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:

\[ \begin{array}{c|c|c} & R & r \ \hline r & Rr & rr \ \hline r & Rr & rr \ \end{array} \]

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.