Km on – you can’t really think I could talk about enzymes without Menten-ing ENZYME KINETICS, did you? From the sluggish lysozyme, which takes a whole 2 seconds to cut a protein, to the mind-blowing-ly fast carbonic anhydrase, which can combine CO2 with water to make carbonic acid – or do the reverse – 1 MILLION TIMES PER SECOND (10 million times faster than it would happen on its own), I love them all. And no love letter for these amazing biochemical reaction-speeder-uppers could be complete without a discussion of how fast they can go, what determines that speed, and how we can measure it. Today we review, then go forth from thermodynamics to look at the age-old question: how many sticks could a stick-snapper snap if a stick-snapper could snap sticks? And to do this we need kinetics!

You can learn a LOT more about enzymes in yesterday’s post: http://bit.ly/2r7x40l

But basically – enzymes are biochemical catalysts – catalysts are things that speed up reactions without being used up in the process (so once they’re through speeding up 1 reaction they can speed up another, then another, then another) – you can find catalysts all over the place – even in your cars (catalytic converter ring a bell?) – and “enzyme” is the name we give to the catalysts we find in our bodies. 

Enzymes are usually proteins, sometimes protein/RNA complexes, and sometimes just RNA. They help biochemical reactions happen – everything from splitting up molecules (e.g. breaking down sugar into smaller parts for energy) to stitching together molecules (e.g. DNA ligase sealing breaks in DNA strands) to just “shifting molecules” (e.g. chromatin remodeling complexes shifting the histone proteins DNA is coiled around (for space-saving) in order to reveal regions to be read). 

Although these processes can be super complex and require a lot of coordination, they all *could* happen without enzymes – because they have the biochemical “drive” to do so. They would just be super unlikely to happen because you’d have to have freely roaming molecules collide at just the right orientation with just the right conditions, etc. 

This biochemical “drive” is a negative change in free energy (G). I like to think of G as molecular “comfiness” – imagine you’re getting a picture taken – it takes energy to hold an awkward pose (uncomfortable molecules have a high G) whereas you can relax if they just want a candid pic (comfy molecules have a low G). I don’t know about you, but I hate those say cheese-y pics – I would much rather be comfortable. 

And molecules would too. So they react and interact in ways that get them to a more comfy position (lower G)-> reactions are energetically favorable if the product(s) P have LESS free energy (lower G) than the reactants (R), and we can call this difference in free energy between the products and the reactants ΔG (Δ is pronounced “delta” and it means “change in”). With enzyme-catalyzed reactions, we usually call the reactants SUBSTRATES (S) – as in, “thing” is a substrate for “cool enzyme.” So we can write the overall reaction as 

E + S ⇌ E + P

and ΔG = GP-GS

Not all reactions have a negative ΔG, but you can couple unfavorable ones to favorable ones to get the job done. This is one reason some enzymes use ATP – splitting ATP releases a lot of energy – and if the ATP splitting & the unfavorable reaction-doing are both happening together in the same enzyme, that energy can be used to power the unfavorable one.

Even when a reaction is really favorable – the products are way comfier than the reactants, so there’s a highly negative ΔG, it often has to go a really uncomfy state to get there – we call this most uncomfy (highest energy) state the transition state (⧧). Imagine you’re trying to snap a stick in half (for the sake of the analogy pretend it’s perforated in the center so there’s only one possible breakpoint). So you start bending the stick and it gets harder and harder & the transition point would be that really tense moment right before the stick breaks (if you think your arms are sore, imagine how the stick feels!). It is at this transition state (TS) that the enzyme hugs the substrate tightest – thus it kinda lures the substrate into adopting that awkward position – the uncomfiness as the substrate “bends” is offset by positive interactions with the enzyme that bending allows for. 

We can figure-ize the changes in energy over the course of a reaction with a REACTION COORDINATE DIAGRAM. It reminds me of a candy-cane shaped rainbow with a pot of gold at the bottom – you have to get over the rainbow before you can get to the gold. The top of the rainbow is the energy of the transition state and the energy required to get there is the activation energy. Enzymes provide an alternative transition state with a “shorter rainbow” – a lower transition state energy means a lower activation energy, so there’s less of a barrier for the reaction to occur. But since the starting and ending points are the same, the overall ΔG for the reaction is the same regardless of whether it’s enzyme-catalyzed or not. 

When you have a quantity like this where you only care about the start and finish “states and not the route taken, we call it a “state property.” And it’s this overall ΔG of the reaction that provides the “drive” for the reaction – so the enzyme doesn’t change that drive – it doesn’t choose whether or not the reaction will happen – and when you lower the barrier you make it easier to “go backwards” too, so enzymes don’t change the equilibrium constants – if you start with the same amount of reactant you’ll eventually reach the same amount of product formed (even if you have to wait billions of years) but they do affect reaction rates – so you don’t have to wait billions of years – but how long do you have to wait? To figure out this we go from the thermodynamic stuff we’ve been talking which deals with reaction favorability to kinetics, which deals with speeds (reaction velocity, v).

A reaction’s only as fast as its slowest step, and you can split a reaction up into a few steps: bind (E+S ⇌ ES), change (ES⇌EP), & release (EP⇌E+P). For simplicity’s sake, let’s say that the reaction’s really favorable so it’s really unlikely to go in the reverse direction (you’re not gonna change product into substrate). (Even if the reaction does go backwards, if you measure kinetics at the very beginning, when there isn’t any product to get turned back into substrate you can ignore the reverse). Then we can remove a couple of those backwards arrows so we have 

bind (E+S ⇌ ES), change (ES->EP), & release (EP->E+P)

You can see that I didn’t take away that first backwards arrow because the binding step is still reversible – and which way the equilibrium lies (is E + S or ES more favored) depends on the affinity (stickiness) of the 2 for one another. And even though we’re assuming you *can* only go ES->EP (and not the other way) – that doesn’t mean you *will* make that change. If you think of things chance-wise (probabilistically), each time a substrate collides with an enzyme, it has a chance of sticking and each time it sticks it has a chance of converting to product. How fast the reaction will occur depends on:

  • how likely E & S are to collide (depends on their concentrations & how much energy they have)
  • how likely E & S are to stick (depends on their stickiness “affinity” for one another) 
  • how likely S is to convert (depends on how high the activation barrier is & how long the substrate’s stuck there so it can “keep trying”)

You also have the releasing step, but, except for some weird cases, that’s usually not a hold-up, so we’ll combine the change and release steps into ES -> E + P. So now we have 2 steps, bind (E+S ⇌ ES) & change/release (ES -> E + S) and  each of these can happen at different speeds

We can give each step a rate constant, k. It’s “constant” in terms of it not being affected by concentrations because it’s an inherent property of the molecules, but it *is* affected by differences in the environment (pH, etc.) so it’s specific for a specific reaction under specific conditions. 

So, how many sticks could a stick-snapper snap if a stick-snapper could snap sticks? To “answer” this let’s turn to Michaelis-Menten kinetics. 

Say you want to figure out how fast a snapper can snap sticks – if you took a single stick-snapper, give her some sticks, set a timer for 60 seconds, then counted how many sticks she snapped, you could divide that by the time to get sticks snapped per second per snapper. This is equivalent to a kinetic constant called kcat (aka “turnover number”) which tells you how many substrate molecules a single enzyme can convert to product per unit time (usually seconds) – as long as there were enough sticks!

If you have a huge excess of sticks, (S >>>> E), you don’t have to worry about running out, so the speed you observe isn’t affected by [S] (concentration of substrate) and each enzyme molecule can work at kcat (each time the snapper snaps a stick there’s another stick there to snap). But if you don’t have a big excess, the substrate all gets used up, so it can only work its fastest in the very beginning of the reaction, right after you mix E & S.

Well, not quite *right* at the beginning. At the very very beginning (we’re usually talking the first couple milliseconds) you have to fill up the enzymes – the snapper has to grab their first stick, so you have a burst of ES formation, but they haven’t had time to snap the sticks yet so product formation starts slow thanks to this lag as the snappers snatch sticks. We call this brief start-up phase the PRE-STEADY STATE.

As the name implies, it’s followed by the STEADY STATE – and this is where kinetic measurements are usually taken. In fact, in Michaelis-Menten kinetics we make a “steady state assumption” – basically we assume that ES is at equilibrium – you know how I said E+S⇌ES was reversible – well, equilibrium is where the *rates* of the forward (E+S -> ES) & reverse (ES -> E+S) reactions are the same – this doesn’t mean that you have the same amount of S bound as unbound, it just means that the amount that’s bound vs. unbound isn’t changing (for every snapper that drops a stick another snapper picks one up).

Initially none was bound and in the pre-steady state the ES vs. E+S was changing as S now had the option of binding. But eventually you reach a dynamic equilibrium where, for every S that gets released or converted, another one binds. This is usually reached within microseconds & is what you measure when you measure the “initial velocity” (vo)

But it can only bind if there’s still S left to bind! And S isn’t just binding – it’s also “dissappearing” because it gets converted into products (and you can’t snap an already snapped stick!). So you enter the POST-STEADY STATE where the substrate starts getting used up, fewer ES complexes can form (poor snappers left stickless) and less product can form

kcat was looking at a a single snapper – but it’s not easy to measure a single enzyme molecule because these guys are really tiny and you usually have a ton of them – so instead of dealing with single molecules you’re dealing with mols of molecules. The mol is the biochemist’s “dozen” – it just means 6.02×10^23 of something.

So now we can imagine a ton of identical snappers, and the amount of sticks snapped depends on how well each snapper can snap sticks (kcat) AND how many snappers there are (enzyme concentration, [E]). 

Instead of finding the maximum rate per snapper, you first find the maximum velocity of product formation (Vmax) for a group of snappers, and then you take the # of snappers into account.

Vmax = kcat[E]o, where [E]o is enzyme concentration

So kcat = Vmax/[E]o

But “turnover rate” isn’t all you care about. How good of a grip does the snapper have on the stick? Is it really a good match? When you have way more substrate than enzyme, the enzyme’s constantly getting bombarded with substrate, so even if it doesn’t like the substrate that much it’s still bound most of the time because every time it lets go there’s another substrate molecule waiting to jam itself in. So you can’t tell if the substrate is actually a really good match for the enzyme.

But if you have lower substrate concentrations, if an enzyme releases a substrate molecule it’s less likely to quickly find another molecule to collide with. So, when those rarer collisions do occur stickiness will matter more – if they don’t stick they’ll have to wait for another one – and you’ll have to wait for product. 

How do we take this into account when “scoring” enzymes – enter the MICHAELIS-MENTEN CONSTANT (Km). I’m gonna derive this in the pics for you so you can see how it comes from the rate constants and concentrations (sorry for sloppy formatting – I’m tired). Km is the substrate concentration at which product formation is at 1/2 it’s maximal speed (so 1/2 Vmax). 

You usually find it by measuring Vo (that initial rate of product formation) at multiple concentrations of the substrate (another chance to use those serial dilutions!). Then you plot substrate concentration on the horizontal axis & reaction rate on the vertical axis. Do this and (often) you’ll get a ,- like curve – a parabola that starts steep and then plateaus. Enzymes that give curves like this obey Michaelis-Menten kinetics. 

It starts steep (but slow) because at there’s more than enough enzyme to quickly change all the substrate into product but then substrate runs out (too many snappers, too few sticks) But as the substrate concentration increases, the enzyme gets saturated – each enzyme is working its hardest, but it can only go so fast – the enzyme becomes limiting (too many sticks, too few snappers). The height at which the curve plateaus is called the maximum velocity (Vmax) and the substrate concentration at which the curve gets halfway to Vmax is called the Km.

This Vmax depends on enzyme concentration (which is why we adjust it to get kcat), but Km doesn’t depend on enzyme concentration, just on how well the enzyme likes the substrate. It’s not quite this simple, but, in general terms & typical cases

lower Km -> better binder (higher affinity for substrate)

higher Km -> worse binder (lower affinity for substrate)

and 

higher kcat -> better changer 

lower kcat -> worse changer

If you want a good idea about how “good” an enzyme is overall for a particular substrate, you need to know how well it binds substrate & how well it changes it. So you combine those 2 measures to get another value, the “specificity constant” (aka “catalytic efficiency”) which = kcat/Km

You know how I said “sometimes” you’ll get a parabola? Well, other times you’ll have enzymes where when you make that graph you’ll get an S-shaped curve. This usually implies that you have an “Allosteric enzyme” – allostery is where something happens somewhere on the molecule that changes something elsewhere on the molecule. Allosteric enzymes usually have multiple active sites and when substrate binds one of them it makes it easier for the other active sites to bind substrate, so you get cooperativity and see a switch-like transition. 

Maud Menten & Leonor Michaelis published their formula in 1913. Maud was a pretty amazing scientist – Born in Canada in 1879, Maud became one of that country’s first female medical doctors, but there were limited opportunities in Canada for a woman to pursue a research career, so she moved to the US, where she worked at the Rockefeller Institute, researching the effects of radium on tumors, before emigrating to Germany to work with Michaelis. After her groundbreaking work there, she moved back to the US to obtain a PhD at the University of Chicago and went on to work as a professor and pathologist at the University of Pittsburgh and a research fellow at the British Columbia Medical Research Institute. Menten also had a full life outside of the lab – in addition being a brilliant biochemist and histologist, she was also a skilled musician and artist and she spoke 6 languages! Menten passed away in 1960, but her name will forever be synonymous with enzyme kinetics.

more on topics mentioned (& others) #365DaysOfScience All (with topics listed) 👉 http://bit.ly/2OllAB0

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