Bio-Rad CHEF-DR® III Variable Angle System - Manual (Page 18)

For Example: 300 bacteria in 5 squares = average of 60 bacteria/square x 25 (squares) x 20
(dilution factor, yeast use 100 for dilution factor) x 10 4 = 3 x 10 8 bacteria per ml. So for 5 ml of
plugs you need 5 ml x 5 x 10 8 cells final concentration ÷ 3 x 10 8 actual cells concentration = 8.33
ml of cell suspension is required.
3.8 Estimation of Agarose Plug DNA Concentration
Two pieces of information are needed to determine DNA concentration:
1. The size in base pairs of the genome. This information is readily available for most organ-
isms, otherwise a best guess is necessary. We use 6 x 10 9 for mammalian, 4.5 x 10 6 for
Escherichia coli and 1.5 x 10 7 for Saccharomyces cerevisiae in the following examples.
2. You need to determine the number of genomes per cell. For example, for stationary growth
phase in yeast or bacterial cells or confluent growth in tissue culture cells, assume one genome
per cell. However, for exponential phase growing cells there is more than one genome per cell.
Make a best guess or assume one per cell which will give the minimum concentration of
DNA. In the below examples we use a value of 1.2 genome equivalents (20%) for mam-
malian cells, 2.5 genome equivalents for bacteria, and 2 genome equivalents for yeast
217 .
Equations for Estimation of DNA Concentration in Agarose Plugs:
(Genome Size bp)(660 g/mole)
= grams DNA/cell (A)
6.02 x 10 23 bp/mole
(grams DNA/cell)(cell/ml) = (grams DNA/ml)(1 x 10 6 µg/ml) = µg DNA/ml (B)
(µg DNA/ml)(genome equivalents)
≅
(µg DNA/ml) (C)
(µg DNA/ml)
≅
(µg DNA/plug)
≅
µg DNA/lane (D)
10 plugs/ml 2 lanes/plug
Example Calculations:
Mammalian:
(6 x 10 9 bp)(660 g/mole)
= 6.578 x 10 -12 g DNA/cell (A)
6.02 x 10 23 bp/mole
(6.578 x 10 -12 g/cell)(5 x 10 7 cell/ml)
=
(B)
(3.289 x 10 -4 g/DNA/ml)(1 x 10 6 µg/ml)
=
329 µg DNA/ml
(3.29 µg DNA/ml)(1.2 genome equivalents)
≅
394 µg DNA/ml (C)
(394 µg DNA/ml)
≅
(40 µg DNA/plug)
≅
20 µg DNA/lanes (D)
10 plugs/ml 2 lanes/plug
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