Enzymes

Noteworthy

Always keep in mind the following equations:

Protein-ligand binding

1) K_a = [PL] / [P] [L] also = K_a = [PL] / [P], when [L] is a lot higher than [P]

2) Fraction of occupied binding sites = [L] / [L] + K_d

3) log (occupied binding sites / empty binding sites) = n log [L] - log K_d (Hill equation)

Enzymes

4) V₀ = V_max [S] / K_m + [S]

5) V_max = K_cat [E_t]

6) Specificity constant = K_cat / K_m

Protein-ligand binding problems

An antibody binds to an antigen with a K_d of 5 X 10^-8 M. At what concentration of antigen will the fraction of occupied binding sites be 0.2?

I know the value of K_d and the value of the fraction of occupied binding sites.

The question is asking for the concentration of antigen (or ligand) = [L]

Then I would use the second equation

Fraction of occupied binding sites = [L] / [L] + K_d

Plugging the values in the equation I got

0.2 = [L] / [L] + 5 X 10^-8 M; rearranging I got. [L] = 5 X 10^-8 M X 0.2 / 1-0.2 = 1.25 X 10^-8

Antibodies usually contains two binding sites per molecule, considering this factor what would be then concentration required to get 0.2?

In this case several equations could be used, I would used the third one

3) log (occupied binding sites / empty binding sites) = n log [L] - log K_d (Hill equation)

This equation could be arranged as follow

(occupied binding sites / empty binding sites) = [L]ⁿ / K_d

Plugging the provided data I got

0.2/1-0.2 = [L]² / 5 X 10^-8 M ; rearranging I got [L] = √ o.25 X 5 X 10^-8 M = 1.1 X 10^-4

Interesting fact is that increasing the number of binding sites per protein increases the need for more ligand to obtain the same number of occupied sites. In this case, 10,000 more molecules (1.1 X 10^-4 / 1.25 X 10^-8).

Enzymes Problems

Problem 1

A research group discovers a new version of studentase, which they called Rogerase, that catalyses the chemical reaction:

Student ------- Professional

The researchers begin to characterize the enzyme.

a) In the first experiment, with [E_t] at 4 nM, they find that the V_max is 1.6 microM s^-1. Based on this experiment, what is the K_cat for Rogerase?

I got [E_t] = 4 nM and V_max = 1.6 microM s^-1, I need K_cat.

Using the equation number five; I got K_cat = V_max/[E_t] or K_cat =1.6 microM s^-1/ 4 X 10^-3 microM = 400 s^-1

b) In another experiment with [E_t] at 1 nM and [Student] at 30 micfroM, the researchers find the V_o = 300 nM s^-1. What is the measurement K_m of Rogerase for its substrate Student?

Now try to do a and b parts of problem 10 from your book.

Problems 2

1. The following data were obtained from an enzyme kinetics experiment. Graph the data using a Lineweaver-Burk plot and determine, by inspection of the graph, the values for K_m and V_max.

[S] (µM) V (nmol/min)

_______ ___________

2. Use the Michaelis-Menten Equation to calculate the missing values of [S] given below if V_max = 2.5 mmol/min. Plot [S] versus V (NOT the reciprocals!). Draw line parallel to the x-axis at V_max and extend your plotted line to show its approach to V_max.

[S] (mM) V (mmol/min)

_______ ___________

[S]₁ 1.7

[S]₂ 2.1

[S]₃ 2.2

[S]₄ 2.5

3. The effect of an inhibitor on an enzyme was tested and the experiment gave the results below. Plot the data and determine, by inspection of the graph, what type of inhibition is involved.

[S] µM V (µmol/min) V (µmol/min) V (µmol/min)

with 0.0 nM with 25 nM with 50 nM

Inhibitor Inhibitor Inhibitor

______ ___________ ___________ ___________

0.4 0.22 0.21 0.20

0.67 0.29 0.26 0.24

1.00 0.32 0.30 0.28

2.00 0.40 0.36 0.32