Question

When XxYyZz is test-crossed, out of a total of 718 offspring, we obtain the following numbers: XYZ 280 xyZ 30 XYz 50 XyZ 3 xYz 5 xYZ 42 Xyz 38 xyz 270 Total= 718 A. Which gene is in the middle? (2 points) B. Calculate map distances. Show your work! (4 points) C. Calculate crossover interference. Show your work! (4 points)

Answer 1

Answer: The test cross between XxYyZz and xxyyzz yield 718 offsping which are as follows:

XYZ : 280 (parental genotype, shows no recombination); xyZ : 30 (single cross over between locus Y and Z); XYz : 50 (single cross over between locus Y and Z); XyZ : 3 (double recombinant); xYz : 5 (double recobinant); xYZ : 42 (single cross over between locus X and Y); Xyz : 38 (single cross over between locus X and Y); xyz : 270 (parental genotype, shows no recombination).

The recombination frequency of the parental genotype =%

The recombination frequency of the recombinants, Xyz and xYZ formed from single cross over between locus X and Y =%; and the recombination frequency of the recombinants, XYz and xyZ formed from single cross over between Y and Z =%; recombination frequency of the double recombinant formed from double cross over =%

Distance between gene locus X and gene locus Y= (% single cross over in the region) + (% double cross over in the region)= (1.11+11.14)=12.25 map units; similarly, distance between gene locus Y and Z= (11.14+1.11)=12.25 map units. Hence, distance between gene X and Z= (12.25+12.25)=24.5 map units

A. Therefore gene Y is in the middle.

B. Distance between gene locus X and gene locus Y= (% single cross over in the region) + (% double cross over in the region)= (1.11+11.14)=12.25 map units; similarly, distance between gene locus Y and Z= (11.14+1.11)=12.25 map units. Hence, distance between gene X and Z= (12.25+12.25)=24.5 map units. Hence, the map distance between gene X and Y is 12.25 map units and distance between gene Y and Z is 12.25 map units, while distance between gene X and Z is 24.5 map units.

C. Recombination frequency in cross over between gene X and Y (both single and double cross over)=12.25% and recombination frequency in cross over between gene Y and Z (both single and double cross over)=12.25%.

Expected double cross over = and observed cross over= Therefore coefficient of coincidence, c.o.c=

Crossover interference =