The distance between gene C and gene B is 8 map units In a c

The distance between gene C and gene B is 8 map units. In a cross between c B/C b and C B/C B individuals, what will be the percentage of CB/CB offspring? a. 4% b. 8% c. 42% d. 46% e. 50%

Solution

c. 42% because mapunit is 8 it means the recombination percentage will be 8% and if the cross is in between cB/Cb and CB/CB individuals happens then the possibility of getting CB/CB offspring is 50% but recombination is 8% then the final percentage of CB/CB offspring will be 42%