The results below were obtained in a three-point test cross. Determine the order of the genes on the chromosomes and the 3 different distances between them.

Question

v w z 592

v + + 61

+ w + 12

v + z 9

+ + z 110

+ + + 616

+ w z 58

v w + 105

Answer 1

Brief explanation:

*In the prompt given, we can assume that vwz and +++ are parental classes (genotype) since they were most abundant in the progeny population (592 and 616, respectively).

*We can also observe that there are single-crossover (v++, +wz, vw+, and ++z) and double-crossover (v+z and +w+) events occurring based on parental classes above.

1) To calculate the distance between v and w, select the progeny genotypes that display appropriate crossover events between v and w --> v++ (its pair +wz) and v+z (and its pair +w+).

*genotypes excluded are vw+ and its pair ++z since they do not display crossover between v and w.

Next, add all counts of v++ (61), +wz (58), v+z (9), and +w+ (12) and divide by total population (1,563) = 0.08957 Morgan (or Map unit).

2) Similarly, we select genotypes that display crossover events between w and z --> vw+ (and its pair ++z) and v+z (and its pair +w+).

*We exclude the genotypes v++ and +wz for reasons indicated above.

All counts of vw+ (105), ++z (110), v+z (9), and +w+ (12) are added and divided by total population (1,563) = 0.15099 Morgan (or Map unit).

3) Last, to calculate the distance between v and z, we add all possible crossover events from 1 (140) and 2 (236) and divide by the total (1,563) = 0.24056 Morgan (or Map units).

*We include the double crossover events ( v+z and +w+) twice in the calculation (from 1 and 2) since they represent two recombination events.

In conclusion, the distances between vwz are v--0.09--w--0.15--z, with v--0.24--z.