In maize, the genes Tu, j2 and gl3 are loacted on chromosome 4 at map positions

Question

In maize, the genes Tu, j2 and gl3 are loacted on chromosome 4 at map positions 101, 106 and 112. If plants homozygous for the recessive alleles of these three genes are crossed with plants homozygous for the dominant alleles, and the F1 plants are crossed to triply recessive plants, what genotypes would you expect and in what proportions?

My question is, how did they get 0.893, 0.047, 0.057, and 0.003???? Please explain in detail. I know all of these numbers add up to be 1, I just don't know how they got them.

Also I had someone answer this yesterday and told me how they got .5 and .6. I need to know how they got 0.893, 0.047, 0.057 and 0.003.

Explanation / Answer

If you look at the map position of genes

Tu (101)

J2 (106)

Gl3 (112)

In other words the above number represents the map distance between them i.e.

Tu and J2 are 5 map units apart.

J2 and Gl3 are 6 map units apart.

Tu and Gl3 are 11 map units apart

Map units represent the recombination frequency in percentage or frequency of recombinants.

The values ( 0.893, 0.047, 0.057, and 0.003) are nothing but rembination events in which the 0.893 is parental number, .047 and .057 are recombinants from single cross over and the lowest one .003 is double recombinant.

We know the distance between genes is calculated on the basis of recombination frequency (in percentage), so

0.047 x 100 % = 4.7 map units

0.057 x 100 % = 5.7 map units

But if you look actual distance between the recombinants is 5.0 and 6.0 map units. This inturn implies that a double recombination event is also taking place.

The frequency of double recombinant in percentage = 0.3 %

Add 0.3 % to each single cross over recombinant, hence

4.7 % + 0.3 % = 5.0 % or 5.0 map units

5.7 % + 0.3 % = 6.0 % or 6.0 map units

Therefore, parental number = 100 - [4.7 + 5.7 + .3] = 89.3 % or 0.893

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