Module 6 question biology

pure-breeding long-wing red-eyed fly and a pure-breeding short-wing white-eyed fly were crossed. All the F1 offspring were long-wing and red-eyed. When members of the f1 generation were crossed the F2 generation included 27 flies with long wings and white eyes.
Which of the options, A to D, shows thelobserved results that most closely match the expected results for the number of long-wing red-eyed flies and short-wing red-eyed flies?
A. 92 long-wing red-eye and 31 short-wing red-eye
B. 27 long-wing red-eve and 29 short-wing red-eve
C. 86 long-wing red-eye and 11 short-wing red-eye
D. 27 long-wing red-eye and 88 short-wing red-eve

The correct answer is A .

Could someone pls explain to me why A is the correct answer?

Reply 1

AriTem

Well it's pure breeding flies and from the 1st generation cross all offspring are long winged and red eyed so you know that the dominant gene is the one that codes for long wings and red eyes.

Therefore all the short wing white eyed flies must be homozygous for the sets of recessive genes (call this WWSS). The long wing red eyed flies must also be homozygous for both sets of dominant genes (call this RRLL) because if any of these flies were heterozygous the offspring would include some short wing white eyed flies.

After that you have to construct a matrix (see below) for potential combination of genes for Gen 2. Gen 1 flies must all be heterozygous for the genes (so all must be RWLS) that codes for long wings and red eyes. So the gametes they produce will be RL/RS/WL/WS (25% each)