# Why is the units of kcat 1/s?

I understand that $k_\text{cat}$ measures the turnover number of an enzyme. This measure is therefore a quantity of molecule conversions per unit of time. I suspect that my problem is more that of a lack of maths but why is the unit expressed in $\mathrm s^{-1}$? Why not just seconds?

The turnover frequency is a rate. (It's actually the kinetic rate of the reaction in saturating substrate concentration, normalized to the amount of enzyme.)

It might be helpful to think about something like a factory that makes a certain number of widgets in a certain number of time. How would you naturally describe how fast the factory could make widgets? Most people would express how fast the factory works in terms of "widgets per hour". It's the same with enzymes - it's most natural to describe how fast the enzyme works in terms of "turnovers per second". It's just that the "turnover" is implied and is viewed as "dimensionless", so it's left off, leaving you simply with "s-1" as a unit.

(Note that even though the "turnover" is left off, it's still very important to know it's there, and to know what, exactly, one "turnover" is. In complicated reactions there is sometimes different reasonable ways to count what "one turnover" is, and different definitions could result in a 2, 3, 4, 5, etc. fold difference in value. Those vexing numeric differences become easy to explain once you realize this researcher is using the formation of one molecule of this product as "one turnover", but that researcher is using the disappearance of one molecule of that substrate as "one turnover".)

Naively, one could invert the ratio to come up with a value in time, rather than in 1/time. For example, if a factory makes 5 widgets per hour, it takes them 0.2 hours to make a widget ... sort of. One issue you run into is that that's the time on average. In a stochastic process like enzyme turnover, it's highly unlikely that any single turnover will take exactly the time specified. Some will take longer, some will be quicker. Only for a large number of turnovers in aggregate does that time value have meaning. Saying that "a turnover takes 5 s" gives a false impression about individual turnovers in a way that saying "this enzyme works at 0.2 turnovers per second" doesn't.

Additionally, in the case of multi-step reactions, saying that "a turnover takes 5 s" may be incorrect. The entire pathway may take much more that that. It's just that once the pathway has reached steady state it's only the rate limiting step which takes 5 s. For example, take a pathway of 20 steps, 19 of which take 4 s each, but one which takes 5 s. It's going to take (19*4 + 5) = 81 seconds for any given substrate molecule to start at the beginning of the pathway and make it to the end. However, once the pathway is full and each step is handing off products to the next step, the pathway will be producing product at a rate of 1 every 5s. (This, of course, doesn't apply to enzymes where all reactions take place at the same active site, but it's a consideration in the general case, which explains why 1/s is the preferred unit for reaction rates.)

• This is a good answer and gets a +1 from me. But "(It's actually the kinetic rate of the reaction in saturating substrate concentration.)" doesn't actually seem to be correct. Somewhere in there we'd need to multiply by a concentration, presumably the concentration of catalyst, to get a rate. Also, the idea in the last paragraph is sound but the math doesn't work out. The "rate" of a 20-step process with 19 4-second steps and 1 5-second step is given by $\nu_{total} = \frac{1}{\sum{\frac{1}{\nu_i}}} = \frac{1}{19\frac{1}{4} + \frac{1}{5}}=\frac{20}{99}$. – Curt F. Nov 16 '15 at 18:51
• @CurtF. You're right that there is a factor related to the amount of enzyme - I'll edit to clarify. On your second point, I don't quite see where you're getting that equation from. It's likely being misapplied if it's saying that adding additional steps in series to a reaction that takes 5 seconds per turnover can somehow make the combined reaction (all of which passes through the obligate 5 per second step) take only 4.95 seconds per turnover. – R.M. Nov 16 '15 at 19:53

You were actually describing the turnover frequency of a catalyst, and frequency has the unit hertz which is 1/s. Turnover number is a dimensionless value: number of molecules of reagent that a catalytic site can convert before becoming inactivated.

• Is the description on wikipedia, that it has different meanings in enzymology and other chemical fields, wrong then? – John H. K. Jan 5 '15 at 15:56
• I think that it is. I don't see why it would be differently defined in enzymology. – RBW Jan 5 '15 at 17:41
• In enzymology, the turnover number has no relationship to the enzyme becoming inactivated. Instead, it is the reaction rate per mole of enzyme, in the extrapolation of infinite substrate concentration. – DavePhD Jan 5 '15 at 18:24

As you say this represents "conversions per unit of time".

The mathematical expression for "per time unit" is 1/(time unit), and seconds is a particular time unit.

• Okay, but why does it have to be 1/(time unit). I.e. why does it have to be (eg.) 5 conversions per second = 5 1/s. Why not simply 5 s. I don't understand the requirement for the '1 divided by'. Thanks – CiaranWelsh Jan 5 '15 at 16:39
• Compare to a situation where you are paid 5 dollars per hour for your work. You work 8 hours and expect 40 dollars at the end of the day. But your boss says, "no your pay for today is 5 hours, not \$40". – DavePhD Jan 5 '15 at 16:55

Vmax=kcat * ([E total]) If the units for Vmax are M/s i.e. s-1 and you divide Vmax by the concentration of total enzyme which has units in M, the Ms will cancel out, leaving you with 1/s or s-1