Question

Suppose that you are studying is a rare recessive condition in humans whereby affected individuals are unable to metabolize a sugar found in certain foods. A woman and her husband are both heterozygous for the allele causing the condition.

A) If the woman is pregnant with twins and the twins are dizygotic (non-identical), what is the probability that both twins will be girls and be affected?

B) If the couple has four affected children, what is the probability that a fifth child will also be affected?

C) What is the probability if the girl twins are identical (monozygotic) and both are affected?

D) If the woman gives birth to four children, what is the probability that none of them will be affected?

E) If the couple has four children, what is the probability that one of will be affected?

F) If the couple has four children, what is the probability that the first two will be affected while the second two will be unaffected?

G) If the couple has three children, what is the probability that two will be affected while one will be unaffected, regardless of birth order?

Answer 1

Solution:

It is a autosomal recessive disease. Ratio when 2 heterozygous are crossed is 1:2:1 in which three are normal and one is affected.

A) Probability of 1st twin to be a girl = 1/2.

Probability of 2nd twin to be a girl = 1/2

Probability of both the girls to be affected = 1/2×1/2×1/4=1/16

B).  Probability that fifth child is affected = 1/4

C). Probability of identical twins to be girls= 1/2.

Probability of both the identical twins to be affected= 1/2×1/4=1/8

D) .

Probability that none of them is affected

i.e all four of them are normal is= 3/4×3/4×3/4×3/4= 81/256

E). Probability that one child willbe affecd= (4!/3!×1! )× (3/4)³×(1/4)¹= 27/64

F) . Probability that first two are affected and second two are unaffected= 1/4×1/4×3/4×3/4 = 9/256

G). Probability that 2 will be affected and one will be Normal = (3!/2!×1!)×(3/4)×(1/4)²= 9/64