Characterisation of enzyme kinetics can play an important role in drug discovery. A good understandi — Physical Chemistry — Kinetics Chemistry Question
Enzyme kinetics
Characterisation of enzyme kinetics can play an important role in drug discovery. A good understanding of how the enzyme behaves in the presence of its natural substrate is necessary before the effect of potential drugs can be evaluated. Enzymes are typically characterised by two parameters, Vmax and Km; these are determined by analysing the variation of the initial rate of reaction with substrate concentration.
Many enzymatic reactions can be modelled using the scheme:
E + S → ES rate constant k1
ES → E + S rate constant k–1
ES → E + P rate constant k2
where E is the free enzyme, S is the substrate, ES is a complex formed between the enzyme and substrate and P is the product.
10.1 Assuming that the system is in steady state and that [S] >> [E] obtain an expression
i) for the rate of production of ES in terms of [E], [S], [ES] and the appropriate rate constants.
ii) for the rate of production of P in terms of [ES] and the appropriate rate constants.
Model Answer
10.1 i) d[ES]/dt = k1[E][S] – (k–1 + k2)[ES]
ii) d[P]/dt = k2[ES]
When doing the experiment [E] is not known, however the total amount of enzyme present is constant throughout the reaction, therefore:
[E]0 = [E] + [ES]
where [E]0 is the initial enzyme concentration.
Also, in enzyme kinetics the Michaelis constant, Km, is defined as:
Km = (k–1 + k2) / k1
10.2 Obtain an expression for [ES] in terms of [S], [E]0 and Km.
Model Answer
10.2 [ES] = ([S][E]0) / (KM + [S])
10.3 Hence obtain an expression for the rate of production of P in terms of [E]0, [S] and the appropriate constants.
Model Answer
10.3 d[P]/dt = (k2[E]0[S]) / (KM + [S])
The maximal rate of reaction, Vmax, occurs when all of the enzyme molecules have substrate bound, i.e. when [ES] = [E]0, therefore:
Vmax = k2 × [E]0
10.4 Obtain an expression for the rate of production of P in terms of Vmax, [S] and the appropriate constants.
Model Answer
10.4 d[P]/dt = (Vmax[S]) / (KM + [S])
The enzyme GTP cyclohydrolase II catalyses the first step in riboflavin biosynthesis in bacteria:
The absence of this enzyme in higher organisms makes GTP cyclohydrolase II a potential target for antimicrobial drugs.
Protein samples were rapidly mixed with different concentrations of GTP. The change in absorbance with time was measured at 299 nm in a 1 ml cell with a 1 cm pathlength. A solution with a concentration of 100 µmol dm–3 of the purified product gave an absorbance of 0.9 in a 1 cm pathlength cell at 299 nm.
10.5 Calculate the initial rate of reaction at each of the GTP concentrations.
Model Answer
10.5 The extinction coefficient of the product is calculated to be 9000 mol dm3cm–1 at 299 nm. The product concentrations at each time point and the initial rate of production at each concentration of GTP are given in the table below:
10.6 Express the equation obtained in part (d) in the form y = mx + c.
Model Answer
10.6 There are a number of different linear forms for the equation obtained in part 10.4. Writing d[P]/dt as V, the simplest form is:
1/V = (KM / Vmax) × 1/[S] + 1/Vmax
10.7 Hence determine Vmax and Km for this enzyme (you may assume that the kinetic scheme outlined above is valid for this enzyme)
Model Answer
10.7
Plotting 1/Vo against 1/[GTP] gives:
Therefore the value for Vmax is 0.114 µmol dm–3 s–1 whilst that for KM is 50.5 µmol dm–3.