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.
  • 3
    $\begingroup$ related $\endgroup$
    – Alan Boyd
    Commented Sep 3, 2017 at 7:34
  • $\begingroup$ For 2. Think A. about the importance of MM kinetics, which was to establish that the reactions were catalysed by a single active site on the enzyme. Simple non-enzymic chemical kinetics would, instead, predict an order proportional to the product of concs of reactants. B. About the interaction of a substrate with the active site and why this has a hyperbolic shape. Consider collisions between entities (here a substrate and active site) to understand the hyperbola. 1 is trivial. $\endgroup$
    – David
    Commented Sep 3, 2017 at 15:03
  • 2
    $\begingroup$ @David 'trivial' is a subjective comment, which in itself is trivial according to what BioSE is for ('... biology researchers, academics, and students.') When there's a Q there's either an answer or it is yet to be unraveled. $\endgroup$
    – Tyto alba
    Commented Sep 3, 2017 at 15:28
  • 2
    $\begingroup$ Hint. Read sections 8.3.1 and 8.4 (start) of Berg online. Oh, and I don't know why you have a fixation on 10%. It's the initial velocity while the rate is linear, whatever that % may be. $\endgroup$
    – David
    Commented Sep 3, 2017 at 16:43
  • 3
    $\begingroup$ For a user with a high reputation you have been rather quiet in reponse to the answers suggested. Did either of them solve your problem? If so you might consider accepting the one that did. If not, perhaps you could explain the problem you still have. $\endgroup$
    – David
    Commented Sep 24, 2017 at 22:20

2 Answers 2


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 Km 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 eo << So (the initial enzyme concentration is very much less than the initial substrate concentration), we may take the So 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 Km and Vmax 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 Km values and at 'physiological' concentration of substrates do obey (to all intents and purposes) second-order kinetics. Catalase, for example, has a Km 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:

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

  • Cornish-Bowden (2004) Fundamentals of Enzyme Kinetics 3rd Edn. Portland Press, London.
  • 2
    $\begingroup$ This answer is far too complex to be comprehensible to a student who is obviously doing a first course in enzymology as part of a general biological sciences degree and is asking very basic of questions. Recommending she read Tipton or Cornish-Bowden is absurd. The answer to her question at the level she requires can be found in the Berg reference in my comment. It would have been better to let her read that (it is on line) and learn for herself. $\endgroup$
    – David
    Commented Sep 4, 2017 at 20:56
  • 1
    $\begingroup$ No, she's just starting enzyme kinetics and her question, though understandable, is a poor addition to SE Biology. I've seen the subjects she has been asking questions on over the last year and have a shrewd guess as to her background. I think it's wise to try to establish first what the poster is asking before writing over 700 words with two equations in which the terms are not even defined. I surmise that you have never taught university students if you imagine that they will even buy a biochemistry text such as Berg when doing a broad biology course, never mind advanced texts. $\endgroup$
    – David
    Commented Sep 4, 2017 at 22:51
  • 2
    $\begingroup$ Her answer is correct but shows only that she has learned the reactions of glycolysis and gluconeogenesis, not that she knows anything of kinetics or even the thermodynamics involved in the irreversible reactions. What Berg is trying to do is provide a visual way for the beginner to think about simple kinetic constants and, most important, that the saturation curve indicates the existance of an ES complex, the kinetics of which validate this. Many students, lacking chemical kinetics, do not understand the significance of the hyperbola, which is the point that needs explaining to them. I may. $\endgroup$
    – David
    Commented Sep 5, 2017 at 10:51

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).

Reaction orders

[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.)

V v. S for simple MM enzyme

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 — KM and kcat — 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.


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.)

  • $\begingroup$ Having given the poster time to think about it, I decided to post an answer to the conceptually interesting part of the question because the second question recently that does not seem to understand it, and it is generally taken for granted in texts. I suppose it needs a numerical explanation, but I didn't want to get bogged down in whether it was a hyperbola or not. I just wanted to get over the idea of why this kind of phenomenon has this sort of curve. $\endgroup$
    – David
    Commented Sep 6, 2017 at 21:49

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