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Help with real-life biochemistry lab math problem.

  1. Nov 19, 2007 #1
    I have this practical problem in my laboratory and would like to see it formulated in a formal calculus way, but my math skills were never very high and they are very rusted now and do not know how to formulate (much less solve) the problem.

    A biological sample containing a virus is run through a chromatographic column and an ultraviolet detector is connected to the column outlet. The virus absorbs U.V. light and produces a signal that is recorded by a pen in a roll of paper chart. The paper advances at a constant speed measured in mm/minute. The sample comes out of the column at a constant rate measured in mL/minute. The deflection of the pen is orthogonal to the direction of paper motion and proportional to the concentration of the virus coming out at a given time. The pen traces a curve that is shaped more or less like a Gaussian bell curve. After all virus passes through the detector, the area under the plotted curve is approximated inscribing triangles in it, measuring them with a ruler and calculating the area of the triangles.

    The problem is to formally derive an equation that relates the area under the curve (measured in mm^2) and the concentration of the virus in the original sample, measured in micrograms per milliliter (ug/mL).

    The “input” data needed in the formula is:

    1) SENSITIVITY = 0.1 AU at full scale. This means that when the detector “see” that the “absorbance” of the liquid coming through is 0.1 it will deflect the pen all the way from zero to the full width of the paper chart, i.e. 100 mm.

    2) SPECIFIC ABSORTIVITY = 36. This means that a *theoretical* solution of the virus containing 1 gram of virus in 100 mL will produce an “absorbance” of 36 AU (“Absorbance Units”). [Please note that in practice there will never be such a concentrated solution, because one gram per 100 mL means *one million* micrograms in 100 mL and typical solutions are about 500 micrograms per 100 mL.]

    3) SAMPLE VOLUME. It is the volume of sample loaded into the chromatographic column and containing the virus and other things that preclude direct measurement. After passing through the column the virus comes out “purified” and can be detected.

    4) CHART SPEED. Measured in millimeters per minute (mm/min).

    5) FLOW RATE. Measured in milliliters per minute (mL/min).

    I have solved the problem in a “practical” way assuming an ideal example where the virus comes out as a hypothetical “rectangular” peak of CONSTANT concentration of the same area as the (real) Gaussian peak, but I would like to have a more formal solution based on the integrated area.

    Thank you in advance.
  2. jcsd
  3. Nov 21, 2007 #2


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    For starters, F(z) = Probability(x < z) = p where z is a specific concentration level and 0 < p < 1. F(z) is the area under the curve. The traced curve is f(z) = F'(z) and by the fundamental theorem of calculus, [itex]F(z) = \int_{-\infty}^z f(x) dx[/itex].

    Also, [itex]F(-\infty) = 0 \text{ and } F(+\infty) = 1[/itex]. So you need to normalize the area under the traced curve to sum up to one. The easiest way is to think of each individual measurement as a percentage of your total measurements. Then by definition they will add up to 100%.

    Next, you need to express the z as a function of elapsed time: z = Z(t). Also time itself is a function of the chart speed, t = T(vc). These transformations are mere "formalities" to translate the concentration level to elapsed time and elapsed time to chart speed; they will not substantially affect the calculations described above.
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