Question

In corn, Zea mays, mutant genes for anther ear (an) and fine stripe (f) are carried at linked loci separated by approximately 20 map units. These mutant genes are recessive to their counterpart alleles, an⁺ and f⁺, which control the wild-type expressions of these traits. The gene for colorless aleurone (c) is recessive to its dominant allele for colored aleurone (C) and is found on a chromosome different from that carrying the an and f loci.
If you selfed a phenotypically wild type corn plant with the following genotype:
an⁺ f /an f⁺, Cc

a.) What percentage of the progeny would have anther ear, fine stripe and colored aleurone?

b.) What percentage of the progeny would have anther ear, fine stripe and colorless aleurone if the wild type parent that was selfed had the genotype:
an+ f+ / an f; Cc ?

Answer 1

Data

A- parent genotype an f+/ an+ f ; Cc

Genotypes of gametes considering the linkage between ear and strpe traits

an f+, C

an f+ ,c

an+ f, C

an + f, c

Product of the self cross of this genotypes there is no chance to obtain a progeny with anther ear, fine stripe, colored aleurone = possible genotypes

an f/ an f ; Cc.

an f/ an f ; cc   

To obtain this genotypes, it is necessary that one of the gametes would have the 2 alleles in mutant form of both traits that are segregated in linked way , in the same dna strand .

Answer= the are 0% possibility of obtain a progeny with this phenotype

B- genotype of parent. an+ f+/ anf. ; Cc

Possibles gametes genotypes

an+f+, C

an+f+, c

an f, C

an f, c

From the self cross of this gametes , the % offspring with the following genotype

an f/ an f, cc. is 1/16. = 6.25%

This is the result of Punnet square crosses