Friday, 18 December 2009

Biochemistry Exams - 2007/2008 3rd year Undenominated Science Semester I - Question 5

Yo, yo, yo!

I said I am back:) This evening together with Question 4 we expose Question 5 of the Sample Exam Paper for 3rd year Undenominated Science (to go to paper click here).

Ok, let's go team.

Question 5 For a competitive inhibitor of an enzyme:

(a) Draw a Lineweaver-Burk (double reciprocal) plot for the enzymatic reaction
in the presence and in the absence of the inhibitor
(b) Explain why the intersection of the lines, plus or minus inhibitor, occurs where
it does
(c) If the inhibitor is present at a concentration equal to its Ki, by how much does
the slope of the plot increase?

Part A) and B) Lineweaver-Burk is a double reciprocal of Michaelis-Menten plot, where [1/V] vs 1/[S] values are plotted. Picture below:

OK, now depending on the nature of inhibitor Km (Michaelis constant, a specific substrate concentration where V = 0.5Vmax) and Vmax (maximum velocity) values behave differently. When there is a presence of:

- competetive inhibitor, Vmax do not change but Km increase. This is because of the competition between inhibitor and substrate for active site of the enzyme. If more of the substrate has to be added to achieve the same result, it means that Km has to increase as well. We can observe that at the L-B plot, see below:

- non-competetive inhibitor interacts with enzyme in a difference place than active site. Its action changes conformation of the protein, what leads to deacrease in substrate processing. In this case adding more substrate will not increase velocity. Km stays constant Vmax decrease, see below.

There are also other inhibitors: reversible, irreversible, mixed inhibitions etc. Their effect on the Km and Vmax might be highly specific or a combination of the above.

Part C) Hmm it is a shame but I am not sure about the answer. I tried to solve it this way:

Remember about the red text:).



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