Question

In humans, six fingers (f) is the dominant trait; five fingers (f) is the recessive trait. if the father is heterozygous for six fingers and the mother has five fingers, what is the probability of their offspring having five fingers

Answer 1

The probability of the offspring of a heterozygous father and homzygous mother having five fingers is 50%.

How to calculate genotype of a cross?

According to this question, a gene coding for the number of fingers in humans is involved. The allele for six fingers (F) is the dominant trait while the allele for five fingers (f) is the recessive trait.

If a cross between a heterozygous father that posseses a genotype of Ff and a homzygous mother that posseses a genotype of ff, the following offsprings will be produced:

• Ff
• Ff
• ff
• ff

This shows that the probability of the offspring of a heterozygous father and homzygous mother having five fingers is ½ (50%).

Learn more about genotype at: https://brainly.com/question/12116830

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