In fruit flies, red eyes (R) are dominant to sepia eyes (r), and the presence of wings (A) is dominant to wingless (a) flies. If two flies heterozygous for each trait are crossed (RrAa × RrAa), how many different phenotypes would be produced in the offspring?
A)2
B)3
C)4
D)5
E)9

Question

Quizplus · Accepted Answer

To solve the problem, we need to use the principles of Mendelian genetics and the Punnett square. 
First, we need to determine the possible gametes of each heterozygous parent: 
RrAa: RA, Ra, rA, ra 
Then, we can use the Punnett square to combine the gametes of both parents:

|   | R  r | 
|---|------| 
| A | RA   | Ra   |
| a | rA    | ra   |

From the Punnett square, we can see that the offspring would have the following genotypes:
- RRAA, RRAa, RRaa, RrAA, RrAa, Rraa, rRAA, rRAa, rrAA, rrAa, rraa 
To determine the phenotypes, we need to apply the principles of dominance and recessiveness: 
- RRAA, RRAa, RRaa, RrAA, and RrAa would all have red eyes and wings. 
- rRAA and rRAa would have sepia eyes and wings. 
- rrAA, rrAa, and rraa would have sepia eyes and be wingless.

Therefore, there are four different phenotypes that could be produced in the offspring of the cross: red eyes and wings, sepia eyes and wings, sepia eyes and no wings, and red eyes and no wings. Thus, the answer is C.

In Fruit Flies, Red Eyes (R) Are Dominant to Sepia