Question

Three linked traits in garden pea plants are each controlled by two alleles. Yellow pods are recessive to green pods, bluish green seedlings are recessive to green seedlings, and creeper is recessive to normal. A plant heterozygous for all three traits was crossed to a creeper plant with yellow pods and bluish green seedlings. The following F2 results were obtained: 171 green pods, bluish green seedlings, creeper 307 yellow pods, green seedlings, creeper 13 green pods, green seedlings, normal 2067 green pods, green seedlings, creeper 165 yellow pods, green seedlings, normal 10 yellow pods, bluish green seedlings, creeper 2033 yellow pods, bluish green seedlings, normal 316 green pods, bluish green seedlings, normal a. Determine the allele arrangement in the heterozygous parent. b. Determine the gene order of these three linked genes. c. Construct a linkage map showing the map distances between the three genes. d. Calculate an interference value.

Answer 1

a). Heterozygous parent allele arrangement = YcB/yCb

b). Order of genes= ycb

c). y----------12.71cM----------c-------7.06 cM--------b

d). Interference = 0.45

Explanation:

Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.

Hence, the parental (non-recombinant) genotype is YBc / ybC.

Yellow pods (y) are recessive to green pods(Y), bluish green(b) seedlings are recessive to green seedlings(B), and creeper(c) is recessive to normal(C)

171 green pods, bluish green seedlings, creeper --Ybc

307 yellow pods, green seedlings, creeper---yBc

13 green pods, green seedlings, normal—YBC

2067 green pods, green seedlings, creeper—YBc

165 yellow pods, green seedlings, normal—yBC

10 yellow pods, bluish green seedlings, creeper—ybc

2033 yellow pods, bluish green seedlings, normal—ybC

316 green pods, bluish green seedlings, normal--YbC

1).

If single crossover occurs between Y & B..

Normal combination: YB / yb

After crossover: Yb/yB

Yb progeny= 171+316=487

yB progeny = 307+165=472

Total this progeny = 959

The recombination frequency between Y&B = (number of recombinants/Total progeny) 100

RF = (959/5082)100 = 18.87%

2).

If single crossover occurs between B & c..

Normal combination: Bc / bC

After crossover: BC/bc

BC progeny= 13+165=178

bc progeny = 10+171=181

Total this progeny = 359

The recombination frequency between B&c = (number of recombinants/Total progeny) 100

RF = (359/5082)100 = 7.06%

3).

If single crossover occurs between Y & c..

Normal combination: Yc / yC

After crossover: YC/yc

YC progeny= 13+316=329

yc progeny = 10+307=317

Total this progeny = 646

The recombination frequency between Y&c = (number of recombinants/Total progeny) 100

RF = (646/5082)100 = 12.71%

Recombination frequency (%) = Distance between the genes (cM)

y----------12.71cM----------c-------7.06 cM--------b

Expected double crossover frequency = (RF between h & w) * (RF between w & b)

= 12.71% * % 7.06= 0.009

The observed double crossover frequency = 13+10 /5082 = 0.005

Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency

= 0.005 / 0.009

= 0.55

Interference = 1-COC

= 1-0.55 = 0.45