61. The RFLP pattern observed for two pure parental lines (P1 and P2) and their F1 progeny is represented below. Further, the P1 plant had red flowers while the P2 had white flowers. The F 1progeny was backcrossed to P(2)The result Obtained, showing the number of progeny with red and white flowers and their RFLP patterns is also represented below.
Which one of the following conclusions made is correct?
(1) The DNA marker and the gene for the flower colour are 10 cM apart.
(2) The marker and the gene for the flower colour are 5 CM apart.
(3) The marker and the phenotype are independently assorting.
(4) The marker and the gene for the colour segregate from one another.
Answer: The DNA marker and the flower-color gene are 10 cM apart, not independently assorting. The correct option is (1).
Problem setup
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Two pure parental lines: P1 (red flowers) and P2 (white flowers). The F1 is backcrossed to P2 (testcross). The RFLP patterns shown indicate two parental marker bands and recombinant patterns in progeny. The progeny counts shown are 45 red, 5 white with one marker pattern, and 45 red, 5 white with the alternate marker pattern, totaling 100 progeny. This yields 10 recombinants out of 100, giving a recombination frequency of 10%, which corresponds to 10 cM genetic distance between the marker and the color locus. Testcross logic linking recombination frequency to map distance is standard in classical genetics texts.
Why option (1) is correct
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Recombination frequency = number of recombinant progeny / total progeny. The data show 5 + 5 = 10 recombinants among 100, so RF = 10/100 = 10%. In mapping, 1% recombination ≈ 1 centiMorgan, so distance = 10 cM between the RFLP marker and the flower color gene.
Why others are incorrect
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Option (2): 5 cM would require 5% recombination, i.e., 5 recombinants in 100, but there are 10 total recombinants (5 in each reciprocal class). Therefore 5 cM is not supported by the counts.
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Option (3): “Independently assorting” implies unlinked loci with 50% recombination, which would yield about a 1:1:1:1 distribution in a testcross, not the strong excess of parental types (45 + 45) over recombinants (5 + 5). Thus the marker and trait are linked, not independently assorting.
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Option (4): “Segregate from one another” is another way of saying unlinked/independent assortment, which again contradicts the parental-excess and 10% recombination observed.
Quick method
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In a backcross/testcross, excess parental classes over recombinants indicates linkage. Sum the two smaller classes to get recombinants, divide by total, convert percent to cM (10 recombinants/100 total = 10 cM). This mapping convention is widely used in linkage analysis.
