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.

Explanation / Answer

if the position of the genes are 101,106, 112 then the distancein map unit between the genes would be,

Tu and j2 = 106-101 = 5 mu, likewise,

j2 and gl3 = 112 - 106 = 6 mu

Also,

parental genotype = Tu+----j2+----gl3+ X Tu-----j2-----gl3   

progeny genotype (single crossover)=

Tu+ -----j2-----gl3, = result of cross over between Tu and j2 , frequency would be = 5/100 = 0.05

Tu----j2+----gl3+ , = result of cross over between Tu and j2 , frequency would be = 0.05

Tu+----j2+----gl3 , = result of cross over between Tu and j2 , frequency would be j2 and gl3 = 6/100 = 0.06

Tu-----j2-----gl3+ = result of cross over between Tu and j2 , frequency would be j2 and gl3 = 0.06

The double cross ove results in the genotype same as parental, expected frequency would be = the product of frequency of single crossover i.e. 0.05 x 0.06 = 0.003

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