(1) How to separate a gene of interest from the great mass
of genomic DNA, and
(2) how to purify enough of the target DNA for analysis.
The b globin
gene contains ~2000 bp.
Each base pair has an average molecular
mass of 635 daltons, so the gene has a total mass of 1.27 X 106
Since there are two copies of the b
globin gene per cell, and the average number of cells in the adult human
body is about 1 X 1013, there is a total mass of about 0.000042
g (42 mg)
of DNA per individual.
If biochemical analysis requires ~1
mg of DNA, it would take about 24 human cadavers to extract this much.
globin gene represents only 1/1,500,000 of the ~3 X 109 bp of
DNA composing a haploid set of human chromosomes. Extraction of 15.1
kg of human genomic DNA would yield only 1 mg b
Since this gene is composed of only
4 repeating nucleotides (as is all DNA), any 2000 bp fragment would look
biochemically identical to any other. Separating the individual gene
from all of the genomic DNA would be virtually impossible.