In corn, the mutant alleles; br (brachytic), an(anther ear), and f(fine stripe)

Question

In corn, the mutant alleles; br (brachytic), an(anther ear), and f(fine stripe) are all recessive to their wild type alleles br+, an+, f+. Corn lines were bred and crossed to produce corn that was heterozygous for all three traits. A farmer then fertilized a heterozygous corn with homozygous recessive pollen. He planted the seeds and observed the progeny.

wildtype                    brbr+anan+ff+              399

fine striped                  brbr+anan+ff                  21

brachytic                      brbranan+ff+                 2

brachytic and               brbranan+ff                   88

fine striped

anther ear                    brbr+ananff+                55

anther ear and brbr+ananff 2

fine striped

brachytic and               brbrananff+                  17

anther ear

brachytic, anther          brbrananff                    355

ear, fine stripped                                                

a. Are the genes linked, clearly explain your reasoning. What hypothesis would you test, and calculate expected numbers only?

b. Determine the sequence of the genes.

c. Determine the map distance between the genes.

d. Determine the interference, I.

Explanation / Answer

a. All 3 genes are linked to each other.

Hypothesis: all 3 genes are not linked i.e. the recombination frequency for each pair of genes is more than 0.5.

Recombination frequency for=( double cross overs + single cross overs)/total no. Of progeny

For anbr, = (2+2+55+88)/939 =0.15

For brf, = (2+2+17+18)/939 = 0.04

For anf, = (2+2+17+55+18+88)/939 = 0.2

All are less than 0.5 and hence are linked genes.

B.sequence of the genes = anbrff

Because the genotypes with the lowest numbers(2 and 2) are the double cross over i.e. brbranan+ff+ and brbr+ananff. The gene which has altered its position with respect to the parental type is be and hence it should be lying between the other two genes.

C. Map distance = %recombination = recombination frequency *100

For anbr, =0.15*100= 15cM

For brf, = 0.04*100= 4cM

D. Interference = 1 - coefficient of coincidence

= 1 - (observed double cross overs/ expected double cross over a)

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