Enzyme kinetics at the chemical level

Question

I am stumped by two questions:

1. Why do we take only the initial 10%(or may be 9.99999....%) of S conversion as the rate of the enzyme reaction. why not more than 10%?
2. Why doesn't the velocity keep on increasing linearly until the Vmax is reached (as all the S molecules get exhausted)

Research and attempts to answer:

Why v₀'s 10%?

1. Steady state kinetics apply when the [ES] is constant, on reaching 10% [P] (for a uni uni reaction) the backward reaction increases, increasing the [ES]?

2. On reaching 10% conversion the rate decreases, so we measure the optimum activity.

Why doesn't the velocity increase linearly until all the molecules are used up?

1. Since an enzyme's turnover number (kcat) is constant...

2. ...

Disclaimer :

1. 'have put two Qs together.
2. Didn't find relevant Q on BioSE.

Answer 1

The answer to the first part of your question is that we don't take the initial 10% of a progress curve (velocity vs time) as a measure of activity, but we measure the initial rate of the reaction. We do this by drawing a tangent at the origin.

It is merely a 'rule of thumb' that progress curves are practically linear provided that not more than 10% of substrate has been used up. If there is significant slowing down before 10% is used up - and there may be if the substrate concentration is way below K_m value - then it is not valid to draw a line through the points: we must draw the tangent to the curve. That is, it is not acceptable to take the average rate during the first 10% of the progress curve.

For this reason it is desirable to use as sensitive a method as possible when measuring an enzymic rate, and to never rely on a single time point (as is common in radioactive assays) without sufficient controls to justify that it it is the initial rate, and not the average rate (or something else) that is being measured.

The reason we measure initial rates is that is makes things simpler on two fronts.

Firstly, the decrease in rate with time of an enzymic reaction may be due to many factors: product may be inhibiting the reaction, the reverse reaction may become important, the enzyme or substrate may be unstable. It is only during the initial rate period that conditions are accurately known, and such effects may be legitimately ignored. It was Michaelis and Menten that first realized the importance of measuring initial rates, and of the great simplification principle that ensued, in enzyme kinetics. (see Cornish-Bowden, 2004).

Secondly, the rate laws themselves are greatly simplified. Provided that e_o << S_o (the initial enzyme concentration is very much less than the initial substrate concentration), we may take the S_o as being equal to the substrate concentration and 'plug in' this value into our rate law (such as the Michaelis-Menten equation), and ignore the effect of time on the value of S. In addtion, we can set all product terms to zero. This can greatly simplify things.

For example, the rate law for a reversible single-substrate mechanism (see here) is the following:

$$ v = { {{{V_{max}^f}\over{K_{m}^s}}\ S\ -{{V_{max}^r}\over{K_{m}^p}}\ P }\over{1 + {{S}\over{K_{m}^s}} + {{P}\over{K_{m}^p}}}}\ \ \ \ \ (1)$$

This one is not as complicated as it looks: we have just defined a K_m and V_max for the forward and reverse directions. But let's set product to zero:

$$ v_i = { {{{V_{max}^f}\over{K_{m}^s}}\ S_o\ }\over{1 + {{S_o}\over{K_{m}^s}} }}\ \ \ \ \ = \ \ \ {{V_{max}^f S_o}\over{{K_{m}^s} + S_o}}\ \ \ \ (2)$$

We now obtain the familiar Michaelis-Menten Equation (where $v_i$ is the intial velocity).

The second part of your question is more fundamental. I'll take it to mean the following. Why does an enzymic reaction not follow second-order kinetics at all substrate concentrations? Why doesn't doubling the substrate always double the rate (as is often the case in chemical kinetics)? A. J. Brown in 1902 suggested that the reason for this is the formation of an enzyme-substrate complex and that at sufficiently high substrate concentration all the enzyme would exist in such a complex and the enzyme would become 'saturated'. The formation of an enzyme-substrate complex, of course, is now a cornerstone of enzyme kinetics. One of the early key pieces of evidence for an enzyme-substrate complex was the observation that an enzyme is much more heat-stable in the presence of its substrate than in its absence.

Finally, it is probably worth pointing out that some enzyme operate way below K_m values and at 'physiological' concentration of substrates do obey (to all intents and purposes) second-order kinetics. Catalase, for example, has a K_m for hydrogen peroxide of about 1 Molar (Ogura, 1955), and one assumes that physiological concentrations never reach this level. (We need to be a little careful about catalase, however: it is one of the few enzymes that is inactivated by substrate).

A great reference on enzyme assays is the following:

• Principles of enzyme assay and kinetic studies by K. F. Tipton in Enzyme Assays: A practical approach

For an introduction to enzyme kinetics, and the early history of enzymology, I like the following:

• Cornish-Bowden (2004) Fundamentals of Enzyme Kinetics 3^rd Edn. Portland Press, London.

Answer 2

The kinetics of enzyme catalysed reactions is a specialized area of biochemistry. It was historically important when much less was known about enzymes and the tools for studying them were limited. Why, then does it retain a place in contemporary text books? I would suggest that it illustrates how many of our ideas of enzymes and their reactions were developed, and that on a more mundane level it provides a cheap basis for laboratory experiments where students can encounter some of the pitfalls of experimental science and also learn to test ideas numerically.

For the average innumerate contemporary student of biology struggling with the Michaelis–Menten equation it can be easy to miss what I regard as the most important concept that this illustrates. This is that the kinetic analysis of simple enzymes supported…

…a model of the enzyme in which the substrate bound and the reaction occurred at a specific site — the active site.

How does it do this? To understand that one must first contrast these enzyme kinetics with those of uncatalysed chemical reactions. It is assumed that chemical reactions occur due to the collisions of the reacting molecule. Depending on the reaction, the rate (r) of a chemical reaction and will be proportional to some power (n) of the concentration of the reactant(s) (S):

r = k[S]ⁿ

and the ‘order of the reaction’ is n — often 1 or 2 (first and second order). (k is the rate constant, which reflects how this proportionality is manifested in different cases.)

What this means is that if one plots r (product per time) against the concentration of the reactant one gets a line of a particular shape (linear or curvilinear — see below) in which the rate continues to increase as the concentration of reactants increase (assuming the formation of product is small in comparison).

[Taken from Askiitians]

In contrast, the rate of enzymic reactions (velocity v) reach a maximum (Vmax), approaching this asymptotically as shown. (This is for a fixed amount of enzyme, as shown for an enzyme obeying simple Michaelis–Menten kinetics.)

The existence of a Vmax is consistent with the idea of a single (or small finite number of) active sites on the enzyme — when all enzyme molecules are saturated (all functioning) increasing the concentration of reactant (substrate) can provoke no further increase in rate.

Finally to the poster’s question (or at least the part that is most interesting conceptually): Why is this maximum velocity approached asymptotically? To appreciate this one should be aware of the disparity in numbers of molecules of reactant and enzyme — for glucose at 5mM and hexokinase at 20µM this is about 250 to 1. As the concentration of reactant is increased the number of ‘free’ enzyme molecules will decrease and a corresponding greater number of reactant molecules will be needed to produce a successful collision with, and binding to, the active site. This effect is small initially, but becomes successively more severe, so that in practice Vmax is never reached and has to be calculated by extrapolation.

And however much the purists may object, the two kinetic constants that emerge from numerical analysis — K_M and k_cat — illuminate for the student two key aspects of the catalytic site. These are that it has a particular binding affinity (which may vary for different substrates) and that it has an intrinsic catalytic capability. So for those who wish to learn to walk before they can run, I recommend section 8.4 of Berg et al., which can be read online.

Addendum

And the answer to the first ‘boring’ part of the question is shown in the insert added to the second figure.The kinetic analysis is only valid if you consider the initial velocity (for which you have presumably been given the arbitrary figure of 10%) while the rate is linear with time. (There are various practical reasons linearity is lost with time — removal of substrate, back reaction as products build up, denaturation of the enzyme etc.)