We work with a membrane protein system where measuring the affinity between the enzyme and the upstream activating protein has been difficult, and when measured in detergent solution, it is almost 100 fold lesser (ie ~100nM) whereas the EC50 in an enzymatic assay using vesicles in ~1-2nM. Would it be reasonable to say that the "real" affinity is ~1nM than 100s of nM based on the biochemical assay?

Alternately, is there a documented system where a huge discrepancy exists between measurements from direct binding and biochemical assays?


2 Answers 2


You can certainly get massive differences between EC50 and affinity. This is especially true for cell-based assays and membrane protein systems.

The reason why is because the appropriate time scales to achieve binding equilibrium (hrs for nM affinity, days for picomolar, feptomolar affinity according to back of the envelope calculations) may be and likely will be different from the appropriate time scales of activation, endocytosis, degradation, etc. Furthermore, you can have avidity effects that would not be apparent in a dilute membrane assay compared to a cell's surface. It could very much be that the dominant terms in mass transport change depending on the geometry and antigen density and while I'm not entirely sure bout vesicles vs. emulsions it is true for cancer models. Theoretical analysis of antibody targeting of tumor spheroids: importance of dosage for penetration, and affinity for retention.

Case study number I: hGH:hGHR interactions have been extensively studied and high affinity hGH have been produced. However, none of the variants showed improved EC50. It turns out that the binding was primarily influenced by the off-rates which were so slow, that affecting those values did little to the kinetics of the system. Growth Hormone Binding Affinity for Its Receptor Surpasses the Requirements for Cellular Activity

Case study number II: CEA-antibodies have been affinity matured to have pM activity compared to the wild-type nM affinity. However, the antibody had minimal effect on tumor uptake. As CEA gets endocytosed on the order of 30 minutes, all of the antibody was simply getting endocytosed rather than acting as an inhibitor. Directed evolution of an anti-carcinoembryonic antigen scFv with a 4-day monovalent dissociation half-time at 37°C.

(edit) I realized that this answer is more applicable to IC50s vs. the EC50 experiments described above. However, many of the points remains. The Km between a substrate and its enzyme is a thermodynamic property where as the EC50 also has dynamic factors that may influence its measurement.

I would also add that simply adjusting the pH 2 units may result in a 2 orders of magnitude change in activity.

(edit2) A further point to make about enzymatic assays is that from an observation of Michaelis-Menten kinetics, the KM of the reaction isn't the KD of the enzyme but alternatively the effective binding constant.

Recall that:

enter image description here

If there is a non-trivial kcat comparable to the dissociation rate kr, the enzyme will have an effective half maximal concentration that is significantly higher than the Kd. There are multiple ways how your assay can result in the discrepancies that you're seeing. The first would your strategy in measuring your EC50. If one is an enzymatic readout and the other is a binding readout, you would naturally have very different results. Alternatively, the enzyme in the detergent system may be catalytically more active than the vesicle system resulting in a higher effective KM. Again, the pH thing.

Finally, there is always that awkward moment when you realized that you might accidentally have measured a titration curve rather than a binding curve.

  • 1
    $\begingroup$ bobthejoe: I like the detail in your answer, especially the hGH example is very surprising. However, my question pertains to comparing affinities and EC50 measurements, both being derived from in vitro assays with purified proteins. I would think the avidity argument might hold, especially when the same system were compared in a vesicular environment versus detergent solution. $\endgroup$
    – gkadam
    Aug 16, 2012 at 20:28
  • $\begingroup$ bobthejoe, just thought I'd say this: You've laid out really well why interpreting Km is not always trivial. Exactly why I chose to focus on EC50 (with the trivial(?) assumption that substrate far exceeds enzyme concentration). I'm going to accept your answer, given the richness of detail that you've laid out! $\endgroup$
    – gkadam
    Aug 20, 2012 at 4:49

In what follows I am going to attempt to answer your question using a specific example of (competitive) reversible activation, and I hope to show what a misleading parameter EC50 can be. (Rate law derivation is an area of interest, hence the long-winded answer).

It can of course give information, but IMO it needs to be used with extreme care.

I am restricting my comments to the case where A is a reversible activator and where [A] >> eo and [S] >> eo. (All constants are defined below).

In short, I am not treating irreversible activation, and I am not treating the case where A is considered a tight-binding activator (so that [A] ≈ eo). In other words, I am making the steady-state approximation (Briggs & Haldane, 1925)

Consider the following simple mechanism for activation in a single-substrate enzyme:

A Mechanism for Competitive, Reversible Activation

Competitive Activation DIAGRAM

The substrate (S) cannot bind to the free enzyme (E) unless the activator (A) is present. That is, only the activator can bind (reversibly) to E, and the substrate binds (reversibly) to the EA complex. Reaction occurs within the EAS complex (giving EAP). Dissociation of product (P) from EAP gives EA.

The (steady-state) initial-rate law for the above mechanism is the following:

Basic Rate law in three parts eqn1

As stated, all constants are defined below and follow the usual definitions in enzyme kinetics. I'll mention just one here: KA is the dissociation constant for the activator, and has units of concentration.

I derived these equations using a computerized version of the King & Altman method (King & Altman, 1956), but it is not difficult to do. It was pure laziness that prevented a 'first principles' derivation. Rate law derivation is treated in depth in most books on enzyme kinetics and good accounts may be found in Segel (1975) and Cornish-Bowden (2004).

One can now make the following points

  • There is no activity without A (as one would expect).
  • When [A] >> KA, the rate law becomes identical with the Michaelis-Menten equation (as one would expect).
  • The equation is similar to the case of competitive inhibition, but (note subtle change) with (1 + KA / [A]) substituted for (1 + [I] / Ki)
  • One can consider this mechanism an example of competitive activation


  • νiexc     =     the initial velocity when A is in vast excess [ie setting [A] equal to ∞ in eqn (1) ].
  • EC50    =     the concentration of A that gives νi equal to νiexc / 2 (half-maximal activation)

allows the following two equations to be derived:

Ratio of velocities eqn3

Setting the left-hand side of the above equal to 2, and solving for [A], gives the following all-important equation

EC50 definition eqn3

One can now draw the following conclusions

  • When [S] = KSm

    Conclusion1 S equals to Km

    In words, when the substrate is present at a concentration equal to the Michaelis constant the value of EC50 is half the value of KA!.

  • When [S] >> KSm

    Conclusion 2 S greater than Km

    When [S] = 10 x KSm, for example, EC50 is one-tenth the value of KA, and when [S] = 100 x KSm the figure is one-hundreth.

  • When [S] << KSm

    Conclusion 3 S less than Km

Thus, for a competitive activator, it is only when [S] << KSm that EC50 gives a reasonable estimate of KA.

The substrate concentration in a 'normal' enzyme assay will probably be equal to or greater than the Michaelis constant (in order to 'see' activity), and thus EC50 will be an underestimate of KA.

If the substrate is present at a 'saturating' concentration (say 10 x KSm) then one is in real trouble.

However, this is not the only consideration. As well as being substrate-concentration dependent, the manner in which EC50 depends on [substrate] depends on mechanism.

As briefly discussed below, the competitive activation case is analogous to that of competitive inhibition. However for uncompetitive inhibition the Ki (inhibition constant) is only equal to IC50 (the concentration of inhibitor giving half-maximal inhibition) at high substrate concentration! (see Naqui, 1982). By analogy, one can conclude that the same is true of uncompetitive activation.

It is very difficult to see how EC50 can be of any practical use, except perhaps when the mechanism of activation is known. At the very least, conclusions drawn from EC50 data should otherwise not be considered definitive.

In addition, I have discussed the only the case of a single-substrate reaction. Two-substrate reactions have the potential for even greater complexity.

Illustrative Example (Competitive Activation)

Let's take an example, and check some conclusions:

  • kfcat   =   1000 s-1
  • eo   =   1
  • KSm    =   10 µM
  • KA      =   50 µM

It may be calculated that:

  • When [S] equals 10 µM (ie when [S] is equal to KSm) and when [A] is 'saturating' (for the sake of argument set equal to ∞) the velocity is 500 µM s-1 [from eqn (1) ].

    From eqn (2), the concentration of A required to achieve half this value (half maximal activation) is 25 µM (EC50 equals 25 µM)

    This value can be directly calculated from eqn (3)

  • When [S] equals 100 µM (ie [S] equals 10 x KSm) and when [A] is 'saturating' the velocity is now 909 µM s-1 [from eqn (1) ].

    From eqn (2), the concentration of A required to achieve half this value is now only ~4.5 µM

    This value can be directly calculated from eqn (3) [EC50 equals (50 / 11) µM]

  • When [S] equals 1 µM (ie [S] equal to one-tenth the Michaelis constant) and when [A] is 'saturating' the velocity is now only ~91 µM s-1 [from eqn (1) ].

    From eqn (2), the concentration of A required to achieve half this value is ~45.5 µM (now almost equal to KA)

    This value can be directly calculated from eqn (3) [EC50 equals (500 / 11) µM]

Reversible Inhibition & IC50. A Brief Comment

There is nothing much new in any of the above. In the field of enzyme inhibition, the inadequecies of IC50 (the concentration of inhibitor giving 50% inhibition) and the relationship to the Inhibition constant are well-known [see, for example, Chou (1974), Segel (1975), Naqui (1982) & Tsou (1987)], but (it seems to me) often ignored.

The paper of Naqui gives a very succinct analysis of the situation and the pdf is freely available to all.

Reversible inhibitors may be divided into four classes, depending on the pattern of inhibition: competitive, uncompetitive, non-competitive and 'mixed' (see Naqui, 1982, and references therein).

In all cases except non-competitive inhibition, Ki ≠ IC50, and IC50 is substrate- dependent.

(Non-competitive inhibition is the 'fluky' one where it just so happens that the catalytic constant and the specificity constant (kfcat / KSm) are changed to the same extent or, to put it another way, the slope and intercept of a Lineweaver-Burk plot (it just so happens) are changed to the same extent)

For an uncompetitive inhibitor (as stated) the IC50 only equals Ki at high substrate concentration (see Naqui, 1982).

Like the case of competitive activation, IC50 equals Ki for a competitive inhibitor only at low susbstrate concentrations (see Naqui, 1982).

'Mixed' inhibition is even more complex, and the inhibition constant may be greater than or less than IC50 depending on experimental conditons (see Naqui, 1982).

Great question, BTW.


As I have derived the full form of the rate law (as opposed to the initial-rate law), it is probably worth posting it. The initial-rate law is obtained from this equation by setting [P] equal to zero.

Full Rate Law for Competitive Activation Kinetic Constant form

Kinetic Constants defined in terms of Rate Contants

Rate constants are numbered (as subscripts) from species, to species, with numbering as in diagram.

Thus the rate constant k1,2 is that from E (species 1) to EA (species 2).

  • Catalytic Constants

    enter image description here

    kcatreverse rate constant form

  • Michaelis Constants

    KmS Rate Constant Form

    KmP Rate Constant Form

  • Specificity Constants

    Specificity Constant for S rate constant form

    Specificity Constant for P rate constant form

  • Activation (Dissociation Constant)

    Ka Rate Constant Form

Kinetic Constant Definition

  • ν velocity

  • νi   the initial velocity

  • KSm    the Michaelis Constant for S

  • KPm    the Michaelis Constant for P

  • kfcat   the catalytic constant (in the forward direction)

  • krcat   the catalytic constant (in the reverse direction)

  • Vmax    the maximum velocity, equal to kfcat eo

  • eo   the total enzyme concentration

    enter image description here

  • KA      the dissociation constant for (activator), equal to k2,1 / k1,2, with units of concentration (Molarity, say) [rate constants are numbered from species, to species, numbering as in diagram].

  • νiexc    the initial velocity when A is in excess ([A] = ∞).

  • EC50   the concencentration of A that gives νi equal to νiexc / 2 (half-maximal activation)

  • [A]    The initial concentration of activator.

  • [S]    The initial concentration of substrate.

The rate law and the diagrams were obtained using Mathematica. For Matematica at SE see here


  • Briggs, G. E. & Haldane, J. B. S. (1925). A note on the kinetics of enzyme action. Biochem. J. 19, 338 - 339.[pdf]

  • Chou, T.C. (1974) Relationships between inhibition constants and fractional inhibition in enzyme-catalyzed reactions with different numbers of reactants, different reaction mechanisms, and different types and mechanisms of inhibition. Mol. Pharmacol., 10, 235-247 [pubmed]

  • Cornish-Bowden, A. (2004). Fundamentals of Enzyme Kinetics. 3rd edn. Portland Press Ltd, London.

  • King, E. L. & Altman, C. (1956). A schematic method of deriving the rate laws for enzyme-catalyzed reactions. J. Phys. Chem. 60, 1375 - 1378. [ACS site]

  • Naqui, A. (1982). What does I50 mean? (1983) Biochem. J. 215, 429-430 [pdf]

  • Segel, I. H. (1975). Enzyme Kinetics. Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems. John Wiley & Sons, Inc., New York.

  • Tsou, C. L. (1987) The screening of enzyme-targeted drugs. Bioessays, 6, 237-238. [pubmed]


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .