You are carrying out the enzyme kinetics experiment with purified Alkaline Phosp

ID: 1009482 • Letter: Y

Question

You are carrying out the enzyme kinetics experiment with purified Alkaline Phosphatase (AP). You are provided with the following stocks solutions, (a) 0.4M Tris-HCl, pH 8.0 (b) 100 uL of AP (1mg/mL) (c) 1 M PNPP and (d) ddH2O AP works optimally in presence of 0.2M Tris-HCl, pH 8.0 , and 1 µg AP. You need to perform AP assay at two different substrate concentrations (shown in table below) in a 2 mL reaction volume. Remember, the smallest volume that you can measure with a micropipette is 2µL. Fill in the table below with the volumes of all of the reagents that you will need to perform the assay. Be careful, you may need to make new stocks from your concentrated reagent stocks.

Final Concentration PNPP | 100 mM 100 mM mM | 0 mM PNPP Name of Working Stock solution (include concentration) V PNPP volume of volume of Working Stock solution V V Working Stock solution solution

Explanation / Answer

100 L AP solution contains (1mg)*(100 L/1 mL)*(1 mL/1000 L) = 0.1 mg = 100 g AP (since 1 mL = 1000 L and 1 mg = 1000 g).

So we have 100 L AP solution containing 100 g AP.

For optimum work, we need to put in 1 g AP in the final solution. The total volume of the final assay is 2 mL = 2000 L.

We will employ the dilution relation

V1*S1 = V2*S2 where V1 and V2 are the volumes of the stock and the assay solutions and S1 and S2 are the concentrations of the stock and assay solution.

Thus, for AP, we have,

V1*(100 g AP) = (2000 L)*(1 g AP)

===> V1 = (2000 L)*1/100 = 20 L AP

Thus, we need to put in 20 L AP into the test tube.

The concentration of Tris HCl is 0.4 M, pH = 8 and we need the final concentration to be 0.2 M, pH = 8 for optimum work.

Thus we have,

V1*(0.4 M, pH = 8) = (2000 L)*(0.2 M, pH = 8)

===> V1 = (2000 L)*0.2/0.4 = 1000 L = 1 mL

We must add 1 mL Tris-HCl, pH = 8 to the test tube.

The assay solution in test tube 1 has 100 mM = 0.1 M PNPP and the aupplied stock is 1 M PNPP. Therefore, using the dilution equation,

V1*1 M = (2000 L)*(0.1 M)

===> V1 = 200 L = 0.2 mL

Total volume of the three reactants = (20 + 1000 + 200) L = 1220 L = 1.22 mL. The rest must be double distilled water which is (2.0 – 1.22) mL = 0.78 mL = 780 L.

For the second part, we shall employ the same dilution equations as above and obtain the same volumes of Tris-HCl, pH = 8 and AP. We have 0 mM PNPP here, which means that we do not add PNPP to this solution. Thus, the total volume of Tris-HCl, pH = 8 and AP = (1000 + 20) L = 1020 L = 1.02 Ml. Hence, volume of dd water to be added = (2.00 – 1.02) mL = 0.98 mL = 980 L.

Final Concentration PNPP -->

100 mM PNPP

0 mM PNPP

Name of working stock solution (with concentration)

Volume of working stock solution (L)

Tris-HCL, pH = 8 (0.4 M)

1000

AP (1 mg/mL)

PNPP (1 M)

200

dd water

780

980

Total (L)

2000 = 2 mL