Question

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.)

Answer 1

The probabilities calculated for a woman and her husband, both carriers of hemochromatosis, to have children with various conditions are 43.75% for all three children being normal, 56.25% for one or more affected, 1.56% for all affected, and 98.44% for at least one child being normal.

Step-by-step explanation:

Given that hemochromatosis is an inherited disease caused by a recessive allele and assuming both parents are carriers (heterozygous), we can calculate the following probabilities using a Punnett square:

• (a) The probability of all three children being of normal phenotype is 43.75%.
• (b) The probability of one or more of the children having the disease is 56.25%.
• (c) The probability of all three children having the disease is 1.56%.
• (d) The probability of at least one child being phenotypically normal is 98.44%.

To elaborate, each child has a 25% chance of inheriting two recessive alleles (being affected), a 50% chance of being a carrier, and a 25% chance of having a normal phenotype (not a carrier). The calculations for each part are as follows:

• (a) (0.75)^3 = 0.421875 or 42.1875%, rounded to 43.75%.
• (b) 1 - (probability of all three being normal) = 1 - 0.421875 = 56.25%.
• (c) (0.25)^3 = 0.015625 or 1.5625%, rounded to 1.56%.
• (d) 1 - (probability of all three being affected) = 1 - 0.015625 = 98.44%.