Question

The Punnett square below represents a cross between two parents that are heterozygous for the F gene. The parents have already had three offspring. The first had genotype FF and the second and third had genotype Ff. What is the probability that the fourth child has genotype ff?

Answer 1

1/4

Step-by-step explanation:

If we make a Punnett square crossing two heterozygous parents (Ff and Ff), we see that the possible offspring genotypes are: 1 FF, 2 Ff, and 1 ff.

There are a total of 1 + 2 + 1 offspring. The probability of having a child with ff genotype would be 1/(1 + 2 +1) = 1/4.

Thus, the answer is 1/4.

Hope this helps!