biochem exam 3 written study (michaelis menten + inhibition)

what is the 1st order part of the michaelis menten curve?

-this is at low amounts of [S], the amount [S] is much smaller than Km

-this linear line, this V0 = Vmax [S] / Km

what is the zero order part of the michaelis menten curve?

-high [S] that is much higher than Km

-V0 = Vmax

what is Km?

Km is the concentration of [S] where V0 = ½ max

when does Km = Kd? Recall Kd is concentration at which half bound

-Km = Kd when binding association/disassociation (k1, k-1) is much faster than k2, much faster than the reaction itself

what is a double reciprocal plot? (lineweaver-burk plot)

-instead of having substrate and V0, it is 1/[S] and 1/V0 which linearizes the data

-it is the same curve in a different view

what can you do using a double reciprocal plot (lineweaver burk plot)

-you can isolate Vmax and Km

what is the lineweaver burk equation?

-it is linear, y = mx + b format

-1/V0 = Km / Vmax [S] + 1 / Vmax

-1/V0 is y, m is Km/Vmax, x is 1/[S] and 1/Vmax is b

what can we use to get Vmax on a double reciprocal plot?

-we can use 1/Vmax, the y intercept to get Vmax

what can we use to get Km on a double reciprocal plot?

-we can use slope, Km/Vmax to get Km

-can also get Km from x intercept, -1/Km

-these should be positive values, you can’t have a negative Km, negative concentration

what does Km actually mean? Recall Km = [S] when V0 = ½ Vmax

-Km is reflection of E+S interactions

what does Vmax actually mean?

-Vmax is how fast does substrate → product conversion occur in enzymes

when is Km a measure of substrate affinity?

-Km varies depending on mechanism (number of steps and rates)

-for 2 step mechanisms it is true that Km = Kd, only true if the reaction is slow compared to binding and k2 is the rate limiting step

what is Vmax proportional to in the two step mechanism where product formation is rate limiting?

-Vmax - k2 [E]T

-Vmax is proportional to the rate limiting step times the amount of enzyme

what would Vmax be if product release was rate limiting? E + S → ES → EP → E + P

-there is a third rate limiting constant, k3

-Vmax = k3 [E]T

-basically Vmax equals kcat x amount enzyme

what is Kcat?

-Kcat is the rate limiting step of an enzyme catalyzed reaction

-this can vary, it could be k1, k2, k3

Why is kcat a first order rate constant?

-1st order rate constant doesn’t rely on collisions

what is the units of Kcat?

-units of s-1 (reciprocal time)

-kcat is called turnover #, the number of substrate molecules converted to product in a given time by a single enzyme molecule

why can’t we tell enzymatic efficiency from kcat alone?

-we don’t know if this rate only occurs at a lot of [S] or low [S]

-could have really high Vmax but if concentration is low enough in the cell, we may never access it

what is catalytic efficiency?

-catalytic efficiency is whether enzymatic rate is limited by creation of product or amount of [S] in environment

what is the specificity constant?

-Kcat/Km is the rate constant for the conversion of E+S to E+P

is Kcat/Km a first or second order rate constant?

-Kcat/Km is a second order rate constant (M-1 s-1)

consider the scenario where k-1 is much greater than k2 (falling off of enzyme is much faster than going forward, k2 is rate limiting step). if kcat/km is high what is kcat?

if kcat/km is high, kcat is much larger than km

-substrate binds tightly (low km) or higher proportion of productive events (large kcat)

consider the scenario where k-1 is much greater than k2. if kcat/km is low what is kcat?

if kcat/km is low, kcat is much smaller than km

-substrate binding weakly (high km) or low proportion of productive binding events (small kcat)

what is catalytic perfection?

-even if kcat is greater than k-1, it will still be limited by k1

-catalytic perfection is the diffusion limit

-the upper limit is the rate at which E and S can diffuse to one another (10^8 M-1 s-1) and achieving this limit is catalytic perfection

-some enzymes aren’t random walk and are faster than the rate of diffusion

can enzymatic systems have more than one substrate?

yes, there can be two substrates for one enzyme

what are the enzyme reaction mechanisms involving a ternary complex?

-ordered: 1 substrate binds 1st, 2nd substrate can bind

-random order: more common, it doesn’t matter which binds first

what is the enzyme reaction mechanism in which no ternary complex is formed?

-no ternary complex: sequential reactions, new conformation and intermediate product

-all of these mechanisms can still be evaluated with kcat/km because there is still a rate limiting step for these

what can you gather from the pH-activity profile for an enzyme?

-you can guess what might be present in the active site, what amino acids are involved in a rate limiting step

-ex: aspartic acid with pepsin because pepsin has an optimal pH of ~2

-ex: glucose-6-phosphatase with histidine because the enzyme has an optimal pH of ~8

at what [S] would an enzyme with a kcat of 30.0 s-1 and a km of .005M operate at ¼ of its maximum rate?

-maximum rate is Vmax and we know Vmax = kcat [E]T

-1/4 Vmax = ¼ kcat [E]T = kcat [E]T [S] / [S] + km

-we can cross out the term kcat [E]T to get just ¼ [S] / [S] + km

-[S] = .00022M

-make sure your km and [S] units match

determine the fraction of Vmax that would be obtained at the following substrate concentrations: ½ Km, 2Km, 10Km

-use michaelis menten equation, and plug in ½ Km where there is [S]

-1/2 Km: 1/3 Vmax

-2 Km: 2/3 Vmax

-10 Km: 10/11 Vmax (things are considered effectively vmax at 5 Km)

an enzyme catalyzes the reaction of Y to Z. the enzyme is present at a concentration of 1 nM, and the Vmax is 2uM/s. The Km for substrate Y is 4uM. calculate Kcat

-all you need to get kcat is amount of enzyme and vmax

-1000nM = 1uM

-divide vmax by molar concentration of active enzyme sites: 2000/1 = 2000

what is the equation for catalytic efficiency?

vmax/Km, and vmax should be proportional to kcat because the enzymes being compared have the same [E]. so kcat/km

what is an irreversible inhibitor?

-irreversible inhibitor: very slow dissociation of inhibitor from enzyme

-inhibitor very tightly bound

-noncovalent or covalent

what is a reversible inhibitor?

-reversible inhibitor: rapid dissociation of enzyme-inhibitor complex

what occurs during competitive inhibition?

-reversible

-inhibitor binds to the active site of the enzyme, in the same place where the substrate typically binds

what occurs during uncompetitive inhibition?

-first substrate binds to enzyme forming ES complex

-then inhibitor binds to ES complex outside of the active site

what occurs during noncompetitive or mixed inhibition?

-inhibitor binds at a separate site from active site and may or may not prevent substrate from binding

-binds to either E or ES complex

how does a competitive inhibitor affect the reaction rate, V0?

-the inhibitor is competing with substrate for binding to the active site

-substrate can only → product when enzyme is bound

-reduced proportion of enzyme bound, so reduced observed rate of reaction (V0)

how can competitive inhibition be overcome?

-as [I] increases, more [S] are required to maintain rxn rate

-competitive inhibition can be overcome by outcompeting I with S

how does vmax change with competitive inhibition?

-enzyme will have the same Vmax with competitive inhibitor

-at high enough [S] the inhibitor is ineffective, all E is bound to S

how does competitive inhibition change Km?

-apparent Km will be increased by inhibitor

-more substrate is needed to obtain ½ Vmax

how does competitive inhibition look on a lineweaver burk plot?

-same y intercept because vmax is unchanged

-change in slope because inhibitor increases Km

what is alpha? what is Ki?

-alpha: degree of inhibitor effectiveness

-Ki: affinity of enzyme for inhibitor, it is the inhibitor concentration at half inhibition

-a lower ki means a more potent inhibitor

what do competitive inhibitors look like as drugs?

-competitive inhibitors look like substrate but react different

say it one more, time, what is occurring with uncompetitive inhibition?

-inhibitor binds to ES complex after substrate has bound to enzyme

-binding site for inhibitor is created after S interacts with enzyme

-does NOT bind to active site

how does uncompetitive inhibition affect Vmax?

-uncompetitive inhibitor reduces the number of ES complexes that can convert into product

-ESI complex cannot react

-less enzyme available reduces Vmax, recall Vmax = kcat [E]T so less total enzyme will decrease Vmax

can you outcompete uncompetitive inhibitors?

-no, more [I], more ESI forms

-cannot outcompete by adding more substrate

how do uncompetitive inhibitors change Km?

-apparent Km will be decreased by inhibitor

-this is unexpected, you would think km might increase. but the thing is any amount that km changes is offset by the decrease in vmax

how does uncompetitive inhibition look on a lineweaver burk plot?

-change in y intercept as vmax decreases from uncompetitive inhibitor

-no change in slope because if vmax is changing, km also changes in a way that cancels out

-km and vmax decrease by the same factor

what is alpha prime and Ki prime?

-alpha’ determines degree to which binding of inhibitor changes affinity for ES

-Ki’ is the affinity of the ES for inhibitor

recall, what happens in mixed and noncompetitive inhibition?

-inhibitor binds to either E or ES

how does mixed and noncompetitive inhibition change vmax and kcat?

-ESI cannot form product

-Vmax is decreased

-decreased turnover number (kcat)

what specifically happens in noncompetitive inhibition?

-halfway between competitive and uncompetitive

-inhibitor has equal affinity for both E and ES

-substrate can still bind enzyme-inhibitor or enzyme, both prevent ES to product

-Ki = Ki’

what specifically happens in mixed inhibition?

-substrate binding to EI and E are different

-inhibitor has different affinity for E vs ES

-Ki does not equal Ki’

what is the effect of more inhibitor when it is mixed/noncompetitive inhibition? how is Vmax changed?

-more inhibitor, more EI and ESI

-ESI and EI cannot form product

-Vmax will be lower

how does apparent Km change for noncompetitive and mixed inhibition?

-noncompetitive: no change, cancels out

-mixed: can increase or decrease by inhibitor

• prefer E, Ki’ > Ki → more like competitive inhibition

• prefer ES, Ki’ < Ki → more like uncompetitive inhibition

how does mixed inhibition look on a lineweaver burk plot?

-change in y intercept because Vmax decreases with inhibitor

-change in slope because Km is changing

how does noncompetitive inhibition look on a lineweaver burk plot?

-Km is same so x intercept is where the line converges

-y intercept changes because Vmax is decreasing with inhibitor

what is alpha? what is alpha’?

-alpha: binding to enzyme degree of inhibitor affinity

-alpha’: binding to ES degree of inhibitor affinity

summarize competitive inhibition

-alpha: binding to enzyme

-inhibitor competes for active site

-graph: same y intercept

-Km increased

-Vmax stays the same

-x int: -1/ alpha Km

-y int: 1/Vmax

-slope: alpha Km/ Vmax

summarize noncompetitive inhibition

-alpha = alpha’ : same binding enzyme or ES

-graph: same x intercept (same Km)

-Km is unchanged

-Vmax is decreased

-x intercept: -1/ Km

-y intercept: alpha’/Vmax

-slope: -alpha’ Km / Vmax

summarize uncompetitive inhibition

-alpha’: binds to ES

-graph: parallel lines

-Km is decreased by same factor as Vmax

-Vmax is decreased

-x int: -alpha’/ Km

-y int: alpha’/ Vmax

-slope: Km/Vmax

summarize mixed inhibition

-alpha’ does not equal alpha: binding for ES or E is in diff amounts

-graph: different x and y intercepts

-Km changes

-Vmax is decreased

-x int: -alpha’ / alpha Km

-y int: alpha’ / Vmax

-slope: alpha Km/ Vmax

why does Km not change for noncompetitive inhibition?

Non-comp inhibition is where the inhibitor will bind the ES and E equally. Binding the ES only will show an apparent increase in enzyme affinity for the substrate and thus a decrease in Km. Binding the E only will show an apparent decrease in enzyme affinity for the substrate and thus an increase in Km. If E and ES are bound equally, Km is simultaneously increased and decreased by equal amounts, and thus does not change

what is the mnemonic MUNC?

-M: mixed is many changes

-U: uncompetitive is unchanged slope

-N: noncompetitive no x change

-C: competitive, change Km only

if you are investigating inhibitors, which is the most promising compound?

-the drug with the lowest Ki, you don’t want to have to use a lot of inhibitor to drug an enzyme

what do irreversible inhibitors do?

-irreversible inhibitors form strong noncovalent associations

-or covalent link with an enzyme

-can destroy a functional group necessary for catalysis

-useful in studying enzymatic mechanisms

what are the three categories of irreversible inhibitors?

-group (side chain) specific reagents

-affinity labels → reactive substrate analogs

-suicide inhibitors

what is the basic mechanism of irreversible inhibitors? how is Vmax changed?

-can interact with E or ES, but once it interacts enzyme is removed permanently

-Vmax is decreased permanently

what are suicide (mechanism-based) inhibitors?

-most specific way to modify an enzyme’s active site

-modified substrates that bind enzyme like a normal substrate

how do suicide inhibitors work?

-processed initially by normal catalytic mechanism

-a chemically reactive intermediate is formed that inactivates the enzyme through covalent modification → enzyme now chemically inactive

-because enzyme is using its normal catalytic pathway, modified group is likely important for catalysis

-can design, if you know mechanism

how does DMFO, the suicide inhibitor to treat African sleeping sickness work?

-first step happens, but can’t continue further

-instead of forming saturated and electrophilic carbon, unsaturated carbon is formed and forms different product

-covalent modifications tend to be suicide inhibitors

what are transition state analogs?

-noncovalent, irreversible

-structural mimic of the transition state

-binds to active site with higher affinity (lower Ki) than substrate to the enzyme

how did we find uninhibited vmax and km from the lineweaver burk practice problem?

-Vmax is 1/y intercept so we just did that

-Km is the x intercept and the x intercept is where y is 0, so we plugged 0 into the y=mx+b equation

-but once we solve for x, the value for x isn’t Km. Km is -1/x

how did we find inhibited Vmax and Km apparent from the lineweaver burk practice problem?

-Vmax is 1/y intercept so we did that

-we did the same procedure as uninhibited to find the inhibited Km, making sure to take the negative reciprocal of x (-1/x)

how did we find Ki?

-Km apparent = alpha Km

-in this example Km apparent was 30 and Km was 10 and [I] was 2

-this mean alpha is 3

-we then used the equation alpha = 1 + [I]/Ki to find Ki