# Forensic use of ABO blood groups and VNTR allele frequencies

The forensic use of ABO blood groups and VNTR allele frequencies

Frequencies of ABO and Rh blood group types in the United States

(see attached file)

a) What is the probability that a B+ blood stain in the US lefta at the scene of a crime randomly matches a person of the following ancestry

Caucasian

Black

Asian

A genetic sytem that is far more variable than the simple blood types consists of atretches of DNA called VNTRs (variable number of tandem repeats). Each of thes VNTR (also known as minisatellites) loci contain many alleles, most of which are very rate.

(see attached)

b) Calculate the probability that an individual picked at random from the Caucasian population has the following genotype (allele 1/allele 2)

5/22 heterozygote at D1S7

6/8 heterozygote at D2S44

5/9 heterozygote at D17S79

12/12 homozygote at D4S139

c) What is the probability that an individual has full multilocus DNA profile in b)?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Question a

Caucasian

Frquency of B= 0.11 (From the tables)

Frequency of Rh+ = 0.85 (From the tables)

Therefore frequency of B+ = 0.0935 =0.11x0.85

Black

Frquency of B= 0.21 (From the tables)

Frequency of Rh+ = 0.9 (From the tables)

Therefore frequency of B+ = 0.189 =0.21x0.9

Asian

Frquency of B= 0.24 (From the tables)

Frequency of Rh+ = 0.99 (From the tables)

Therefore frequency of B+ = 0.2376 =0.24x0.99

Question b

Caucasian

5/22 heterozygote at D1S7

Freuency at 5= 0.011 (From the tables)

Freuency at 22= 0.077 (From the tables)

Therefore frequency of genotype= 0.000847 =0.011x0.077

6/8 heterozygote at D2S44

Freuency at 6= 0.034 (From the tables)

Freuency at 8= 0.106 (From the tables)

Therefore frequency of genotype= 0.003604 =0.034x0.106

5/9 heterozygote at D17S79

Freuency at 5= 0.015 (From the tables)

Freuency at 9= 0.200 (From the tables)

Therefore frequency of genotype= 0.003 =0.015x0.2

12/12 homozygote at D4S139

Freuency at 12= 0.190 (From the tables)

Freuency at 12= 0.190 (From the tables)

Therefore frequency of genotype= 0.0361 =0.19x0.19

c)

Probability that the individual has full multilocus profile in b)= 0.000000000330595

=0.000847x0.003604x0.003x0.0361

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