You cross truebreeding melonberries with the genotype XX YY

You cross true-breeding melonberries with the genotype XX; YY; zz and true-breeding melonberries with the genotype xx; yy; and ZZ. You then take some of the F1 offspring and perform a testcross. The resulting F2 offspring have the following genotypes in the following numbers:

Xx; Yy; Zz 15

Xx; Yy; zz 1048

Xx; yy; Zz 70

Xx; yy; zz 248

xx; Yy; Zz 254

xx; Yy; zz 76

xx; yy; Zz 1069

xx; yy; zz 10

a. Draw a map showing the three genes. Include the calculated map distance between each pair

b. Calculate the frequency of double recombinants, the coefficient of coincidence, and the interference

Solution

Answer.

Arrange the observed genotypes in decreasing order.

Observed

Genotype

Type of Gamete

1069

xx; yy; Zz

Parental

1048

Xx; Yy; zz

Parental

254

xx; Yy; Zz

Single-crossover between gene wx and cn

248

Xx; yy; zz

Single-crossover between gene wx and cn

76

xx; Yy; zz

Single-crossover between genes b and wx

70

Xx; yy; Zz

Single-crossover between genes b and wx

15

Xx; Yy; Zz

Double-crossover

10

xx; yy; zz

Double-crossover

To determine the gene order, we need the parental genotypes as well as the double crossover genotypes.

From the above table,

parental genotypes: xx; yy; Zz and Xx; Yy; zz

double crossover genotypes: Xx; Yy; Zz and xx; yy; zz

From the first double crossover, the y allele is associated with the x and z alleles. Therefore, y is in the middle, and the gene order is x y z.

Determining the linkage distances:

x - y distance calculation. This distance is derived as follows: 100*((254+248+15+10)/1000) = 52.7 cM

y - z distance calculation. This distance is derived as follows: 100*((76+70+15+10)/1000) = 17.1 cM

Interference:

To measure interference, we first calculate the coefficient of coincidence (c.o.c.) which is the ratio of observed to expected double recombinants.

Interference is then calculated as 1 - c.o.c.

As calculated above, the recombination frequency is 0.527 between genes x and y, and the recombination frequency between y and z is 0.171. Therefore, expected double recombinants will be [100*(0.527 x 0.171)] = 9.0%.

With a sample size of 1000, this would amount to 90 double recombinants. However, we actually detected only 25.

c.o.c. = 25/90 = 0.28

Interference = 1 – coc

= 1- 0.28 = 0.72

Observed	Genotype	Type of Gamete
1069	xx; yy; Zz	Parental
1048	Xx; Yy; zz	Parental
254	xx; Yy; Zz	Single-crossover between gene wx and cn
248	Xx; yy; zz	Single-crossover between gene wx and cn
76	xx; Yy; zz	Single-crossover between genes b and wx
70	Xx; yy; Zz	Single-crossover between genes b and wx
15	Xx; Yy; Zz	Double-crossover
10	xx; yy; zz	Double-crossover