A truebreeding wildtype female fly with a gray body and long

A true-breeding wild-type female fly with a gray body and long bristles is crossed to a true-breeding male fly with an ebony body and short bristles. Then, a test cross is conducted in which F1 heterozygous females are crossed to true-breeding male flies with ebony bodies and short bristles. The ebony (e+) and short (s+) genes are linked. The progeny of this cross are described in the table below. Calculate the distance between the ebony (e+) and short (s+) genes. Show your calculations. Gray body, Long bristles 447 Ebony body, Short bristles 453 Gray body, Short bristles 52 Ebony body, Long bristles 48

Solution

More number individuals are parents: 447 & 453; Less number individuals are recombinants: 52 & 48

Then % of Recombination = No. of Recombinant individuals / Total no. of progeny x 100

= 100 / 1000 x 100

= 10%

Here % of Recombination is 10%, and is equal to the map distance between genes. So the distance between e+ and s+ is 10 map units or centi Morgans or Morgan units

One map unit is equal to 10⁶ base pairs. So the distance between e+ and s+ is 10 x 10⁶ base pairs = 10⁷ base pairs

According to Watson Krick, each base pair length is 3.4 A^o or 0.34 nm. Then the distance between e+ and s+ is 0.34 x 10⁷ nm

= 34 x 10⁵ nm

= 3400000 nm

= 34/10000 meters

= 0.0034 meters