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 over 2
B

1 over 4
C

3 over 4
D
4 over 4

Answer 1

To determine the likelihood that their first child will have freckles, we can use a Punnett square.

The mother has two recessive genes for freckles (rr), and the father has one recessive gene for freckles and one dominant gene for clear facial skin (Rr).

Let’s set up the Punnett square:

• The mother's alleles: r (from rr)
• The father's alleles: R and r (from Rr)

The possible combinations of alleles from the parents are as follows:

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

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

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

Now, we can calculate the probabilities:

• Rr: 2/4 (or 1/2)
• rr: 2/4 (or 1/2)

Since only the genotype rr will result in freckles, the likelihood that their first child will have freckles is 2 out of 4, which simplifies to 1 out of 2.

Thus, the answer is: A 1 over 2.