Question

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 0.61, determine the frequency of each phenotype in the testcross progeny. Round each answer to 5 decimal digits.

Color

Body Type

Disposition

6cM

21cM

What is the frequency of gray, sleek testcross progeny?

What is the frequency of pudgy, mean testcross progeny?

What is the frequency of gray, friendly testcross progeny?

What is the frequency of pudgy, friendly testcross progeny?

What is the frequency of black, friendly testcross progeny?

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

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

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

Answer 1

So the distance between the genes is:

Color --- 6 cM --- Body type --- 21 cM --- Disposition

Coefficient of coincidence of 0.61 means 61% of expected double crossovers actually occur.

Parents are crossed as given below:

gg SS ff × GG ss FF

All F1 are Gg Ss Ff

F1 testcross cross: Gg Ss Ff × gg ss ff

This gives parental phenotype of: gray, sleek, friendly and black, pudgy, mean.

All other phenotypes are recombinants.

Frequencies:

Gray, Sleek (G,S) :

Recombination frequency = 6%

Hence 100 - 6 = 94% are parental. Half these are gray and sleek and half are black and pudgy. Hence 94/2 = 47% are gray and sleek.

Pudgy, mean (s, f):

Recombination frequency = 21%

Hence 79% are parental. Of these half are sleek and friendly and half are pudgy and mean. Hence 79/2 = 39.5% are pudgy and mean.

Gray, friendly (G, F):

Distance between the two genes = 6 +21 = 27cM

Great, friendly are parental. Hence 73% are parental. 73/2 = 36.5% are gray, friendly.

Pudgy, friendly (s, F):

This would be a recombinant.

Recombination frequency = 21%

Half of recombinants are pudgy friendly, half are sleek, mean. Hence answer is 21/2 = 10.5%

Black, friendly (g, F):

Recombination frequency = 27%

Half are black, friendly and half are gray, mean. Hence answer is 13.5%

Black, pudgy, mean (g, s, f):

This is are parental phenotype.

Recombinants includes single and dinner crossovers.

Expected double crossovers = 0.06 × 0.21 = 0.0126 = 1.26%

But coefficient of coincidence = 0.61

Hence only 61% of the 1.26% double crossovers actually occur. That is 0.61 × 1.26 = 0.7686%

Total recombinants = 6 + 21 + 0.7686 = 27.7686%

Hence parentals = 100 - 27.7686 = 72.2314%

Half of these are black, pudgy, mean and half are gray, sleek, friendly.

Hence answer is 72.2314/2 = 36.1157%

Gray, sleek, mean (G, S, m):

This will include parental at gray, sleek and recombinant and sleek, mean.

Parentals at great sleek = 100-6 = 0.94

Recombinant at sleek, mean = 0.21

0.94 × 0.21 = 0.1974 = 19.74%

Gray, pudgy, friendly (G, s, F):

This requires a double crossover.

As calculated above, double crossovers are 0.7686%

Half are gray, pudgy, friendly and half are black, sleek, mean.

Hence answer is 0.7686/2 = 0.3843%