Question

Suppose a mother carries two recessive genes for freckles (rr) and a father carries one recessive gene for freckles and one dominant gene for clear facial skin (Rr). What is the likelihood that their first child will have freckles?

Answer 1

The answer is 50%.

This is an example of Mendelian monohybrid cross:
Parents: rr x Rr
Offspring: Rr Rr rr rr

So, in this cross, 2 out of 4 offspring will be heterozygous (Rr), and other 2 out of 4 (which is 50%) offspring will be recessive homozygous (rr) and will have freckles.

Answer 2

Answer: The correct answer is 50%.

Explanation-

Genotype of the mother with freckle trait is rr. This means that it is a recessive trait.

Genotype of of the father having clear facial skin is Rr, showing that it is a dominant trait. There can be two genotypes for this trait that is Rr and RR.

When these parents are crossed, they produce 50% offsprings with the freckles trait (rr) and 50% with clear facial skin ( Rr).

Thus, likelihood of the first child with freckles is 50% ( refer punnett square)