Exam 4

• Rate is proportional to transition state activation energy = Delta-G-dagger
• Likelihood the reaction occurs = Delta-G of the reaction
• Rate enhancement is proportional to Delta-Delta-G-double-dagger
• Equilibrium is proportional. To Delta-G of the reaction

Catalytic Strategies for Enzymes

• acid / base

Rnase A

• Has 2 histidines in the active site

  • Histidine at 118 Acts as an acid ( gives proton )
  • Histidine at position 12 acts as a base

• PH has to be near the $p{K}_a$$pK_a$ of the histidine side chain

• Histidine is a “popular” one for acid / base reactions

  • $p{K}_a$$pK_a$ of side chain = 6.0

Covalent Catalysis

• good at binding and stabilizing the transition state

  • Lowers activation energy

Metal Ion Catalysis

• amino acid chemistry stinks at redox reactions?

  • Good at acid base ?

• Help with acid base chemistry

  • You don’t have to change the pH

• Zinc changes the “polarizability” of the water around it

Proximity and Orientation

• closer in space = “more probable the bump into each other”

  • 5x enhancement

• Proper orientation = 100x enhancement

• Remove background movement

  • Enzyme “calms” the natural vibration of the substrates
  • $1\ast {10}^7$$1*10^7$ enhancement
  • Aka stabilizes the transition state

Geometry

• molecular orbital theory
• Van Der Waal Radii

Problem - Which of the following amino acid residues …

• B , histidine , aspartate , glutamate

Problem 2 - The figure to the left shows a proposed …

• Zinc = changes polarizability of water

• Arg 145 = helps to bind substrate

  • During transition state it helps stabilizes through ionic interactions

    • Hydrogen bonds

• Glu 270 = acts as a base , and then as an acid

Chemical Kinetics

• Kinetics helps understand reaction mechanism

• Concentration Changes :

  • Substrates
  • Products
  • Intermediates

• Concentration Doesn’t Change :

  • Catalyst

  • Solvent

  • Anything with extremely high concentration

    • Little change is negligible

• Changes in concentration can change the reaction order

• Zero Order Reactions = doesn’t change with respect to concentration

• First Order = changes with respect to the concentration of 1 thing

• Second Order = changes with respect to the concentration of 2 things

• Third Order = ….

Zero Order Reactions

• $\mathrm{A}⟶\mathrm{P}$$\ce{A -> P}$
• $\frac{d[A]}{dTime}=-k\ast [A{]}^0$$\frac{d[A]}{dTime} = -k * [A]^0$
• $[A{]}^0=$$[A]^0 =$ concentration at time zero
• $[A{]}^t$$[A]^t$ = concentration at time $t$$t$

• $[A{]}_t=[A{]}_0-(k\ast t)$$[A]_t = [A]_0 - (k * t)$

• $A$$A$ vs $T$$T$ should be linear , negative

• Slope = $-k$$-k$

  • $k$$k$ units = concentration / time

  • Negative because you are subtracting out anything that takes [A] away

    • Forward reaction

First Order Reactions

• $\mathrm{A}⟶\mathrm{P}$$\ce{A -> P}$

• $\frac{d[A]}{dT}=-k\ast [A{]}^1$$\frac{d[A]}{dT} = -k * [A]^1$

• $ln\left(\frac{[A{]}_t}{[A{]}_0}\right)=-k\ast t$$ln\left(\ \frac{[A]_t}{[A]_0}\ \right) = - k * t$

• Or

• $[A{]}_t=[A{]}_0\ast e^{-k\ast t}$$[A]_t = [A]_0 * e^{-k * t}$

• Plot = x-axis = time

  • Y-axis = $ln\left(\frac{[A{]}_t}{[A{]}_0}\right)$$ln\left(\ \frac{[A]_t}{[A]_0} \ \right)$

• Slope = $-k$$-k$

  • $k$$k$ units = $\frac{1}{T}$$\frac{1}{T}$

Second Order Reactions

• Unimolecular :

  • $2\mathrm{A}⟶\mathrm{P}$$\ce{2A -> P}$

  • $\frac{d[A]}{dT}=-k\ast [A{]}^2$$\frac{d[A]}{dT} = -k * [A]^2$

  • $${\int }_{[A_0]}^{[A_t]}\frac{d[A]}{[A{]}^2}=-k\ast {\int }_0^{T}dT$$

  • = $\frac{1}{[A{]}_t}=\frac{1}{[A{]}_0}+k\ast t$$\frac{1}{[A]_t} = \frac{1}{[A]_0} + k * t$

    • Or $[A{]}_t=\frac{[A{]}_0}{1+[A{]}_0\ast k\ast t}$$[A]_t = \frac{[A]_0}{1 + [A]_0 * k * t}$

  • Plot = linear

  • X-axis = time

  • Y-axis = $\frac{1}{[A]}$$\frac{1}{[A]}$

  • Slope = $+k$$+k$

    • $K$$K$ units = $\frac{1}{\text{concentration}\cdot \text{time}}$$\frac{1}{\text{concentration} \cdot \text{time}}$

• Bimolecular :

  • $\mathrm{A}{\phantom{A}}^{+}\mathrm{B}⟶\mathrm{P}$$\ce{A+B -> P}$
  • Harder than unimolecular
  • You can view this as change in product formation over time
  • $\frac{dP}{dT}=k\ast [A]\ast [B]$$\frac{dP}{dT} = k * [A] * [B]$
  • Integral Result = $\frac{1}{[A{]}_0-[B{]}_0}\ast ln\left(\frac{[B{]}_0\ast [A{]}_t}{[A{]}_0\ast [B{]}_t}\right)=k\ast t$$\frac{1}{[A]_0 - [B]_0} * ln\left(\ \frac{[B]_0 * [A]_t}{[A]_0 * [B]_t} \ \right) = k * t$

• Having 2 things coming together in the proper position and proximity is harder than 1 thing

• 3 things coming together at the same time is very rare in biology / enzymes

Opposing Reactions

• $\mathrm{A}\underset{\mathrm{k}{\phantom{A}}_{-1}}{\stackrel{\mathrm{k}{\phantom{A}}_1}{\stackrel{-\rightharpoonup }{\leftharpoondown -}}}\mathrm{B}$$\ce{A <=>[k_1][k_{-1}] B}$

• $\frac{d[A]}{dT}=-k_1\ast [A{]}^1+k_1\ast [B{]}^1$$\frac{d[A]}{dT} = -k_1 * [A]^{1} + k_1 * [B]^{1}$

  • Subtracting anything that takes $A$$A$ away

    • Then multiply by reactions

  • Add anything that makes it

    • Multiple by its reactants ??? See transcript @@ 10:48

       

• If you have some 3rd order reaction $\mathrm{A}+\mathrm{B}{\phantom{A}}^2⟶\mathrm{P}$$\ce{A + B^2 -> P}$

• But you don’t know which reactant is second order and which reactant is first order

• You can add so much concentration of one of the products to create a “pseudo “ order reaction

  • Then the rate no longer depends on it
  • Makes it “zero” order aka “pseudo” order

Half-Life

• time required to reduce the concentration by half
• Substitute in $\frac{[A{]}_0}{2}$$\frac{[A]_0}{2}$ for $[A{]}_t$$[A]_t$
• Zero Order = $\frac{0.5\ast [A{]}_0}{k}$$\frac{0.5*[A]_0}{k}$
• First Order = $\frac{ln(2)}{k}=\frac{0.693}{k}$$\frac{ln(2)}{k} = \frac{0.693}{k}$
• Second Unimolecular = $\frac{1}{k\ast [A{]}_0}$$\frac{1}{k * [A]_0}$

Problems :

• Determine the velocity of the reaction :

  • $\mathrm{X}+\mathrm{Y}⟶\mathrm{Z}$$\ce{X + Y -> Z}$

  • $V=\frac{d[P]}{dT}=k\ast [X]\ast [Y]$$V = \frac{d[P]}{dT} = k * [X] * [Y]$

    • $=4\ast {10}^2{M}^{-1}{S}^{-1}\ast 3\ast {10}^{-6}\ast 5\ast {10}^{-6}=\frac{6\ast {10}^{-9}M}{second}$$= 4 * 10^2\ M^{-1}\ S^{-1} * 3 * 10^{-6} * 5*10^{-6} = \frac{6 * 10^{-9}\ M}{second}$

• The reaction has a rate constant of $0.01s^{-1}$$0.01\ s^{-1}$ :

  • Units for $k$$k$ tell us its first order reaction
  • Half Life = $t_{\frac{1}{2}}=\frac{ln(2)}{k}=\frac{0.693}{0.01s^{-1}}=69.3seconds$$t_{\frac{1}{2}} = \frac{ln(2)}{k} = \frac{0.693}{0.01\ s^{-1}} = 69.3\ seconds$

• $k_{cat}$$k_{cat}$ = catalytic turnover :

  • measures the number of substrate molecules converted to product per enzyme per second when the enzyme is saturated

Transition State and Enzyme Kinetics 1

• Transition State Theory = Enzyme binds best to the transition state ??? Ok

• $\mathrm{A}+\mathrm{B}\stackrel{-\rightharpoonup }{\leftharpoondown -}\mathrm{X}{\phantom{A}}^{†}\stackrel{\mathrm{k}{\phantom{A}}^{{}^{\prime }}}{\to }\mathrm{P}$$\ce{A + B <=> X^{\dagger} ->[k^’] P}$

• K’ = rate of decomposition of transition state to products

• Reactants in the transition state are in equilibrium

• ??? See lecture audio

• $\frac{d[P]}{dT}=k\ast [A]\ast [B]=k^{{}^{\prime }}\ast [x{]}^{†}$$\frac{d[P]}{dT} = k * [A] * [B] = k^{’} * [x]^{\dagger}$

• $K^{†}=\frac{[x{]}^{†}}{[A]\ast [B]}$$K^{\dagger} = \frac{[x]^{\dagger}}{[A]*[B]}$

• $\mathrm{\Delta }{G}^{†}=-R\ast T\ast ln(K^{†})$$\Delta G^{\dagger} = - R * T * ln(K^{\dagger})$

• $\frac{d[P]}{dT}=k^{{}^{\prime }}\ast e^{\frac{-\mathrm{\Delta }{G}^{†}}{R\ast T}}$$\frac{d[P]}{dT} = k^{’} * e^{\frac{- \Delta G^{\dagger}}{R * T}}$

• What is the rate of the reaction proportional to?

  • Transition state energy
  • Rate $\propto \mathrm{\Delta }{G}^{†}$$\propto \Delta G^{\dagger}$

• frequency can bump the transition state form products

• The equation proves why it is proportional to transition state energy

Enzyme Catalysis

• The fastest an enzyme can go is linked to the “fusion”

  • How fast can the substrate get to the enzyme

• $\mathrm{\Delta }\mathrm{\Delta }G$$\Delta \Delta G$ = “rate enhancement”

• Biggest jump = rate limiting step

How to Lower $\mathrm{\Delta }{G}^{†}$$\Delta G^{\dagger}$

• enthalpy has the biggest effect

Imaginary Stickase

• “picture” of transition state theory

• If it prefers to bind to the transition state , then it lowers the energy of activation

• We need to have the number of molecules to reach the transition state to be high

  • We can use temperature to help
  • have to overcome peak of transition state
  • Higher temperature has more molecules able to reach transition state

• Enzymatically , we only have 1 line , because the body does not alter temperatures

  • Enzymes cause left shift on the curve

    • Kinetic energy vs number of molecules graph on the right

Mechaelis-Menten

• not the best , because of assumptions

  • Assumes there is only 1 substrate

  • Zero-order kinetics , doesn’t rely on the concentration of anything

    • This doesn’t accurately depict the situation in biology
    • But gives a good enough approximation to compare and conceptualize

• $\mathrm{E}+\mathrm{S}\stackrel{-\rightharpoonup }{\leftharpoondown -}\mathrm{ES}\stackrel{-\rightharpoonup }{\leftharpoondown -}\mathrm{E}+\mathrm{P}$$\ce{E + S <=> ES <=> E + P}$

  • Bunch of rate constants on each equilibrium position
  • $k_{-2}$$k_{-2}$ is very unlikely ( transition of Enzyme + Product back into the Enzyme-Substrate complex ) , and makes the math difficult , so it is often disregarded

• Velocity of product formation = $velocity=\frac{d[P]}{dTime}=k_2\ast [ES]$$velocity = \frac{d[P]}{dTime} = k_2 * [ES]$

• [ES] is a function of other things

• $\frac{d[ES]}{dT}=(k_1\ast [E]\ast [S])-(k_{-1}\ast [ES])-(k_2\ast [ES])$$\frac{d[ES]}{dT} = (k_1 * [E]*[S]) - (k_{-1}*[ES]) - (k_2*[ES])$

• There are a couple assumptions you can make

  • Assume an equilibrium expression $k_s$$k_s$

    • Textbook says it doesn’t work , but you can make this assumption but math is more complicated

    • But instead we use the steady state approximation

      • The change in enzyme-substrate concentration with respect to time is zero

        • Only works if we are operating under “saturation” conditions

          • Start with really high concentration of substrate , binds to the enzyme

            • As soon as the enzyme releases substrate , new substrate binds immediately

              • there is so much substrate around , there there is no chance of the enzyme ever being by itself

• Total amount of enzyme = free enzyme concentration + enzyme-substrate concentration

• Set $\frac{d[ES]}{dT}=(k_1\ast [E]\ast [S])-(k_{-1}\ast [ES])-(k_2\ast [ES])$$\frac{d[ES]}{dT} = (k_1 * [E]*[S]) - (k_{-1}*[ES]) - (k_2*[ES])$ equal to zero , and then rearrange it :

  • $k_1\ast [E]\ast [S]=(k_{-1}\ast [ES])+(k_2\ast [ES])$$k_1 *[ E ] * [S] =(k_{-1} *[ES]) + (k_2 * [ES])$

• Then some other step :

  • $\frac{([E{]}_T-[ES])\ast [S]}{[ES]}=\frac{k_{-1}+k_2}{k_1}=K_m$$\frac{([E]_T - [ES]) * [S]}{[ES]} = \frac{k_{-1} + k_2}{k_1} = K_m$

• High $K_m$$K_m$ = low affinity

• Low $K_m$$K_m$ = high affinity

• $K_m$$K_m$ is the mechaelis constant , roughly equal to binding affinity

  • Equals substrate concentration at $\frac{V_{max}}{2}$$\frac{V_{max}}{2}$

• Now “substitute that in , and solve for [ES]” , ok

• $[ES]=\frac{[E{]}_T+[S]}{K_m+[S]}$$[ES] = \frac{[E]_T +[S]}{K_m + [S]}$

• Now plug this into the original :

  • $\frac{d[P]}{dT}=k_2\ast [ES]$$\frac{d[P]}{dT} = k_2 * [ES]$
  • $=\frac{d[P]}{dT}=\frac{k_2\ast [E{]}_T\ast [S]}{K_m+[S]}$$= \frac{d[P]}{dT} = \frac{k_2 * [E]_T * [S]}{K_m + [S]}$

• $k_2$$k_2$ is the rate of the overall reaction

  • Getting rid of K and K-minus-one

• $V_{max}=k_{cat}\ast [E{]}_T$$V_{max} = k_{cat} * [E]_T$

• $V_0=\frac{V_{max}\ast [S]}{K_m+[S]}$$V_0 = \frac{V_{max} * [S]}{K_m + [S]}$

• If operating under zero-order kinetics : The only thing that dictates $V_{m}ax$$V_max$ is $k_{cat}$$k_{cat}$ and enzyme total

• Why does $K_m$$K_m$ being a ratio of rates make sense ?

  • $k_{-1}$$k_{-1}$
  • Asdfasf see transcription
  • Enzyme’s affinity for substrate dictates rate , $K_m$$K_m$ takes all of this into account? Sure

• $k_{cat}$$k_{cat}$ = catlyitic constant

• Catalytic Efficiency = $\frac{k_{cat}}{K_m}$$\frac{k_{cat}}{K_m}$ = how good the enzyme is

  • “Rate / affinity “

Line-Weaver Burk Plot

• Double-Reciprocal Graph

• Graph of kinetic data

  • Substrate concentration vs velocity

• Line-weaver is a “normalization” , creates a linear curve

• By plotting it this way , we can easily determine variables

• $Y=mx+b$$Y = mx + b$

• Slope = $\frac{K_m}{V_{max}}$$\frac{K_m}{V_{max}}$

• Y-Intercept = $\frac{1}{V_{max}}$$\frac{1}{V_{max}}$

  • Upward Shift on the Y-Axis = smaller $V_{max}$$V_{max}$
  • Downward Shift on the Y-Axis = larger $V_{max}$$V_{max}$

• X-Intercept = Set Y = 0 , solve for X

  • $=-\frac{1}{K_m}$$= -\frac{1}{K_m}$
  • Left Shift on the X-Axis = smaller $K_m$$K_m$ = better binding affinity
  • Right Shift on the X-Axis = larger $K_m$$K_m$ = worse binding affinity

• Better because of Pipetting reasons ?

Problem : An enzyme catalyzed reaction has a $K_m$$K_m$ of $1mM$$1\ mM$ and a $V_{max}$$V_{max}$ of $5nM{s}^{-1}$$5\ nM\ s^{-1}$. What is the velocity of the reaction when the substrate concentration is $1.5mM$$1.5\ mM$ ?

• as long as substrate and enzyme are in the same units , you don’t have to convert

Kinetics and Mechanisms

• 3 major reaction mechanisms

  • random substrate binding

    • intersect at y-axis

  • ordered /sequential substrate binding

    • first enzyme causes conformational change that is needed for the second substrate to bind

  • ping-pong mechanism

    • bind to substrate and release product

      • during this step , you modify the enzyme

        • the modified enzyme-substrate complex can then bind to second substrate

Lineweaver-Burk Plot

• $Y=mx+b$$Y = mx + b$

• Slope = $\frac{K_m}{V_{max}}$$\frac{K_m}{V_{max}}$

• Y-Intercept = $\frac{1}{V_{max}}$$\frac{1}{V_{max}}$

  • Upward Shift on the Y-Axis = smaller $V_{max}$$V_{max}$
  • Downward Shift on the Y-Axis = larger $V_{max}$$V_{max}$

• X-Intercept = Set Y = 0 , solve for X

  • $=-\frac{1}{K_m}$$= -\frac{1}{K_m}$
  • Left Shift on the X-Axis = smaller $K_m$$K_m$ = better binding affinity
  • Right Shift on the X-Axis = larger $K_m$$K_m$ = worse binding affinity

Inhibitors

• Competitive Inhibitors

  • all 3 are reversible

  • binds to the same place as the substrate

  • increases $K_m$$K_m$ , right shift on graph

  • $V_{max} stays the same

  • $K_m^{app}$$K_m^{app}$ or $\frac{-1}{\alpha K_m}$$\frac{-1}{\alpha\ K_m}$ = $K_m$$K_m$ in the presence of the inhibitor

  • cross at y-axis

  • $K_m^{app}=\alpha K_m$$K_m^{app} = \alpha\ K_m$

  • $\alpha$$\alpha$ = “level of inhibition”

    • $\alpha =1+\frac{[Inhibitor]}{K_I}$$\alpha = 1 + \frac{[Inhibitor]}{K_I}$

  • $K_m$$K_m$ = enzyme affinity to substrate

  • $K_m^{apparent}$$K_m^{apparent}$ = enzyme-substrate with inhibitor present

  • $K_I$$K_I$ = enzyme : inhibitor , when its binding to the same place as the substrate

• these equations help us in drug design

  • trying to inhibit a target , use these equations to prove / demonstrate viability

• alcohol dehydrogenase has multiple different substrates it can use

  • ethanol , methanol , propanol

  • $K_m$$K_m$ vs $K_I$$K_I$ ? Which is lower?

    • whichever is lower = higher affinity = the winner of which binds

Uncompetitive Inhibitor

• Binds to ES complex

• $V_{max}$$V_{max}$ = decreased

• $K_m$$K_m$ = decreased

  • means $K_m^{app}<K_m$$K_m^{app} < K_m$

  • binds in a different position

  • has to be enough substrate for enzyme to bind to form ES complex , if there is no ES complex , the inhibitor can’t bind

    • Not enough substrate for it to affect

  • competitive inhibitor is effective a low substrate concentrations

  • uncompetitive is effective at high substrate concentrations

  • perfect drug design is uncompetitive , because it can never be beaten

  • most of the time though drugs are competitive

  • rare in mechaelis-menten kinetics

    • less rare when there is 2 substrates

• Prime = denotes inhibitor is binding to ES

• $K_m^{app}=\frac{K_m}{{\alpha }^{‘}}$$K_m^{app} = \frac{K_m}{\alpha^{‘}}$

• lines are parallel

• ${\alpha }^{‘}=1+\frac{[I]}{K_I^{‘}}$$\alpha^{‘} = 1 + \frac{[I]}{K_I^{‘}}$

Mixed Inhibitor

• binds to the enzyme or enzyme-substrate complex

• has a measurable $K_I$$K_I$ and a $K_I^{‘}$$K_I^{‘}$

• subset of mixed called “non-competitive”

  • $K_I=K_I^{‘}$$K_I = K_I^{‘}$

• Non-Covalent Inhibitors :

  • Competitive = Binds Enzyme

    • $K_I$$K_I$ and $\alpha$$\alpha$
    • effective at low substrate concentrations
    • LWB = crosses @ $y$$y$-axis

  • Uncompetitive = Binds ES complex

    • $K_I^{‘}$$K_I^{‘}$ and ${\alpha }^{‘}$$\alpha^{‘}$
    • effective at high substrate concentrations
    • LWB = does not cross

  • Mixed = binds E or ES

    • effective at low and high concentrations
    • one subset called “non-competitive” , where $K_I=K_I^{‘}$$K_I = K_I^{‘}$
    • LWB = crosses anwhere but at the $y$$y$-axis

• SEE TRANSCRIPT FOR WHAT NOT TO MISS !!!

• $K_I<K_I^{‘}$$K_I < K_I^{‘}$ :

  • mixed

• $K_I^{‘}<K_I$$K_I^{‘} < K_I$ :

  • mixed

• $K_I=K_I^{‘}$$K_I = K_I^{‘}$ :

  • mixed
  • non-competitive

• $\alpha =1$$\alpha = 1$ = no inhibition

Problem : An enzyme has a $K_m$$K_m$ of $8\mu M$$8\ \mu M$

• $K_m=8\mu M$$K_m = 8\ \mu M$
• $K_I=$$K_I =$
• $K_m^{app}=12\mu M$$K_m^{app} = 12\ \mu M$
• $[I]=3$$[I] = 3$

• $K_m^{app}=\alpha K_m$$K_m^{app} = \alpha K_m$

  • $\alpha =\frac{K_m^{app}}{K_m}$$\alpha = \frac{K_m^{app}}{K_m}$
  • $\alpha =\frac{12}{8}=1.5$$\alpha = \frac{12}{8} = 1.5$

• $\alpha =1+\frac{[I]}{K_I}$$\alpha = 1 + \frac{[I]}{K_I}$

• $K_I=\frac{[I]}{\alpha -1}=\frac{3\mu M}{0.5}=6\mu M$$K_I = \frac{[I]}{\alpha - 1} = \frac{3\ \mu M}{0.5} = 6\ \mu M$

Problems Slide Deck

• #2 on homework 6 = Methanol Question with the dog

  • substrate =
  • inhibitor =
  • $V_{uninhibited}=\frac{V_{max}\ast [Methanol]}{K_M+[Methanol]}$$V_{uninhibited} = \frac{V_{max}*[Methanol]}{K_M + [Methanol]}$
  • $V_{inhibited}=\frac{V_{max}\ast [Methanol]}{\alpha \ast K_m+[Methanol]}$$V_{inhibited} = \frac{V_{max} * [Methanol]}{\alpha * K_m + [Methanol]}$
  • we wan’t to make sure its reduced to $5\mathrm{\%}$$5\%$
  • $\frac{V_{inhibited}}{V_{unihibited}}=0.05$$\frac{V_{inhibited}}{V_{unihibited}} = 0.05$
  • make substitutions and then algerbra

• A first order reaction is $25\mathrm{\%}$$25\%$ complete in 39 mins at 298K. What is the rate constant?

  • $ln([A_t])=ln([A_0])-k\ast t$$ln\left([A_t]\right) = ln([A_0]) - k * t$

  • solve for $t$$t$

  • $-k\ast t=\frac{ln[A_t]}{[A_0]}$$-k * t = \frac{ln[A_t]}{[A_0]}$

  • $[A_t]=0.75[A_0]$$[A_t] = 0.75\ [A_0]$

    • substitute this in
    • something cancels

  • $\frac{ln(0.75)[A_0]}{[A_0]}=-k\ast t$$\frac{ln(0.75)[A_0]}{[A_0]} = -k * t$

  • $\frac{ln(0.75)}{39min}=-k$$\frac{ln(0.75)}{39\ min} = -k$

  • $-k=-1.22\ast {10}^{-4}$$-k = -1.22 * 10^{-4}$

  • $k=1.22\ast {10}^{-4}{s}^{-1}$$k = 1.22 * 10^{-4}\ s^{-1}$

• For a simple reaction the $k_1$$k_1$ for the uncatalyzed reaction is $1\ast {10}^{-2}{s}^{-1}$$1 * 10^{-2}\ s^{-1}$ and the $k_1$$k_1$ for the catalyzed reaction is $5\ast {10}^6{s}^{-1}$$5 * 10^{6}\ s^{-1}$

  • Rate Enhancement = $\frac{k_{cat}}{k_{non}}=\frac{5\ast {10}^6{s}^{-1}}{1\ast {10}^{-2}{s}^{-1}}=5\ast {10}^8$$\frac{k_{cat}}{k_{non}} = \frac{5 * 10^{6}\ s^{-1}}{1 * 10^{-2}\ s^{-1}} = 5 * 10^{8}$

• From the following date determine $V_{max}$$V_{max}$ and $K_m$$K_m$

  • scatter plot ( 1 / S ) and ( 1 / V ) in excel

  • find equation of a line

  • Now we can find Vmax and Km

  • $V_{max}=\frac{1}{Y-intercept}=\frac{1}{0.0061}=163.9\mu Mmi{n}^{-1}$$V_{max} = \frac{1}{Y-intercept} = \frac{1}{0.0061} = 163.9\ \mu M\ min^{-1}$

    • same units on $V_0$$V_0$

  • $K_m=$$K_m =$

    • set equation equal to 0

    • $0=0.1961x+0.0061$$0 = 0.1961\ x + 0.0061$

    • $x=\frac{-0.0061}{0.1961}=0.0311$$x = \frac{-0.0061}{0.1961} = 0.0311$

    • $K_m=\frac{-1}{x}=\frac{-1}{0.0311}=32\mu M$$K_m = \frac{-1}{x} = \frac{-1}{0.0311} = 32\ \mu M$

      • same units as substrate concentration

• From the following data determine :

  • The type of inhibitor

    • plot everything as inverses

      • ( 1 / S ) and ( 1 / V )
      • should be positive slope , straight line

    • bottom line = $0$$0$ of the inhibitor ( aka only substrate )

    • the middle line = $3mM$$3\ mM$ of the inhibitor

    • top line = $5mM$$5\ mM$ of the inhibitor

    • lines are not parallel , slopes are different

      • rules out uncompetitive

    • where do they cross ?

      • the intercepts are different by only 0.01 = competitive

• Apparent $K_m$$K_m$ at each concentration of $I$$I$

  • No Inhibitor :

    • y = 0.4843x + 0.1951 ; find $K_m$$K_m$

      • $x=\frac{-0.1951}{0.4843}=-0.4028$$x = \frac{-0.1951}{0.4843} = -0.4028$
      • $K_m=\frac{-1}{-0.4028}=2.48mM$$K_m = \frac{-1}{-0.4028} = 2.48\ mM$

  • $3mMI$$3\ mM\ I$ :

    • y = 0.755x + 0.1969 ; find $K_m$$K_m$

      • $x=\frac{-0.1969}{0.755}=-0.2608$$x = \frac{-0.1969}{0.755} = -0.2608$
      • $K_m^{app}=\frac{-1}{-0.2608}=3.83mM$$K_m^{app} = \frac{-1}{-0.2608} = 3.83\ mM$

  • $5mMI$$5\ mM\ I$ :

    • y = 1.0062x + 0.1861 ; find $K_m$$K_m$

      • $x=\frac{-0.1861}{1.0062}=-0.1850$$x = \frac{-0.1861}{1.0062} = -0.1850$
      • $K_m^{app}=\frac{-1}{-0.1850}=5.41mM$$K_m^{app} = \frac{-1}{-0.1850} = 5.41\ mM$

• uncompetitive uses alpha-prime

• alpha or just $K_I$$K_I$ = competitive

• $K_I^{{}^{\prime }}$$K_I^{'}$ or ${\alpha }^{{}^{\prime }}$$\alpha^{'}$ = uncompetitive = binding at a different site than the substrate

• Estimate $K_I$$K_I$

  • $\alpha =1+\frac{[I]}{K_I}$$\alpha = 1 + \frac{[I]}{K_I}$
  • $\alpha K_m=K_m^{app}$$\alpha K_m = K_m^{app}$
  • $\alpha =\frac{K_m^{app}}{K_m}$$\alpha = \frac{K_m^{app}}{K_m}$
  • $\alpha =\frac{3.83mM}{2.48mM}=1.54$$\alpha = \frac{3.83\ mM}{2.48\ mM} = 1.54$
  • $K_I=\frac{[I]}{\alpha -1}$$K_I = \frac{[I]}{\alpha - 1}$
  • $K_I=\frac{3mM}{1.54-1}=5.56mM$$K_I = \frac{3\ mM}{1.54 - 1} = 5.56\ mM$
  • $\alpha K_m=K_m^{app}$$\alpha K_m = K_m^{app}$
  • $\alpha =\frac{K_m^{app}}{K_m}=\frac{5.41}{2.48}=2.18$$\alpha = \frac{K_m^{app}}{K_m} = \frac{5.41}{2.48} = 2.18$
  • $K_I=\frac{[I]}{\alpha -1}$$K_I = \frac{[I]}{\alpha - 1}$
  • $K_I=\frac{5mM}{2.18-1}=4.24mM$$K_I = \frac{5\ mM}{2.18 - 1} = 4.24\ mM$

• $K_I$$K_I$ is different because :

  • experimental error
  • lineweaver-burk plot is an estimation , creates error

• The catalytic efficiency of many enzymes depends on pH. Chymotrypsin shows a maximum value of $k_{cat}$$k_{cat}$

  • amino acids responsible for $K_{cat}$$K_{cat}$ Are different than the amino acids responsible for binding

    • $K_M$$K_M$ between 6 and 7 = probably histidine
    • 8 and 10 = probably lysine

  • $K_m$$K_m$ increasing = bad for catalytic efficiency

• An enzyme follows simple Michaelis-Menten kinetics

  • $K_m$$K_m$ for substrate $S$$S$ is $1mM$$1\ mM$
  • $K_m$$K_m$ for substrate $T$$T$ is $10mM$$10\ mM$
  • $k_2$$k_2$ for substrate $S$$S$ is $2\ast {10}^{4}se{c}^{-1}$$2 * 10^{4}\ sec^{-1}$
  • $k_2$$k_2$ for substrate $T$$T$ is $4\ast {10}^{5}se{c}^{-1}$$4 * 10^{5}\ sec^{-1}$
  • $S$$S$ is the preferred substrate because it has a lower $K_m$$K_m$
  • enzyme preferentially binds to $S$$S$ but it doesn’t have the greatest catalytic efficiency
  • Cat Eff = $\frac{k_{cat}}{K_m}=\frac{2\ast {10}^{4}se{c}^{-1}}{1mM}=2\ast {10}^{4}m{M}^{-1}se{c}^{-1}$$\frac{k_{cat}}{K_m} = \frac{2*10^{4}\ sec^{-1}}{1\ mM} = 2 * 10^{4}\ mM^{-1}\ sec^{-1}$
  • Cat Eff = $\frac{k_{cat}}{K_m}=\frac{4\ast {10}^{5}se{c}^{-1}}{10mM}=4\ast {10}^{4}m{M}^{-1}se{c}^{-1}$$\frac{k_{cat}}{K_m} = \frac{4*10^{5}\ sec^{-1}}{10\ mM} = 4 * 10^{4}\ mM^{-1}\ sec^{-1}$

• Draw curves that would be obtained when velocity vs $[S]$$[S]$ is plotted

  • $V_{max}$$V_{max}$ doubles

• Initial rate data for an enzyme that obeys Michaelis-Menten kinetics……

  • $V_{max}=\frac{1}{0.00426}=234.7\mu mol\cdot m{L}^{-1}\cdot s^{-1}$$V_{max} = \frac{1}{0.00426} = 234.7\ \mu mol \cdot mL^{-1} \cdot s^{-1}$
  • $k_{cat}=\frac{V_{max}}{[E_t]}=\frac{234.7\mu mol\cdot m{L}^{-1}\cdot s^{-1}}{0.003\mu mol\cdot m{L}^{-1}}=78200s^{-1}$$k_{cat} = \frac{V_{max}}{[E_t]} = \frac{234.7\ \mu mol \cdot mL^{-1} \cdot s^{-1}}{0.003\ \mu mol \cdot mL^{-1}} = 78200\ s^{-1}$
  • $V=\frac{V_{max}\ast [S]}{K_m+[S]}$$V = \frac{V_{max} * [S]}{K_m + [S]}$
  • $K_m=\frac{V_{max}\ast [S]}{V}-[S]=\frac{234.7\ast 160}{132}-160=124\mu M$$K_m = \frac{V_{max} * [S]}{V} - [S] = \frac{234.7 * 160}{132} - 160 = 124\ \mu M$
  • y-intercept = $\frac{{\alpha }^{‘}}{V_{max}}$$\frac{\alpha^{‘}}{V_{max}}$
  • ${\alpha }^{‘}=82.6$$\alpha^{‘} = 82.6$
  • ${\alpha }^{‘}=1+\frac{[I]}{K_I^{‘}}=\frac{1.2\ast {10}^{5}M}{82.6}=1.43\ast {10}^{-7}M$$\alpha^{‘} = 1 + \frac{[I]}{K_I^{‘}} = \frac{1.2 * 10^{5}\ M}{82.6} = 1.43 * 10^{-7}\ M$

Misc

• Office Hour - Next Week :

  • Monday = 10:00 - 13:00
  • Tuesday = 09:00 - 12:00

• Final Exam :

  • 35% from Dr. Leffak
  • Exam 3 = high level overview
  • Exam 4 = practice exam will be posted
  • Time = 10:15 - 12:15

Enzyme Examples

• Chymotrypsin and lysozyme ?

• Chymotrypsin :

  • protease , cleaves peptide bonds. Used in sequencing

  • cleaves after an aromatic on the C-terminus side

  • poster child for how cool enzymes are

  • Catalytic Triad = Aspartate 102 , Histidine 57 , and Serine 195

  • Specificity Pocket = where the amino acid that

    • scissored bond = bond that is cleaved

    • chymotrypsin vs trypsin

      • trypsin doesn’t have the serine in the core of the pocket

        • its swapped out for aspartate 189

    • the serine doesn’t disrupt bonding in the aromatic group

    • glycine is important because it is flexible and it is small enough to allow the aromatic group to fit into the specificity pocket

    • Trypsin cleaves after a positive amino acid

      • so it has a negative charged amino acid in the core of the pocket

• elastase cleaves after small , neutral amino acids

  • has a neutral core , with bulky amino acid side chains. Only allows small amino acids like alanine to fit inside

• Serine Proteases ( serine in active site ) = Chymotrypsin , trypsin , elastase

  • have 40% of their sequences is identical

  • specificity pocket is unique

  • clotting cascade = all serine proteases ( really common mechanism )

  • produced by pancreas , secreted into G.I track

    • produced and stored as the inactive ( pro ) form

    • then something is cleaved off to produce the active enzyme

      • = enzyme control : proteolytic cleavage

        • aka a zymogen

Figure 6-22 : Flow Charge

• what are the 3 amino acids in the catalytic triad doing ?

  • Aspartate 102 = has polarizing effects on the histidine
  • Histidine 57 = acting as a base
  • Serine 195 =

• Mechanism :

  • serine attacks sicissle bond

    • this transfers a proton to the histidine

      • the ability to transfer this proton is aided by the polarizing effects of Aspartate 102

    • this forms a tetrahedral intermediate

  • the intermediate decomposes

    • proton donation

    • this forms the acyl enzyme intermediate

      • this pushes part of the intermediate into the oxyanion hole

  • leaving group leaves

  • water is added

  • reverse of 1st step

    • has to return to original form to be an “enzyme”

  • Reasons Serine Proteases are Classical Examples

    • Because of the attack and forming a tetrahedral intermediate , it forms a covalent bond
    • “transfer of proton” = acid base chemistry
    • oxyanion hole = “preferentially binding to transition state”

  • during formation of intermediate , we are changing how it sits in the enzyme

    • pushed into the oxyanion hole

      • then it hydrogen bonds here

    • the best enzymes preferentially bind to the transition state

Figure 11-29

• Hydrogen bonding an polarizing effects matter

• hydrogen bonding to the “Asp”

• some hydrogen bonds are stronger than others

• low barrier hydrogen bond :

  • the nitrogen on the histidine and the carboxylic acid on the asp ,

    • the hydrogen is shared between the two amino acids

      • less as a “hydrogen bond” , more as a “shared hydrogen”
      • makes the bond stronger ( think covalent bond electron sharing )

• If Asp is mutated and swapped for Glycine , what happens to $k_{cat}$$k_{cat}$ ?

  • it decreases , catalytic rate goes down

    • If it is instead replaced with Glu , it would be more unpredictable

    • you still have carboxylic acid to form hydrogen bond

    • but now its 3d space problem

      • there is a chance it lines up better and increases $k_{cat}$$k_{cat}$

  • which affects $k_{cat}$$k_{cat}$ more ?

    • serine or aspartate ?

      • the serine , because the serine is the main player in the mechanism

Lysozyme

• found in mucous membranes , lacrimal glands

• attacks carbohydrate portion of cell membranes in bacteria

• specifically binds to 6 residues

  • cleaves between $D$$D$ and $E$$E$ site

• similar to chymotrypsin

• C-2 Fluorine = bad leaving group , won’t allow reaction to occur

  • substrate is “locked in place” = allows researcher to determine structure of the enzyme

    • without the substrate their, its harder to pinpoint which are the “catalytic residues”

• Catalytic Mechanism :

  • 2 paths because historically their are two different research groups with different conclusions for how the mechanism actually works

  • Left Mechanism :

    • Glu = acts as an acid ( donates proton ) , then in the next step it acts as a base ( takes hydrogen from water )

      • then we have a break in carbohydrates

    • Asp = sitting there chilling , polarizing the electrostatic interaction

  • Right Mechanism :

    • Glu = acting as an acid in first step. In the next step , the asp forms a covalent bond in the transition state

  • Right side = the winner ?

    • proved via fluorine ?

Lysozyme Activity vs pH

• catalytic mechanism is dependent on two ionizable side chains

  • asp needs to be deprotonated

• their pKas are pretty similar

• using titration of the enzyme , the pKa of the Glu = 6.2 = higher than it is in the table in the book

  • surrounding environment alters pKa values

• pKa of Asp 3.7

• allows for wider range

• when it is acidic < 3.7 = non active = protonated = can’t act as a nucleotide

• at high pH > 6.2 , the Glu will donated a proton = can’t act as an acid anymore

• delicate balance [ 3.7 , 6.2 ] = most active environment for enzyme = highest $k_{cat}$$k_{cat}$

Problem - Predict the effect of mutating ASP 102 of trypsin on substrate binding and catalytic activity

• if mutate Asp 102 , catalytic activity decreases . Binding is harder to predict.

Lysozyme residues Asp 101 ….

• they are part of a hydrogen bonding network
• with Asp is sitting right there
• others are part of a larger hydrogen bonding network
• if you mess up the network , you mess up the chemistry

Are we going to effect enzymes near equilibrium or far from equilibrium ?

• $\mathrm{F}\text{-}6\text{-}\mathrm{P}\underset{\mathrm{FBPase}}{\stackrel{\mathrm{PFK}}{\stackrel{-\rightharpoonup }{\leftharpoondown -}}}\mathrm{F}\text{-}1,6\text{-}\mathrm{BP}$$\ce{F-6-P <=>[PFK][FBPase] F-1,6-BP}$

  • trying to effect this pathway
  • both enzymes are far from equilibrium
  • if they are both far from equilibrium , and you are trying got control , then turn one of the valves……

Final Exam Study Sessions

• pH range around 6.0

• steady state approximation

• Uncompetitive inhibition , line-weaver burk plot = parallel lines

  • with inbititor = x-axis-left-shifted = $\frac{-1}{K_m^{app}}=\frac{-{\alpha }^{‘}}{K_m}$$\frac{-1}{K_m^{app}} = \frac{-\alpha^{‘}}{K_m}$

    • $K_m^{app}=\frac{K_m}{{\alpha }^{‘}}$$K_m^{app} = \frac{K_m}{\alpha^{‘}}$

• Competitive Inhibition :

  • With Inhibitor = x-axis-right-shifted :

    • $\frac{-1}{K_m^{app}}$$\frac{-1}{K_m^{app}}$ or $\frac{-1}{\alpha \ast K_m}$$\frac{-1}{\alpha * K_m}$
    • $K_m^{app}=\alpha \ast K_m$$K_m^{app} = \alpha * K_m$

• Practice Test - #7 :

  • enzyme is active in 6.5 to 8.5

  • competitive inhibitor can only bind at 6.5

  • at pH of 8.5 , the competitive inhibitor is deprotonating , and then no longer able to interact with the enzyme

    • unlikely enzyme would be deprotonating

• Competitive Inhibitor :

  • once you find K_m^{app} , then find alpha
  • solve for alpha
  • $\alpha =\frac{K_m^{app}}{K_m}$$\alpha = \frac{K_m^{app}}{K_m}$
  • then you use th equation : $\alpha 1+\frac{[I]}{K_I}$$\alpha 1+\frac{[I]}{K_I}$
  • then solve for $K_I=\frac{[I]}{\alpha -1}$$K_I = \frac{[I]}{\alpha - 1}$

• Practice Test - #3 :

  • Rate Enhancement = $e^{\frac{\mathrm{\Delta }{\mathrm{\Delta }}^{‡}}{R\ast T}}$$e^{\frac{\Delta \Delta^{\ddagger}}{R * T}}$
  • Rate Enhancement = $e^{\frac{13,000J\cdot mo{l}^{-1}}{\frac{8.314J}{mol\cdot K}\ast 298K}}$$e^{\frac{13,000 J \cdot mol^{-1}}{\frac{8.314\ J}{mol \cdot K}* 298\ K}}$
  • $RE=190$$RE = 190$

• Practice Test - #4 :

  • normally it has a negative charge

    • if we replace it with a positive charge , the $K_M$$K_M$ for substrate $A$$A$ would go up ,

      • $K_m$$K_m$ for substrate $B$$B$ decreases

• Homework 6 - Question #2 :

  • Velocity of Uninhibited : $V_{uninhibited}=\frac{V_{max}\ast [S]}{K_M+[S]}$$V_{uninhibited} = \frac{V_{max} * [S]}{K_M + [S]}$

  • $V_{inhibited}=\frac{V_{max}\ast [S]}{\alpha \ast K_M+[S]}$$V_{inhibited} = \frac{V_{max} * [S]}{\alpha * K_M + [S]}$

  • Get rid of 95%

  • $0.05=\frac{V_{inhibited}}{V_{uninhibited}}$$0.05 = \frac{V_{inhibited}}{V_{uninhibited}}$

  • $0.05=\frac{K_m+[S]}{\alpha \ast K_m+[S]}$$0.05 = \frac{K_m + [S]}{\alpha * K_m + [S]}$

  • Substrate Concentration

    • $[Methanol]=\frac{25mL\ast 0.9}{\frac{32g}{mol}\ast 15L}\dots .=0.0469M$$[Methanol] = \frac{25\ mL * 0.9}{\frac{32\ g}{mol}* 15\ L} …. = 0.0469\ M$

  • Ethanol = Inhibitor , 10 mM = 0.01 M

  • $K_M=$$K_M =$

  • $0.05=\frac{0.01+0.0469}{\alpha \ast 0.01+0.0469}$$0.05 = \frac{0.01 + 0.0469}{\alpha * 0.01 + 0.0469}$

  • $\alpha =109.1$$\alpha = 109.1$

  • $\alpha =1+\frac{[I]}{K_I}$$\alpha = 1 + \frac{[I]}{K_I}$

  • $K_I$$K_I$ = $K_M$$K_M$ for ethanol

  • Inhibitor Concentration = $K_I\ast (\alpha -1)$$K_I * (\alpha - 1)$

    • $[I]=0.001M\ast 108.1$$[I] = 0.001\ M * 108.1$
    • $[I]=0.1081M$$[I] = 0.1081\ M$

  • $\frac{0.1081mol}{1L}\ast 15L=1.623molEthanol\ast \frac{46g}{mol}=74.6gethanol$$\frac{0.1081\ mol}{1\ L} * 15\ L = 1.623\ mol\ Ethanol * \frac{46\ g}{mol} = 74.6\ g\ ethanol$

  • $74.6gethanol\ast \frac{1]mL}{0.9g}=82.9mLethanol\ast \frac{1}{0.12}=691mLwine$$74.6\ g\ ethanol * \frac{1] mL}{0.9\ g} = 82.9\ mL\ ethanol * \frac{1}{0.12} = 691\ mL\ wine$