IBO 2008

[Artush]

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.

[protomandaniel]

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.