Question

In: Biology

In Drosophila melanogaster, black body (b) is recessive to gray body (b+), purple eyes (pr) are...

In Drosophila melanogaster, black body (b) is recessive to gray body (b+), purple eyes (pr) are recessive to red eyes (pr+), and vestigial wings (vg) are recessive to normal wings (vg+). These three genes are linked with pr located between b and vg at a genetic distance of 6 cM from b and 13 cM from vg. Previous experiments have shown that the interference associated with these three genes is 0.5. A fly with a black body, purple eyes and vestigial wings is crossed with a fly with a gray body, red eyes and normal wings. All of their F1 progeny is wild-type. The F1 females are mated with male flies that have black bodies, purple eyes and vestigial wings. Ten thousand progeny are produced from this testcross. How many of the testcross progeny are expected to have purple eyes but be wild-type for body color and wing type?

A. 19.5 B. 39 C. 78 D. 280.5 E. 561

Solutions

Expert Solution

Interference = 1- Coefficient of coincidence

So Coefficient of coincidence = 1-interference

=1-0.5

0.5

Coefficient of coincidence = 0.5

Coefficient of coincidence is given by

The observed frequency of double recombinant/ expected frequency of double recombinant

Map distance between color and body type gene is 10cM

Map distance between body type and disposition is 14cM

expected frequency of double recombinant is given by

Map distance between color and body type /100 * map distance body type and disposition/100

6/100 * 13/100 = 0.0140

Observed frequency of double recombinant we need to find out

Coefficient of coincidence = the observed frequency of double recombinant/ expected frequency of double recombinant

0.5 = X /0.0078

X = 0.0039

The observed frequency of double recombinant = sum of the frequency of both the double recombination

Since nothing is given about the progeny numbers so we will take the double recombinant frequency as half of the total double recombinants that we got.

So the frequency will be 0.0039/2 = 0.00195

black body, purple eyes and vestigial wings and gray body, red eyes and normal wings are the parent type.

purple eyes black body normal wing is the phenotype.

Gene arrangement in wild-type is

b pr vg

b+ pr+ vg+

b pr+ vg

Since the purple gene is the middle gene so it can only be translocated to another chromosome if has undergone double crossover.

So the frequency of a purple eyes black body normal wing is the phenotype. will be 0.00195

Given that the total number of progenies are 10000

so purple eyes black body normal wing will be 10000*0.00195

19.5

So the answer is option A.

gladiator answered 1 year ago

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