An individual homoygous for gene a and b is cross with wild type and `F_1` was back crossed with the double recessive. The appearance of the offspring is as follows
`++ to 39`
`ab to 31`
`a+ to 17`
`+b to 13`
Find out the distance between genes a and b .

An individual homozgous for genes cd is crossed with wild type and F_(1) is crossed back with the double recessive. The appearance of the offspring is a follows: {:(++" "-" "903),(cd" "-" "897),(+d" "-" "98),(c+" "-" "102):} The distance between genes c and d is

An individual homozgous for genes cd is crossed with wild type and F_(1) is crossed back with the double recessive. The appearance of the offspring is a follows: {:(++" "-" "903),(cd" "-" "897),(+d" "-" "98),(c+" "-" "102):} The distance between genes c and d is

An individual homozygous for genes cd is crossed with type and F_1 Crossed back with the double recessive . The appearance of the offspring is as follows {:(++rarr,903),(cdrarr,897),(+drarr,98),(c+rarr,102):} The distance between the genes c and d is

$ A cross between the pure recessive parent and the F-1 offspring is called Test cross. ! Test cross is a type of Back cross.

$ A cross between the pure recessive parent and the F-1 offspring is called Test cross. ! Test cross is a type of Back cross.

Individuals homozygous for 'xy' genes were crossed with wild type '++' . The F_1 dihybrid thus produced was test crossed . It produced progeny in the following in the following ratio '+ + ' 900, ' +y ' 115, 'xy' 880, 'x+' 105. Find out the recombination frequency .

A homozygous recessive with cd genes was crossed with dominant (+ +). If their hybrid is test crossed and the following result was obtained + + - 900 cd - 880 + d - 115 + c - 105 than what will be the distance between genes c and d.

Individuals homozygous for ab genes were crossed with wild type (++). The F_(1) hybrid thus produced was test crossed. It produces progeny in following proportion. Now, calculate the distance between a and b genes.