DNA Profiling Module Writing Assignment

DNA Profiling Part 3

1

Part 3: Analyzing the certainty of DNA profile comparisons

At the end of the DNA data analysis in which two samples are being compared to decide if they match, the analyst concludes one of the following: - “cannot exclude” - “can exclude” - “inconclusive/uninterpretable”

If the answer is “cannot exclude” (I.e. there was a DNA match), a

statistical interpretation is also provided which may be stated similarly to the following:

“There is a 1 in 300 million chance that suspect and crime scene

DNA derive from different persons by chance” This number refers to a calculation of the “random match probability”

Alleles in DNA profiling •  Allele: A specific sequence

of DNA observed in nature. In the top example these would be chocolate, vanilla and strawberry.

•  In DNA profiling, each allele is assigned a number (as in bottom example).

•  The allele’s number normally refers to how many copies of the repeated stretch of DNA has been observed in the STR.

•  Allele range: The entire cohort of different alleles observed at a specific chromosomal locus.

•  The allele range in this example would be 8-14.

Mom Dad Alleles

Allele range

The “CSF1PO locus”

Chr1 copy1 Chr1 copy2

DNA Profiling Part 3

2

Allele frequencies •  Allele frequency: The proportion of all

copies of a locus that is made up of a particular sequence variation.

–  Alleles are not all observed in nature in the same proportions

–  Cataloging which ones are actually observed and how often helps determine how rare they are

–  The rarer the allele, the less likely it will be observed by chance in two unrelated people

•  If Mom and Dad are the only two people in the whole world….the allele frequencies for the chromosome shown would be the following: –  chocolate = 0.5 –  vanilla = 0.25 –  strawberry = 0.25

Mom Dad

Allele frequencies

Chr1 copy1 Chr1 copy2

Allele frequencies: •  Sum to “1” or “100%” at a given locus

(depending on how table shows info) •  Are a mathematical expression of the

likelihood of observing that particular allele in a randomly chosen DNA sample

•  Using the example shown on the right, a frequency of “25.369” means that 25.369% of the time, a scientist testing DNA will observe 10 copies of the tandem repeat of DNA associated with the CSF1PO locus (if all DNA donors are Caucasian)

•  Tables of allele frequencies are used which are broken down according to broad ethnic populations (eg. Caucasian, African American, Hispanic, Asian) Total 99.999

Allele frequencies

Locus

Budowle et al J Forensic Sci, 1999, 44(6), pp. 1277-86.

DNA Profiling Part 3

3

How do they narrow it down to “1 chance in 300 million”?

•  Multiple loci are analyzed •  The overall probability (of two

samples being from different people by chance if they give the same result…) is known as the “random match probability”.

•  The random match probability assumes complete randomness of all the alleles in the human population and is computed using the “basic product rule”:

…so what does THAT mean? P = 2

n A1A2B1B2C1C2....N1N2

The Basic Product Rule •  Overall probability is represented by “P”

•  Each locus being analyzed is represented by the letters A, B, C etc.,

•  The letters A, B, C etc each appear twice because there are two copies of each chromosome

•  The two copies of each chromosome can either bear the same allele or different alleles and are represented by subscripts “1” and “2” ie. A1A2

•  The 2n represents a necessary doubling required per locus… because in the laboratory it cannot be distinguished which allele came from the maternal chromosome and which from paternal and they count as separate possibilities when tabulating all the possible combinations

P = 2 n A1A2B1B2C1C2....N1N2

P = 2 n A1A2B1B2C1C2....N1N2

P = 2 n A1A2B1B2C1C2....N1N2

P = 2 n A1A2B1B2C1C2....N1N2

Mom Dad …was it …or

Mom Dad

DNA Profiling Part 3

4

Example “basic product rule” calculation •  Imagine analyzing two

DNA samples (e.g suspect DNA and crime scene DNA) for the 3 loci shown

•  For the three loci, you obtain the following results for both samples

–  DS1358 = Alleles 16 & 17*

–  VWA = Alleles 17 & 18 –  FGA = Alleles 21 & 22

•  HOW SURE are you that the DNA match is not by chance?

Crime Scene

Suspect *

*Two alleles are typically observed per locus because ~80% of people are heterozygous at these loci I.e. “vanilla, strawberry”.

Allele frequency tables from the literature (same ones FBI uses)

Budowle et al J Forensic Sci, 1999, 44(6), pp. 1277-86.

Example of “basic product rule calculation”

•  Multiply the observed allele

frequencies (expressed as decimal) together according to the equation

•  Convert expression to a probability

..that the Suspect and Crime Scene samples derived from different people by chance (I.e. that there was a “random match”).

P = 2n D3S1358 Allele 16 and 17 frequencies⎡⎣ ⎤⎦× VWA Allele 17 and 18 frequencies⎡⎣ ⎤⎦× FGA Allele 21 and 22 frequencies⎡⎣ ⎤⎦

P = 23 0.23153( )× 0.21182( )⎡⎣ ⎤⎦ × 0.26276( )× .22194( )⎡⎣ ⎤⎦ × 0.17347( )× .18878( )⎡⎣ ⎤⎦ = 7.49 × 10

−4 = 0.000749

0.000749 =

7.49 10,000

= 1

1335 ≈ 1 in 1335 chance

P = 2 n A1A2B1B2C1C2

DNA Profiling Part 3

5

More loci analyzed = greater certainty that DNA samples do

not match by chance •  In previous example with 3 loci we narrowed the

probability to: ~ 1 in 1335

•  A few things to point out: –  Every additional locus tested decreases the

probability of a random match. –  The results of the calculation are VERY

DEPENDENT on the actual alleles observed in the analysis.

–  Allele frequencies vary with ethnic background. –  The random match probability is not an “index

of guilt” and does not even address the likelihood that the defendant is the source of the DNA. Its sole purpose is to help a jury address the hypothesis that a DNA match between crime scene and suspect was observed BY CHANCE!

Budowle et al J Forensic Sci, 1999, 44(6), pp. 1277-86.

0.000749 =

7.49 10,000

= 1

1335 ≈ 1 in 1335 chance

Group work on basic product rule calculation

•  Break into groups of ~4 people •  Work through the following calculation: Imagine analyzing 3 sets of 2-DNA samples

(suspect and crime scene DNA) using 4 loci, and the allele frequencies below: 1.  7,8; 13,14; 14,15; 25,26 2.  10,11;16,17; 18,19; 21,23 3.  11,12; 18,19; 19,20; 18,22

•  Spend ~15 min. discussing within your group the following: –  Why analyzing more loci decreases the probability of a random match. –  Why the calculation is only as good as the table(s). –  What you think the forensic expert probably does in cases in which the ethnic

background of the possible perpetrator isn’t known. –  Are there negative consequences to incorrectly calculating the number? What –  Why does a “match” between crime scene and suspect DNA not necessarily

mean the suspect is guilty?

DNA Profiling Part 3

6

Group work on basic product rule calculation

•  Break into groups of ~4 people •  Work through the calculation provided on the

worksheet •  Spend ~15 minutes discussing within your

group the following: –  Why analyzing more loci decreases the

probability of a random match. –  Why the calculation is only as good as the

table(s). –  What you think the forensic expert

probably does in cases in which the ethnic background of the possible perpetrator isn’t known.

–  Are there negative consequences to incorrectly calculating the number? If so, what might some of them be?

–  Why does a “match” between crime scene and suspect DNA not necessarily mean the suspect is guilty?