Question

Using the information in the first question and the data in the following table, what is the map distances between genes? Hint: It helps to re-write the above NCO in the correct gene order. pr = prawny; wd = wider; and ct = cut wildtype 215 prawny, cut, wider 200 3506 prawny cut, wider wider 3507 28 prawny, cut 34 cut 552 prawny, wider 578 O pr-ct -5.5 mu and ct-wd - 13.8 mu O pr-ct=13.8 mu and ct-wd = 5.5 mu O pr-ct = 13.1 mu and ct-wd = 4.8 mu O pr-wd = 13.8 mu and wd-ct = 5.5 mu

Answer 1

In the case prawy, cut, wides is the recessive type So the genotype = po ct wd Then the wild recessive type is the potcft wat CEPT pot ctt wat 215 ar ct wd - 200 ctt pr wat - 3506 p&t ct wd - 3507 prt wd ett - 28 odt po ct 34 Wat et pot 552 ctt pr wd - 578 Parental The type is 8620 set ct wd ett pa wat and Double crossing over is the so the po cat odt prt ct wd pot wd ett and wat prct. as the Dco single crossing cover. I are the Prett wat pst ct wd Po, ct, ud and pyt ett wat Single crossing over I are the py ctt wat REDMI NOTE 7 AT DUAL CAMERA pot & Wd py ex two and portal wat
RI
So the map distance - he map distance between pr-ct Single crossover i to poco__ x100 Total no of genotype 215 +200 +28+34 x 100 - 5.5 mu The map distance between et-nd - Se. S'ingle crossingoveyti + OCO Total no. of genotype -xoo = 552 +578+28+39 Xloo 8620 - 13.8 mu