You are mapping three linked genes in tomato plants. The recessive alleles for t

Question

You are mapping three linked genes in tomato plants. The recessive alleles for the three genes (when in the homozygous condition) cause absence of anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j), respectively. Two true-breeding parents were crossed, producing trihybrid (heterozygous) F1s, which were then test-crossed. The 3000 offspring were grouped into the following phenotypic classes (a trait not mentioned in a particular phenotypic class can be considered to be dominant or wild-type for that trait):

Phenotype Number Observed

hairless 249

jointless, hairless 40

jointless 931

wild type for all three 270

anthocyaninless, jointless, hairless 278

anthocyaninless, hairless 941

anthocyaninless 32

anthocyaninless, jointless 259

a) What are the original parental phenotypes?

b) Which of the three genes is in the middle?

c) Calculate the map distance between the three genes

Explanation / Answer

Step 1. Define the phenotype and genotype:

Phenotype

Genotype

anthocyanin

A

anthocyaninless

a

hair

H

hairless

h

joint

J

jointless

j

Step 2. Arrange the phenotypes in ascending order of number observed.

Phenotype

Genotype

Observed

anthocyaninless, hairless

ahJ

941

jointless

AHj

931

anthocyaninless, jointless, hairless

ahj

278

wild type for all three

AHJ

270

anthocyaninless, jointless

aHj

259

hairless

AhJ

249

jointless, hairless

Ahj

40

anthocyaninless          

aHJ

32

Step 3. Define the type of gamete.

a)   The genotypes found most frequently are the parental genotypes. From the table it is clear that the ahJ and AHj genotypes are the parental genotypes.

b)   The double-crossover gametes are always in the lowest frequency. From the table the Ahj and aHJ genotypes are in the lowest frequency.

Phenotype

Genotype

Observed

Type of Gamete

anthocyaninless, hairless

ahJ

941

Parental

jointless

AHj

931

Parental

anthocyaninless, jointless, hairless

ahj

278

Single-crossover between genes A and J

wild type for all three

AHJ

270

Single-crossover between genes A and J

anthocyaninless, jointless

aHj

259

Single-crossover between genes H and J

hairless

AhJ

249

Single-crossover between genes H and J

jointless, hairless

Ahj

40

Double-crossover

anthocyaninless          

aHJ

32

Double-crossover

Step 4. Find out the middle gene.

The next important point is that a double-crossover event moves the middle allele from one sister chromatid to the other. This effectively places the non-parental allele of the middle gene onto a chromosome with the parental alleles of the two flanking genes. We can see from the table that the A gene must be in the middle because the recessive a allele is on the same chromosome as the H and J alleles, and the dominant A allele is on the same chromosome as the recessive h and j alleles.       

Hence, A allele is between H and J allele and the order of the gene is HAJ.

Step 5. Calculate map distance.

Distance between genes H and A = [100*((278+270+40+32)/3000)] = 20.67 cM

Distance between genes A and J = [100*((259+249+40+32)/3000)] = 19.3 cM

Phenotype

Genotype

anthocyanin

A

anthocyaninless

a

hair

H

hairless

h

joint

J

jointless

j

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