This has been discussed at length on another thread so I do not intend to repeat it except to refer you to the COA decision, thus...
Seventeen of the twenty bands attributable to Sheila Caffell had been detected in DNA from the internal swabbings. Random chance would have suggested thirteen common bands would be found and hence since there was significantly more than thirteen, it provided some support for the DNA of Sheila Caffell being in the moderator.
Question to you Holly. How many bands in common are accepted as a match under UK law?
Its not the number of bands but the size - see first diagram - A, B and C match and D doesn't. The DNA from the individual needs to match the sample as follows:
-
Inclusions -- If the suspect's DNA profile matches the profile of DNA taken from the crime scene, then the results are considered an inclusion or nonexclusion. In other words, the suspect is included (cannot be excluded) as a possible source of the DNA found in the sample.
-
Exclusions -- If the suspect's DNA profile doesn't match the profile of DNA taken from the crime scene, then the results are considered an exclusion or noninclusion. Exclusions almost always eliminate the suspect as a source of the DNA found in the sample.
-
Inconclusive results -- Results may be inconclusive for several reasons. For example, contaminated samples often yield inconclusive results. So do very small or degraded samples, which may not have enough DNA to produce a full profile
The bands are created by a process called Agarose Gel Electrophoresis. The gel matrix acts as a sieve: smaller DNA molecules migrate faster than larger ones, so DNA molecules of different sizes separate into distinct bands during electrophoresis - see second diagram.
Example calculation used for bands:
DNA concentration was determined by electrophoresis in 8% agarose gel. 1µl, 0.5µl and 0.25µl of ?phage/Bst EIIDNA were loaded in the gel to assist the quantification.
1. Add up the sizes of all ? DNA bands to calculate total size of ?phage genome
? genome= 48161 bp
2. Choose a band with a similar size to the plasmid DNA
3.675bp
3. Calculate the percentage of ? DNA genome represented by that ?band.
%= 3.675*100/48161= 7,63%
4. Calculate relative concentration of the selected band:
Stock solution of ? phage DNA contains 500 ng genomic DNA/µl
500 ng genomic DNA/µl* 7.63/100= 38,15ng of 3675 bp band/µl
See third diagram for above calculation