In humans, albino individuals are homozygous recessive for an autosomal gene. If
ID: 3201959 • Letter: I
Question
In humans, albino individuals are homozygous recessive for an autosomal gene. If two normal parents have an albino daughter and a normal son. What is the probability that their son is homozygous? a. 1/4 b. 1/2 c. 1/8 d. 2/3 e. 1/32 Referring to question 12, what is the probability that the father is heterozygous? a. 1/4 b. 1/2 c. 1 d. 1/16 e. 0 Again referring to question 12, what is the probability that out of 4 children, 2 are albino? a. 1/256 b. 67/256 c. 5/24 d. 54/256 e. 54/128 Again referring to question 12, what is the probability that out of 4 children, 2 or more are albino? a. 1/256 b. 67/256 c. 5/24 d. 54/256 e. 54/128Explanation / Answer
Answer:
Since albinism is caused by homozygous recessive mutation for an autosomal gene
Parent
A
a
A
AA
Aa
a
Aa
aa
AA: Homozygous dominant with probability 1/4 – no albino
Aa: Heterozygous genotype with probability 2/4 = 1/2 – no albino
Aa: homozygous recessive with probability 1/4 – albino
1.Probability that son is homozygous = 1/4 (probability of AA)
2. Probability that father is heterozygous = 1
3. Probability that out of 4 children, 2 are albino = this can be solved with the help of binomial expression, where p = probability of success 1/4 (albino) and q = probability of failure 3/4 (no albino) – Formula 4C2 x p2 x q(4-2) = 6*(1/4)2*(3/4)2 = 54/256
4. Probability that out of 4 children, 2 or more are albino – this is possible when 2 children are albino, 3 children are albino or all 4 children are albino
4C2 x p2 x q(4-2) + 4C3 x p3 x q(4-3) + 4C4 x p4 x q(4-4)
=6*(1/4)2*(3/4)2 + 4*(1/4)3*(3/4)1 + 1*(1/4)4*(3/4)0
= 54/256 + 3/64 + 1/256 = 67/256
Parent
A
a
A
AA
Aa
a
Aa
aa