Question

2. (6) An experiment was done to determine the linkage relationship of three genes in Arabidopsis, one affecting leaf shape, another flower col our and one affecting sepals. Homozygous plants that have smooth leaves, yellow flowers and fused sepals are were crossed with homozygous plants that have serrated leaves, white flowers and unfused sepals. The FI plants resulting from this cross were all phenotypically serrated, yellow and unfused. These were test-crossed with homozygous plants that were phenotypically smooth, white and fused to give the following F2 phenotypes and numbers. Petals Sepals Total Serrated Serrated Serrated Serrated Smooth Smooth Smooth Smooth White White Yellow Yellow White White Yellow Yellow Fused Unfused Fused Unfused Fused Unfused Fused Unfused 17 535 58 56 507 23 Based on this information, draw a map indicating the position ofthe loci on the chromosome relative to each other and the distance between them based on recombination frequencies.

Answer 1

Answer:

Chromosome map = s--------9.68mu------y----3.51mu----f

Trihybrid = SyF / sYf

The order of genes = syf

Explanation:

2. (6) An experiment was done to determine the linkage relationship of three genes in Arabidopsis, one affecting leaf shape, another flower col our and one affecting sepals. Homozygous plants that have smooth leaves, yellow flowers and fused sepals are were crossed with homozygous plants that have serrated leaves, white flowers and unfused sepals. The FI plants resulting from this cross were all phenotypically serrated, yellow and unfused. These were test-crossed with homozygous plants that were phenotypically smooth, white and fused to give the following F2 phenotypes and numbers. Petals Sepals Total Serrated Serrated Serrated Serrated Smooth Smooth Smooth Smooth White White Yellow Yellow White White Yellow Yellow Fused (Syf) Unfused (SyF) Fused (SYf) Unfused (SYF) Fused (syf) Unfused (syF) Fused (sYf) Unfused (sYF) 17 535 58 56 507 23 Based on this information, draw a map indicating the position ofthe loci on the chromosome relative to each other and the distance between them based on recombination frequencies.

Smooth = s

Serrated = S

Yellow = Y

White = y

Unfused = F

Fused = f

Hint: Non-recombinant (parental) combinations are produced more in number than recombinants. The parental (non-recombinant) genotypes is SyF / sYf

1).

If single crossover occurs between s&y

Normal combination: Sy/sY

After crossover: SY/sy

SY progeny=0+58 = 58

sy progeny = 2+56 = 58

Total this progeny = 116

Total progeny = 1198

The recombination frequency between s&y = (number of recombinants/Total progeny) 100

RF = (116/1198)100 = 9.68%

2).

If single crossover occurs between y&f

Normal combination: yF/Yf

After crossover: yf/YF

yf progeny= 17+2=19

YF progeny = 23+0=23

Total this progeny = 42

The recombination frequency between y&f = (number of recombinants/Total progeny) 100

RF = (42/1198)100 = 3.51%

3).

If single crossover occurs between s&f

Normal combination: SF/sf

After crossover: Sf/sF

Sf progeny=17+58=75

sF progeny = 23+56 = 79

Total this progeny =154

The recombination frequency between s&f = (number of recombinants/Total progeny) 100

RF = (154/1198)100 = 12.85%

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

s--------9.68mu------y----3.51mu----f