Question

The following F2 phenotypic data was obtained from a Drosophila testcross using an F1 offspring. Assume red eye and brown body are dominant wild-type phenotypes, and white eye and yellow body are mutant phenotypes.White-eye, brown body 670Red-eye, yellow body 650White-eye, yellow body 38Red-eye, brown body 561. What is the genotype of the F1? How do you know this?

Answer 1

The genotype of the F1 was wy+/w+y.

Step-by-step explanation:

One of the given options has a typo: the red eye-brown body offspring count should be 56 instead of 561.

We have two genes with two alleles each:

Red eyes (w+) is dominant over white eyes (w).

Brown body (y+) is dominant over yellow body (y).

The phenotypes of the F2 tesulting from a test cross (F1 x wy/wy) are:

• wy+/ey (white-eye, brown body): 670
• w+y/wy (red-eye, yellow body): 650
• wy/wy (white-eye, yellow body): 38
• w+y+/wy (red-eye, brown body 56

If the genes w and y are linked, two phenotypes in the F2 will be much more abundant than the other two. Recombination during meiosis is a rare event, so the most abundant phenotypes are the parentals (the ones present in the F1 parent).

Every individual in the offpsring has a wy chromosome, as this was the gamete inherited from the test cross individual.

In this case, the most abundant gametes are wy+ and w+y, so the genotype of the F1 was wy+/w+y.

Notice how when recombination occurs in the F1 parent, the recombinant gametes appear: wy and w+y+, which are the less abundant in the F2 progeny.