Question

4. Assume the database that was used to generate Table 1 consists of 36 individuals. Using this information, what is (show equations and calculations, the probability (report the RMP and combined frequency) that two individuals have the same profile of 15/16; 8/8; 12/13 if (and no, allele 13 is not a typo): a) they are unrelated, but from a smaller demographic, explain use of equation (4 marks) b) there could be a parent-offspring relationship possible, explain use of equation (4 marks) c) there could be a full-sibling relationship possible, explain use of equation (4 marks) Table 1: Observed allele frequency per locus D3S1358 TH01 D135317 D3S1358 Frequency TH01 Frequency | 0.01 D135317 Frequency 0.5 0.4 0.2 10 15 16 0.4 0.4 17 8 0.35 0.64 12 0.1

Answer 1

a) They are unrelated, but from a smaller demographic, explain use of equation

• Since, the alleles were heterozygous,
• Locus A: 15/16 = 2pq = 2 (0.4)(0.4) = 0.32

• Since, the alleles were homozygous and θ = 0.03 (smaller demographic)
• Locus B: 8/8 = P2 +p (1-p)θ = (0.64)2 + (0.64)(1 - 0.64)0.03 = 0.4165

• 13 does not exist on the table because it showed up less than 5 times
• So we can use the equation 5/2N must be used.
• 13 = 5/2N = 5/2 *(36) = 0.069
• Since, the alleles were heterozygous,
• Locus C: 12/13 = 2pq = 2*(0.1)(0.069) = 0.0138

• Pro1le frequency: (0.32)(0.4165)(0.0138) = 1.839 x 10-3
• Combined frequency: 1/1.839 x 10-3 = 544
• The probability is 1 in 544

b) There could be a parent-offspring relationship possible, explain use of equation

• Probability of a same genotype, and the alleles are heterozygous
• F = 0.25 because the relative is a parent
• Locus A: 15/16 = 2pq +2(p+q - 4pq)F = 2 (0.4) (0.4) + 2((0.4+0.4) - 4 (0.4)(0.4))(0.25) = 0.4
• Locus B: 8/8 = p2+4p (1-p)F = (0.64)2 + 4(0.64)(1-0.64)(0.25) = 0.64
• Locus C: 12/13 = 2pq +2(p+q-4pq)F = 2(0.1)(0.069) + 2((0.1+0.069)-4(0.1)(0.069))(0.25) = 0.08455
• Pro1le frequency: (0.4)(0.64)(0.08455) = 0.0216
• Combined frequency: 1/ 0.0216= 46
• The probability is 1 in 46

c) There could be a sibling relationship possible, explain use of equation

• Locus A: 15/16 = (1+p+q+2pq)/4 = (1+0.4+0.4+2(0.4)(0.4))/4 = 0.53
• Locus B: 8/8 = (1+2p+p2)/4 = (1+2(0.64)+(0.64)2)/4 = 0.6724
• Locus C: 12/13 = (1+p+q+2pq)/4 = (1+0.1+0.069+ 2(0.1)(0.069))/4 = 0.2957
• Pro1le frequency: (0.53)(0.6724)(0.2957) = 0.1054
• Combined frequency: 1/ 0.1054 = 9.4
• The probability is 1 in 9