- What is the variable in our ELISA?
In our assays, the
**variable**is the concentration of "competitor" rabbit IgG. Everything else is held constant (or omitted in some controls). - A
**standard curve**graph (see sketch at right) has known concentrations of IgG, micrograms/ml, on the X axis (abcissa). Since the concentration of IgG competitor is increasing, the**slope of the curve is negative**. In the standard curve Eric did, 35% inhibition occurred at 0.65 micrograms rabbit IgG/ml, hence that value on page 111. - Eric chose 35% inhibition as his "
**reference level**" for the graph on page 111 because it is about half way between the maximum A450 1.38 ("no competition" control), and the minimum A450 control 0.56 ("maximum competitor"). So it is at the point of maximum slope; that is,**maximum sensitivity**of the assay to change in concentration of competitor. The best reference level for any given plate is half way between the minimum and maximum A450 values obtained; that is,**half-maximal inhibition**. Half way between 1.38 and 0.56 is 0.97. Eric used 35% inhibition, that is 1.38 - (0.35)(1.38) = 0.90, which is close to half-way between the maximum and minimum values. When you graph your ELISA data, you should calculate the A450 that is**half way between your maximum and minimum A450 control values**. This will be the best "reference level" for comparing your standard curve and unknowns. - How do we
**get the unknown concentration from the standard curve**? In Eric's plate, the standard IgG, at known concentrations, gave 35% inhibition at 0.65 micrograms/ml. So when the unknown is at a dilution that gives 35% inhibition, it must also be at 0.65 micrograms/ml. If you think about it, this should be clear, but if you need help, see page 110 for how to use this to calculate the concentration of IgG in your unknown from your ELISA results. - Page 111 does not show the standard curve (which was done in
the same plate). It shows the unknown (normal rabbit serum), with
**dilution-fold**increasing on the X axis. As dilution-fold increases, concentration of IgG competitor decreases, so the**slope is positive**. - The 3 data points at the top of page 111 labeled "No competition" are just replicate sets of wells (their positions on the X axis don't mean anything). The horizontal line is the average value of the 3 averages.
- In order to
**interpolate from a semi-log plot**, you need to understand what the tick marks mean. For example, in between dilution-folds of 1000 (1K) and 10,000 (10K), the tick marks mean 2K, 3K, 4K, 5K, 6K, 7K, 8K, and 9K fold-dilutions. Drawing the line between observed data averages, find where it crosses 35% inhibition, and draw a vertical line connecting that crossing point to the X axis. There you can read from the tick marks the approximate fold-dilution that gave half-maximal inhibition. - Interpolating the curve on page 111 shows that 35% inhibition occurred at a dilution of 1/7,000. Already, because of the tiny divisions on the graph, we can't get more significant digits, that is, we can't tell 1/6,500 reliably by interpolation from 1/7,500. However this source of uncertainty is smaller than the uncertainty inherent in the ELISA as we are doing it.
- Thus the unknown sample tested on page 111 (normal rabbit serum) had (7,000 times 0.65 micrograms/ml) equals 4.5 milligrams/ml. (Is this a reasonable value? Textbooks often quote "typical" values of 10-12 mg/ml, so it is within 3-fold of the textbook value.)
**Error bars**on all points are SEM's calculated with the SEM program available on the lab computers. (SEM is freeware downloadable from the class home page -- look for "SEM Free Software Download").- As you know, +/- twice the SEM is the 95% confidence interval. By interpolating from +/- twice the SEM bars, you can interpolate the minimum and maximum dilutions within the 95% confidence interval that give 35% inhibition. For the graph on page 111, these were 1/4,000 and 1/10,000. Using these, you can calculate the 95% confidence interval for your result. In Eric's result on page 111, normal rabbit serum contains 4.5 mg IgG/ml with a 95% confidence interval from 2-9 mg/ml. Remember the textbook value of 10-12 mg/ml? So 9 mg/ml is close enough to that to be perfectly reasonable.