lecture 11 - notes

how quickly reacitons take place. this is about enzyme connetics. this is the enzyme discipline focused on dtemrinign the rate of a reaction and also how i changes in response to changes in expreimental paramters . yesterday we described a general enzyme catalyzed reaction so rfree enzyme plus free substrate come together to forma. complex, then that gets convere3ted to a enzyme product complex and then the product and then enzyme gets released. this is a three step description of a general enzyme catalyzed reaction. This is a two step process . E plus S is ES. the second step in the 6-1 equaion is difficult is to detect or measure in the lab. the second step is measured direclty by the amjount of product that is formed. ,not all enzyjme catlyzed reactions can be similfied like this. for now this is how we describe a typical enzyme catalyzed reaction. e plus S going to enzyme substrate complex then going to E plus P.

what happenes to the concentraiton of the components over time.

look at the ocmpoentns one by one. the conentraitono fht esubstrat ein your test tube and as epextcted u have a certain amount of substrate. you decide to start witha. certain amjount of substrate and then as soon as the reaction start, the substrate concnetraiton goes down. the straight line is the concentration of P product . before the reaction starts it is at zero . over time it goes up. those two are expected, the concnetration of E and ES something else happens. So the E starts at a certain value , something u detmine , over time it will decrease… why? becuase some of that enzyme over time is bindign a substrate molvuel and beocming an ES. so the concntraiton fo the fre enzyme goes down over time. opposite the conentraiton of the ES complex at the bginning is zero because u hadnt added substrate as soon as the reaction starts it goes up but after a certain time period.. both the free enzyme conentraiotn and the enzyme concnetration stops chaingingfor a brief period. soon after the reactions tarts the Es conentraion and E concentraion will remain steady and during this time the S concnetraion keeps going down and the P concnetraion is going to keep going up and the Es and E concnetraion stays the same. the intial hase where the concentraion of E and ES do change is the presteady state and the period that comes after that which ES and other intemreidiates remain constatnt is the steady state. the concnetraion of ED and ES remain constnat. speed of chemical reaction during presteady state is difficutl to study. so we focus on what happens at the steady state because it is easier to study. (look at grpah at page 4)

Now plot at page 5)

now the plot shows plot vs product concentraiotn. there are three curves bc the experiment was done at three different substrate concentraions. the lowest one shows s concnetration when it was .2 uM the other one was .5uM and the last one was .10uM. immediatly u can see the more substrate that u start the reaciton with the mroe product u ar going to form over time. the enzyme concnetraion is not changing we are using same concentraion of enzyme for all three cases. but even in that case we are formign more product over time. what about the speed of the reaction…. speed of chemical reaction is measured by product formed per unit time or amount of substrate molcuel disappearing per unit of time. so we look at product ofmraiton. u can cualte the speed by cualting the change in product conentraion over itme. the difficulty with studying enzyme kinetics, as time goes on the reaciton velocity changes. its. HYPERBOLIC cruve. over time reaction velocity changes. that complicates things. We are not going to stuyd v we study V^0 which is the intial reaction velocity. taken at time close to zero. this is the reaction, its the rate of the reacation at the very beginning of the process. as u cans ee from the black curves the vlocity is greatest at the very beginning and over time it slows down.

the tanget lines at time zero, they are the dottoed straight lines. slop of those dotted lines represent vo for the indivual three cases. once you do like an expeirment with 20 different enzyme substrate concentration( in the graph example there are 3) u can make a plot (pg 6) and the x axis is the substrate concnetraion and on the y u have the intial velocity Vo. the intail volceity (Vo) depeonds on the substrate concnetraiton. it is not a stright line, it is a hyperbolic curve. at a fixed enzyme conentraiotn just by increasing the concention of S. u can make the reaction go faster and faster. if u have a low substate concnetrion ur intail volcity is small and thats a slow reactioin but u can increase the substrate and the reactio will go faster. but ther is a limit to this u cant achieve the high velocity but increaisn the substrate concentration a lot. so u reach a certain substrate cocnentrion. the curve is almost plateu, adding more substrate form that point on will by u a teen tiny but more speed but there is a limtie to the max rate of the chemcial reaction and that is Vmax. theoretically u would reach vMAX when u put the conentration of S as infifiintiy but there is not such thing as inficnat substrate concnetraion in a lab. every mocleul has a solubility limit u can dissolve something indefinetly. the plateau like Vo region is close tpo the max vMAX. the curve is ur experimental data, that is what michaelis mentien did in 1913. they measured the V naught of their enzymes and plotted over a substrate concnetrioan dn they got a curve. they measure the v naught of their enzyme and plotted over a substrate concentration and they got a curve. you can do this too in the lab and once u have this plot u can esitmate what the VMAX value is for your system. estimate and not dtmerine becase waht vmax is trying to guesss when the curve ocmpeltely platues and u can do that precisley u can only estimate. vMAX is observed when all the enzyme is present as the ES complex. that is why adding even more substrate will not icnrease the reaction. velocity. this is BC all of the enzymes already have a substrate bound and theyre busy converting that enzyme bound susbtrate into product. so there is no free enzyme left and adding more substrate does not help u. further increase in S have no effect on the reaciton rate.

Michelasis marten they looked at their exprimental data and they were wondering why its shaped like that. more than 110 years ago like 50 years before ekdre and perutz solved the first structure of ap rotie myoglobin. they had no idea what enyzkme looked like and how it worked. they were going by experimental data and trying to figure out why it is hyperbolic. so equation 6-7 and 6-8 we broken down the equaiton up and into it indiaul compoennts of step one and step 2. step one enzyme and substrate combine to beocmes a complex in a revrsible and relatively fast step. so its a reverisble reaction. there would be a rate constatt with the reverse and forward. it is k1 and k-1 which is forward and rever4se. the complex breaks down to yield to E and P in a slower stpe. Es goes to E plus P this proces si s reveirsible so k2 and k-2 associated with that equaiton. so mikhail and menten went oon to use those two expression to analyze their data. and 6-9 this is the onclusion that they came up with and this equation is called the michaelis meten equaiton. Vo = vmax times substrate concnetriaon over km plus substrte concnetriaon . this equation exaclty matched thei expiremntal observation, this euqaiton u can plug in the various values and itll match what u get expiermentally, u alreayd know three of the four ocmpeontns. intail velocity is the value on the y axis, vmax is the max velocity and uc an estimate using a plot, substrate concnetraion is on the x axis, and Km is michealis constnat. u can look at this value on a table. this euqaiton describes the changing behavior of the chemail reaction depending on the the substrate concentraiton .

what they did is they got rid of the reverse reaction fo the second step . what we are trying to detmein is a equaiton that gives u Vo . not the genral vleocty but the intial Vo. so tthis is the speed of the reaciton that imeediate time point when u just start the reaciton. and when u start ur reaction there would be neglible amount of product P in the test tube because u just started the reaction. If p is small the reverse reaction E plus P going to US is going to be neglibly small. so u can ignore the reverse reaction of hte second step . so in that early time period of the reaction u only have to cocner ourselves with these three preocess. E plus S going to Es . ES going back to E plus s and the third E plus S going to E plus P

we can assume that of the three compoentn reacitons. Es plus E plus P is the slowest step. this is an assumption not all enzyme catlyzed reactions are like this. one ten years later that assunmption is valid. there are acceptions but it holds for most enzyme reactions. enzyme convering into substrat and produce and relasing is is most likely slower than free enzyme comign together with substrat to form a ES complex. the first stpe is a binding reaction and the second step is a chemcial transomfriaon so in most cases binding is fast and chem transomfraiton is slower. then what is the rate for the overall process. the overall rate of the reaction is equal to the rate of the rate limiting step. it can be any faster than the slowest step which is the second step

vO equal k2 times enzyme substrate complex concnetraion. 6-11

this is not that usual. bc u need the Es concentraion and that is hard to measure. so we need to change the equation 6-11. we susbtitue the Es as something that can be measured. we bringin another expression.. ET stands for total. this is the total concentrion enzyme in your system so u have to add two thigns together , the concentrion of the free enzyme and the concnetrion of hte ES. u combing those two and u ahve the total enzuyme concnetraion.

u can also do

E = et - ES

es - et -e

in a biolgical system the concnetaion of substrate is larger than the concentraion of enzyme. there are exceptions but in most cases ther eis more substrate than enzyme. hemoglbin there is more o2 molcules in the lungs than hemlgobin protien bc S is larger than ET. the amount of substrate bound by the enzyme at any give them is negliglble compared with the total S. that means we can treate S as a consntat under bio conditions. michealis and metond made a second assumption

it says that when u reach a stead systate when u reach steayd state of the Es complex concentratoin stays consnat. for ES to remain consnat the rate of the ofmration of ES has to equal the rate of dissocaiton of the ES that the only way that the Es concentraion will remain consant. the rate that it get sfomed must match the rate that it gets disscoated in order for es to remain a constant. so u write the expression for es formation and es is fomed through only one process e plus s going to ES> if u write that mathemativelly its k1 rate consant times concnetriaon of e times S.

the rate of ES breakdown. it can go back to free E plus substrate or it can become free E plus product. so the rate of breakdown is the addition of those two processes. so k-1 times ES + k2times ES. expressions 12 and 13. now we can apply that assumption. that those two rates are the same under steady state condition.

so steady state means that there is going to be a constant making of ES = constant breaking of ES. This is going to be rate of formation = k1(E)(S) ( k1 because it is the forward reaction of E+S=ES)

rate of Es breakdown = k-1(ES) + k2(ES)

k-1 (ES) becasue we are using the first step reverse reaction so ES = E + S

k2(ES) because we are using the second step forward reaction to break down ES into E + P

so the full rate of formation = rate of breakdown is

k1(E)(S) = k-1(ES) + k2(ES)

bg assumption rate of fomation of es equal the rate of es breakdown.

by equating both expression we get equation 6-14 . next we take that same equation and we solve for ES. we rearrange this so we hav ES. on the left and everything else on the right .

6-18 - idk u have abunch of rate consatns this is consant plus consatnt dvidie dby a constant so the whole equation is the michealis constant Km = (k-1 + k2)/k1.

another subsitution is 6-11 which says Vo = k2 times enzyme susbtrat concnetaion and we encoutnred this concept before. if this second step is hte rate limitng step then the rate for the overall reacito must match the rate for this second step which can be expressed as k2 times concentaion of ES. this is a unimolar reactio so the rate gets detmined by concnetaion of that single reactant mocluels times the rate consant.

so u plug this expression into 6-19 and y get 6-20. we start to get something looking like the ifnal michealis meten equation. looking at 6-20. we usbsutie Es by Et that means all of the enzume .

stopped at 37mins cuz wtf

OK, but instead of ES complex,

if we substitute that with total enzyme concentration,

that means all of the enzymes, 100% of your enzymes

are in the enzyme substrate complex form, okay?

And under that condition, you must have reached Vmax,

right? All of your enzymes are in the complex form.

There is no more free enzyme in your system,

okay? Therefore, that's the theoretical maximum

velocity you've reached, which we called Vmax early

on. so we just substitute that term k2 times et

with v max and then we finally get this expression

v naught equals v max times s plus km plus s.

This is the final Michaelis -Menten equation okay

and we've just now derived it over 15 or so steps.

okay for for exam two again I'm going to give you

all the all the equations that appear on all of

the slides and all of the papers okay but it's not

if you don't understand that you're not going to be

able to answer all of the questions right so the

derivation that we went through you should be able

to do it on your own okay and by that I mean you

start from 6-1 okay and you make all

the arguments that we just made and

you should be able to derive 6-9 okay you you will

be given equation 6-9 but if you're able to derive

it that means you fully understand each of the

steps all of the assumptions that's being applied,

okay, and you'll be ready for any

question I throw at you, okay?

Okay, so what does, what is the meaning

of the Michaelis constant Km, okay?

For now,

don't ask why,

just see what happens when you make Km equals

substrate concentration, okay? So you're going

to substitute that term there, Michaelis constant,

with simply substrate concentration, okay?

Then the Michaelis equation simplifies to V naught

equals one-half of V max, okay? So that means

Km is the substrate concentration at which your

system will achieve one -half of maximum velocity.

That's sort of the English definition of Km.

So again, Km is something that you can determine

experimentally by plotting this curve. Okay? Substrate

concentration versus initial velocity V naught. Okay? How?

Well, first, you draw that curve. Okay? From

your curve, you determine what V max is.

You can't determine V max. You can estimate

V max. Okay? So let's, for argument's sake,

your estimated V max is

50 micromolar per minute,

okay? So that is your best

estimate of the Vmax value,

okay? Then with that information, you can

calculate Km because what's one-half Vmax?

That would be 25 micromolar per minute,

okay? And then you look at the curve and you

determine what the substrate concentration must be,

that gives you one half of Vmax, or 25 micromolar

p minute per minute, okay, and that substrate

concentration value is your Michaelis constant, Km,

all right.

But I kept saying that you can't determine

Vmax, you can only estimate it. That

is true if you have a hyperbolic curve.

Well nowadays you can use math CAD and you

know software to pretty accurately determine

VMAX, but 110 years ago you couldn't,

okay?

Well, but you could, okay? So what

did Michaelis-Menten do, well,

Line River Burke do after Michaelis-Menten?

They come, they came up with this really elegant

method for accurately determining Km, okay?

To accurately determine Km, you need to

first accurately determine Vmax, right?

How do you do this? Okay,

so here's the Michaelis -Menten equation, here

6-2 This is just the reciprocal of both sides,

okay?

And then you expand this. You get equation 6-26,

which is called the Leinweaver -Burk equation, okay?

And look at this.

1 over V naught equals Km over V max

over substrate concentration. So V

max and Km, these are all constants,

okay? So looking at this equation, if you were

to plot 1 over S versus 1 over V0, okay? Remember

before here you were plotting simply S

versus V0, okay? But you take the exact same data

points you got experimentally, you plot 1 over

S versus 1 over V0, you get a straight line.

And we like straight lines, right?

So once you have a straight line like this,

you can very accurately determine the

slope of that line, or the y-intercept of

that line, or the x -intercept of that line.

And from those values, you plug

it in back to the equation 6-26,

okay? And then you can very accurately

determine both Km and Vmax values, okay?

With the hyperbolic curve, you can get a good estimate

of those with the Lineweaver -Berg plot, much easier.

Okay, here are some real-life Km

values out there for real enzymes,

okay? KM

can vary for different substrates of the same enzyme.

Let's look at an enzyme called hexokinase.

You are going to learn about this when

you study glycolysis with Dr. Kulsatra.

It's an enzyme that phosphorylates sugar molecules,

and it gets the phosphate moiety from ATP.

It turns out that that enzyme can do this on

glucose sugar as well as fructose sugar, okay? But

no matter which sugar it's phosphorylating, it

needs to bind two substrate molecules, right?

It's a bimolar reaction.

You have to bind both ATP and a sugar

molecule, and then you produce ADP plus

the phosphorylated sugar, okay? Okay,

so it's just showing you that for

this phosphorylation reaction,

okay, the Km value determines on the substrate,

okay?

When you're using glucose as the substrate,

Km is 0.05. When you're using fructose as the

substrate, it's 1.5, okay? ATP doesn't change

because it's the same molecule no matter which

sugar it's phosphorylating, okay? So here, obviously,

glucose has a much smaller Michaelis constant

compared to fructose. What does this tell you?

What's the English definition of Km?

It's the concentration of the substrate at

which you reach one half of Vmax. OK,

so hexokinase, even though it can accept

both glucose and fructose as a substrate,

it'll reach one half Vmax much more easily when

you use glucose compared to when you use fructose.

Okay? I think these are,

the units here are micromolar.

I'll have to double check, but for now, let's say the

unit is micromolar. That means when you use glucose,

the concentration of glucose in your test tube

only needs to be 0.05 micromolar, and then

the system will reach 1 .5 Vmax. But if you use

fructose, you need to have your sugar concentration

at a much higher value, 1.5 micromolar.

So this is why KM values are useful.

Carbonic anhydrase, we encountered this

many times. KM value is higher now.

So there are a lot of tables like this where you can

go and simply grab the Km values. You don't have to

always do this experiment yourself because scientists

have already done it for you. But you need to understand

what it is, how it was obtained, and how to use it.