Albinism,lack of pigmentation in humans,results from an autosomal recessive gene.Two parents with normal pigmentation have an albino child.What is the probability that their next three children will be wild type?
A)1/32
B)0%
C)3/64
D)1/4
E)3/8

Question

Quizplus · Accepted Answer

When two parents with normal pigmentation (who are heterozygous carriers for the recessive gene) have an albino child, it means that both parents have one copy of the recessive gene. Therefore, the parents' genotypes are both Aa (where "A" represents the dominant allele for normal pigmentation and "a" represents the recessive allele for albinism).
The Punnett square for the parents' cross would be:
   | A   | a  
---|---|---
A | AA | Aa 
a  | Aa | aa 
So, the probability of a child being wild type (having normal pigmentation) is 3/4 or 0.75. 
The probability of having three consecutive wild-type children is the product of their individual probabilities, since each birth is an independent event. Therefore, 
P(wild type child) x P(wild type child) x P(wild type child) = (3/4) x (3/4) x (3/4) = 27/64 or 0.4219.
However, we also have to consider the probability of having three children in general, regardless of their pigmentation. The probability of having three children is (1/2) x (1/2) x (1/2) = 1/8. 
Therefore, the overall probability of having three consecutive wild-type children is:
P(wild type child) x P(wild type child) x P(wild type child) x P(three children) = (27/64) x (1/8) = 3/64 or 0.0469. 
So the answer is C: 3/64.

Albinism,lack of Pigmentation in Humans,results from an Autosomal Recessive Gene