LEVEL I
Materials
Procedure
| Tube # |
Final Concentration of Dopachrome (mM) |
|---|---|
| 1 |
0 |
| 2 |
0.125 |
| 3 |
0.25 |
| 4 |
0.5 |
| 5 |
1.0 |
| 6 |
2.0 |
| 7 |
4.0 |
| 8 |
8.0 |
With these dilutions, you have prepared tubes containing concentrations from 0 to 8 mM dopachrome (tubes 1-8). Tube 1 contains no dopachrome and is used for blanking the spectrophotometer.
The units of concentration are millimolar (mM). A 1.0 mM solution contains .001 moles per liter or .000001 moles per ml. Thus, with a volume of 3.0 ml, there are .000003 moles of dopachrome, or 3 micromoles. Correspondingly, tubes 2-8 contain 1 to 24 micromoles of dopachrome. For the remainder of this exercise, be sure to distinguish between concentration (mM) and total amount of substance present (micromoles).
| Tube |
Concentration of Dopachrome (mM) |
Absorbance |
A/C |
| #1 |
0 |
0 |
---- |
| #2 |
0.125 |
||
| #3 |
0.25 |
||
| #4 |
0.5 |
||
| #5 |
1.0 |
||
| #6 |
2.0 |
||
| #7 |
4.5 |
||
| #8 |
8.0 |
You can more accurately determine the extinction coefficient by performing a linear regression analysis of your data, and computing the slope and y intercept. The slope of the linear regression will represent the extinction coefficient for your sample.
Notes
Since tyrosinase catalyzes the conversion of L-DOPA to dopachrome, this exercise measures the conversion of colorless DOPA to the dark orange dopachrome. Substrate and product are in a 1:1 ratio for this reaction, thus the amount of product formed equals the amount of substrate used. The optical density of dopachrome @475 nm is directly proportional to the intensity of orange color formation in solution (Beer-Lambert Law).