A Fruit Fly Cross
This article explores the genetic cross of a black female and a gray male fruit fly (Drosophila) to predict the F1 phenotypic ratio. The black color is governed by a recessive mutant allele (bb), while the gray color is dominant (B). We examine the genetic backgrounds of the parents: the black female is homozygous recessive (bb), and the gray male is heterozygous (Bb) due to having a black father. By setting up a Punnett square for the cross, we determine the expected ratios of gray and black offspring, revealing a balanced ratio of 50% gray and 50% black flies.
A Fruit Fly Cross
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Presentation Transcript
Background information • A recessive mutant allele, black, causes a very dark body in fruit flies, Drosophila, when homozygous. • The wild-type color is described as gray.
So the Question is… • What F1 phenotypic ratio is predicted when a black female is crossed to a gray male whose father was black? • Hint: Use B for the dominant trait
Time to Think…5 Steps • Review the data • Set up the problem • Make your punnett square • Look at your results • Do they make sense?
And the Answer is… • Since the female parent is black, she must be ________ for the mutant, recessive color (bb).
That was the answer because… • The male parent is gray and thus he must have at least one dominant allele (B). • Since his father was black (bb), and since he received one of the chromosomes from his father, the male parent must be heterozygous (Bb).
Look at the facts… • So the female is bb (black) • And the male is Bb (gray) • Put them into a punnett square bb x Bb
And the results are… _______ % Gray _______ % Black 50 50
