Question

F1 females from a cross of AABBCC to aabbcc were backcrossed to aabbcc males. In the progeny of this cross there were 150 AaBbCc, 145 aabbcc, 55aabbCc, 52 AaBbcc, 31 aaBbcc, 29 AabbCc, 5 Aabbcc and 3 aaBbCc.

Is there evidence of interference? What is the coefficient of coincidence?

Answer 1

Yes, there is evidence of interference and its value is 0.51

The coefficient of coincidence = 0.49

Explanation:

For the sake of understanding we can divide genotype as chromosome wise.

150 AaBbCc = ABC / abc

145 aabbcc = abc/ abc

55 aabbCc = abC / abc

52 AaBbcc = ABc / abc

31 aaBbcc = aBc / abc

29 AabbCc = AbC / abc

5 Aabbcc = Abc / abc

3 aaBbCc = aBC / abc

Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.

Hence, the parental (non-recombinant) genotypes is ABC/abc

1).

If single crossover occurs between A&B..

Normal combination: AB/ab

After crossover: Ab/aB

Ab progeny= 29+5=34

aB progeny = 31+3=34

Total this progeny =68

Total progeny= 470

The recombination frequency between A&B = (number of recombinants/Total progeny) 100

RF = (68/470)100 = 14.47%

2).

If single crossover occurs between B&C..

Normal combination: BC/bc

After crossover: Bc/bC

Bc progeny= 52+31=83

bC progeny = 55+29 = 84

Total this progeny = 167

The recombination frequency between B&C = (number of recombinants/Total progeny) 100

RF = (167/470)100 = 35.53%

3).

If single crossover occurs between A&C..

Normal combination: AC/ac

After crossover: Ac/aC

Ac progeny= 52+5=57

aC progeny = 55+3=58

Total this progeny = 115

The recombination frequency between A&C = (number of recombinants/Total progeny) 100

RF = (115/470)100 = 24.47%

Recombination frequency (%) = Distance between the genes (cM)

B----------14.47cM--------A-----------24.47cM--------------C

Expected double crossover frequency = (RF between B & V) * (RF between V&S)

= 14.47% * 24.47% = 0.035

The observed double crossover frequency = 5+3 / 470 = 0.017

Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency

= 0.017 / 0.035

= 0.49

Interference = 1-COC

= 1- 0.49 = 0.51