Question

In an 'Enzyme Kinetics' experiment, the following calibration data, product (p-nitrophenol, PNP) concentration (M) versus UV Absorbance 405 nm, was obtained If the UV absorbance 405 nm within the 1.5 mL-cuvette was measured as 0.144 in one of the trials during the experiment, determine the amount of pNP (in mg) produced. Assume total volume of reaction mixture stays at 1.5 mL at all times. (Round up to two decimal places) Molecular Weight of PNP: 139.11 g/mol PNP Concentration UV Absorbancem (MM) 405 nm 7.19E-04 0.375 3.59E-04 1.80E-04 0.125 1.44E-04 0.075 7.19E-05 0.05 0.2

Answer 1

This problem can be solved by using the Beer Lambert's Law.

According to Beer Lambert's Law;

A = E C L

where;

A = Absorbance

E = Molar absorptivity

C = Concentration of solution

L = Length of cuvette

The plot of A vs C is a straight line with slope = E L.

The plot is given below:

pNP Concentration	Absorbance
0.000719	0.375
0.00035	0.2
0.00018	0.125
0.000144	0.075
0.0000719	0.05
0	0

Absorbance vs PNP Concentration Absorbance - 515*x + 0.0118 0.0 + 0.0000 0.0002 0.0004 0.0006 C(MM)

The equation of the fitted line is :

A = 515 C + 0.0118

Given;

In one of the measurements;

A = 0.144

0.144 = 515 C + 0.0118

which on solving gives;

C = 0.00026 mM

Volume of solution = 1.5 mL = 0.0015 L

Number of moles = 0.00026 X 0.0015 = 3.9 X 10^-7 mmol = 3.9 X 10^-10 mol

1 mole = 139.11 g

Therefore;

Amount of pNP produced = 3.9 X 10^-10 X 139.11 = 5.425 X 10^-8 g

Amount of pNP produced = 5.425 X 10^-5 mg