Solutions to DNA Mixture Practice Problems
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Consider the following DNA mixture
| Marker |
Sample |
Victim |
| 1 |
11, 12, 13 |
13, 13 |
| 2 |
8, 9, 10, 11 |
10, 11 |
- 1: Assuming a two-person 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 two-person 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.