=
=
=
=
=
=
-2
1.
10 mM
1 µm
1 liter
0.1 ml
10 mg
10 µg
0.010 M (or 10 M)
-6
10 M
3
1,000 ml (or 10 ml)
100 µl (or 102 µl)
0.010 g (or 10-2 g)
10-5 g
2.
2.0 g NaOH dissolved in sufficient water to give 200 ml.
3.
8.5 g NaCl dissolved in sufficient water to give 1,000 ml.
4.
40 ml
5.
Weigh 1.0 g agarose. Combine with 6.25 ml of 20X TAE. Add sufficient water to give
125 ml.
6.
0.01 M
7.
Before dilution, the number of antibody molecules in a 1.5 ml sample of the solution
would be:
⎛
moles ⎞ ⎛ 1.5 liters ⎞
molecules
23 molecules ⎞ ⎛
2 X 10-9
= 1.8 X 1012
⎜⎝ 6 X 10
mole ⎟⎠ ⎜⎝
liter ⎟⎠ ⎜⎝ 1,000 sample ⎟⎠
sample
The number of antibody molecules would be many times greater then the number of cells
in the 30 µl sample.
After dilution, the calculated number of antibody molecules in a 1.5 ml sample of the
solution would be:
⎛
moles ⎞ ⎛ 1.5 liters ⎞
molecules
23 molecules ⎞ ⎛
2 X 10-109
= 1.8 X 10-88
⎜⎝ 6 X 10
⎟
⎜
⎟
⎜
⎟
mole ⎠ ⎝
liter ⎠ ⎝ 1,000 sample ⎠
sample
This number is far, far less than one. In other words, there are ZERO antibody molecules
present in the assay.
There is no known physical basis for this phenomenon. Because the results has not been
consistently repeated by other laboratories, it is not accepted by the scientific community.