Question

R and Y genes of maize lie very close to each other. When RRYY and rryy genotype are hybridized, F${}_2$ generation will show

• A. Segregation in 12 : 3 : 1 ratio
• B. Segregation in 3 : 1 ratio
• C. Higher number of dominant parental types
• D. Higher number of recombinant types

Answer 1

Answer: (C)

Higher number of dominant parental types

Independent assortment is segregation of factors for a trait independent of other factors during gamete formation followed by their random rearrangement in progeny thereby producing both parental and new combinations. The linkage is the tendency of closely placed genes on a chromosome to stay together during inheritance; it produces more parental combination and less/no new combinations of the gene.

Genes present together on a chromosome show linkage and do not follow independent assortment resulting in more parental types and fewer recombinants in the progeny.

Since the R and Y genes are closely placed on the chromosome and show linkage; they will not assort independently and more parental types (RRYY and rryy) and fewer recombinants (Ry, rY) will be formed. Two pairs of genes showing independent assortment give 9 : 3: 3: 1 ration in F${}_2$ progeny. Phenotypic ratio 3 : 1 is obtained for an allele of a gene showing segregation (monohybrid cross, the question is about dihybrid cross). Thus, the correct answer is option C.