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Fruit flies can either have a brown body or a black body. The allele for a brown body (B) is dominant to the allele for a black body (b).

A brown fruit fly was crossed with a black fruit fly as shown below. What is the probability of a resulting offspring being black?
B b
b ? ?
b ? ?
A.
50%
B.
100%
C.
75%
D.
25%

1 answer

To determine the probability of the resulting offspring being black when crossing a brown fruit fly with a black fruit fly, we need to analyze the genotype of each parent.

  1. The brown fruit fly can be either homozygous (BB) or heterozygous (Bb). The black fruit fly must be homozygous recessive (bb) since it expresses the black body phenotype.

  2. If we assume the brown parent is heterozygous (Bb), we can set up the following Punnett square:

       B   |   b
     ----------------
  b |  Bb  |  bb
     ----------------
  b |  Bb  |  bb

From the Punnett square, we have:

  • 2 offspring with genotype Bb (brown body)
  • 2 offspring with genotype bb (black body)

This means:

  • Probability of brown body offspring = 2 out of 4 = 50%
  • Probability of black body offspring = 2 out of 4 = 50%

Thus, if the brown parent is heterozygous (Bb), the probability of getting a black offspring (bb) is 50%.

  1. If we assume the brown parent is homozygous (BB), the Punnett square looks like this:
       B   |   B
     ----------------
  b |  Bb  |  Bb
     ----------------
  b |  Bb  |  Bb

From this square, we have:

  • 4 offspring with genotype Bb (all brown body)
  • 0 offspring with genotype bb (no black body)

Thus, if the brown parent is homozygous (BB), the probability of getting a black offspring is 0%.

Since we do not know whether the brown parent is homozygous or heterozygous, the best answer considering the brown parent is likely to be heterozygous is:

A. 50%