Question

In fruit flies the presence of wings (W) is dominant to the absence of wings (w), and red eyes (R) are dominant to brown eyes (r). A wingless fly that is heterozygous for eye color is crossed with a fly that is heterozygous for wings and has brown eyes. What is the probability that the offspring would be homozygous recessive for wings and eye color?

Answer 1

The probability that offspring from the described cross will be homozygous recessive for both wings and eye color is 25%.

Step-by-step explanation:

You are asking about the probability that offspring will be homozygous recessive for wings and eye color when a wingless fly that is heterozygous for eye color is crossed with a fly that is heterozygous for wings and has brown eyes. To calculate this, we can set up a Punnett square. The wingless, heterozygous-eyed fly has the genotype wwRr, while the fly that is heterozygous for wings with brown eyes has the genotype Wwrr.

To find the probability of homozygous recessive offspring (wwrr), we look at the probability for each trait separately and then multiply them together. For wings, the probability of a wingless (ww) offspring is 1/2. For eye color, the probability of brown eyes (rr) is also 1/2. Therefore, the probability of an offspring being homozygous recessive for both traits is 1/2 (for wings) × 1/2 (for eyes), which equals 1/4 or 25%.