Question

Two pure lines differ in 6 characters (1-6) whose pattern of inheritance seems to be Mendelian (i.e. they are determined by a single gene). Two individuals, one from each line, are crossed giving rise to progeny M1m1 M2m2 M3m3 M4m4 M5m5 and M6m6, where M and m denote the alleles of the parental lines and the relationship of dominance (M being dominant over m). A hybrid individual of the F1 is then used in a test cross. The recombination frequencies of the co-segregating characters are:

	M1	M2	M3	M4	M5	M6
m1	-	0.11	0.31	0.51	0.49	0.53
m2		-	0.22	0.46	0.52	0.48
m3			-	0.25	0.51	0.50
m4				-	0.49	0.52
m5					-	0.33
m6						-

Work out the genetic map for the six markers under study. Hint: because of experimental error, use 0.45 instead 0.5 as the threshold frequency to deem two loci as being part of the same linkage group, that is part of the same chromosome.

Answer 1

The markers present in the same chromosome or in different chromosomes but which are far apaert will have a 50:50 probability of co-segregation.
The following inference is drawn from the table:

• Locus 1 is close to locus 2.So they a part of same linkage group.
• Locus 1 is less closer to locus 3 as compared to locus 2.
• Locus 2 is closer to locus 3 as compared to locus 1.
• The order must be: locus 1-Locus 2- Locus 3
• Locus 3 is linked to locus 4.Locus 4 must be present with loci 1,2 and 3.
• Locus 4 is far from loci 1 and 2.
• Locus 5 and 6 are of the same linkage group.The linkage group encompassing loci 1,2,3 and 4 is different from the linkage group of 5 and 6.

So loci 1,2,3 and 4 are present on one chromosome in the order 1-2-3-4.
Loci 5 and 6 are present on a different chromosome in the order 5-6.