Morgan crossed true breeding red eye (W+), Long leg (SL+) females with true breeding white eye (w), short leg (sl) males. F1 females from this cross were then backcrossed to true breeding white eye (w), short leg (sl) males.

From this cross the following flies were collected:

348 white eye (w), short leg (sl) flies

356 red eye (W+), Long Leg (SL+) flies

68 white eye (w), Long Leg (SL+) flies

55 red eye (W+), short leg (sl) flies

How far apart are the red eye and long leg genes on the X chromosome? (Display your answer in centimorgans, to 2 decimal places.

Question

Biology

Saunders

2 November, 10:02

Morgan crossed true breeding red eye (W+), Long leg (SL+) females with true breeding white eye (w), short leg (sl) males. F1 females from this cross were then backcrossed to true breeding white eye (w), short leg (sl) males.

From this cross the following flies were collected:

348 white eye (w), short leg (sl) flies

356 red eye (W+), Long Leg (SL+) flies

68 white eye (w), Long Leg (SL+) flies

55 red eye (W+), short leg (sl) flies

How far apart are the red eye and long leg genes on the X chromosome? (Display your answer in centimorgans, to 2 decimal places.

+2

Answers (1)

Know the Answer?

Hayden Lowe · Answer 1

The F1 parents is : W + SL+ / w sl

The genetic distance is calculated by the (recombinat/total) * 100 = { (68+55) / (68+55+348+356) }*100 = (133/837) * 100 = 15.89 cM.

Here recombinat are lower in number and parental are greater in number.

15. 89 is the right answer.