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?

Question

Responses
A




1 2
B




1 4
C




3 4
D




4 4

Answer 1

To determine the likelihood that their first child will have freckles, we can use a Punnett square to analyze the possible gene combinations from the mother and father.

The mother's genotype is $rr$ (homozygous recessive), and the father's genotype is $Rr$ (heterozygous).

Let's set up the Punnett square:

Parent genotypes:

• Mother: $rr$
• Father: $Rr$

The possible gametes from each parent:

• Mother can only contribute $r$ (since she has two recessive alleles).
• Father can contribute either $R$ or $r$.

Now we fill in the Punnett square:

$$\text{Misplaced hline}$$

From the Punnett square, we see the possible genotypes of the children:

• $Rr$ (clear skin) - 2 out of 4 squares (50%)
• $rr$ (freckles) - 2 out of 4 squares (50%)

Therefore, the probability that their first child will have freckles (genotype $rr$) is:

$$\frac{2}{4}=\frac{1}{2}$$

So, the correct answer is:

A 1/2.