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?

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.

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StaceeLichtenstein StaceeLichtenstein · Answer 2 · 2018-11-26T03:02:47-05:00

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)