Help with real-life biochemistry lab math problem.

[mspittel]

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.

 

[EnumaElish]

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, $F(z) = \int_{-\infty}^z f(x) dx$.

Also, $F(-\infty) = 0 \text{ and } F(+\infty) = 1$. 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(v_c). 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.

 

[tacman]

Hello,

Thank you for reaching out for help with your biochemistry lab math problem. I understand that you are looking for a formal calculus equation that relates the area under the curve to the concentration of the virus in the original sample. I am happy to assist you with this.

First, let's define some variables for the equation:

A = area under the curve (measured in mm^2)
C = concentration of virus in the original sample (measured in ug/mL)
S = sensitivity (0.1 AU at full scale)
SA = specific absorbivity (36 AU for 1 gram of virus in 100 mL)
V = sample volume (measured in mL)
CS = chart speed (measured in mm/min)
FR = flow rate (measured in mL/min)

To begin, we can use the definition of specific absorbivity to relate the absorbance of the virus (A) to the concentration (C):

A = SA * C

Next, we need to relate the absorbance (A) to the sensitivity (S) and the chart speed (CS). We can use the fact that the chart speed is measured in mm/min to convert it to mm^2/min:

A = S * CS * t

Where t is the time in minutes.

Now, we need to relate the time (t) to the sample volume (V) and the flow rate (FR). We can use the definition of flow rate (FR = V/t) to get:

t = V/FR

Substituting this into the previous equation, we get:

A = S * CS * (V/FR)

Finally, we can substitute the value of A from the first equation into this one to get the final equation:

SA * C = S * CS * (V/FR)

Solving for C, we get:

C = (S * CS * V)/(SA * FR)

This is the formal equation that relates the area under the curve to the concentration of the virus in the original sample. I hope this helps and that you are able to use this equation to solve your problem. Good luck with your lab work!