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Genetics as it applies to evolution, molecular biology, and medical aspects.

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schnupps12
Garter
Posts: 2
Joined: Thu Oct 08, 2009 1:43 am

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Last edited by schnupps12 on Mon Oct 12, 2009 10:36 pm, edited 1 time in total.

zami'87.
Coral
Posts: 257
Joined: Sat Mar 05, 2005 6:56 pm
Location: Serbia

### Re: 3 point test cross

Let me see..if I remember well genetic classes

Write given phenotypes in form of received genotypes from female gametes:

Dy+*ct+*w+ 18
dy*ct+*w+ 153
dy+*ct*w+ 6
dy+*ct+*w 76
dy+*ct*w 150
dy*ct*w+ 72
dy*ct*w 22
dy*ct+*w 3
total: 500
Parental class is the biggest.So dy*ct+*w+ 153 and dy+*ct*w 150 don’t have recombinations. Note classes with least individuals 3 and 6. Those are double crossing overs.Gene in the middle switches. So w gene is in the middle.So correct orders on two X chromosomes are Dy*w+*ct+ and dy+*w*ct . Next calculate recombinant frequency by dividing (number of recombinants for each gene pair) with (total number). For example distance in m.u between genes w and ct is..(search w*ct+ and w+*ct pairs in classes..forget dy gene for now) 100*(6+76+72+3)/500=100* 0.314=31.4 m.u and by calculating in similar fashion distance between dy+ and w(or dy*w+) is 9.8 m.u. If you check distance between dy+ and ct you’ll notice that you’ll get smaller number than by summing 9.8+31.4 . Why?Because you forgot about double crossingovers that in this case you add twice. Next calculate interference by using formula I=1-(observed doublerecombinants/expected recombinants).Observed double recombinants are 6 + 3 as I said. Expected are..product of frequencies in individual regions..thus (0.314*0.098)*500=15.386..I=1-9/15.386=0.4151 or 41.5 %..hope this helped.Cheers!

ps feel free to correct me..
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