Question

The female has a wood stick-like body and short antenna, T The male has a leaf-like...

The female has a wood stick-like body and short antenna, T

The male has a leaf-like body and fanned antenna.

All the flies in the F1 generation have a wood stick-like body and a fanned antenna.

Associate the alleles with the appropriate letter, with “B” an “b” for body type and “A” and “a” for antenna type. The uppercases “B” and “A” indicate the dominant allele, the lowercases “b” and “a” the recessive allele.

You next cross F1 females with true-breeding stick body, short antenna males. There are 1600 flies in the F2 generation.

If the body type and antenna type traits were linked and no recombination occurred,

1. what phenotypes and corresponding genotypes would you expect?

2. What would be the number of individual for each phenotype? (1 point)

3. When you count the F2, you really get:87Stick body, fanned antenna ; 714 Stick body, short antenna; 726 Leaf body, fanned antenna; 73 Leaf body, short antenna

What is the genetic distance between the body type and antenna type genes?

You conducted another crossing to determine the distance between the body type and the wing shape. You found that the recombination frequency between these two genes was 4%.

4 Draw all possible positions of the wing shape gene compared to the body type and antenna type on a chromosome?

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Answer #1

BBaa x bbAA

BbAa all f1 are wood type body and fanned antenna

BbAa x BBaa

BBAa, BBaa, BbAa, Bbaa

When it is unlinked

If they are linked then

BA/ba x Ba/Ba

BA/Ba, BA/Ba, ba/Ba, ba/Ba

If they are linked no recombination occurs or only two phenotype were produced that is woodstick and fanned antenna,,, woodstick and short antennae in equal proportion

3.sticky body fanned antennnae=87

Stick body and short antenna =714

Leafy body and fanned antennae=726

Leafy body and short antennae=73

Map distance=no of recombinant prigeny/total offspring

=726+73/1600

=799/1600

=0.49*100

=49.9cM

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