## Related questions with answers

Galactosemia is a recessive human disease that is treatable by restricting lactose and glucose in the diet. Susan Smithers and her husband are both heterozygous for the galactosemia gene. a. Susan is pregnant with twins. If she has fraternal (nonidentical) twins, what is the probability both of the twins will be girls who have galactosemia? b. If the twins are identical, what is the probability that both will be girls and have galactosemia? For parts c–g, assume that none of the children is a twin. c. If Susan and her husband have four children, what is the probability that none of the four will have galactosemia? d. If the couple has four children, what is the probability that at least one child will have galactosemia? e. If the couple has four children, what is the probability that the first two will have galactosemia and the second two will not? f. If the couple has three children, what is the probability that two of the children will have galactosemia and one will not, regardless of order? g. If the couple has four children with galactosemia, what is the probability that their next child will have galactosemia?

Solution

Verifieda. Susan and her husband are both heterozygous (Aa) for galactosemia, which means that they don't have this disorder, but they are both carriers of the defected gene. The probability of having a girl is $\dfrac{1}{2}$, while the probability for a homozygous genotype is $\dfrac{1}{4}$ (AA, Aa, Aa, and aa). The possibility for an affected girl is $\dfrac{1}{2} * \dfrac{1}{4} = \dfrac{1}{8}$. Thus, the probability for this couple to have twin girls (fraternal) where both of them have galactosemia is $\dfrac{1}{8} * \dfrac{1}{8} = \dfrac{1}{64}$.

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