lecture 10.105 Kinetics _ Lipids HO

pdf

School

University of California, Davis *

*We aren’t endorsed by this school

Course

BIS105

Subject

Biology

Date

Oct 30, 2023

Type

pdf

Pages

Uploaded by CommodoreLightningBarracuda37

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 1 We will finish the derivation of the Michaelis-Menten equation 1. The initial rate assumption. Since we are using initial rates, v 0 , P is never present in significant quantities. Therefore we can assume that the conversion of P and E to the ES complex can be ignored. That is, can ignore the term k -2 . Now, E + S →  − k k 1 1 ES → k cat P + E and v 0 = d P dt [ ] = k cat [ES] 2. The steady-state assumption. Assume that k cat is small compared to k 1 and k -1 , meaning that the formation of and dissociation of the ES is fast compared to the conversion of ES to P. If so, then [ES] can be considered a constant . d ES dt [ ] = − d ES dt [ ] or formation = break down k 1 ([E t ] - [ES])[S] = (k -1 + k cat )[ES] Step 3. Solve for [ES]: Multiply out and distribute. [ES] = k E S k k k S t cat 1 1 1 [ ][ ] [ ] − + +

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 2 [ES] = [ ][ ] [ ] E S k k k S t cat − + + 1 1 Step 4. Now the rate constants are all collected together. Define this term as K M , the Michaelis- Menten constant. K M = k k k cat − + 1 1 Step 5. Get rid of enzyme concentrations: From initial rate assumption, v o = dP dt = k cat [ES] v o = k cat × ] [ ] ][ [ S K S E M t + = ] [ ] ][ [ S K S E k M t cat + but when is v 0 = V max ? When all the enzyme is tied up in the ES complex. That is, when ES = E t . v 0 = V max when k cat [ES] = k cat [E t ]. Substituting from above, v 0 = V S K S M max [ ] [ ] + This is the Michaelis-Menten equation . Note that it has the same form as the equation for oxygen binding by myoglobin, i.e., c x x y V v + = = max 0 A rectangular hyperbola.

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 3 Recall that K M was defined as a collection of rate constants K M = k k k cat − + 1 1 What else can we discover about K M ? Examine the situation where v 0 = ½ V max , at [S] ½ . ½V max = V S K S M max [ ] [ ] 1 2 1 2 + ½ = [ ] [ ] S K S M 1 2 1 2 + ½K M + ½[S] ½ = [S] ½ ½K M = [S] ½ − ½ [S] ½ so, K M = [S] ½ K M = the concentration of S that gives ½ V max Look at the plot again.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 4 How do we determine K M and V max ? Initially, at [S] << K M , [S] + K M  K M v 0 = V S K S M max [ ] [ ] + = V K S M max [ ] , which is a straight line with a slope = V K M max . What are the units on this parameter? Reciprocal time. What does that suggest to you? This is a first order rate constant. At the end, when [S] >> K M , v 0 = V S K S M max [ ] [ ] + = V S S max [ ] [ ] = V max . Could use these positions in the curve to determine V max and K M , except 1. v 0 never really reaches V max 2. initial slope hard to pick out accurately

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 5 Better to use a linear conversion of the Michaelis-Menten curve, and make a Lineweaver-Burke plot. From v 0 = V S K S M max [ ] [ ] + create the reciprocal, 1 0 v K S V S K V S S V S M M = + = + [ ] [ ] [ ] [ ] [ ] max max max ] [ 1 1 1 max max 0 S V K V v M  + = Features: When [S] is very large, 1/[S] is very small. So at 1/[S] → 0, 1/ v0 → 1/V max . When 1/v 0 → 0, 1 1 1 0 v K V S V M = + max max [ ] = 0 1 1 0 1 V K S K S M M max [ ] [ ] +       = = + K S M [ ] = − 1 1/[S] = − 1/K M

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 6 Another look at K M Recall that K M = k k k cat + − 1 1 If the steady-state assumption is to hold, k cat << k -1 . Then K M  k -1 /k 1 , which is the dissociation constant for the ES complex. Written this way, ES →  − k k 1 1 E + S, K d = [ ][ ] S E ES At equilibrium, k -1 [ES] = k 1 [S][E], so, k k E S ES − = = 1 1 [ ][ ] [ ] K d = K M Under these conditions, K M is a measure of the affinity of S for E. High affinity means high [ES] means low K M . In many cases in the cell, [S] is near the enzyme's K m . Reasonable efficiency for the enzyme at this [S] level.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 7 Let’s now talk about V max or k cat V max or k cat Intuitively, this tells you something about the absolute potential activity of the enzyme, as it is the maximum rate achievable when [S] approaches infinity for a given enzyme concentrations. Recall that V max = k cat  [E t ]. So, k cat = V max / [E t ] Thus, k cat = a first order rate constant, units of s -1 . It is also the number of substrate molecules converted to product molecules per unit time on a single enzyme when the enzyme is saturated with substrate . This is the enzyme turnover number . You might expect a good enzyme to have a high turnover number. What is the best measure of the efficiency of an enzyme? Look at low [S], i.e., when [S] << K m . v = V max  [S] / (K m + [S]) = (V max /K m )  [S] Here V max /K m is a first order rate constant. Homeworks require you to recognize this fact. Further, v = k cat  [E t ]  [S] / (K m + [S])

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 8 so at [S] << K m v = (k cat /K m )  [E t ]  [S] Now (k cat /K m ) is recognized as a second order rate constant. If this number (k cat /K m ) is big, that means the rate of ES  P is high (k cat ), and the affinity of E for S is also high (low K m ). So, big k cat /K m , efficient enzyme. This provides a good way to compare the efficiency of the same enzyme for different substrates. For example, chymotrypsin. This enzyme cleaves peptide bonds immediately after aromatic amino acids. X - Gly - X; k cat /K m = 1.3 x 10 -1 M -1 s -1 X - Phe - ; k cat /K m = 10 5 M -1 s -1 So chymotrypsin cuts after F 10 6 x better than after G. Look again at v = (k cat /K m )  [E t ]  [S]. There is a theoretical limit as to how fast two molecules can collide by simple diffusion. This imposes an upper theoretical limit to the value of k cat /K m . This theoretical diffusion-controlled limit is on the order of 10 8 - 10 9 M -1 s -1 .

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 9 Many enzymes have k cat /K m values close to the diffusion-controlled limit. For example, acetylcholinesterase. k cat = 1.4 x 10 4 s -1 K m = 9 x 10 -5 M k cat /K m = 1.6 x 10 8 M -1 s -1 Thus, this enzyme operates a nearly the maximum that can be expected, at the diffusion controlled limit. Let's spend a minute and look at how the data might really appear, and what you do with it. Plot [P](t) vs t at 6 different [S]. Get 6 plots Now take initial slope of each plot. Units are  M per time; V o .

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 10 Say the numbers are: [S] (mM) = 1, 2, 5, 10, 20, 40 V 0 (  M/min) = 10.9, 20, 40, 60, 80, 96 Now we can plot V 0 vs [S], This looks like a hyperbola. What is K m ? V max ? Hard to tell. We want a Lineweaver-Burke plot. The numbers are: 1/[S] = 1, 0.5, 0.2, 0.1, 0.05, 0.025 1/V 0 = 0.092, 0.05, 0.025, 0.0167, 0.0125, 0.0104

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 11 Now we can see the parameters better. 1/V max = 0.0083, so V max = 120  M/min 1/[S] = -0.1, so K m = 10 mM I really recommend that you do the homework problems. Plot out the graphs. Don't do them for the first time at the exam. Enzyme inhibition How do you control the activity of an enzyme? Can control by changing the concentration of active enzyme . How? 1. Change the actual concentration of the enzyme Rate of synthesis Rate of degradation 2. Change the activity of the enzyme with inhibitors .

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 12 We will restrict our discussion to modes of action of reversible inhibitors. Enzyme kinetics can tell us about the type of inhibition. Three types of inhibition we will consider are: Competitive Noncompetitive Uncompetitive Competitive inhibition Substrate and inhibitor bind to the same site on the enzyme. What do you expect will be the effect on the v 0 vs. [S] curve? 1. As [S] →  , S competes out I, so no change in V max . 2. At [S] = K M , some S displaced by I, so it requires more S to reach ½ V max . So, K M,app > K M . Graphically

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 13 It is more obvious when plotted as a Lineweaver-Burke plot. This plot behavior is diagnostic of competitive inhibition. Formally,

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 14 K I is the dissociation constant for EI. ] [ ] ][ [ EI I E K I = eqn. 1 We also earlier showed that ] [ ] ][ [ 1 1 1 1 ES S E K k k k k k K D cat M = = = + = − − Given the similar nature of K I and K M and what you know about K M , what would you say K I tells you about the affinity of the enzyme for the inhibitor? We’re going to skip the derivation of the form the Michaelis-Menten equation takes when a competitive inhibitor is present, and simply write it down. For the curious I’ll leave it in the notes. Now we will derive an expression between the rate and [S] when an inhibitor, I, is present. ] [ 0 ES k v cat = Divide by [E] t = E + ES + EI

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 15 EI ES E ES k E v cat t + + = ] [ 0 Substitute [ES] = [E][S]/K M (from eqn. 2) and [EI] = [E][I]/K I (from eqn. 1), then factor out [E]. I M M cat I M M cat t K I K S K S k E E K I E K S E E K S E k E v ] [ ] [ 1 ] [ ] [ ] [ ] ][ [ ] ][ [ ] ][ [ 0 + +  = + + = Now cancel [E], divide both sides by k cat , and set the entire denominator over K M . I M M M I M M M M M I M M MAX t cat K I K S K S K K I K K S K K K S K I K S K S V v E k v ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ 1 ] [ [ 0 0 + + = + + = + + = = Resolve denominator by factoring out K M from 1 st and 3 rd term in denominator.         + + = I M MAX K I K S S V v ] [ 1 ] [ ] [ 0 or         + + = I M MAX o K I K S S V v ] [ 1 ] [ ] [ See that         + = I M app M K I K K ] [ 1 , , so that app M MAX o K S S V v , ] [ ] [ + =

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 16 The reciprocal form is MAX I MAX M V S K I V K v 1 ] [ 1 ] [ 1 1 0 + •                 + = So you can see that what has changed is the slope, not the y-intercept. The properties are these: As [I] increases, the reciprocal plot pivots about the y- axis intercept counterclockwise. The slope, K M /V max , increases, while the x-axis intercept decreases, as Km,app increases. How do you find K I ? A number of ways. 1. You can look at the x-axis intercept for – 1/K M,app , and calculate K I from there. Use         + = I M app M K I K K ] [ 1 , , so K I = 𝐾 𝑀 [𝐼] 𝐾 𝑀,𝑎𝑝𝑝 −𝐾 𝑀 2. Given v 0 with I present, V MAX , KM, [I] and [S], solve for K I . I M M I M K I K K S S V K I K S S V v ] [ ] [ ] [ ] [ 1 ] [ ] [ max max 0 + + =         + + =

Biological Sciences 105 Lecture 10, November 1, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 17 ] [ ] [ ] [ max 0 0 0 S V K I K v K v S v I M M = + + M I M K v S v S V K I K v 0 0 max 0 ] [ ] [ ] [ − − = ( ) M M I K v S v V I K v K 0 0 max 0 ] [ ] [ − − = Now simply plug in the numbers. There other more complicated graphical ways.