homozygous recessive condition in their genes.

There is a possibility that Beth and Tom can be the carriers of the disease due to a recessive allele. The probability of each parent(Tom) compared to their sibling is 2/3. Both parents are the carriers, so the possibility is ¼.

The probability of a child inheriting this disease is calculated as follows:$=\frac{2}{3}\times \frac{2}{3}\times \frac{1}{4}=\frac{1}{9}$

Probability

Hence, both the parents have to be the carriers to cause the disease in the child.