you and your spouse are carriers (you have the allele, but you don\'t show the d
ID: 52176 • Letter: Y
Question
you and your spouse are carriers (you have the allele, but you don't show the disease) for maple syrup urine disease ( an autosomal recessive disease) and you want to start a family)A) what is the probability of having 10 children, the first 6 are normal and the last 4 are affected?
B) what is the probability of having 10 children, the first 2 are normal and of the remaining children only 2 are affected with this disease) you and your spouse are carriers (you have the allele, but you don't show the disease) for maple syrup urine disease ( an autosomal recessive disease) and you want to start a family)
A) what is the probability of having 10 children, the first 6 are normal and the last 4 are affected?
B) what is the probability of having 10 children, the first 2 are normal and of the remaining children only 2 are affected with this disease) you and your spouse are carriers (you have the allele, but you don't show the disease) for maple syrup urine disease ( an autosomal recessive disease) and you want to start a family)
A) what is the probability of having 10 children, the first 6 are normal and the last 4 are affected?
B) what is the probability of having 10 children, the first 2 are normal and of the remaining children only 2 are affected with this disease)
Explanation / Answer
Based on the given data, both mother and father are carriers (both are heterozygous); therefore the cross of an autosomal recessive disease between two carriers is as follows:
Thus, 1/4th of children will be affected and 1/2nd of children will be carrier’s and1/4th of children will be normal.
A)
The probability of having 10 children, the first 6 are normal and the last 4 are affected is:
P[N6D4] = [P(N)]6×[P(D)]4
Where,
= (1/4)6 × (1/4)4
= (1/4)10
= 0.000001
Thus, the probability of having 10 children, the first 6 are normal and the last 4 is: 0.000001.
B)
The probability of having 10 children, the first 2 are normal and of the remaining children only 2 is:
NN (82) DD
N2 (82) D2
= (1/4)2 × (82) × (1/4)4
= (1/4)4 × (82)
= 0.0109378
Thus, probability of having 10 children, the first 2 are normal and of the remaining children only 2 is 0.0109378.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.