Question

Problem 2

You cross a true-breeding yellow-bodied, smooth-winged female fly with a true-breeding red-bodied, crinkle-winged male. The red body phenotype is dominant to the yellow body phenotype and smooth wings are dominant to crinkled wings. Use B or b for body color alleles, and W or w for wing surface alleles.

a) What are the genotypes of the P generation flies?

b) What will be the genotype(s) and phenotype(s) of the F1 offspring?

c) You discover that the genes for body color and wing surface are linked. You perform a dihybrid test crossbetween the F1 flies from part (b) with a true-breeding yellow-bodied, crinkle-winged fly. Use the followingF2 results to determine the recombination frequency (%) between the body color and wing surface genes. (Remember that the recombinants are the ones that do not resemble the parental types from the P generation.)

Body Color	Wing Surface	# of Individuals
red	smooth	102
yellow	smooth	404
red	crinkled	396
yellow	crinkled	98

You decide to turn your attention to a different gene, one that controls wing length. This gene has two alleles, "L orl" where long wings are dominant to short wings. Remember that the red body phenotype is dominant to the yellow body phenotype. You again mate two true-breeding flies:

P: red-bodied, short wing male X yellow-bodied, long wing female

F1: All red-bodied, long wing

d) You perform a test cross between the F1 flies above with true-breeding yellow-bodied, short-winged flies. You get the following F2 results. What is the recombination frequency (%) between the genes for body color and winglength?

Body Color	Wing Length	# of Individuals
red	long	45
red	short	460
yellow	long	440
yellow	short	55

e) Based on the information in (c) and (d), what are the two possible arrangements of these three genes: bodycolor, wing surface and wing length? Draw two linkage maps to show the possible arrangements of thesegenes and the map distance between genes.

Answer 1

PROBLEM 2 :-

Answer A :-

The parental phenotypes shows male parent as red body and crinkle wings (BBww) and the female parent as yellow body and smooth wings (bbWW)

Answer B :-

All the F1 individuals will be heterozygous and will show genotype 'BbWw' indicating red body and smooth wings.

Answer C :-

Recombinant frequency = Number of recombinants / Total progeny

The recombinants in this case will be red body, smooth winged and yellow body, crinkle winged (102 + 98 = 200 )

Total progeny is 1000.

So, RF = 200 / 1000

= 0.2 or 20%

Answer D :-

The phenotypes that will be considered recombinants have red body, long wings and yellow body, short wings.

Recombination frequency = Number of recombinants / Total progeny

= 100 / 1000

= 0.1 or 10%

Answer E :-

The gene for body colour is represented as 'B', the gene for Wing surface is represented as 'W' and the gene for Wing size is represented as 'L'.

Recombination frequency between B and W is 20% and that between B and L is 10%

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