Question

2. Recombination frequency is given below for several gene pairs. Create a linkage map for these genes, showing the map unit distance between loci. j, k 12%; k, 1 6%; j, m 9%; 1, m 15% 3. In guinea pigs, black (B) is dominant to brown (b), and solid color (S) is dominant to spotted (s). A heterozygous black, solid-colored pig is mated with a brown, spotted pig. The total offspring for several litters are black solid = 16, black spotted = 5, brown solid = 5, and brown spotted = 14. a. Are these genes linked or nonlinked? b. If they are linked, how many map units are they apart?

Can you explain how to formulate map units in these types of questions?

I have a answered but i would like help with part B.

Answer 1

3) recombination frequency in percentage is the distance between genes in map units.

recombination frequency between the genes are given.

distance between j and k=12 cM, distance between k and l=6 cM, distance between j and m=9 cM, distance between l and m=15 cM

so l is in between J and k and the m is on the other side of the j, so

j to k is 12 cM, and j to l and l to k=6 cM and l to m = m to j + j to l=9+6=15 cM

2) BbSs * bbss

9 см 6 см 6 см

	BS	Bs	bS	bs
bs	BbSs ( black, solid)	Bbss ( black, spotted)	bbSs (brown,solid)	bbss ( brown, spotted)

if the genes are not linked the expected ratio is

black, solid: black, spotted: brown, sold:brown,spotted=1:1:1:1

here the oberved ratio is

black, solid: black, spotted: brown, sold:brown,spotted =16:5:5:14=3.2:1:1:2.8

the ratio differs considerably from the expected ratio so the genes are linked.

b) maximum recombination frequency=50% then the genes are not linked, if the genes are linked recombination frequency is less 50%, recombination frequency is the percentage of recombinant progenies.

here the number of black, spotted and brown, solid is less than the number of other 2 phenotypes so the recombinant progenies are black, spotted and brown, solid

recombination frequency= (number of recombinant progenies/total number of progenies)* 100

= (5+5)/(40)

= 25 %

distance between the genes=25 cM