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Enzyme Kinetics and Ligand Binding [email protected] http://glutxi.umassmed.edu / S1-824 (a.m.), LRB 926 (p.m.) 6-5570 Resources: 1. The plots and data analyses in this lecture http://inside.umassmed.edu/uploadedFiles/gsbs/courses/2013-2014_Core_Course_Materials/ Block_I/CoreKinetics.zip 2. Graphing software (Prism) http://graphpad.com/apps/index.cfm/register/subscription/ID/3970/subscription/XBYHWLGI Monday, September 23, 13
Transcript
Page 2: Grad Kinetics

Objective of Class

• To emphasize that "kinetics" provides a set of tools to understand time-dependent processes and the underlying mechanisms that govern them.

Monday, September 23, 13

Page 3: Grad Kinetics

3

ContextMany of the phenomena we study in biology

are time-dependent phenomena

time for heat to propagate from the base to theends of the chains, as a function of the length hof the alkane molecules.

An 800-nm, 500-fs-duration laser pulse froman amplified titanium-doped–sapphire laser (5)incident on the Au/glass interface (the backside) of the 50-nm-thick Au layer generated hotelectrons within a skin depth of ~15 nm (6).Because the hot electrons have a large diffusioncoefficient, the electron temperatures at the frontand back of the Au layer equalized even beforeelectron-phonon coupling brought the hot elec-trons into equilibrium with the lattice (6). Within~1 ps, the Au layer was in thermal equilibri-um and uniformly heated throughout (6). Toimprove the adhesion of Au to glass it wasnecessary to add a Cr layer beneath the Au.Unfortunately, heat transfer from a Cr layer toAu is relatively slow; to minimize this effect, wemade the Cr layer just 0.8 nm thick. An ultrafastthermoreflectance apparatus (2, 7) was used tocharacterize the temperature rise of the Au layer.As shown in Fig. 1C, there is a fast increase ofthe Au surface temperature to 80% of the finaltemperature within 1 ps. There is also a slower(1.5 ps time constant) rise to the final temper-ature due to the Cr layer. The same transientresponse was observed with either front-side orback-side flash-heating and with or without aSAM. The Au layer remained at an approxi-mately constant high temperature for severalnanoseconds, subsequently cooling by heatdiffusion into the glass. In the SFG experiments,the intensity of the heating pulse was varied tolocate the threshold for melting the Au, and thenthe pulse was attenuated by 20%. Because themelting temperature of Au Tm = 1064°C, thisprocedure resulted in flash-heating of the Aulayer to ~800°C.

SAMs have been studied extensively bySFG since 1991 (8), but ultrafast probing of aflash-heated SAM requires some elaboration. Inthe SFG technique we used, a femtosecond in-frared (IR) pulse at 3.3 mm with a bandwidth of150 cm!1 is incident on the SAM, coherentlyexciting all the alkane CH-stretch transitions inthe 2850 to 3000 cm!1 range, along withelectrons in the Au skin layer, producing anoscillating polarization in both the Au and theSAM layers. At the same time, a picosecond-duration 800-nm pulse (“visible”) with a band-width of 7 cm!1 is incident on the sample. Thevisible pulse interacts with this oscillating polar-ization through coherent Raman scattering tocreate a coherent output pulse at the IR + visiblefrequency. This combined IR-Raman interactionis forbidden (in the dipole approximation) incentrosymmetric media because the second-order susceptibility c(2) vanishes in such media.Because the methylene –CH2- groups of thealkane SAM form a nearly centrosymmetricsolid, the SFG signal that we observed originatedpredominantly from the Au surface and theterminal methyl –CH3 groups. The well-knownSFG spectrum obtained in ppp polarization (4),

from a SAM with n = 17 (i.e., an 18-carbon orC18 SAM), is shown in Fig. 1D. Molecularvibrational transitions appear as dips against abroad nonresonant background from Au. Thesemethyl transitions have a spectral width Dn = 15cm!1, corresponding to a coherence decay timeconstant T2 = 0.7 ps, which indicates that SFGsignals are emitted during an ~1 ps time win-dow. Thus the time resolution of these SFGmeasurements is ~1 ps.

Three intense vibrational transitions wereobserved, originating from the symmetric nsCH3

and antisymmetric naCH3 methyl stretching vi-brations and from the dCH3 bending overtonetransition, which draws intensity from a 2:1Fermi resonance with the CH stretches (4, 8).All methylene transitions are weak, which isindicative of a high degree of order (4). Figure1D shows the spectrum of a C18 SAM ~400 psafter flash-heating, where the SAM is in equi-librium with Au at ~800°C. All three methyltransitions have lost intensity as a result ofthermal disordering of the methyl groups. The2dCH3 band evidences a red shift. The red shiftis caused by thermal excitation of the ~1500 cm!1

v = 1 state, which introduces an additionalcontribution from the anharmonically red-shiftedv = 1 ! v = 3 transition. It is notable thatmethylene transitions remain weak at hightemperature and that the transient intensity lossis reversible once the SAM returns to ambienttemperature. This indicates that chains remainupright and remain bonded to their originalsites. Under ordinary circumstances, alkaneSAMs on Au desorb to form the disulfideCH3-(CH2)n-S-S-(CH2)n-CH3 at 175 to 225°C(9, 10), which displays enhanced methyleneSFG transitions, so the unexpected stability ofthese SAMs at 800°C must be attributed to thebrief duration of the temperature increase.

We performed molecular simulations of aC16 SAM on Au (27 molecules with periodicboundary conditions) to better understand ther-mal disordering of the terminal methyl groups.When the SAM was equilibrated at 300 K,the well-known (11) all-trans structure with achain tilt of ~35° and a zenith angle (anglebetween surface normal and final C-C bond)of ~25° was obtained. The nsCH3 transitionhas an IR transition dipole moment of mag-

Fig. 2. Results of molecular simu-lations of alkanethiol SAMs. (A)Structure of alkanethiol SAM (n =15). Simulations were performedon a unit cell of 27 alkanes withperiodic boundary conditions. WhenT is increased to a high tempera-ture, the methyl head groups be-come orientationally disordered.(B) The SFG intensity for the nsCH3transition is approximately propor-tional to the square of the nor-malized ensemble–average IR dipolemoment (!m"/mIR)

2, which is temper-ature dependent. (C) With aninstantaneous temperature jump to1100 K, the methyl head groupsbecome orientationally disordered in less than 2 ps.

Fig. 3. (A) SFG spectraof C8 (n = 7) and C18(n = 17) SAMs withoutheating pulses (blue) andwith flash-heating to800°C (red). (B) VRF fora C8 monolayer. (C) VRFfor a C18 monolayer.

10 AUGUST 2007 VOL 317 SCIENCE www.sciencemag.org788

REPORTS

Vibrational Response Functions (VRFs) of Self-Assembled Monolayers. B VRF for a C8 monolayer. C VRF for a C18 monolayer

From: Wang et al., SCIENCE VOL 317, pp 787-790, 2007

W.Stuhmer et al.

ARCK1 ==

1 SOpA

RCK3 1

J6pA

RCK4

4 0 0 p A~~~~~~4Op

RCK5 1

,,,J@20pA

20ms

BG/Gm

1.2 T +

-80 -40

Fig. 5. Conductance-voltage relations of RCK channels. (A) Familiesof outward currents in response to depolarizing voltage steps. Fromtop to bottom RCKI, RCK3, RCK4, RCK5. The traces are responsesto 50 ms voltage steps from -50 to 40 mV in 10 mV intervals.Ensemble currents recorded from macro-patches. Sampling at 10 kHz,filtering at 3 kHz low pass. (B) Plots of normalized conductance(G/Gm) versus test potential for different RCK channels (RCK1: opencircles; RCK3: crosses; RCK4: diamonds; RCK5: filled circles). Toobtain the conductance values the current at a particular test potentialwas divided by the driving potential assuming a reversal potential of- 100 mV. The lines showed the results of a non-linear least-squaresfit of a Boltzmann isotherm (see Materials and methods) to theconductance values. The maximal conductance (Gm) obtained by the fitwas used to normalize the data. The half-activation voltages in thisplot are -24 mV (RCK1), -37 mV (RCK3), -30 mV (RCK4) and-40 mV (RCK5).

to voltage steps from -60 to 0 mV are shown in Figure 7.The step size of elementary currents at 0 mV varied between0.46 pA (RCK4) and 1.02 pA (RCK5). The single channelcurrent - voltage relations were measured in cell attachedpatches with normal frog Ringer's solution on the extra-cellular side. For all channels, the current-voltage relationis linear in the voltage range -20 to 20 mV. However, sincethis is a rather narrow range for conductance estimation, wemeasured the average amplitudes at 0 mV membranepotential. While the RCK1, RCK3 and RCK5 channels haverather similar single-channel current amplitudes, that of theRCK4 channel is considerably lower (Table I).

Pharmacology of RCK channelsA profile of the pharmacological sensitivity of the differentRCK channels to the K+ channel blockers 4-aminopyridine(4-AP) and tetraethylammonium (TEA) and several basicpeptide toxins was determined. The concentration

RCK1

5LOOpA

RCK3

lOpA

RCK4

I 200pA

RCK5

300pA

1 s

Fig. 6. Inactivation time course of currents mediated by different RCKchannels. Ensemble currents from macro-patches recorded at 0 mV testpotential from oocytes expressing RCKI, RCK3, RCK4 and RCK5channels at 0 mV test potential. Duration of the test pulse was 3.2 s.

Holding potential was -80 mV. Note difference in the degree ofinactivation at the end of the 3.2 s pulse. Sampling at 62.5 Hz andlow pass filtering at 120 Hz.

dependence of the block of outward currents by a particularsubstance was determined in whole-cell current recordingsat 20 mV test potential and the results are summarized inTable II. A striking difference in the inhibition of K+currents by TEA is observed between channels formed byRCK1 and RCK5 proteins. The RCK4 channels have a lowersensitivity to 4-AP than the other RCK channels. Both slowlyinactivating channels, RCK1 and RCK5, are much moresensitive to DTX than the inactivating channels RCK3 andRCK4. A different profile is observed for CTX, whichblocks RCK 1, RCK3 and RCK5 well, but is much lesseffective on RCK4 channels.

Discussion

Comparison between K+ channels in neurones andRCK channels expressed in Xenopus oocytesAn important question resulting from the molecular andfunctional diversity of RCK proteins is their relation to K+channels in native membranes. To establish the molecularidentity of a K+ channel in its native cell membrane anda particular RCK channel expressed in Xenopus oocytes,properties such as the voltage and time dependence, thesingle-channel amplitude and the susceptibility to blockersshould be compared. This comparison assumes that the K+channels expressed in oocytes accurately reflect thefunctional properties of K+ channels in the nativemembrane. This has not yet been shown and onlypreliminary conclusions can be drawn on the molecularstructure of K+ channels in native membranes.Delayed K + outward currents which inactivate only

slowly, on a time-scale of hundreds of milliseconds, arefound in neurones of different origins (Hille, 1984). Non-inactivating outward currents, e.g. in PC12 cells or frogspinal cord, are mediated by channels with a low (5-15 pS)channel conductance (Harris et al., 1988; Hoshi and Aldrich,1988a,b). Low conductance non-inactivating K+ channelswhich are DTX sensitive were found in rat sensory neurones

(Feltz and Stansfeld, 1988). Non-inactivating K+ channelswhich are DTX sensitive participate in regulating transmitter

3240

Conductance-voltage relations of RCK channels. (A) Outward currents in response to depolarizing voltage steps. From top to bottom RCKI, RCK3, RCK4, RCK5. The traces are responses to 50 ms voltage steps from -50 to 40 mV in 10 mV intervals.

From: Stühmer et al. EMBO Journal vol.8, pp.3235 - 3244, 1989.

Chemical control of metabolically-labeled Shaker channels with CTX–Biotin. (A) CTX–Biotin (10 nM) inhibits Shaker-IR K+ currents in CHO-K1 cells treated with either 50 µM thiol sugar 1 (h) or 0.5% ethanol as a vehicle (s). Only metabolically-labeled (+thiol sugar) channels were irreversibly blocked by CTX–Biotin after a simple washout; inhibition was completely reversed by an application of 1 mM TCEP. Reaction profiles were monitored with a 200 ms, 40 mV pulse every 15s.

From: Hua Z, Lvov A, Morin TJ, Kobertz WR. Chemical control of metabolically-engineered voltage-gated K(+) channels. Bioorg Med Chem Lett 2011.

needed to be chemically reversible. Although disulfide bond forma-tion between CTX and thiol-containing sialic acids on the cell sur-face is an obvious chemoselective and cell friendly reaction, wechose to label CTX with a bismaleimide that had an internal disul-fide bond because maleimides are inherently more stable in waterthan MTS reagents. Moreover, this subtle difference would allowfor delivery of a small molecule probe to the modified K+ channelsubunit after cleavage with reductant, which would be useful insubsequent biochemical, biophysical or imaging experiments. Tosimplify the synthesis of the bismaleimide, derivatization of CTX,and ensure delivery of a molecular probe to a K+ channel subunit,we set out to synthesize a symmetrical bismaleimide (Scheme 1)from cystamine dihydrochloride 2 that would allow for the facileincorporation of a molecular probe in the final step of the synthe-sis. The amino groups of cystamine 2 were capped with 2 equiv of adoubly amino-protected, activated ester of L-lysine 3. Selectivedeprotection of the Fmoc protecting groups gave the symmetricdiamine 4. Addition of 2 equiv of the NHS-ester of 3-(maleimi-do)propionic acid and N-Boc deprotection afforded bismaleimide5, which was subsequently biotinylated with 2 equiv of NHS–Bio-tin to give biotin bismaleimide 6. CTX was then derivatized bylabeling a cysteine mutant of CTX (R19C)32 with 100-fold molar ex-cess of biotin bismaleimide 6 to yield CTX–Biotin, which was puri-fied by reverse phase HPLC as we have previously described.17,19

With CTX–Biotin in hand, we decided to change both the chan-nel (Shaker–IR)33 and the expression system (CHO-K1 cells) todemonstrate that our approach was versatile and could be usedto label glycosylated ion-conducting subunits. Inactivation-re-moved Shaker is similar to Q1 in that it is an archetypical volt-age-gated K+ channel with a pore-forming domain flanked by fourvoltage-sensing domains.1 However, each Shaker voltage sensor isdoubly N-glycosylated on the extracellular loop connecting the firsttwo transmembrane segments (S1 and S2 loop).2 Previous studieshave shown that the N-glycans on Shaker contain sialic acid resi-dues.34 Therefore, each tetrameric Shaker channel will have eightN-glycans and potentially up to 24–32 sialic acids if every antennaon each glycoconjugate is capped. Figure 3 shows the effects of10 nM CTX–Biotin on CHO cells expressing Shaker K+ channelsincubated with either 50 lM thiol sugar 1 or vehicle in perfo-rated-patch whole cell recordings.35 CTX–Biotin inhibition ofShaker in vehicle treated cells (circles) was completely reversibleafter 3 min. In striking contrast to control and Q1 channels( Fig. 2C), metabolically-labeled Shaker channels (squares) wereirreversibly inhibited after identical treatment with CTX–Biotin.Irreversible inhibition of the Shaker K+ currents was reversed with

tris(2-carboxyethyl)phosphine (TCEP) (1 mM for 2 min), whichreduced the disulfide bond, freeing the toxin moiety and leaving abiotin moiety on a Shaker K+ channel subunit. Rewardingly, both

H2NS

SNH2 BocHN COOPfp

NHFmoc

2 HCl

NHBiotin

NH2

NHBoc

TFA

NH

HN

S

R3

ON

OO

O

NH

HN

S

R2

ON

OO

O

H2N

HN

S

R1

O 2

2

R2=

R3=

4

5

6

a, b

c, d

2

R1=

2 3

+

e

Scheme 1. Reagents and conditions: (a) MeOH, Et3N, rt; (b) piperidine, CH2Cl2, rt; (c) 3-(maleimido)propionic acid N-hydroxysuccinimide ester, DMF, rt, 22% for three steps(a–c); (d) TFA, CHCl3, H2O, rt; (e) NHS–biotin, Et3N, rt, 32% for two steps (d and e).

Time (s)0 200 400 600 800 1000 1200

Nor

mal

ized

cur

rent

0.0

0.5

1.0

TCEP

CTX-Biotin

Washout

A

BInitial

Washout

After TCEP

1 nA

50 ms

+ Thiol Sugar

- Thiol Sugar +Thiol Sugar

1 nA

50 ms

- Thiol Sugar

InitialWashout

C

V, mV

I/Imax

AfterTCEP

Initial

1.0

0.5

+ Thiol Sugar

V, mV-80

I/Imax

1.0

0.5

- Thiol Sugar

InitialWashout

-60 -40 -20 0 20 40 60 -80 -60 -40 -20 0 20 40 60

Figure 3. Chemical control of metabolically-labeled Shaker channels with CTX–Biotin. (A) CTX–Biotin (10 nM) inhibits Shaker-IR K+ currents in CHO-K1 cellstreated with either 50 lM thiol sugar 1 (h) or 0.5% ethanol as a vehicle (s). Onlymetabolically-labeled (+thiol sugar) channels were irreversibly blocked by CTX–Biotin after a simple washout; inhibition was completely reversed by an applicationof 1 mM TCEP. Reaction profiles were monitored with a 200 ms, 40 mV pulse every15 s. (B) Raw current traces before (black) and after (light gray) CTX–Biotintreatment. Traces of the recovered current after TCEP treatment are shown in thedark gray. (C) I–V relationship obtained from the same cell before inhibition withCTX–Biotin and after the current recovery by either washout (!thiol sugar) orreduction with TCEP (+thiol sugar).

Z. Hua et al. / Bioorg. Med. Chem. Lett. xxx (2011) xxx–xxx 3

Please cite this article in press as: Hua, Z.; et al. Bioorg. Med. Chem. Lett. (2011), doi:10.1016/j.bmcl.2011.04.099

Monday, September 23, 13

Page 4: Grad Kinetics

Time course of 3-0-methylglucose uptake in isolated muscle cells of Balanus nubilis. Ordinate: ratio of intracellular activity to extracellular activity of 3-0-methylglucose per equivalent volume of bulk external solution. Abscissa: time in hours. External sugar concentration, 1 mM. Uptake was measured using conventional (filled squares) and scintillator probe (open circles) methods. The water content of isolated fibres (70%) is shown by the continuous line above the points. The time at half-equilibration is shown by the dashed line. Mean fiber diameter, 1352 µm ; 21 ºC.

From: Carruthers, A. J. Physiol. VOL 336, pp 377-396, 1983

where Jt is uptake at time t , Jm is uptake a t complete equilibration and k is the rate

constant. The experiment'al results of Fig. 2 are wnsisknt with a value of k of

0.195 x s-l. Fig. 1 also shows that uptake is essentially linear during the initial

45 min of exposure to sugar. 3feasurements of sugar uptake in subsequent experiments

were made using an incubation period of 30 min in order to obtain accurate

measurements of t2he initial rate of sugar uptake.

Time (h)

Fig. I . Time course of 3-0-methylglucose uptake in isolated muscle fibres. Ordinate: ratio of i n t r a ~ l l u l a r activity t o extracellula~ activity of 3-0-methy1glucose per equixra1ent volume of bulk external solution, .4bscissa: time in hours. External sugar concentration, 1 m M Uptake was measured using rc)nventional and scintillator probe (0) methods. Kumber of points per conventional determination, five or more. The water content of isolated fibres (70%) is shown by the continuous line above the points. The time a t half-equilibration is shown by thedmhed line. Mean fibre diame*r, 1352 p m ; k m p r a t u r e . 21 OC.

The calculated rate of sugar uptake a t 30 min is 2 pmol . cm-*. s-I. Assuming

uptake is not saturated, the permeability of the barnacle muscle fibre, 2' (em. s-I),

to 3-0-methy~g~ucose is related to the sugar flux, J (mol. ernv2. s-I), by

where So is the external sugar concentration (in01 . emp3). This corresponds to a value

of P of 2 x cm. s-I which is some 3 4 orders of magnitude larger than the

permeability of artificial lipid bilayers to sugars (Jung & Snell, 1968 ; Lidgard & Jones,

19'75).

At equilibrium, the 3-0-methylglucose space of the fibre is 70 yo, which is in close

sgreement with estimates of the xra%er contents of barnacle muscle (71 & 1 7; ; R = 5 ) .

Assuming this water is not bound, these results shour that 3-0-methylglucose is not

accumulated by barnacle muscle and that the transfer of the sugar across the

sarco1emma is mediated by a passive, facilitated process. EJat ofphloretia on mqzr uptuke. Re18tively low concentrations of phloretin inhibit

The tools used to analyze the data obtained, however, are invariant and fall under the general umbrella of “kinetic analysis”.

First-order decay analysis of Drosophila embryonic total RNA during normal and slow larval growth conditions. The percentage of embryonic RNA remaining at various times during a chase under normal growth conditions (solid circles) or during slow growth conditions (open circles) is plotted. The regression lines indicate that the stability of embryonic RNA increases from 48 h 115 h if the larval growth rate is reduced.

From: Winkles et al., J. Biol. Chem., Vol 269, pp 7716-7720, 1985

7718 Ribosomal R N A Stability and Growth Conditions

Fraction

FIG. 1. Stability of embryonic total RNA during normal larval growth conditions. Adult females were fed a density-labeled Chlorella paste containing [3H]uridine for 84 h. Embryos were col- lected during the final 14 h of labeling, and some were transferred into a yeast medium containing ['*C]uridine for subsequent develop- ment. Larval chase times were estimated by assuming that the average age of collected embryos was the length of the collection period divided by two, that the duration of embryonic development is 23 h (Poulson, 1950), and that larvae begin feeding immediately after hatching. RNA was prepared from embryos and larvae and centri- fu.ged in CsHC02 equilibrium gradients. The arrow in panel A denotes the position of a ["CIRNA light buoyant density marker which was added to the embryonic RNA sample. A , embryonic RNA; B , 64-h larval RNA; C, 84-h larval RNA.

The autoradiogram is shown in Fig. 3C. The signals evident in the total RNA samples reflect hybridization to rRNA; hybridization to the tRNA carrier is negligible. The signal from sample 2, reflecting the relative amount of rRNA in the stable RNA population whose half-life we have measured above, is approximately 18% higher than the signal from sample 1, representing embryonic RNA. Another experiment using dense RNA from a 50-h chase point gives a similar result. The apparent increase in signal in the RNAs from the chase points is probably not significant, however, given slight variations seen in this kind of hybridization experiment. The results indicate that the fraction of RNA which is rRNA is

Length of chose, h

C""

FIG. 2. First-order decay analysis of embryonic total RNA during normal and slow larval growth conditions. The per- centage of embryonic RNA remaining a t various times during a chase with light yeast (normal growth conditions, solid circles) or dense algae (slow growth conditions, open circles) has been plotted. The values in solid circles are derived from the experiment shown in Fig. 1 as well as four other independent experiments (not shown). The values in open circles are derived from the experiment shown in Fig. 4 as well as one other independent experiment (not shown). The regression lines indicate that the stability of embryonic RNA in- creases from 48 h ( r2 = 0.915) to 115 h ( r2 = 0.892) if the larval growth rate is reduced.

not depleted during the chase. The dense RNA sample may be slightly enriched in rRNA, as would be expected if non- rRNA species decay faster than rRNA. Therefore, the half- life of total RNA is a good estimate of rRNA half-life.

To investigate whether the stability of embryonic rRNA varied under different larval growth conditions, we then mea- sured the half-life of this RNA using the L + D protocol as described under "Materials and Methods." A representative experiment using a dense chase is shown in Fig. 4. Adult females were fed a yeast paste containing [3H]uridine for 46 b. Some of the embryos laid during the last 12 h of this radioactive labeling period were transferred into a medium containing 50% 13C,15N-labeled Chlorella cells, 50% 13C,2H,'5N-labeled Chlorella cells, and [14C]uridine. First-in- star larvae were collected after 31 and 67 h of development. Total RNA was prepared and centrifuged to equilibrium in KI/NaI gradients. Since larvae develop and grow more slowly in algal medium, we might expect that the incorporation of dense isotopic precursors into the originally light embryonic RNA pools would be a gradual process. In the extreme case, this would result in an ineffective chase and an inaccurate estimate of stable RNA. As seen in Fig. 4B, after a chase of 31 h the newly synthesized larval [14C]RNA had a wide band width, indicating an RNA population of heterogeneous buoy- ant density. However, at this early chase time, the chase was still effective enough so that the buoyant density of the new larval RNA was greater than that of the inherited embryonic RNA and the amount of stable RNA could be calculated. During the next 36 h of larval development, continued incor- poration of stable isotope produced larval [14C]RNA with a more homogeneous buoyant density distribution and a greater density shift (Fig. 4C).

The half-life of embryonic rRNA under slow larval devel- opment conditions was calculated, as previously discussed, by quantitating the percentage of embryonic RNA remaining a t various times of larval development. A decay curve was gen- erated assuming first-order decay kinetics; experimental val- ues from two independent experiments were used in this analysis (Fig. 2). This analysis indicates that the half-life of embryonic rRNA inherited by slowly developing larvae is

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Because reactions may be measured over intervals of psec (10-12 s) to days (105 s), the methods of observation must be very different to resolve these different reactions.

These are the types of analyses you will be undertaking in your own research, in class and when reading the work of other researchers.

Monday, September 23, 13

Page 5: Grad Kinetics

What does kinetics tell us?

• Kinetic analysis determines the intermediates (however transient) in a reaction pathway and the roles they play in the reaction.

• This is done by:

• Analysis of reaction time courses (this is called chemical or reaction kinetics)

• Analysis of how rates of reactions at fixed [enzyme] vary with [substrate] (this is called steady-state kinetic analysis).

• From these data we can describe intermediates, predict mechanisms and test predicted behavior experimentally

Monday, September 23, 13

Page 6: Grad Kinetics

6

Assumed Knowledge/Glossary of terms

velocity or rates - amount of material (substrate, product, # cells etc) consumed or produced or number of events occurring per unit timecatalyst - an agent that accelerates a chemical reaction but which is unchanged in amount or chemistry at the end of the reaction.enzyme - a biological catalyst

substrates - molecules that are acted upon by enzymes

products - molecules produced by the action of enzymes on substratescatalysis - the process by which an enzyme or catalyst accelerates a reaction.

equilibrium - a state in which opposing forces are balanced.

chemical equilibrium - a reaction in which forward and reverse reactions continue to proceed but are quantitatively balancedthermodynamic equilibrium - Keq = [Product]e/[Substrate]e i.e. the ratio of product formed : substrate remaining at equilibriumthermodynamics of reaction rates - enzymes introduce alternative reaction pathways in which the Gibbs free energy of activation is reduced reaction order - the dependence of the reaction rate on [substrates]n when n is the number of substrates which must interact to form a single molecule of product.

Monday, September 23, 13

Page 7: Grad Kinetics

Specific Goals

This  lecture  reviews:

1. Theory of reaction order and rate constants2. Ligand binding

i. Theory of binding.ii. How to measure binding constantsiii. Inhibition of binding

3. Enzyme catalyzed reactionsi. Theory of binding.ii. How to measure binding constantsiii. Inhibition of binding

4. Interpretation of binding and Michaelis constants5. Limits on reaction rates.

Monday, September 23, 13

Page 8: Grad Kinetics

The Order of a Reaction

The rate or velocity, v, of a reaction or process describes how fast it occurs.

Usually, the velocity is expressed as a change in concentration per unit time,

v = dcdt

but it may also express the change in a population of cells with time, the increase or decrease in the pressure of gas with time or the change in absorption of light by a colored solution with time.

Monday, September 23, 13

Page 9: Grad Kinetics

describes how the velocity of the reaction depends upon the concentration of reactants.

The Order of a Reaction

molarity per secmolarity

= k1 = per sec

In the isomerization reaction

the theory of chemical kinetics tells us

Because the exponent, m =1, this is first-order with respect to A and, since A is the

only independent (concentration) variable in the rate equation, the reaction is first-

order overall.

The units for this first order reaction are derived from moles of product formed per

second per mole of reactant or,

molarity per sec = k1.molarity

A Bk1

[ ][ ] [ ]dt

d Bk A k Am1 1= =

Monday, September 23, 13

Page 10: Grad Kinetics

The units of this second order reaction are derived from moles of product

formed per second per mole2 of reactants or

molarity per sec = k1 (molarity)2

molarity per secmolarity2 = k1 = per molarity per sec

In a reaction of the type

The rate of the reaction is proportional to [E].[S].

E + S k1⎯ →⎯ E ⋅ S

Because m = 1 for both species, the reaction is first-order with respect to E or

S but is second order overall as one single step is involved in the reaction of

two species.

Monday, September 23, 13

Page 11: Grad Kinetics

Understanding the stoichiometry of a reaction is insufficient to predict the rate law of the reaction. This is illustrated on the next slide.

The Order of a Reaction must be determined experimentally

Monday, September 23, 13

Page 12: Grad Kinetics

If the concentration of a reactant remains unchanged by the reaction, it is frequently omitted in the rate-law expression. For example, with the first reaction, a more complete rate law is:

v = k[sucrose][H+][H2O]

(the reaction is 3rd order overall). However, H+ is a catalyst and its [] is constant during the run; [H2O] (solvent) is little changed because it is present in vast excess (55.5 M). Thus the terms [H+] and [H2O] are omitted in the rate law. If the reaction were carried out at different [H+] or in an inert solvent, a first-order dependence of the reaction on [H+] and on [H2O] is seen.

Stoichiometric�  Reaction Rate�  Law Kinetic�  Order

v�  =�  k[sucrose] 1

L-­isoleucine�  ⇄�  D-­isoleucine v�  =�  k[L-­isoleucine] 1

2N2O5�  ⇄�  4NO2�  +O2 v�  =�  k[N2O5] 1

2NO2�  ⇄�  2NO�  +O2 v�  =�  k[NO2]2 2

Hemoglobin.3�  O2�  +�  O2 ⇄�  Hb.4O2�  �   v�  =�  k[Hb.3�  O2][O2] 2�  (overall)

H2�  +�  I2 ⇄�  2HI�  �   v�  =�  k[H2][I2] 2�  (overall)

2�  D-­glucose�  ⇄�  maltose�  +�  H2O v�  =�  k[D-­glucose]2 2

C2H5OH�  �  �  �  ⇄�  �  �  �  CH3CHO v�  =�  constant 0

Sucrose+�  H2O ⇄�  Fructose�  +�  glucoseH+

DISTINGUISHING ORDER AND STOICHIOMETRY

Monday, September 23, 13

Page 13: Grad Kinetics

N2O5 NO2 + NO3

NO3 NO2 + O

2O O2

Reaction 3 Dinitrogen pentoxide decomposition

Hence, the first step is first order. 2N2O5 was indicated to balance the equation. It could just as easily have been written as: N2O5 ⇌ 2NO2 + ½O2

2NO2 ONOONO 2NO + O2

Reaction 4 Nitrogen dioxide decomposition to nitric oxide and oxygen.This involves formation of an intermediate thought to be the one shown in the reaction below

The first step is therefore second order.

The first column in the table indicates only stoichiometry, NOT reaction mechanism or order.

Monday, September 23, 13

Page 14: Grad Kinetics

The next few slides illustrate 4 types of reactions you may observe in the research setting.

Examples of reaction orders in nature

• Zero-order kinetics• First order kinetics• True Second order kinetics• Second order kinetics characterized by pseudo-first order behavior.

The reactions we illustrate in this lecture are available in the file: “CoreKinetics.pzf” This file is a “GraphPad Prism file that contains the data for each type of plot we will make and the types of analysis we will perform.

If you wish to plot these data yourself, you should download the file from the Core Curriculum website:

http://inside.umassmed.edu/uploadedFiles/gsbs/courses/2012-2013_Core_Course_Files/CoreKinetics.pzf%20-%20for%20students%20only.zip

And download the “GraphPad Prism” software from:http://inside.umassmed.edu/content.aspx?id=41754

Monday, September 23, 13

Page 15: Grad Kinetics

Zero-order reaction

0 2 4 6 8 100

2

4

6

8

10

TIME

[Sub

stra

te] o

r [P

rodu

ct]

SubstrateProduct

A zero-order reaction.

Note that [substrate] decreases linearly with time and [product] increases linearly with time. This observation suggests that we should perform “Linear Regression” analysis of the data to obtain constants (slopes) for substrate loss and product formation.

zero-order kinetics

0 100 2000

20

40

60

80

100

[S] µM

v (

d[P

]/d

t)

Monday, September 23, 13

Page 16: Grad Kinetics

e.g. ethanol conversion to acetaldehyde by the liver enzyme, alcohol dehydrogenase (ALDH). The oxidizing agent is nicotinamide adenine dinucleotide (NAD+) and the reaction can be written:

CH3CH2OH + NAD+⇌ CH3CHO + NADH + H+ALDH

The negative sign is used with reactant - ethanol - because its concentration decreases with time. The concentration of its product, acetaldehyde, increases with time.

At saturating [alcohol] (about 2 beers) and with NAD+ buffered via metabolic reactions that restore it rapidly, the rate of this reaction in the liver is zero-order over most of its course (ALDH is saturated by ethanol & NAD+).

v dt dtd[acetaldehyde] k[ ]d ethanol

o=- = =

t00

C0

C

CH3CH2OH

CH3CHO

Monday, September 23, 13

Page 17: Grad Kinetics

Theory of Zero-order ReactionsA zero-order reaction corresponds to the differential rate law

dCdt

= k0

The units of k0 are molarity per sec. This is a “zero-order reaction because there is no concentration term in the right hand of the equation

Defining C0 as the concentration at zero time and C as the concentration at any other time, the integrated rate law is:

C = C0 + k0t

ory = y-intercept + slope * x

Monday, September 23, 13

Page 18: Grad Kinetics

Zero-order reaction

0 2 4 6 8 100

2

4

6

8

10

TIME

[Sub

stra

te] o

r [P

rodu

ct]

SubstrateProduct

Best-­‐fit  values ! ! ! ! Substrate!! Product! ! Units          Slope! ! ! ! ! -­‐1  ±  0  ! ! 1  ±  0       mols/sec          Y-­‐intercept  when  X=0.0! ! 10  ±  0! ! 0 ± 0!! ! mols          X-­‐intercept  when  Y=0.0! ! 10.00!! ! 0! ! ! sec      Goodness  of  Fit! !          R  squared! ! ! ! 1.000!! ! 1.000

y = x-intercept + slope * x

Monday, September 23, 13

Page 19: Grad Kinetics

A first-order reaction.

Note here that [substrate] decreases in a curvilinear fashion with time and [product] increases in a curvilinear manner with time. This observation indicates that the reaction is NOT zero-order. How can we analyze this further?

1stOrder

0 2 4 6 8 100

1

2

3

4

5

TIME

[A] o

r [B] [Substrate]

[Product]

first-order kinetics0 100 200

0

20

40

60

80

100

[S] µM

v (

d[P

]/d

t)

Monday, September 23, 13

Page 20: Grad Kinetics

Theory of First-order Reactions

A first-order reaction corresponds to the differential rate-law:

dCdt

= k1C

The units of k1 are time-1 (e.g. s-1). There are no concentration units in k1 so we do not need to know absolute concentrations - only relative concentrations are needed.

The reaction:A ➝ B

has the rate law:

where k1 is the rate constant for this reaction.

The velocity may be expressed in terms of either the rate of disappearance of reactant (-d[A]/dt) or the rate of appearance of product (d[P]/dt).€

v = −d[A]dt

=d[B]dt

= k1[A]

k1

Monday, September 23, 13

Page 21: Grad Kinetics

First Order reactions - loss of substrate (see Appendix 1)

y = slope x + intercept

Theory

[ ] [ ]d A k A dt1 0- =

Defining [A]o as [A] at [A] at zero-time and integrating between A at time 0 and time t gives

ln[ ] ln[ ]kA t A1 0=- +

[A] = [A]0 e−k1t

Integrated rate law

Half-life

.lnt k k2 0 693

/1 21 1

= =

Defining [A] at t1/2 as [A]0/2

Monday, September 23, 13

Page 22: Grad Kinetics

Linear plot

[P] = [P]∞ 1− e−k1t{ }

First Order reactions - product formation (see Appendix 1)

Log plot

ln([P]∞ − [P]t ) = −kt + ln[P]∞

y = slope x + intercept

ln([P]∞ − [P]t ) = −kt + ln[P]∞

Half-life

Monday, September 23, 13

Page 23: Grad Kinetics

The raw data suggest that [substrate] falls from 5 mM to an equilibrium value of 0 mM.

Let us now make a log plot of log [substrate] vs time (or show the y-axis data on a log scale).

This produces a linear plot which is consistent with 1st order kinetics!

1stOrder

0 2 4 6 8 100.01

0.1

1

10

TIME

[A]

[Substrate]

Monday, September 23, 13

Page 24: Grad Kinetics

A second clue comes from the measurement of half-times. As [Substrate] declines from 5 - 2.5 mM, from 2.5 - 1.25 mM and from 1.25 to 0.625 mM, the time required for each 50% reduction is unchanged at ≈1.4 sec.

This is characteristic of “first-order decay” as observed with radioactive decay.

Constant decay times and the linear relationship between log {[S]t - [S]∞} vs time indicate a first order process. Let us check this by applying a first-order analysis to the data.

1st Order

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

TIME

[Sub

stra

te]

1.4 sec

1.4 sec

1.4 sec

Monday, September 23, 13

Page 25: Grad Kinetics

The goal of nonlinear regression is to fit a model (an equation) to XY data.

The model is expressed as an equation that defines Y as a function of X and one or more parameters.

Nonlinear regression finds the values of those parameters that generate the curve that comes closest to the data. Those best-fit values of the parameters are the best possible estimate of the values of those parameters.

To use nonlinear regression, therefore, you must choose a model or enter one. GraphPad Prism offers a model for first-order reaction called “One-Phase Decay”

The equation is:

Y=(Y0 - Plateau)*exp(-k*X) + Plateau

In which the parameters are defined as:

Non-linear regression analysis of untransformed data

Y0 is the Y value when X (time) is zero.

Plateau is the Y value at infinite time.

k is the rate constant (per unit time).

Tau is the time constant (the reciprocal of k).

Half-life is computed as ln(2)/k.

Span is the difference between Y0 and Plateau

URL: http://www.graphpad.com/help/Prism5/Prism5Help.html?reg_exponential_decay_1phase.htm

Monday, September 23, 13

Page 26: Grad Kinetics

1. Start with initial estimated values for each parameter in the equation.

2. Generate the curve defined by the initial values. Calculate the sum-of-squares -- the sum of the squares of the vertical distances of the points from the curve.

3. Adjust the parameters to make the curve come closer to the data points -- to reduce the sum-of-squares. There are several algorithms for adjusting the parameters - Prism uses the Marquardt algorithm.

4. Adjust the parameters again so that the curve comes even closer to the points. Repeat.

5. Stop the calculations when the adjustments make virtually no difference in the sum-of-squares.

6. Report the best-fit results. The precise values you obtain will depend in part on the initial values chosen in step 1 and the stopping criteria of step 5. This means that repeat analyses of the same data will not always give exactly the same results.

URL: http://www.graphpad.com/help/Prism5/Prism5Help.html?how_regression_works2.htm

The basic idea of nonlinear regressionEvery nonlinear regression method follows these steps:

Monday, September 23, 13

Page 27: Grad Kinetics

One  phase  decay!Perfect  fit! ! [Substrate]! [Product]! ! UnitsBest-­‐fit  values ! !          Y0!! ! ! ! ! 5.000!! ! 0       mM          Plateau! ! ! ! ! 0! ! ! 5.000       mM          k! ! ! ! ! ! 0.5000! ! 0.5000     per  sec          Half  Life!! ! ! ! 1.386  ! ! 1.386       sec          Tau  =  1/k! ! ! ! 2.000!! ! 2.000       sec          Goodness  of  Fit! !          Degrees  of  Freedom! ! 48! ! ! 48          R  square!! ! ! ! 1.000!! ! 1.000

Y=(Y0 - Plateau)*exp(-k*t) + Plateau

1stOrder

0 2 4 6 8 100

1

2

3

4

5

TIME

[Sub

stra

te] o

r [Pr

oduc

t]

[Substrate][Product]

Monday, September 23, 13

Page 28: Grad Kinetics

General rules for 1st order reactions

1. Plot of log(St-S∞) vs time produces a straight line with slope = -k2. The half-time (t1/2) and k are invariant of the starting value of St chosen.3. t1/2 = 0.693/k4. Plot of log(P∞-Pt) vs time produces a straight line with slope = -k

5. The units of k1 are time-1 (e.g. s-1). There are no concentration units in k1 so we do not need to know absolute concentrations - only relative concentrations are needed.

6. k may be obtained by direct curve fitting procedures by nonlinear regression7. The appropriate equation for loss of substrate is

[S]t = {[S]0 - [S]∞} e-(k.t) + [S]∞8. The appropriate equation for product formation is

[P]t = [P]∞ (1 - e-(k.t))

9. When a first order reaction is reversible (as most are), e.g.

A ⇌ B The overall rate constant is k1 + k2

k1

k2

Monday, September 23, 13

Page 29: Grad Kinetics

Second Order Reactions

Class 1 (A+A ⇌ P)

The differential rate law is v=k2[A]2

Although one or more reactants may be involved, the rate law for many reactions depends only on the second power of a single component. e.g.

Fall into 2 categories in which the rate law depends: 1) upon the second power of a single reactant species, or 2) the product of the concentrations of two different reagents.

2 proflavin [Proflavin]2

2[ ]2

2 A–A–G–C–U–UA–A–G–C–U–U

U–U–C–G–A–A

2[ 2 2]2

Monday, September 23, 13

Page 30: Grad Kinetics

2nd Order class 1

0 2 4 6 8 100

1

2

3

4

5

TIME

[A] o

r [B]

[Substrate]

[Product]

[substrate] decreases and [product] increases in a curvilinear fashion with time. This indicates that the reaction is NOT zero-order. How can we analyze this further?

The curves drawn through the points were computed by nonlinear regression assuming first order kinetics (one-phase decay equation). Note the systematic deviations from the fit. This strongly suggests that this reaction does not follow first order kinetics.

We can investigate this further by plotting the residuals of the fit (how each point deviates from the calculated fit) vs time.

Monday, September 23, 13

Page 31: Grad Kinetics

Nonlin fit of 2ndOrderIrrev:Residuals

0 1 2 3 4 5 6 7 8 9 10-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

TIME

[Substrate][Product]

The non random scatter of deviations of fit from data around the origin confirms the poor fit and that we should consider either an error in data sampling or another model for the data.

Monday, September 23, 13

Page 32: Grad Kinetics

Class 1, 2nd order Transform of data

0 2 4 6 8 100

2

4

6

8

TIME

[A] 0/

[A]

2nd order data

Linear  regression  analysisBest-­‐fit  values !          Slope! 0.66  ±  0          Y-­‐intercept  when  X=0.0! 1.0  ±  0          X-­‐intercept  when  Y=0.0! -­‐1.515        Goodness  of  Fit!          R  square!1.000

slope = [A]0 k2

Thus one expects a linear relation between the reciprocal of the reactant concentration and time.

Defining [A] at zero-time = [A]0, it can be shown that the integrated rate law is

1[A]

−1

[A]0

= k2t

1[A]

= k2t +1

[A]0

multiply both sidesby [A]0

[A]0

[A]= [A]0 k2t +1

Theory of Class 1 Second-order Reactions

[A]0/[A] versus time for normalized 1st and 2nd order kinetics with identical t½

Class 1, 2nd order Transform of data

0 2 4 6 8 100

2

4

6

8

TIME

[A] 0/

[A]

1st order data

2nd order data

slope = [A]0 k2

Monday, September 23, 13

Page 33: Grad Kinetics

A0k (slope) vs A0 second order

0 2 4 6 8 100.0

0.5

1.0

1.5

[A]o

[A] 0

k pe

r sec

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

[A]0 k per sec

0.1320 ± 2.842e-009-3.974e-009 ± 1.763e-0083.010e-0087.576

half-time vs [A]

0 2 4 6 8 100

2

4

6

8

[A]

t1/2

sec

How starting [A] affects rate of 2nd order reaction

0 2 4 6 8 100

5

10

15

TIME

[A] 0/

[A]

12345678910

Increasing[A]0

Monday, September 23, 13

Page 34: Grad Kinetics

General rules for 2nd order reactions (Class 1)

1. Standard 1st order analysis does not work2. Plotting [A]0/[A] vs time produces a straight line with

slope [A]0 k3. Plotting slope vs [A]0 produces a straight line with

slope k and y-intercept 0.4. The half-time (t1/2) falls with increasing [A].5. The units of k are concentration-1.time-1.6. This analysis breaks down when the reaction is

reversible (i.e. when kr ≥ kf/10

Monday, September 23, 13

Page 35: Grad Kinetics

2nd order reactions Class 2 (pseudo-first order; A+B⇌P)

A reaction that is 2nd order overall is 1st order with respect to each of the two reactants.

E+S ESk1

k2

For example, in the reaction

If the enzyme E were maintained at a constant low [] (e.g. [E] < [S]/100) and the substrate were varied, the reaction differential rate law is:

v= [E]k1[S]

Let us review this by examining ligand (L) binding to a receptor (R).

Upon rapid mixing of R and L, the receptor may undergo a fluorescence change allowing measurement of ligand binding. Alternatively, it may be possible to measure ligand binding by use of radiolabeled Ligand and filter-bound receptor. Either way, the time course of ligand binding may be examined to determine whether it displays first or second order kinetics.

R L LRkr

kf+

Monday, September 23, 13

Page 36: Grad Kinetics

The data were fitted with the one-phase association equation and the fit is excellent in each case (the residuals < [LR]/100)

You can also see that the reaction becomes faster at higher [L] - k increases and t1/2 falls with increasing [L].

Pseudo 1st Order

0 5 100.0000

0.0002

0.0004

0.0006

0.0008

0.0010

TIME

[LR

] µM

.167

.278

.1

.464

.774

1.292

2.154

3.594

5.995

10

[L]

At zero-time, various concentrations of L (µM) were mixed with 1 nM R. The time course of LR formation was monitored at each [L].

Monday, September 23, 13

Page 37: Grad Kinetics

We will show below that:

1. The slope is kf

2. The y-intercept is kr

3. The x-intercept is -kr/kf

kobs vs L

0 2 4 6 8 100

5

10

15

20

25

k obs

per s

ec

[L] µM

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

kobs per sec

1.999 ± 0.00018610.5012 ± 0.0007352-0.25070.5002

1. We obtain k or kobs from the one-phase association equation fits.

2. If we then plot kobs versus [L], you can see that the plot is linear with 2 constants - slope and y-intercept.

Monday, September 23, 13

Page 38: Grad Kinetics

For our reaction

The rate of LR formation is given by:

1. The time dependent component of this expression is e-t(kr+kf[L]) = e-t kobs.

2. Thus kobs = (kr+kf[L])

3. In a plot of kobs versus [L], kobs increases linearly with [L] (slope = kf) and the y-intercept = kr.

4. The x-intercept (when kobs = 0) = -kr/kf. Why?

5. 0=kr + kf[L]; thus -kf[L] = kr; thus -[L] = kr/kf

6. Analysis of the time course of L binding to R at varying [L] permits computation of kf, kr and kf/kr = Keq for the reaction.

7. This is ONLY true when [L] >> [R]. Here, first-order kinetics are observed because [L] does not change significantly. If [L] ≈ [R] the system will behave like a class 1 second order reaction.

Theory for pseudo first-order reactions

If [R]0 is the amount of receptor at t= 0, it can be shown that the integrated rate law is:

Monday, September 23, 13

Page 39: Grad Kinetics

What is the difference between a first order reaction and a second-order reaction that behaves like a first order reaction?

• A second-order reaction that behaves like a first order reaction (e.g. page 36) is called a pseudo-first-order reaction. Its rate constant, kobs, increases linearly with [S] (i.e. kobs = kr+kf[S]). t1/2 falls with increasing [substrate].

• A true first-order reaction is characterized by a rate-constant, k, that is independent of [substrate] or [product]. t1/2 is independent of [substrate].

What is the difference between a class 1 second order reaction and a class 2 second-order (pseudo-first-order) reaction?

• kobs for a “class 1” 2nd order reaction is k[S]0 and when [S]0 is 0, kobs = 0

• A “class 1” 2nd order reaction is not described accurately by first order equations but when 1/[S] is plotted vs time, the plot is linear.

• kobs for a “class 2” 2nd order reaction is kf[S]0 + kr and when [S]0 is 0, kobs = kr.

• A “class 2” 2nd order reaction is described accurately by first order equations.

Monday, September 23, 13

Page 40: Grad Kinetics

Summary: How to determine the Order and rate constant of a reaction.Use direct data plots

• If a curved plot results, other plots such as log [] versus time (first order) or 1/[] versus time (2nd order) may be tried. One looks for a linear plot indicating order and whose slope is proportional to the rate constant.

• First order: v = [S] k1 = molarity/sec

Once reaction order is determined, and rate constants computed by linear or nonlinear regression analysis, velocities (mol/sec) are obtained as:• zero-order: v = k0 = molarity/sec

• Second order: v = [S] [S] k2 = molarity/sec, or v = [E] (kr + kf[S]) = molarity/sec,

These experiments are then repeated at different [S] to obtain the relationship between [S] and velocity. These data then allow to determine reaction mechanism - the topic covered by the remainder of this lecture.

• A plot of [] versus time gives a clue to the order. If the plot is linear, a zero-order reaction is indicated.

Monday, September 23, 13

Page 41: Grad Kinetics

We will analyze the kinetic behavior of a simple enzyme and examine inhibitions of enzymes by various types of inhibitor.

We ask:

1. How does the activity of an enzyme vary with substrate concentration and why?

2. What are kcat, Vm, Km, Kd and Ks and how are they related and how can we measure them?

3. How do certain chemicals inhibit enzymes and how can we determine these mechanisms in practice?

Part 2 - Steady-state kineticsGoals

Monday, September 23, 13

Page 42: Grad Kinetics

CATALYTIC PRINCIPLESA key step in enzyme function is the formation of the enzyme (E)/substrate (S) complex, ES.

A number of observations point to the existence of ES prior to the release of product (P) and free E. These are:

1. ES complexes have been directly visualized by EM +X-ray crystallography.

2. The physical properties of an enzyme can change upon binding S.

3. The spectroscopic characteristics of E and S can change upon ES formation.

4. High specificity for ES formation is observed.

5. The ES complex may be isolated in pure form.

Jack Griffith developed techniques that let scientists see the finer details of DNA. In 1971 he and Arthur Kornberg published this photo, the first electron microscope image of DNA bound to a known protein - DNA polymerase.

Monday, September 23, 13

Page 43: Grad Kinetics

6. At constant [E], increasing [S] results in increased product formation to a point where product formation no longer increases. This saturation is presumed to reflect the fact that all E is now in the form ES. This is illustrated below.

Vm is a theoretical, maximum value for v.

Km is that concentration of [S]

producing a v of Vm/2.

Vm

0.5 Vm

Km

0 10 20 30 40 500

20

40

60

80

100

120

[S] mM

v (d

[P]/d

t), r

ate

of r

eact

ion

0.01 0.1 1 10 100 1000

0

25

50

75

100

[S] µM

v (

d[P

]/dt)

[S] mM

Monday, September 23, 13

Page 44: Grad Kinetics

The reaction velocity curve is a section of a rectangular hyperbola which in this instance takes the generic form

This equation is called the Michaelis-Menten equation.(note in this handout the terms Vm and Vmax may used interchangeably)

How can this phenomenon be rationalized in terms of enzyme-mediated catalysis?

v =

Vm * [S]Km + [S]

Monday, September 23, 13

Page 45: Grad Kinetics

Steps in enzyme-mediated catalysis

Let us examine an enzyme-mediated reaction.

Here enzyme E reacts with substrate S to form the complex ES. ES is then converted to EP which dissociates to E and product (P). The rate constants kf, kr, k2, k-2, kp and k-p describe the rates of the various steps.

Monday, September 23, 13

Page 46: Grad Kinetics

! !The solution (equation describing the rate of reaction in terms of substrate and product levels and rate constants) is quite complex to derive. However, we can make a number of simplifying assumptions in order to more readily obtain a solution to this scheme. How do we do this?

1.! !Assume that the reverse reaction (P→ S) is negligible. While this reaction is thermodynamically feasible, we (acting as the biochemists working with this enzyme) can establish experimental conditions that preclude or minimize the reverse reaction. e.g. adding an additional enzyme which converts P into another species Q which cannot react with our enzyme. Or, we can measure the rate of reaction at very early time points where the reverse reaction is insignificant.

2.! !Assume only a single central complex (ES) exists. i.e. ES breaks down directly to E + P.

3.! ! [S] >> [E] i.e. S interaction with E to form ES does not significantly affect [S].

Monday, September 23, 13

Page 47: Grad Kinetics

The overall scheme is visualized as follows.

There are two parts to this reaction:

1)! Formation of ES2)! ES breakdown to product P and free enzyme E

Monday, September 23, 13

Page 48: Grad Kinetics

For our reaction,

The formation of ES is a second order process and the breakdown of ES to E + S or to E + P are first order processes.

The units are kr = kp = per sec.

kf = per M per sec. You will see later why these units are important.

Monday, September 23, 13

Page 49: Grad Kinetics

Enzyme/substrate interactions ≡ Receptor/ligand interactions

The first step in enzyme function is the formation of the enzyme (E)/substrate (S) complex, ES.

Consider the reaction

Defining Et as E + ES

The derivation of this solution is included as Appendix 2

what is kr/kf?

Monday, September 23, 13

Page 50: Grad Kinetics

The ratio kr/kf has units of:

per secper M per sec

= M

This means that one-half of [Et] = [ES] when [S] = kr/kf

When [S] = kr/kf,

0 25 50 75 100

0.0

0.2

0.4

0.6

0.8

1.0

[S] µM[E

S]/[

E t]

kr/kf

Monday, September 23, 13

Page 51: Grad Kinetics

What is the significance of kr/kf?

kf/kr = Keq = equilibrium constant or association constant for E-S interaction.

kr/kf = 1/Keq = Ks or Kd = dissociation constant for the ES complex

Monday, September 23, 13

Page 52: Grad Kinetics

When Kd is 1 µM, 50% of Et = ES when [S] = 1 µM = 1 x 10-6M.

When Kd is 1 nM, 50% of Et = ES when [S] = 1 nM = 1 x 10-9M.

A low value for Kd means that the ES complex is more stable (less dissociates to E + S) thus at any [S], there is a higher probability that ES is formed. Less S is required to occupy one-half of the available binding sites. The enzyme shows high affinity for S.

A low value for Kd (high affinity for S) results from energetically favorable interactions between E and S (e.g. H-bonding and multiple van der Waals).

Monday, September 23, 13

Page 53: Grad Kinetics

∆Go∆Go

Keq per M Kd M kcal/mol kJ/mol1,000,000 0.000001 -8.2 -34.210,000 0.0001 -5.5 -22.8100 0.01 -2.7 -11.410 0.1 -1.4 -5.71 1 0.0 0.00.1 10 1.4 5.70.01 100 2.7 11.40.0001 10000 5.5 22.80.0000001 10000000 9.5 39.9

You may recall from “Thermo” that Keq = e-∆Gº/RT

Thus at 20ºC, an equilibrium constant of 10 corresponds to a ∆Gº of -1.36 kcal/mol (see Table below)

Note: H-bond energies range from 3 to 7 kcal/mol. van der Waal’s bond energies are approximately 1 kcal/mol.

Monday, September 23, 13

Page 54: Grad Kinetics

Fit these data in Prism by nonlinear regression using “One-site binding - specific binding”

Y=Bmax*X/(Kd + X)

How do we measure Kd and Maximum binding?In this experiment we continue our analysis of ligand binding started on slide 36. Here we measure the amount of ligand bound to the receptor R at 10 sec (where [L], [R] and [LR] are in equilibrium; see slide 36).

[L]  µM! [LR]  µM0.100! 2.848e-0040.167! 3.995e-0040.278! 5.262e-0040.464! 6.496e-0040.774! 7.557e-0041.292! 8.377e-0042.154! 8.959e-0043.594! 9.349e-0045.995! 9.599e-00410.000! 9.756e-004

equilibrium binding

0 1 2 3 4 5 6 7 8 9 100.0000

0.0002

0.0004

0.0006

0.0008

0.0010

[L] µM

[LR

] µM

One site -- Specific bindingBest-fit values

BmaxKd

[LR] µM

0.00100.25 µM

µM

Monday, September 23, 13

Page 55: Grad Kinetics

Lineweaver Burk

-5 0 5 10

1000

2000

3000

4000

1/[L] per µM1/

[LR

] per

µM

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

1/[LR] per µM

251.0 ± 0.09620999.2 ± 0.3800-3.9800.003983

slope = Kd/Bmax

x-cept = -1/Kd y-cept = 1/Bmax

There are other ways of computing Kd and Bmax. These are shown below and involve linearization of nonlinear data. The derivations are included as Appendix 3. You should

note that linearization greatly amplifies errors of analysis!

Hanes-Wolf Transform of equilibrium binding

0 5 10

2000

4000

6000

8000

10000

[L] µM

[L]/[

LR]

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.0

[LR] µM

999.9 ± 0.03448250.5 ± 0.1362-0.2505

slope = 1/Bmax

y-cept = Kd/Bmax

x-cept = -Kd

Scatchard Transform of equilibrium binding

0.0000 0.0005 0.0010 0.00150.000

0.001

0.002

0.003

0.004

0.005

[LR] µM

[LR

]/[L]

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

[LR] µM

-3.984 ± 0.0027590.003986 ± 2.095e-0060.001000-0.2510

slope = -1/Kd

y-cept =Bmax/Kd

x-cept = Bmax

Monday, September 23, 13

Page 56: Grad Kinetics

If Bmax is measured as 10 µM bound ligand and you know that [Et] is 10 µM, the number of ligand binding sites per enzyme is 1 and Bmax = [Et].

If Bmax is measured as 10 µM ligand and you know that [Et] is 5 µM, the number of ligand binding sites per enzyme is 2 and Bmax = 2[Et].

If Bmax is measured as 10 µM ligand and [Et] is 20 µM, the number of ligand binding sites per enzyme is 0.5 (or 1 if one-half of the enzyme is denatured!)

and Bmax = 0.5[Et].

When you perform a binding assay (measuring the concentration of bound ligand as a function of free [ligand]), you typically measure the maximum binding capacity (Bmax) of your system and Kd. Your challenge is then to understand how Bmax is related to [Et]

How is [Et] related to Maximum Binding or Bmax?The plots on the previous page gave us maximum [ligand bound] or Bmax under the prevailing experimental conditions

Monday, September 23, 13

Page 57: Grad Kinetics

With some enzymes, the functional unit comprises only 1 enzyme molecule which binds 1 substrate or ligand. Here, [Et] = Bmax.

If 2 subunits are required for the functional complex, Bmax = [Et]/2. If 3 subunits are required for the functional complex, Bmax = [Et]/3.In general, if n subunits are required for the functional complex, Bmax = [Et]/n.

SubunitA

Other enzymes form a complex that contains 2 or more functionally non-interacting enzyme molecules - each of which binds 1 substrate or ligand. Again, [Et] = Bmax.

SubunitA

SubunitB

Yet other enzymes are complexes comprising 2 or more functionally interacting enzyme molecules. Here, two enzyme molecules may be required to form a single binding site. Defining [Et] as the concentration of enzyme monomers:

Subu

nit A

Subu

nit B

Monday, September 23, 13

Page 58: Grad Kinetics

Competitive antagonists displace substrates from the enzyme

Ligand binding in the presence of a competitive antagonist I

1 10 100 1000 100000.0

0.2

0.4

0.6

0.8

1.0[I]=0[I]=5[I]=25[I]=50[I]=100

[S] µM

[liga

nd] b

/[Enz

yme]

t

S I

enzyme

E + I EIES +SKiKs

The curve is shifted to the right (Kd increases) but the maximum amount of binding (Bmax) is unchanged.

Monday, September 23, 13

Page 59: Grad Kinetics

Binding in the presence of a competitive antagonist is described by:

[ ]{ [ ] } [ ]

[ ] [ ]substrateK K

Inhibitor substrate

Enzyme substrate

1b

di

t=+ +

( [ ]/ )

[ ]

[ ]

K K I K

KK

KI

K

KKI

1

1

1

dapp d i

d

dapp

i

i

d

dapp

= +

- =

=-

Defining Kdapp as the Kd measured in the presence of Inhibitor

Monday, September 23, 13

Page 60: Grad Kinetics

Some physiologic antagonists can result in down-regulation of a receptor or enzyme. When ligand binding binding to the receptor is measured, there are fewer receptors. This is a noncompetitive type of inhibition.

Ligand binding in the presence of a noncompetitive antagonist I

1 10 100 1000 100000.0

0.2

0.4

0.6

0.8

1.0

[I]=0[I]=5[I]=25[I]=50[I]=100

[S] µM

Rel

ativ

e bi

ndin

g

Kd

Bm

Bm 5

Bm 25

Bm 50

Bm 100

Monday, September 23, 13

Page 61: Grad Kinetics

Let us now consider the breakdown of ES to form regenerated E and product, P.

The overall scheme is visualized as follows.

Monday, September 23, 13

Page 62: Grad Kinetics

Defining the rate of product formation, v as

v = kp [ES] (1

It can be shown that:

v =Vm[S]Km + [S] (2

The derivation of this solution is included as Appendix 4

Km =k2 + k3k1

(4

where

Vm = [Et]kp (3

Monday, September 23, 13

Page 63: Grad Kinetics

Properties of the Michaelis Menten Equation

However, Vm is rarely measurable because in many instances it is not possible to add sufficient quantities of substrate to saturate the enzyme . You may then ask, if Vm is not measurable, how does one determine Vm and Km for a reaction?

at low [S],

i.e. v increases linearly with [S] v ≈

Vm [S]Km

first-order kinetics

i.e. v does not increase with increasing [S]

At very high [S], ! ! v ≈ Vm ! !

zero-order kinetics

0 100 2000

20

40

60

80

100

[S] µM

v (

d[P

]/d

t)

Monday, September 23, 13

Page 64: Grad Kinetics

How do we measure Vm and Km?In this experiment we have measured the rate of product formation in an enzyme catalyzed reaction and express this rate, v, versus [S].

Non-linear regression (using Prism) using the Michaelis-Menten equation yields Vm and Km.

There are other methods of calculating these parameters by linearization of the data and by using linear regression. These methods and their derivations are included as Appendix 5.You should note that linearization greatly amplifies errors of analysis!

Enzyme

0 20 40 600

20

40

60

80

VmKm

M/min100.025.00

[S] M

rate

of r

eact

ion

M/m

in

Lineweaver Burk Enzyme

-0.1 0.0 0.1 0.2 0.3 0.4

0.05

0.10

0.15SlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

1/v min/ M0.2500 ± 1.824e-0090.0100 ± 2.711e-010-0.04004.000

1/[S] per M

1/v

min

/M

slope = Km/Vm

x-cept = -1/Km

y-cept = 1/Vm

Hanes Wolf Enzyme

-20 20 40 60

-0.2

0.2

0.4

0.6

0.8

1.0

[S]

[S]/v p

er

min

Slope

Y-intercept when X=0.0

X-intercept when Y=0.0

1/slope

[S]/v

0.0100 ± 2.320e-010

0.2500 ± 7.331e-009

-25.00

100.0

slope = 1/Vm

y-cept = Km/Vm

x-cept = -Km

Monday, September 23, 13

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ENZYMES ARE INHIBITED BY SPECIFIC MOLECULESReversible Inhibitors - May occur naturally or may be synthesized as drugs.

King-AltmanRepresentation

A competitive inhibitor, I, competes for binding with S. I is not transformed into product. The inhibitor, I, resembles the substrate, S. I reduces the rate of reaction + S by reducing the proportion of enzyme in the form ES

Competitive InhibitionE + I EI

ES E + Pkp

+S

Ki

Ks

S I

enzyme

Monday, September 23, 13

Page 66: Grad Kinetics

Noncompetitive Inhibition

S and I are not mutually exclusive but ESI is catalytically inactive. When S binds, the enzyme undergoes a conformational change which aligns the catalytic center, C, with the susceptible bonds of S; I interferes with the conformational change, but has no effect on S binding.

E

E + P

E + I EI

ES + I ESIE

I

I

I

I

S S

S S

E

C C

C C

kp

E + P

kp

+S

+S

Ki

Ks

Monday, September 23, 13

Page 67: Grad Kinetics

Competitive inhibitionS and I compete for binding to the same site

i.e. when [I] increases ! Km(app) increases because Km(app) includes an [I] term ! Vm(app) is unaffected because = k3[E t] (i.e. includes no [I] terms)

Defining Vm(app) and Km(app) as Vm and Km in the presence of inhibitor

v[Et ]

=

k3 [S]KS

1+ [S]KS

+ [I]Ki

v =[Et ]k3 [S]

KS 1+ [I]Ki

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+ [S]

[ ][ ]

K SV S( )

( )

m app

m app=+

[ ]K K

KI

1( )m app Si

= +' 1

Monday, September 23, 13

Page 68: Grad Kinetics

Competitive inhibition

Vm is unaffected but Km(app) increases with [I].

Km(app) = KS(1 + [I]/Ki)

0

2

4

6

8

10

12

0 20 40 60 80 100

control+ inhibitor

v, ra

te o

f rea

ctio

n

[S] mM

Vm

Km

Km(app)

0

0.2

0.4

0.6

0.8

1

1.2

-0.1 0 0.1 0.2

Control+ inhibitor

1/V

1/[S] per mM

1/Vm

-1/Km

-1/Km(app)

Monday, September 23, 13

Page 69: Grad Kinetics

Examples  of  Compe--ve  InhibitorsNaturally  occurring  (source  and  targets)

Digitalis  (fox  glove;  Na,KATPase)

Tetradotoxin  (puffer  fish;  Na-­‐Channel)

Cytochalasin  B  (fungi;  glucose  transporters)

Atropine  (deadly  nightshade;  Acetylcholine  receptor)

Synthe5c  (their  targets  and  use)

Ibuprofen  (cyclo-­‐oxygenase  inhibitor  -­‐  anH-­‐inflammatory)

Sulfanilamide  (dihydropteroate  synthetase;  anHbacterials  -­‐  inhibit  folate  synthesis)

NeosHgmine  (acetylcholinesterase;  prolong  neuromuscular  transmission  -­‐  treat  myasthenia  gravis)

Indinavir  (HIV  protease  II  inhibitor;  prevent  HIV  transmission  and  full-­‐blown  AIDS)

see http://www.emdbiosciences.com/docs/docs/LIT/ISB_USD.pdfMonday, September 23, 13

Page 70: Grad Kinetics

thus the inhibitor I reduces Vm(app) but has no effect on Km(app)

Noncompetitive inhibition

[ ][ ]

or vK SV S( )

( )

m app

m app=+

Monday, September 23, 13

Page 71: Grad Kinetics

Noncompetitive inhibition

Km(app) is unaffected but Vm(app) is decreased.

Vm(app) = Vm/(1 + [I]/Ki)

where Ki is the dissociation constant for the EI complex, Ki = [E][I]/[EI]

0

2

4

6

8

10

12

0 20 40 60 80 100

control+ inhibitor

v, ra

te o

f rea

ctio

n

[S] mM

Vm

Vm(app)

Km

0

0.2

0.4

0.6

0.8

1

1.2

1.4

-0.1 0 0.1 0.2

Control+ inhibitor

1/V

1/[S] per mM

1/Vm

1/Vm(app)

-1/Km

Monday, September 23, 13

Page 72: Grad Kinetics

Noncompetitive inhibition is a common theme in feedback inhibition. For example, the biosynthesis of isoleucine from threonine in bacteria involves 4 steps mediated by 4 different enzymes. The first reaction is catalyzed by threonine deaminase (TD). This enzyme is noncompetitively inhibited by isoleucine. Thus as product increases, the first step in the reaction decreases. As P falls due to shut down of TD, so TD is released from inhibition due to dissociation of P from the TD.P complex and the reaction cycle proceeds again.

Threonine Isoleucine

threonine

deaminase

A B D

-

Monday, September 23, 13

Page 73: Grad Kinetics

Examples  of  Non-­‐Compe--ve  Inhibitors

Naturally  occurring  (source  and  targets)

Caffeic  Acid  (tomato;  lipoxygenase)

Caffeine  (tea,  coffee;  cAMP  phosphodiesterase,  glucose  transporters)

Synthe5c  (their  targets  and  use)

Haloperidol  (brain  &  endothelial  cell  Nitric  Oxide  Synthase  inhibitor  -­‐  anH-­‐psychoHc)

TrichostaHn  A  (Histone  DeAcetylase;  anH-­‐cancer)

mycophenolic  acid  (Inosine  monophosphate  transferase;  Dengue  virus)

see http://www.emdbiosciences.com/docs/docs/LIT/ISB_USD.pdf

Monday, September 23, 13

Page 74: Grad Kinetics

Significance of Km and Vm

Km is the concentration of [S] at which 1/2 of the active sites are filled with substrate. The fraction of filled sites [ES]/[Et]!

Hence one-half of the enzyme is in the form of ES and the rate of the reaction, v, is

0.5 [Et] kp or v = 0.5 Vm

Thus when [S] = Km

ES⎡⎣ ⎤⎦Et⎡⎣ ⎤⎦

=Km

2Km

= 0.5

ES⎡⎣ ⎤⎦Et⎡⎣ ⎤⎦

=S⎡⎣ ⎤⎦

S⎡⎣ ⎤⎦ + Km

00.20.40.60.81

0 20 40 60 80 100

[ES]/[E

t]

[S]

Monday, September 23, 13

Page 75: Grad Kinetics

We also saw above that! ! !! ! ! Thus if kr >> kp

(ES dissociation is much faster than catalytic steps)

where Kd is the dissociation constant or Kd for S interaction with E.

Thus in some reactions where kp << kr, Km = Kd while in other enzyme mediated reactions where kp ≥ kr, Km > Kd.

Monday, September 23, 13

Page 76: Grad Kinetics

Vm reveals the turnover number (sometimes called kcat) of an enzyme if [Et] is known because !

Vm = kcat [Et]

if [Et] = 1 µM and Vm = 600 mmols/L/sec, kcat = 6 x 105 sec-1 - Each round of catalysis is 1/kcat = 1.7 µsec.

Monday, September 23, 13

Page 77: Grad Kinetics

“Kinetic Perfection” and the kcat/Km CriterionWhen [S] << Km,

We also know that Vm = kcat [Et]

In the enzyme mechanism we are studying kcat = kp, thusVm = kp [Et]

v is therefore directly proportional to and [S] at fixed [Et]

[ ]v

KV S

m

m=

since Vmax = kp [Et], this means

Kk

m

p

Monday, September 23, 13

Page 78: Grad Kinetics

Are there limits upon kcat/Km?

Examination of this eqn indicates that k1 is limiting.

kf is the second order rate constant that describes association of E and S to form the ES complex. kf includes terms that describe the collisional frequency of E and S.

Thus the highest value that kp/Km can achieve is kf.

kr is the rate of ES dissociation to E + Skp is the rate of ES breakdown to E + P

When k2 << k3 (i.e. catalysis is much faster than ES dissociation)

Kk

k kk k

m

cat

f r

p f= +

§k kk k

kk k

kf r

p f

f

p ff+ =

Monday, September 23, 13

Page 79: Grad Kinetics

For uncharged molecules in solution, the encounter rate constant, ke can be calculated as

ke =4πN(DA + DB )(rA + rB )

1000 M-1.sec-1

note,DA =kT6πrAη

D and r refer to the diffusion constants and reaction radii of molecules A and B and N is Avagadro’s number. For an enzyme molecule of radius = 30 Å and a substrate molecule of radius = 5 Å, the encounter rate constant is 109 M-1.sec-1.

Monday, September 23, 13

Page 80: Grad Kinetics

Thus diffusion limits the rate of encounter of E and S and an upper limit of kp/Km is therefore 109 M-1.sec-1. Even this requires that the substrate can encounter the enzyme surface in any orientation.

kcat/Km ratios of some enzymes, e.g. acetylcholinesterase and carbonic anhydrase are between 108 - 109 M-1 sec-1 indicating they have achieved kinetic perfection.

For these enzymes, because kp/Km ≈ 108 M-1 sec-1, this means that kp is so fast relative to kr that the ES complex breaks down faster to form product and E than it dissociates back to S and E.

Their activity is limited only by the rate at which they encounter substrate in solution. Any further gain can only be achieved by decreasing diffusion times. This can be achieved by sequestering substrates and products in the confined volume of a multi-enzyme complex, e.g. mitochondria.

Biochemistry 1988, 27, 11 58-1 167 Triosephosphate Isomerase Catalysis Is Diffusion Controlled.Stephen C. Blacklow, Ronald T. Raines T. Wendell A. Lim, Philip D. Zamore, and Jeremy R. Knowles

Monday, September 23, 13

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Summary

3.! We examined receptor & enzyme inhibitions (Competitive, noncompetitive) and showed that these are distinguishable. Competitive inhibitors increase Kmapp and noncompetitive inhibitors reduce Vmapp.

5.! We examined the significance of kcat and saw that when kcat ≥ kr/10, Km > KS. !! When kcat is very low, Km approaches KS.

4.! We discussed the significance of Km, Ks and Vm

2.! We reviewed methods of computing Kd, Bmax, # binding sites/enzyme, Km and Vmax. We saw that non-linear regression using some form of the Michaelis Menten equation is the modern way to analyze binding/[ligand] & activity/[substrate] data.

1.! Making simplifying assumptions we derived the Michaelis-Menten equation and expressed Km and Vm in terms of rate constants and [E]total

Monday, September 23, 13

Page 82: Grad Kinetics

APPENDIX

1. Analysis of first order reactions2. Substrate binding to an enzyme3. Methods for linearization of substrate

binding data4. Enzyme catalyzed reactions5. Methods for linearization of [substrate]

velocity data.

Monday, September 23, 13

Page 83: Grad Kinetics

[A] = [A]0 e−k1t

when t = t1/2 , [A]= 0.5[A]0

12

[A]0 = [A]0 e−k1t1/2

12= e−k1t1/2

ln12= −k1t1/ 2 ∴−0.693 = −k1t1/ 2

t1/ 2 =0.693

k1

and because τ=1k1

, t1/ 2 = 0.693τ

y = slope x + intercept

Linear plot Log plot

Appendix 1 First Order reactions - loss of substrate

ln[ ] ln[ ]A kt A 0=- +

ln[ ] ln[ ]A kt A 0=- +

Monday, September 23, 13

Page 84: Grad Kinetics

Linear plot Log plot

when t = t1/2 , [P]= 0.5[P]∞12

[P]∞ = [P]∞ 1− e−k1t{ }12= 1− e−k1t{ }−

12= −e−k1t

12= e−k1t

ln12= k1t1/ 2 ∴0.693 = k1t1/ 2

t1/ 2 =0.693

k1

and because τ=1k1

, t1/ 2 = 0.693τ

[P] = [P]∞ 1− e−k1t{ } ln([P]∞ − [P]t ) = −kt + ln[P]∞

y = slope x + intercept

ln([P]∞ − [P]t ) = −kt + ln[P]∞

Appendix 1 First Order reactions - product formation

Monday, September 23, 13

Page 85: Grad Kinetics

The rate of ES formation is given by:

At equilibrium:

The rate of ES breakdown is given by:

Thus:

Enzyme/substrate interactions ≡ Receptor/ligand interactions

The first step in enzyme function is the formation of the enzyme (E)/substrate (S) complex, ES.

Consider the reaction

Appendix 2 - Derivation of Michaelis-Menten equation for ligand binding

Monday, September 23, 13

Page 86: Grad Kinetics

thus

If we are interested in how much ES is formed (how much substrate is bound) at any given [S] and [E], we can express [ES] as as a fraction of total enzyme [Et] as:

[ES][Et ]

= [ES][E]+ [ES]

and, substituting for [ES],

Canceling [E] thus gives us:

Appendix 2

Monday, September 23, 13

Page 87: Grad Kinetics

How do you measure Ks (Kd)? - 1) Lineweaver Burk method

Here we assume that each enzyme molecule binds 1 substrate molecule. Taking the reciprocal of this expression:

1[Sb ]

= KS[Et ]

1[S]

+ [S][S]

1[Et ]

= KS[Et ]

1[S]

+ 1[Et ]

Appendix 3 - Linearization of ligand binding data

Monday, September 23, 13

Page 88: Grad Kinetics

+ ycept* xslopey =

1[Sb ]

= KS[Et ]

1[S]

+ 1[Et ]

Is a familiar linear expression of the type

y = slope*x + y-intercept. Thus

The expression,

1[Sb ]

= KS[Et ]

1[S]

+ 1[Et ]

Appendix 3

Monday, September 23, 13

Page 89: Grad Kinetics

When [S] = -Ks (a theoretical state because in life, [S] ≥ 0),

1[Sb ]

= KS[Et ]

1−KS

+ 1[Et ]

= −1[Et ]

+ 1[Et ]

= 0

Thus the x-intercept = 1/-Ks

Appendix 3

Monday, September 23, 13

Page 90: Grad Kinetics

1[Sb ]

= KS[Et ]

1[S]

+ 1[Et ]

-0.1 0.0 0.1 0.20

2

4

6

8

1/[S] per µM

1/m

ol b

ound

Lineweaver Burk analysis

1/-KS

1/[Etot]

y = 1+ 25 x

Appendix 3

Monday, September 23, 13

Page 91: Grad Kinetics

How do you measure Ks or Kd? - 2) Hanes Woolf method

Multiply by [S]

1[Sb ]

= KS[Et ]

1[S]

+ 1[Et ]

When [S] = -Ks

Thus the x-intercept = -Ks

[S][Sb ]

= KS[Et ]

+ −KS[Et ]

= 0

y = y-cept + x * slope

[S][Sb ]

= KS[Et ]

[S][S]

+ [S][Et ]

= KS[Et ]

+ [S] 1[Et ]

Appendix 3

Monday, September 23, 13

Page 92: Grad Kinetics

[S][Sb ]

= KS[Et ]

+ [S] 1[Et ]

-25 0 25 50 75 1000

40

80

120

[S] µM

[S]/[

S]b

Hanes-Woolf analysis

-KS

KS/[Et]

y = 20 + 1 x

Appendix 3

Monday, September 23, 13

Page 93: Grad Kinetics

y y-cept - slope*x=

∴ [Et ]Ks

= [Sb ][S]

+ [Sb ]Ks

[Sb ][S]

= [Et ]Ks

- [Sb ]Ks

= [Et ]Ks

- 1Ks

[Sb ]

How do you measure Ks or Kd? - 3) Scatchard Analysis

[Sb ] = [Et ][S]KS + [S]

∴ [Sb ] Ks + [S]{ }[S]

= [Et ] = [Sb ] Ks[S]

+ [Sb ]

[Sb ][S]

= [Et ]Ks

- [Et ]Ks

= 0When [Sb] = [Et],

Appendix 3

Monday, September 23, 13

Page 94: Grad Kinetics

[Sb ][S]

= [Et ]Ks

- 1Ks

[Sb ]

0.0 0.2 0.4 0.6 0.8 1.00.00

0.01

0.02

0.03

0.04

[Sb] µM

[Sb]

/[S]

Scatchard Analysis

[Et]

slope = -1/KS

y = 0.04 - 0.04 x

Appendix 3

Monday, September 23, 13

Page 95: Grad Kinetics

Defining the rate of product formation, v as

v = kp [ES] (1

We must again express [ES] in terms of known quantities!

Rate of [ES] formation

+d[ES]/dt = kf [E][S] (2

Rate of breakdown of [ES]

-d[ES]/dt = kr[ES] + kp [ES]or! ! ! !

-d[ES]/dt = (kr + kp) [ES] (3

Appendix 4 - Deriving the Michaelis-Menten equation

Monday, September 23, 13

Page 96: Grad Kinetics

In the "steady state" the concentrations of intermediates (e.g. ES) are unchanged, whereas [S] + [P] can change. If we limit measurements of v to early stages, [ES] does not change (there is no reverse reaction)! ! !

d[ES]/dt = 0 (4

! ! i.e.kf [E] [S] = (kr + kp) [ES] (5

hence

(6

(7

Appendix 4

Monday, September 23, 13

Page 97: Grad Kinetics

The following steps are algebraic tricks

[Et] = [E] + [ES] (8

where Et is total enzyme

if we divide the velocity equation (1 by Et we obtain

! (9

Appendix 4

Monday, September 23, 13

Page 98: Grad Kinetics

We can rearrange this to

(10

(11

then substitute for [ES] from equation (6 to give

Appendix 4

Monday, September 23, 13

Page 99: Grad Kinetics

Define !Vm as:! ! [E]tkp = Vm! ! (12

vVm

=

[S]Km

1+ [S]Km

(14

Km =k2 + k3k1

(13

v =Vm[S]Km + [S] (15or,

Appendix 4

Monday, September 23, 13

Page 100: Grad Kinetics

A simple, algebraic trick solves this problem! The reciprocal of the Michaelis-Menten equation is

1v=

Km

Vm [S]+

[S]Vm [S]

1v=

Km

Vm

1[S]

+1

Vm

thus making: 1/v = y and, 1/[S] = x

a plot of 1/v versus 1/[S] will produce a straight line of:positive slope = Km/Vm and, y-intercept = 1/Vm

This is a linear equation with the familiar form, + intercepty = slope * x

Appendix 5 - Linearization of [substrate] velocity data

Monday, September 23, 13

Page 101: Grad Kinetics

When [S] = -Km (a theoretical condition because [S] ≥ 0 in the real world), this means that

(9

1v=

Km

Vm

1−Km

+1

Vm

=1v=

1−Vm

+1

Vm

= 0

thus when [S] = -Km, 1/v = 0. The plot known as a Lineweaver-Burk plot looks like

0.00 0.05 0.10 0.15 0.20 0.25

0.1

0.2

0.3

0.4

0.5

0.6

1/[S] per µM

1/vsecperµmol

slope = Km/Vm

1/Vm

-1/Km

y = 0.1 + 2.5 x

Appendix 5

Monday, September 23, 13

Page 102: Grad Kinetics

If we multiply the reciprocal form of the MM equation by [S] we obtain

The Hanes-Woolf Plot

[S]v

=1

Vm

[S]+Km

Vm

-20 0 20 40 60 80 100

4.0

8.0

12.0

[S] µM

[S]/v

pers

ec

slope = 1/Vm

Km/Vm-Km

y = 2.5 + 0.1 x

Thus a plot of [S]/v versus [S] will produce a single straight line with slope 1/Vm and y-intercept of Km/Vm.

[S]v

=−Km

Vm

+Km

Vm

= 0

When [S] = -Km, the equation becomes

Thus the x-intercept = -Km

Appendix 5

Monday, September 23, 13


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