for q2 that answer is 4/16 and q 3 the answer is 1/3 only show me the calculatio

Question

for q2 that answer is 4/16 and q 3 the answer is 1/3

only show me the calculation because i keep getting different value

Problem 1 Wild type fruit flies have gray bodies and straight wings. Consider two pure-breeding lines of fruit flies. One has red bodies and straight wings and the other has gray bodies and curved wings. Reciprocal crosses give the same result: all the Fl were wild type in appearance. Assume that body color and wing shape are Mendelian traits that assort independently and use this information to help you answer the questions below. Note that a chi-square table and Pascal's triangle are provided on th e figure page 1) Which is the most correct genotype of the F1? a) males and females are gray with straight wings b) males are red with curved wings and females are gray with straight wings c) males and females are r r c+c d) males are r Y c+ c and females arer r c+ c' e) males are r Y c+ c , and females are r r c+ c 2) If F1 are mated to produce F2 flies, what proportion of F2 progeny are expected to be heterozygous for the gene controlling body color and homozygous for the gene controlling wing shape a) 9/16 b) 3/16 c) 1/16 d) 2/16 e) 4/16 3) If Fl are mated to produce F2 flies, what is the probability that a red-bodied, straight-winged F2 fly will be pure- breeding for both traits? a) 1/3 b) 1/4 c) 1/9 d) 2/3 e) 1/16

Explanation / Answer

1. Two pure breeding lines of fruit fly were crossed.

One has red bodies and straight wings and the other has gray bodies and curved wings. Wild type flies have gray bodies and straight wings. So, the dominant Color allele is gray and the dominant wing shape is straight.

So, let the genotype of red bodies and straight wings flies be ggSS and the genotype of gray bodies and curved wings be GGss.

A cross between these two pure-breeding flies results in a F1 generation that all has gray bodies and straight wings (GgSs). So, the correct option is option A, males and females are gray with straight wings.

2. Gametes that can be formed by a F1 generation fly include GS, Gs, gS, gs.

Punnett square for the cross between two F1 generation flies is as follows:

From the Punnett square, the proportion of F2 progeny expected to be heterozygous for the gene controlling body color and homozygous for the gene controlling wing shape is 4/16 (GgSS, Ggss).

Number of flies with GgSS genotype = 2

Number of flies with Ggss genotype = 2

So, the correct option is option E, 4/16.

3. From the Punnett square, the number of F2 flies with red body and straight wings is 3 (ggSs, ggSS).

Number of flies with ggSs genotype = 2

Number of flies with ggSS genotype = 1

Therefore, the probability that a red-bodied, straight-winged F2 fly will be pure-breeding (ggSS) for both the traits is therefore 1/3.

So, the correct option is option A, 1/3.

GS Gs gS gs GS GGSS GGSs GgSS GgSs Gs GGSs GGss GgSs Ggss gS GgSS GgSs ggSS ggSs gs GgSs Ggss ggSs ggss

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