Can any one reaction in a cell be at equilibrium?
I think there are a few principles that we need to consider before answering your question.

(a) When a reaction is at equilibrium, the rate of any elementary reaction is exactly balanced by that of the reverse process.  This is an important one.

The above principle follows from transition-state theory, which holds that the activated state for the reaction in one direction is that same for that in the reverse direction.  It also follows from the principle of microscopic reversibility at equilibrium or, more correctly, the principle of detailed balance at equilibrium, which states that “in a system at equilibrium each collision has its exact counterpart in the reverse direction, and that the rate of every chemical process is exactly balanced by that of the reverse process” (my emphasis) [Laidler, 1987, p 130]. It is important to realise that these two points of view are equivalent.  To again quote Laidler (p130), if one is working within the framework of TST “the principle of microscopic reversibility presents nothing new”. 

To put it bluntly: a system where any elementary reaction, either explicitly or implicitly, is not exactly balanced by the reverse process is not at equilibrium. By this criteria, the actin example alluded to above cannot, even loosely, be considered at equilibrium. 

This principle has many important consequences, even when a system is not at equilibrium.  One is that the the product of the ratios of rate constants around a cycle (closed loop) equal the equilibrium constant for the cycle (see Cornish-Bowden, 2004, p 104). [see Addendum for an example of where this consequence was not adhered to].

(b) When a species is in a steady-state, the rate of formation equals the rate of breakdown. The steady-state concentration may differ markedly from the equilibrium concentration.

(c) As @WYSIWYG has pointed out, an enzyme (catalyst) does not change the equilibrium constant for the uncatalyzed reaction.

(d) When we are dealing with equilibria, we are usually interested in Gibbs free energy changes.  For example (Silbey &amp; Alberty, 2001, p 277).

&#916;Go' = -RT ln K’ 

(In the above equation K’ is the apparent equilibrium constant, that is the equilibrium constant at specified pH, and &#916;Go' is the standard transformed Gibbs free energy of a biochemical reaction.  More on this distinction, if you are interested, in the Silbey and Alberty reference quoted above). 

The important point is that there is a relatively simple relationship between Gibbs free energy and the equilibrium constant. The Gibbs free energy, of course, gives a measure of the spontaneity of a reaction and also its capacity to do work.  

When the transformed Gibbs free energy is of a reaction (&#916;G') is zero, the reaction (at specified pH) is at equilibrium and cannot perform useful work, and reactions with negative &#916;G' may be considered ‘spontaneous’.  It is also important to realize that equilibrium is a dynamic state: chemical bonds are continuously being broken and energy is continuously being redistributed (with no loss of energy). [I am dealing here with the transformed free energy, &#916;G', that is the free energy change at specified pH, rather than &#916;G (the pH independent value), as this is the most useful when considering a biochemical reaction.  However, similar conclusions apply to &#916;G]. 



Having got those out of the way, I can now attempt to answer your question (slightly rephrased). Can a reaction in a metabolic pathway be at equilibrium?  

To be extremely pedantic, if there is a flux through the pathway (net conversion of first substrate to end product) then the answer is no (Newsholme &amp; Start, 1973, p 11). That is, if there is a flux through the pathway, &#916;G' cannot be exactly zero for any individual reaction.   However, reactions in a metabolic pathway may be very close to equilibrium (Newsholme &amp; Start, 1973, chapter 1).

Let’s (once again) rephrase your question.  Are there any examples of reactions in metabolic pathways that are close to equilibrium, and how can we determine this?

To again quote Newsholme &amp; Start (1973, p11)  “In a series of reactions that constitute a metabolic pathway, a few may be displaced far from equilibrium, whereas the majority of reactions may be close to equilibrium”. 

So how can this be determined? One way would be to measure the ratio of products to substrates (or the ratio of product to substrate pairs) in the cell, and compare this with the equilibrium constant.  Note that it is only the ratio of substrate/product pairs we are interested in, not the absolute concentrations. We might be interested in the NAD+/NADH ratio in the cell, for example.  

That is, we measure the mass action ratio and compare this with the equilibrium constant.  

Such measurements are fraught with difficulties, but let’s agree that they can be made. We could rapidly freeze the tissue sample to -190&deg;C (using liquid nitrogen), for example, and then measure the ratio of metabolites.  Finally, it should be pointed out that comparison of mass-action ratio with the equilibrium constant is not the only way of deducing that a reaction is near equilibrium, and agreement between alternative methods is highly desirable before any firm conclusions are drawn. 

Let’s consider glycolysis as an example.  It is generally agreed that the reactions catalyzed by phosphoglucoisomerase, phosphoglycerate mutase and enolase are all close to equilibrium: the mass action ratios and the equilibrium constants are about the same (see Newsholme &amp; Start, 1973, p 98).  

It is also generally agreed that the reactions catalyzed by phosphofructokinase and pyruvate kinase are far from equilibrium (see Newsholme &amp; Start, 1973, p 98).  

What is the rationale? In general, it is control reactions catalyzed by regulatory enzymes that are displaced from equilibrium. (A key property of a regulatory enzyme is that the activity is controlled by factors other than substrate concentration). 

The enzyme phosphofructokinase is a good example.  This enzyme plays a key role in the regulation of glycolysis. Measurement of the mass-action ratio for this enzyme gives a value of 0.029, where the equilibrium constant for the reaction is about 1000 (see Newsholme &amp; Start, 1973, p31). It is clear that physiologically this enzyme catalyzes a reaction that is far from equilibrium. 

In conclusion, many reactions in metabolic pathways may be close to equilibrium, but regulatory enzymes almost always catalyze reactions that are displaced from equilibrium.

Finally, if anyone wishes to provide more recent references or to otherwise improve the answer, feel free to edit. 

Addendum 

For a well-known controversy in the enzyme kinetics field where the requirement that product of the ratios of rate constants around a closed loop equal the equilibrium constant for the cycle was not adhered to, see Selwyn (1993) and Topham &amp; Brocklehurst (1992).  These authors were critizing the work of Varon et al (1992), who made a bit of a hash of things. If you are interested in this area, you may wish to read up on this controversy, which is very informative. (Selwyn’s paper is a great start). [All the above papers are freely available in PubMed Central].

References

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

Laidler, K. J. (1987) Chemical Kinetics 3rd Edn. Harper &amp; Row.

Newsholme, E.A. &amp; Start, C (1973) Regulation in Metabolism. John Wiley &amp; Sons.

Silbey, R. J. &amp; Alberty, R.A (2001) Physical Chemistry. 3rd Edn. John Wiley. 



Eric Arthur Newsholme (1935–2011) Obituary