Answers/help to problems 1-3 1) To calculate the probability for a DNA profile u

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Answers/help to problems 1-3
1) To calculate the probability for a DNA profile using allele frequencies, one has to assume that the alleles follow the Hardy-Weinberg Principle, meaning that the alleles will remain constant over time. This is an example of a Locus (gene) with only two alleles (A and B). If we have allele A (with allele frequency p) and allele B (with frequency q) in a population, the genotype frequency for genotype AA will be p, genotype AB will be 2pq, and genotype BB will be q. The frequency p+ 2pq+q-1 will remain this way over generations. 2) Using the table below, if you have a homozygous genotype (two of the same alleles), such as TPOX (6,6), the genotype frequency will be p 0.006 x 0.006. 3) If you have a heterozygous genotype (two different alleles), such as TPOX (6, 8), the genotype frequency will be 2pq 2 x 0.006 x 0.506. 4) Once you have obtained all of the genotype frequency for those loci, all you have to do is use the product rule and multiply those genotype frequencies together to get the frequency of the full DNA profile. 5) Note that all answers that involve calculation of a frequency should be given in scientific notation. Therefore, a frequency of 0.000092 would be 9.210 (or 9.2*10-5). Use the allele frequencies in the following table. Allele Frequency 0.006 0.506 0.094 0.063 0.294 0.038 Allele 6 Locus TPOX 10 12

Explanation / Answer

1. Calculate the frequency of the following DNA profile: TPOX (6,12) vWA (15,21) THO1 (9, 9.3)

Solution: Since TPOX (6,12) vWA (15,21) THO1 (9, 9,3) are heterozygous genotype, the genotype frequency would be 2pq.

TPOX (6,12)- 2pq = 2 * 0.006 * 0.038

                          = 0.000456

                          = 4.56 * 10^-4

vWA (15,21)- 2pq = 2 * 0.67 * 0.001

                          = 0.00134

                          = 1.34 * 10^-3

THO1 (9, 9.3)- 2pq = 2 * 0.232 * 0.026

                            = 0.102064

                            = 1.2064 * 10^-2

Therefore, frequency of the DNA profile TPOX (6,12) vWA (15,21) THO1 (9, 9,3) = (4.56 * 10^-4) (1.34 * 10^-3) (1.2064 * 10^-2) =7.37 * 10^-9

2. Calculate the frequency of the following DNA profile: TPOX (8, 9) vWA (17, 17) THO1 (6, 10)

Solution: Here TPOX (8, 9) and THO1 (6, 10) are heterozygous genotypes whereas, vWA (17, 17) is a homozygous genotype which will follow the p2 formula for calculation.

TPOX (8, 9) vWA (17, 17) THO1 (6, 10)

= (2pq) (p2) (2pq)

= (2 * 0.506 * 0.094) (0.300 * 0.300) (2 * 0.295 * 0.116)

= (9.5128 * 10^-2) (9 * 10^-2) (6.844 * 10^-2)

= 585.950429 * 10^-6

~ 586 * 10^-6 = 5.86 * 10^-4

3. Of the three loci (TPOX, vWA and THO1) and alleles, which DNA profile can be considered to be the most rare. In other words, which profile would occur at the lowest frequency?

Solution:

Frequency of TPOX (6,12) vWA (17,18) THO1 (9.3, 10) = 3.383 * 10^-7

                    TPOX (6,6) vWA (20,21) THO1 (9.3, 9.3) = 1.94 * 10^-14

                    TPOX (6,6) vWA (21, 21) THO1 (9.3, 9.3) = 2.4336 * 10^-14

                    TPOX (6, 12) vWA (17,17) THO1 (9.3,10) = 2.4755 * 10^-7

The lowest frequency calculated is 1.94 * 10^-14 which belongs to the profile TPOX (6,6) vWA (20,21) THO1 (9.3, 9.3)

                         

                           

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