Question

Diagrams of a bent pinkie finger and a straight pinkie finger are shown below. The allele for having a bent pinkie finger is dominant to the allele for having a straight pinkie finger.

If two people who are both heterozygous for the trait of a bent pinkie finger have a child, what is the percent probability that the child will have straight pinkie fingers?

Answer 1

25%

Step-by-step explanation:

This question involves a single gene coding for pinkie finger shape in humans. The allele for a bent pinkie finger (B) is dominant over the allele for a straight pinkie finger (b). This means that the 'B' allele will always be expressed over the 'b' allele in a heterozygous state.

An heterozygous individual will contain a combination of both alleles i.e. Bb. According to the question, two people that are heterozygous for the bent pinkie finger trait are crossed i.e. Bb × Bb. Each parent will produce B and b gametes. Using these gametes in a punnet square, 4 possible offsprings will be produced with genotypes: BB, Bb, Bb, bb

Bb, Bb - phenotypically bent pinkie finger

bb - phenotypically straight pinkie finger

Out of the possible four offsprings, 1 of them will possess a straight pinkie finger. This means that 1/4 × 100 = 25% is the percent probability of producing an offspring with straight pinkie finger by crossing two heterozygous individuals.