Question

Females of genotype AaBbCc were crossed with males of genotype aabbcc.This resulted in 1000 progeny of the following phenotypes:

A B C	3
A B c	365
A b C	84
A b c	44
a B C	65
a B c	98
a b C	340
a b c	1

a) Calculate the distance between genes A and B.

b) Calculate the distance between genes B and C.

c) Calculate the observed frequency of double crossovers.

d) Calculate the expected frequency of double crossovers.

e) Draw the map of these loci.

Answer 1

Answer-

According to the given question-

Here we have a female which are heterozygous , having genotype AaBbCc was crossed with homozygous recessive male having genotype AaBbCc then we get total 1000 offspring out of which-

ABc = 365

abC = 340

aBc = 98

AbC = 84

aBC = 65

Abc = 44

ABC = 3

abc= 1

Total number of offspring = 1000

Here parental are ABc = 365 and abC = 340 while double cross over are ABC = 3 and abc = 1.

Total number of observed double cross over = 3+ 1= 4.

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

Distance between gene A and Gene B = 98+84+3+1= 186 = 186/1000 = 0.186 or 18.6cM.

Distance between gene B and gene C = 65+44+3+1= 113 = 113/1000= 0.113 or 11.3 cM

Distance between gene A and Gene C = 18.6 + 11.3 = 29.9.

Observed double cross over= 3+1 = 4 out of 1000 offspring, so frequency = 4/1000= 0.004.

Expected double cross over= frequency of cross over between A and B × frequency of cross over between B and C × number of total offspring.

Expected double cross over= 0.186 × 0.113 × 1000. = 21.018 or 21.

So number of expected double cross over = 21 out of 1000 so frequency = 21/1000= 0.021.

So the observed double cross over frequency is 0.004 and expected double cross over frequency is 0.021.

So distance between gene A and B = 18.6 cM

Distance between gene B and gene C = 11.3 cM