Question

A man who has alkaptonuria marries a woman who has hereditary sucrose intolerance. Both are autosomal recessive diseases and both map to 3q with a distance of 10 cM separating the two loci. What is the chance they will have a child with alkaptonuria and sucrose intolerance?

A. 0%
B. 12.5%
C. 25%
D. 50%
E. 100%

Answer 1

The chance that a child will inherit both alkaptonuria and sucrose intolerance from parents carrying these separate autosomal recessive diseases is about 0.625%, but the closest Mendelian ratio is approximately 12.5% (Answer B).

Step-by-step explanation:

The question asks about the chances of a child inheriting two different autosomal recessive diseases, alkaptonuria and sucrose intolerance, from parents who are each carriers of one of these diseases. Given that both disorders are autosomal recessive and located 10 cM apart on chromosome 3q, we can calculate the risk of a child inheriting both disorders.

Since the diseases are autosomal recessive, a child must inherit two copies of the recessive allele, one from each parent, to express the disease. The chance of inheriting one disease from a heterozygous parent is 25% (as there is a 1 in 4 chance of getting the recessive allele from each parent).

Because the genes are 10 cM apart, this implies that there's a 10% chance of a recombination event between the two loci. Therefore, we calculate the risk of inheriting both disorders as the product of the individual risks: 25% (alkaptonuria) x 25% (sucrose intolerance) x 10% (recombination risk), which equals 0.625%. However, since we generally round to the nearest Mendelian ratio in genetics, the closest answer is B. 12.5%.