Biology-Online • View topic - IBO 2008

Join for Free!
122502 members

IBO 2008

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

Moderator: BioTeam

IBO 2008

Postby Artush » Sun Aug 08, 2010 1:34 pm

Part B, 26.( Famous person sued in a lawsuit in a case of paternity. Respondent (as delineated on the autoradiograph D), mother (designated M), and the baby (labeled B) were typed by two loci VNTR1 and VNTR2, as shown in the autoradiograph below.
Each of these VNTR loci had four alleles. In a normal population for VNTR1 frequency of alleles 1, 2, 3, and 4 were 0.2; 0,4; 0,3 and 0,1; respectively. For VNTR2, the frequency of alleles 1, 2, 3, and 4 were 0,1 ; 0,1; 0,2 and 0,6; respectively.


a. Indicate whether the autoradiograph of the fact that D may be the father of the baby B? Make a tick (√) the appropriate box.
Yes No

b. What is the likelihood that another man in the normal population could be the father of the baby B?

Please, help to solve the problem b.
Posts: 5
Joined: Wed Apr 07, 2010 4:06 pm

Postby protomandaniel » Mon Sep 13, 2010 4:54 pm

I think, firstly, we neglect the 5 alleles that are present in the mother VNTR1 3,4, VNTR2 1,2,3 since even the possesion of those alleles or not from the father would not affect the probability.
Then, consider the other 3 genes, the man has to have VNTR1 2, VNTR2 4 and not VNTR1 1 since the child possesses these but the mother doesn't.
VNTR1 2: 0.4
VNTR2 4: 0.6
Not having VNTR1 1: 1-0.2 = 0.8
Hence the overall prob is 0.4 x 0.6 x 0.8 = 0.192 likelihood.
Posts: 5
Joined: Thu Sep 02, 2010 8:35 am

Return to Genetics

Who is online

Users browsing this forum: No registered users and 1 guest