In peas, the genes a, b, c, d, e and f are linked on chromosome V and in the order given and the map distances are RF(a-b)=7 cM, RF(b-c) = 17 cM, RF(c-d) = 12 cM, RF (d-e) = 14 cM RF(e-f)= 10 cM. You cross a pure-breeding strain carrying wild-type alleles at all loci to a strain that is homozygous mutant for all six genes and cross the F1 a b c d e f/+ + + + + + plants to the a b c d e f tester line. To your surprise, the measured map distances turn out to be RF(a-b)=5 cM, RF (b-c) = 1 cM, RF (c-d) = 0.5 cM, RF (d-e) = 8 cM, RF (e-f) = 10cM. In aIDition, the F1 plants showed reduced fertility. a.(1 point) Explain these results being as specific as possible. b.(1/2 point) What is the likely origin of the recombinants in the b-c and c-d intervals?
