What was the recombination percent obtained when white eyed yellow bodied Drosophila was crossed with its wild type?

When white eyed and miniature winged Drosophila melanogaster is crossed with its wild type, it produces following percent of recombinants. A) 1.3% B) 37.2% C) 62.8% D) 98.7%

Correct option:(B)37.2%

Solution:

Drosophila, when white eyed and miniature winged (wwmm) female is crossed with wild type, that is red eyed normal winged (w $^{+}$ w $^{+}$ m $^{+}$ m $^{+}$ ) males, 62.8% of the progeny was of parental types .
And only 37.2% were recombinants.
Shows that linkage between genes for white body and miniature wing is loose linkage.
So, recombinant occurs along with parental combinations in offsprings, this type of linkage is called as incomplete linkage.

• According to Mendel, the F2 phenotypic dihybrid ratio is 9:3:3:1.

• The ratios of Morgan’s dihybrid cross between white eyed, yellow bodied female Drosophila and red eyed, brown bodied male Drosophila was significantly different. It produced 1.3% recombinants and 98.7% progeny with parental type combinations.

• Morgan and his group knew that the genes were located on the X chromosome. Hence, they observed that when two genes in a dihybrid cross are located on the same chromosome, the proportion of parental gene is much higher than the non-parental gene.

• He also coined the terms linkage (the physical association of genes on a chromosome) and recombination(when crossing over of genes occurs). When linkage is very tight, recombination is lower and when linkage is very loose, recombination is higher.