Hemochromatosis is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following?

(a) All three children are of normal phenotype.

(b) One or more of the three children have the disease.

(c) All three children have the disease.

(d) At least one child is phenotypically normal.

(Note: It will help to remember that the probabilities of all possible outcomes always add up to 1.)

Inherited disorders are caused by the transfer of defective genes from the parents to the offspring. The recessive allele causes the disease by the transfer of one allele from each parent.

The possibility of the offspring genotypes for the given condition is as follows.

The combination of two recessive alleles (nn) is responsible for hemochromatosis disease. The combinations NN, Nn, and nN are carriers for the trait but have a normal phenotype.

The given condition is all three children have a normal phenotype. The chance that each child would be normal is ¾. Thus, the overall probability is as follows.

$\text{Probability}=\frac{3}{4}\times \frac{3}{4}\times \frac{3}{4}=\frac{27}{64}$

27/64 is the probability that all the children would be normal.

The total of all possible probability outcomes is one. By subtracting, the normal phenotype from one results in the probability of the disease condition.

Probability of the child with disease = $1-\frac{27}{64}=\frac{37}{64}$

37/64 is the probability that one or more of the three children would be diseased.

The given condition is that all three children have the disease condition. The chance that each child would be diseased is ¼. Thus, the overall probability of all three children is as follows.

$\text{Probability}=\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times =\frac{1}{64}$

1/64 is the probability for all three children to be diseased.

The possibility of one child with the disease is obtained by subtracting the overall probability of the three children with the disease from one.

$\text{Probability}=1-\frac{1}{64}=\frac{63}{64}$

The overall probability of one child with the disease is 63/64.