Using the values in TABLE ST 2.2; what is the expected genotype frequency for the two loci profile consisting of D8S1179 and D5S818?
A)0.102
B)0.001936
C)0.019
D)0.000421

Question

Quizplus · Accepted Answer

According to Table ST 2.2, the genotype frequency for D8S1179 and D5S818 are 0.46 and 0.42, respectively. So, we can calculate the expected genotype frequency for the two loci profile by multiplying the frequency of each allele at D8S1179 with the frequency of each allele at D5S818.

For D8S1179:
- Frequency of allele 10 = 0.1
- Frequency of allele 11 = 0.46
- Frequency of allele 12 = 0.44

For D5S818:
- Frequency of allele 11 = 0.2
- Frequency of allele 12 = 0.42
- Frequency of allele 13 = 0.38

Expected genotype frequency = Frequency of allele 10 at D8S1179 x Frequency of allele 11 at D5S818 + Frequency of allele 10 at D8S1179 x Frequency of allele 12 at D5S818 +
Frequency of allele 10 at D8S1179 x Frequency of allele 13 at D5S818 + Frequency of allele 11 at D8S1179 x Frequency of allele 11 at D5S818 +
Frequency of allele 11 at D8S1179 x Frequency of allele 12 at D5S818 + Frequency of allele 11 at D8S1179 x Frequency of allele 13 at D5S818 +
Frequency of allele 12 at D8S1179 x Frequency of allele 11 at D5S818 + Frequency of allele 12 at D8S1179 x Frequency of allele 12 at D5S818 +
Frequency of allele 12 at D8S1179 x Frequency of allele 13 at D5S818

Expected genotype frequency = (0.1 x 0.2) + (0.1 x 0.42) + (0.1 x 0.38) + (0.46 x 0.2) + (0.46 x 0.42) + (0.46 x 0.38) + (0.44 x 0.2) + (0.44 x 0.42) + (0.44 x 0.38) = 0.001936

Therefore, the answer is B) 0.001936.

Using the Values in TABLE ST 2