1 In dolphins gray G is dominant to black g Sleek S is domin

1. In dolphins, gray (G) is dominant to black (g), Sleek (S) is dominant to pudgy (s) and friendly (F) is dominant to mean (f). All loci are autosomal. Pure-breeding black, sleek , mean dolphins were mated to pure-breeding gray, pudgy, friendly dolphins. The F1 were testcrossed. Based on the map below and a coefficient of coincidence of 0.8, determine the frequency of each phenotype in the testcross progeny. Round each answer to 5 decimal digits.

ROUND EACH ANSWER TO 5 DECIMAL PLACES!

(a) What is the frequency of gray, sleek testcross progeny?

(b) What is the frequency of pudgy, mean testcross progeny?

(c) What is the frequency of gray, friendly testcross progeny?

(d) What is the frequency of pudgy, friendly testcross progeny?

(e) What is the frequency of black, friendly testcross progeny?

(f) What is the frequency of black, pudgy, mean testcross progeny?

(g) What is the frequency of gray, sleek, mean testcross progeny?

(h) What is the frequency of gray, pudgy, friendly testcross progeny?

I have asked this question 3 times already and nobody on this website seems to know the correct answer or how to go about doing this reverse mapping problem.

This is a reverse mapping questions, there is no need for total amount given.

Solution

The recombination between three genes results in the formation of different combination of genes in different frequencies viz, parental combination, single recombinant combination between first and second gene, single recombinant between second and third gene and double recombinants.

The distance between the three genes are

Colour gene and body type gene = 16 cM (centiMorgen).

Colour gene and Disposition gene = 16 cM + 24 cM = 40 cM.

Body type gene and Disposition gene = 24 cM.

The gametic frequencies of F1 geneation with gSF/Gsf genotype will be as under

gSF= 1/2 (0.84) (0.76) = 0.3192

Gsf = 1/2 (0.84) (0.76) = 0.3192

gsf = 1/2 (0.16) (0.76) = 0.0608

GSF = 1/2 (0.16) (0.76) = 0.0608

gSf = 1/2 (0.84) (0.24) = 0.1008

GsF = 1/2 (0.84) (0.24) = 0.1008

gsF = 1/2 (0.16) (0.24) = 0.0192

GSf = 1/2 (0.16) (0.24) = 0.0192

Here the coefficient of coincidence is 0.8, that means 80% of the expected double recombinants are formed and 20% shows interference.

Therefore double recombinants ( gsF and GSF) will show the frequency of 0.0192 x 0.8 = 0.01536.

The F1 progeny will form gametes with different gene combination depending upon the distance between thase genes.

a) Here gray, sleek combination will be present in 7.61600% gametes and when test crossed give rise to 7.61600% test cross progeny as gray and sleek.

b) Here pudgy, mean combination will be present in 38% gametes and when test crossed give rise to 38% test cross progeny as pudgy, mean.

c) Here gray, friendly combination will be present in 16.160% gametes and when test crossed give rise to 16.160% test cross progeny as gray, friendly.

d) Here pudgy, friendly combination will be present in 11.6160% gametes and when test crossed give rise to 11.6160% test cross progeny as pudgy, friendly.