Solutions to DNA Mixture Practice Problems
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 since 29 March 2008

Consider the following DNA mixture
Marker 
Sample 
Victim 
1 
11, 12, 13 
13, 13 
2 
8, 9, 10, 11 
10, 11 
 1: Assuming a twoperson mixture, what are the possible
genotype(s) of the second contributor?
 11, 12 at marker 1; 8, 9 at marker 2.
 2: Now suppose the genotype of the victim is 11, 13 instead of 13, 13. Under a twoperson mixture, what are the possible
genotype(s) of the second contributor?
 11, 12 or 12, 13 at marker 1; 8, 9 at marker 2.
Suppose the allele frequencies at these two markers are
as follows:
Marker 1 
Freq 
11 
0.1 
12 
0.2 
13 
0.3 


Marker 2 

8 
0.05 
9 
0.01 
10 
0.2 
11 
0.1 
 3: Assuming a two person mixture (and the victim is 13, 13 are marker 1), what is the probability that a random person would
fail to be excluded as a contributor to the mixture?

Pr(11, 12 at 1)*Pr(8, 9 at 2) = (2*0.2*0.2)*(2*0.05*0.01) = 0.00005 or 1/25,000
 4: Redo this problem, but now assume the victim is 11, 13 are marker 1.

Pr(11, 12 or 12, 13 at 1)*Pr(8, 9 at 2) = (2*0.2*0.2+ 2*0.2*0.3)*(2*0.05*0.01) = 0.00005 or 1/25,000 = 0.00016 or 1/6250.
 5: Now make no assumptions as to the number of contributors.
What is the probability that a random person would fail to be excluded as a contributor to the mixture?

Pr(11 or 12 or 13)^{2} * Pr(8 or 9 or 10 or 11)^{2} = 0.047 or 1/22.