You cross true-breeding melonberries with the genotype XX; YY; zz and true-breed

Question

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

Explanation / Answer

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

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