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Question: You have two pure-breeding lines of fruit flies. One line (A) has stubby wings and cannot fly and the other line (B) has normal wings. – Free Chegg Question Answer

Transcribed text From Image: You have two pure-breeding lines of fruit flies. One line (A) has stubby wings and cannot fly and the other line (B) has normal wings. You also know the genotypes of four different molecular markers (M1-M4) that differ between the two strains. You cross flies from line A to flies from line B and all of the offspring have normal wings. You then cross the F1 flies back to flies from line A. You observe the wing phenotypes of the F2 generation and determine the genotypes at the 4 markers. The data is presented in the attached table Which of the following can you conclude from the data?
M2 B/B b/b b/B b/B B/B b/B B/B B/B b/B b/B b/B B/B b/B B/B M3 M4 D/D d/d d/D d/D D/D d/D D/D D/D d/D D/D d/D D/D d/D D/D Phenotype M1 Stubb Normal CIC CIC CIC Al A Line A Line B a/a F2 fly 1Normal F2 fly 2Normal F2 fly 3Stubb F2 fly 4 Normal F2 flv 5 Stubb F2 flv 6Stubb A/A A/A CIC C/C A/A A/A Normal F2 flv 8Stubb F2 fly 9Normal F2 fly 10 Stubb F2 fly 11 Normal F2 fly 12 Stubb A/A A/A C/C C/C
The gene that determines wing stubbiness in line A is most closely linked to M1 True False Reset Selection
The recombination frequency between the gene that determines wing stubbiness in line A and M3 is 33%. True False
The gene that determines wing stubbiness in line A is most closely linked to M4 True False
If the F1 flies were crossed back to flies from line B, we would get similar results to the data above. True False

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Answer:

Answer


1. False, F2 fly shoes Normal wings even with A/A genotype

2. True, 2 recombinants out of 6 Stubby in M3 = 2/6 = 0.33

3. True, All D/D are stubby

4. False, They would be more biased towards normal wings


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