In corn (Zea mays) a cross is made between a homozygous red kernel and a homoygo

Question

In corn (Zea mays) a cross is made between a homozygous red kernel and a homoygous white e kernel strain. The F1 individuals all have white kernels only, and the F2 express both phenotypes in a distriution of 1285 white kernel: 303 red kernel.

a. What ratio is best represented by the outcome in the F2? What type of epistatic interaction does this ratio represent?

b. Using the 3 step process, what is the likely genotype of the F1 parents?

c. How many genes are segregating? In other words, how many genes are directly responsible for the observed ratio?

Explanation / Answer

The phenotypic ration of the white kernel to red kernel is 1285:303

this is approximately 3:1 . This ratio is for monohybrid cross

in F2 generation the ration will become 9:3:3:1 i.e it will become a dihybrid cross.

In context with epeistatis the allele present on one locus afects the gene present at the other locus so the ratio which comes in case of independent assortment as 9:3:3:1 will become 9:3:4 of 9:7 in case of epistasis becomes genes gets corelated

answer b:- genotype of f1 parents with be a hetrozygous red and white kernel foer example if white is represented by W and red by R then the genotype of F1 will be Rr and Ww

answer c:- two genes are segregrating four genes R,r,W,w are responsible

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