Question

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 10kb Probe А 5kb -UI III TIT II Probe B 11kb 7kb — color pink white red pink red pink pink white pink white red pink red white pink pink red pink white pink red pink c. probes A and B: list the animals that show recombination between these markers and then calculate CM distance between them. Which other rabbits have genotypes that could only have been generated by recombination between A and B? Count these and then calculate the distance in cM by computing the ratio #recombinants/# meiosis d. Color and probe B: list the animals that show recombination between the Color gene W and marker B. Calculate the CM distance between W and B (Hint: look only at Color (W/w) and B probe RFLP patterns - ignore RFLP with probe A)

Answer 1

ANSWER C :-

From the above mentioned data, it is clear that the A gene will be having two alleles namely (A¹⁰) and (A⁵) and the B gene will be having two alleles namely (B¹¹) and (B⁷). The homozygous combinations will be (A¹⁰B⁷) and (A⁵B¹¹) whereas combinations apart from this would be considered as recombination events.

In case of chromosome number 2, it can be seen that the genotypes are (A⁵B¹¹) and (A⁵B⁷) respectively. The second combination is a recombinant and hence this progeny can be considered to be an output of recombination and in this manner individuals (progeny) 2, 3, 4, 9, 10, 17, 18 and 21 are recombinants corresponding to a total of 8 individuals.

Recombination frequency = Number of recombinants / Total number of individuals (progeny)

= 8/22 x 100

= 0.3636 x 100

= 0.36 or 36.36 %

= 36.36 cM

= 36cM

The formula indicates the progeny although in the formula listed above, there is indication of denominator in terms of number of meiosis which accounts to 44. In such a case, the distance will be calculated to be (8/44 x 100 = 18.18% or 18 cM).

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