In corn, a strain homozygous for the recessive alleles a (green), d (dwarf) and

Question

In corn, a strain homozygous for the recessive alleles a (green), d (dwarf) and rg (normal leaves) was crossed to a strain homozygous for the dominant alleles of each of these genes, namely A (red), D (tall) and Rg (ragged leaves). Offspring of this cross were then crossed to plants that were green, dwarf and had normal leaves. The following phenotypic classes were observed. Propose a linkage map with distances between the three genes. What is the coefficient of coincidence for these genes? What degree of interference occurred in this experiment?

Explanation / Answer

a.) Steps to find linkage map :-

The genotypes found most frequently are parental genotypes.

So, Here, Parental genotypes are:-

Red tall ragged ADRg/ adrg (265) &

green dwarf normal adrg/ adrg (275)

The genotypes found least frequently are double cross overs (DCO).

So, here, DCOs are :-

red tall normal ADrg/adrg (24) &

green dwarf tagged adRg / adrg (16).

A DCO event moves the middle allele from one sister chromatid to the other. So, the two genes found together in DCO and parental will be placed at extremes in the linkage map and the gene present in a DCO from other chromatid will be present in middle in the linkage map.

Here, A and D are present together ,so at ends and Rg in middle.

Linkage map is-

A------- Rg-------D

Steps to find map distance-

1st, map distance between A and Rg =

= [(no. of SCO between A and Rg + no. of DCO)/ total no. Of progeny] × 100

Here, SCO between A and Rg are-

red dwarf normal Argd/ argd(90) and

green tall ragged aRgD/argd (70)

Total no. of offsprings = sum of all = 1000

So, map distance between A and Rg =

= [{(90+70)+(24+16)}/1000]×100= [(200/1000)×100] =

=20 mu

2nd, map distance between Rg and D

= [(no. of SCO between Rg and D + no. of DCO)/ total no. of progeny]× 100

Here, SCO between Rg and D are-

green tall normal argD/argd (140) &

rer dwarf ragged ARgd/ argd (120)

So,map distance between Rg and D is-

=[{(140+120)+(24+16)}/1000]×100 = [(300/100)×100] =

=30 mu

So, A---(20 mu)-----Rg-----(30mu)-------D.

2.) Coefficient of coincidence (COC) is calculated as-

COC = observed DCO/ expected DCO

and expected DCO = frequency of SCO in 1st region × frequency of SCO in 2nd region × total no. of progeny

Here, expected DCO = 0.20 × 0.30× 1000 = 60

observed DCO = 24+16= 40

So COC = 40/60 =0.667

3.)Since, here COC is between 0& 1, it indicates partial interference.

Interference = 1- COC = 1- 0.667 =0.333.

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