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?

Answer 1

To determine the likelihood of the child having freckles, we need to look at the genetic makeup of the parents.

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

When we set up a Punnett square to analyze the possible genotypes of their offspring, we can list the possible alleles contributed by each parent:

• The mother (rr) can only contribute r.
• The father (Rr) can contribute either R (dominant) or r (recessive).

Now we set up the Punnett square:

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

From the Punnett square, we can see the possible combinations of alleles for their offspring:

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

So, out of the four possible genotypes:

• 2 will be Rr (not have freckles)
• 2 will be rr (will have freckles)

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

\[ \text{Probability} = \frac{\text{Number of rr outcomes}}{\text{Total outcomes}} = \frac{2}{4} = \frac{1}{2} = 50% \]

Thus, there is a 50% chance that their first child will have freckles.