Question

You performed this test cross and obtained these numbers of progeny gamete as follows:

Trihybrid Tester

v+/v. b / b+. p / p+ (♀) x v / v. b / b. p / p (♂)

Progeny

V b+ p+ 580

V+ b p 592

v b p+ 45

V+ b+ p 40

v b p 89

V+ b+ p+ 94

v b+ p 3

v+ b p+ 5

Total 1448

Compute the map distances between the loci and draw a linkage map from this information

Answer 1

Answer-

According to the given question-

Here we have to make a test cross between female of F1 generation and male having all recessive character.

v+ / v. b / b+. p / p+ (♀) x v / v. b / b. p / p (♂)

After cross we get following offspring-

V b+ p+ = 580

V+ b p = 592

v b p+ = 45

V+ b+ p = 40

v b p = 89

V+ b+ p+ = 94

v b+ p = 3

v+ b p+ = 5

Now we have to find the gene present in the middle-

Total number of offspring = 1448.

Here V b+ p+ and V+ b p are parental while v b+ p and v+ b p+ are double cross over.

First arranged the given number of offspring in decreasing order.

V+ b p = 592

V b+ p+ = 580

V+ b+ p+ = 94

v b p = 89

v b p+ = 45

V+ b+ p = 40

v+ b p+ = 5

v b+ p = 3

Genes are present as follow-

Now we have to calculate the map distance-

Distance between V and b = 94 + 89 + 5 + 3 = 191 / 1448 = 0.131 = 13.1 cM

Distance between b and p = 45 + 40 + 5 + 3 = 93 / 1448 = 0.064 = 6.4 cM.