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

Answer 1

To determine the likelihood that the first child of the mother (rr) and the father (Rr) will have freckles, we can use a simple Punnett square.

The mother's genotype is rr, and the father's genotype is Rr.

Here’s how the Punnett square is set up:

    R    |    r
-----------------
r |   Rr   |   rr
-----------------
r |   Rr   |   rr

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

• Rr (clear skin) - 2 squares
• rr (freckles) - 2 squares

Out of the 4 possible combinations:

• 2 combinations result in Rr (clear skin).
• 2 combinations result in rr (freckles).

Therefore, the probability of the child having freckles (rr) is:

\[ \text{Probability of rr} = \frac{\text{number of rr outcomes}}{\text{total outcomes}} = \frac{2}{4} = \frac{1}{2} \]

Thus, the answer is A \( \frac{1}{2} \).