Question

The data below is from a series of two-point crosses.
Use the recombination frequencies to construct a genetic map of all three genes.

cv – ct 34%

cv – vg 50%

vg – ct 30%

a) Draw a genetic map including all three genes and the distance between them.

b) A fourth gene, g, is found to have 12% recombination when tested with cv. Next you plan to test g and ct . Predict the outcome of the cross between g and ct.

Answer 1

Recombination frequency in percentage is the distance between genes in cM.

maximum recombination frequency between genes is 50%, then the genes are not linked, can be far apart to each other on one chromosome or can be on different chromosomes.

so the order of genes is cv-ct-vg

the map is

b) recombination frequency between cv and g=12%

distance between cv and g= 12 cM

g can be on either side on the cv.

if g is on the same side as that of ct.

distance from g to ct= distance between cv and ct- distance between cv and g

=34-12

=22 cM

if g is on the same side as that of ct, cross between g and ct would yield 22% recombinant progenies and 88% parental type progenies, recombination frequency is the percentage of recombinant progenies.

If g is on the other side

distance from g to ct= distance between cv and ct+ distance between cv and g

=34+12

=46 cM

If g is on the other side,cross between g and ct would yield 46% recombinant progenies and 54% parental type progenies, recombination frequency is the percentage of recombinant progenies.