Answer:
a. 81/256
b. 175/256
C. 1/4
D. 9/256
E. 54/256
Explanation:
Since both parents are heterozygous for galactosemia (Gg). The possibility of their children will be
GG Gg Gg gg. The probability for one child to be unaffected is 3/4 and affected is 1/4 since it is a recessive diseases:
A. Probability for none of the four children to be affected is 3/4*3/4*3/4*3/4=81/256
B.
Since the probability of unaffected children is 81/256
The probability that at least one child will be affected is 1-(3/4*3/4*3/4*3/4)=1-(81/256)
= 175/256
C. The probability that only one if the child will be affected is 1/4
D. The probability that the first two will have galactosemia and the second two will not is
1/4*1/4*3/4*3/4=9/256
E. The probability that two will have galactosemia and two will not, regardless of order.
Since the probability that two will have galactosemia and two will not is 9/256.
The total possible outcome that 2 among 4 children will be born regardless of order (first and second, first and third, first and fifth, second and third, second and fourth, third and fourth) is 6
Hence the probability two will have galactosemia and two will not, regardless of order is 9/256*6= 54/256
