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.

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1. Step 1: Understand the inheritance pattern. Hemochromatosis is caused by a recessive allele, meaning a child must inherit two recessive alleles (one from each parent) to express the disease. Each parent being a carrier has one recessive allele (h) and one dominant allele (H), so their genotype is Hh.
2. Step 2: Calculate the probability of each child's genotype. Using a Punnett square, the possible genotypes for each child are: HH (25%), Hh (50%), and hh (25%).
3. Step 3: Address part (a) - Calculate the probability that all three children are of normal phenotype (do not have the disease). This occurs if none of the children have the hh genotype. The probability of a child being either HH or Hh (normal) is 75%. Therefore, the probability that all three children are normal is 0.75 * 0.75 * 0.75 = 0.421875 or 42.19%.
4. Step 4: Address part (b) - Calculate the probability that one or more of the three children have the disease (hh genotype). First, find the probability that none of the children have the disease (all are HH or Hh) which is 0.421875 as calculated in step 3. The probability that at least one child has the disease is 1 - 0.421875 = 0.578125 or 57.81%.
5. Step 5: Address parts (c) and (d) - Calculate the probability that all three children have the disease (all are hh). The probability of one child being hh is 25%, so for three children it is 0.25 * 0.25 * 0.25 = 0.015625 or 1.56%. For part (d), the probability that at least one child is phenotypically normal is the complement of all children having the disease, which is 1 - 0.015625 = 0.984375 or 98.44%.