The autosomal not X- linked gene for brachydactyly,short fingers,is dominant to normal finger length.Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers.What is the probability that their first child will be a brachydactylous girl?
A)1/2
B)2/3
C)3/4
D)1/4
E)1/8

Question

Quizplus · Accepted Answer

First, determine the genotypes of the parents:

Female with brachydactyly in the heterozygous condition: Bb
Man with normal fingers: bb

Write out a Punnett square:

|   | B  | b  |
|---|---|---|
| b | Bb | bb |
| b | Bb | bb |

There is a 50% chance of each possible genotype in their offspring.

To have a brachydactylous girl, she must inherit the dominant B allele from her mother and the recessive b allele from her father.

The probability of this occurring is:

Pr(Bb female) = Pr(B from mother) x Pr(b from father) 
= (1/2)x(1/2) 
= 1/4

Therefore, the probability of their first child being a brachydactylous girl is 1/4, or choice D.

The Autosomal Not X- Linked Gene for Brachydactyly,short Fingers,is Dominant