Respuesta :
Answer:
[tex]\frac{1}{4}[/tex]
Explanation:
The complete question is as follows - 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 -
It is given that Susan and her husband are heterozygote which means they are not affected but they are carrier of the defect galactosemia.
The genotype of both the parents would be “Gg”
The cross between the two parents will result into following offspring
Gg * Gg
GG, Gg, Gg, gg
Thus, out of four, nearly three offspring do not have galactosemia defect while only one offspring with genotype “gg” is the affected one.
Thus, probability of offspring with defect is 1/4
