Geeves 1

Question 1

Specific activity

Answer 1

Activity per mg of protein

This is the activity of one enzyme unit

Question 2

Proof enzymes were proteins

Answer 2

Sumner 1962- discovered that Jack beans used urea
Managed to separate Jack bean urease from samples by size, charge ph etc.
Showed that there is a reaction Urea + H2O -> CO2 + NH3

Question 3

Enzyme acceleration

Answer 3

Accelerate by 2 fold to 10^17 fold
Lower activation energy
Acceleration is the same in both directions- cannot alter eqm

Question 4

Simple transition state theory

Answer 4

Catalyst lowers the activation energy without changing eqm
Energy to break depends on length, angle and neighbour
TS exists for a bond vibration (10s)
Transmission coefficient- between 0-1 determines the outcome either S or P once TS is reached

Question 5

Total activity

Answer 5

u moles of S->P per minute

The activity of the all of the enzymes in a sample

Question 6

Arrhenius equation

Answer 6

Rate (k) = Ae -Ea/RT
Where A is the probability factor, e is natural log, Ea is the activation energy, R is the gas constant 8.314 J/mol and T is temp

Question 7

Dialysis for ligand binding

Answer 7

Semi permeable membrane, chambers filled with ligand
Protein added to one side
Ligand diffuses to equilibrate free ligand

Question 8

Actin tropomyosin sedimentation

Answer 8

Proteins binding to actin will sediment with it
Run the supernatsnt and pellet on gels
The actin will then separate from the tropomyosin
The band intensities correspond to conc so can be used for Kd

Question 9

The steady state assumption

Answer 9

Assumes that the ES concentration is constant

Must be small compared to the rate of change of S or P

Question 10

The 4 key points of MM
3 assumptions of MM
Curve
How Km and S

Answer 10

Initial rates- S is constant, P is 0
Steady state- ES constant
Total E is fixed

Curve saturates gradually

If Km is lower than the S in the cell then it is S independent

If S is low than Km, then half Vmax will never be reached, and there will be a linear line (v= Vmax[S]/Km)

Question 11

Catalytic efficiency

The equation for max rate

Answer 11

Vmax/Km or kcat/Km
The upper limit is k1, where ES would instantly form E and P

v = Vmax X S
K + S

Question 12

Km and the EQ constant

Answer 12

Km = k2 + k-1
k1

Or

K1/ (1+K2)

Question 13

Enzymes are more than simple catalysts

7

Answer 13

Specific
Control flux
Signals
Protecting labile substrates
Energy economy of the cell
Part of the information pathway
Are proteins

Question 14

The lock and key model

Answer 14

Binding site designed to match substrate
Does not explain how TS lowered
But maybe if it matched the TS and introduced strain
Replaced by induced fit model

Question 15

Ordered substrate binding

Answer 15

Both substrates then both products

Lactate + NAD -> Pyruvate + NADH

Question 16

Ping pong model

Answer 16

ATP + creatine -> ADP + phosphocreatine

Question 17

Serine proteases

Answer 17

Chymotrypsin
His-Asp-Ser catalytic triad
Specificity pocket allows greater surface area for binding. Backbone amides or positive charges.
Oxyanion hole- stabilises transition state, negative charge on oxygen or alkoxide (O of alcohol)

Question 18

The lock and key model

Answer 18

Binding site designed to match substrate
Does not explain how TS lowered
But maybe if it matched the TS and introduced strain
Replaced by induced fit model

Question 19

Ordered substrate binding

Answer 19

Both substrates then both products

Lactate + NAD -> Pyruvate + NADH

Question 20

Ping pong model

Answer 20

ATP + creatine -> ADP + phosphocreatine

Question 21

Serine proteases

Answer 21

Chymotrypsin
His-Asp-Ser catalytic triad
Specificity pocket allows greater surface area for binding. Backbone amides or positive charges.
Oxyanion hole- stabilises transition state, negative charge on oxygen or alkoxide (O of alcohol)

Question 22

Limits on enzyme specificity

Answer 22

Discriminate similar size substrates
E.g. Aminoacyl tRNA synthetase
even though Val and Thr are a similar shape, the Val tRNA is 200x more selective as it has an additional OH in the binding site

Question 23

Induced fit model

Answer 23

Replaced lock and key
Preformed site but coalesces around substrate
Substrate induces conformational change
Protein is dynamic
Enzymes stresses and strains the substrate to destabilise it

Question 24

Diffusion limited collisions

Equation

Answer 24

E/t = k [E] [S]

Where S is constant
The binding site only represent about 1% of the protein, so 1% of k

Question 25

Orbital steering

Answer 25

Increases the target zone of ligand binding Might have a negative pocket for a positive ligand Directs ligand towards pocket

Question 26

Multi step docking and catalysis

Answer 26

Proteins structure changes Correct substrate aligns the catalytic residues Allows discrimination Smaller substrate doesn't have the binding energy to position the active site residues

Question 27

Conformational trapping

Answer 27

The enzyme flips between different forms Substrate selects the optimal form Selection and induced fit can occur in same system

Question 28

Protein confirmation and ligand binding | Square diagram

Answer 28

Will be more inactive E | But more of the active E is bound to ligand

Question 29

Competitive inhibitors | Chymotrypsin inhibitor

Answer 29

``` Same site Resembles substrate TS analogues Effectiveness depends on the affinity of the I and S Inhibiter reduces free enzyme ``` Blocking specificity pocket prevents function Benzene, phenol are crude inhibitors ALSO PRODUCT INHIBITION

Question 30

Non competitive inhibition

Answer 30

Different site Slows down catalysis e.g. k'2 is smaller than k2 The Vmax is reduced but the Km doesn't change, as the binding of S to E isn't affected

Question 31

Mixed inhibition

Answer 31

Different site But inhibits S BINDING AND CATALYSIS affects Vmax and Km

Question 32

Un competitive inhibition

Answer 32

Rare I only bind to the ES complex Slows catalysis Km and Vmax affected

Question 33

Activators

Answer 33

Accelerate catalysis or increase binding affinity Always allosteric site One example is calcium- a second messenger used to activate many kinase and proteases etc. Removes the pseudo sequence of PKC Also GAP for G proteins

Question 34

Proteins switches

Answer 34

Kconf gives the equilibrium constant of if the reaction is shifted towards E or E' etc. So large means shifted right This is then reflected in the Kd binding So a x100 higher amount of inactive E means that ligand binding is e.g. X100 less strong

Question 35

Cooperativity

Answer 35

The binding of ligand to one site increases the affinity of the other sites E.g. Haemoglobin Gives an S shaped curve instead of standard hyperbolic

Question 36

Monod wyman changeux Cooperativity

Answer 36

Describes allosteric transitions Each promoter has a R or T state Ligand can bind to either state, conformational change which alters affinity

Question 37

3 examples of cooperative enzymes and their activators

Answer 37

Isocitrate dH - AMP Deoxythymidine kinase - cCDP Pyruvate kinase - fructose DP

Question 38

Strengths of the steady state

Answer 38

``` Minimum enzyme info Specific activity Specificity Inhibitors and activators Small enzyme amounts ```

Question 39

Weakness of steady state

Answer 39

Km and Vmax have limited meaning Can't explore detailed mechanism Doesn't say what is happening on the enzyme

Question 40

Advantages of the pre steady state

Answer 40

More detail Shorter time scales need special equipment High enzyme conc needed

Question 41

pre steady state example

Answer 41

Looked at the nitro phenol production In MM, if there is little S then the v becomes linearly dependent on S If S is constant, then the intercept should be 0,0 But the intercept isn't- suggests that product is being formed before it can be observed The burst phase

Question 42

Burst amplitude | Burst rate constant

Answer 42

Y intercept Amp = [E] x (k'1) (k1 + k2)^2 Rate constant = k'1 + k2 k'1 = k1 [S]

Question 43

Stopped flow technique

Answer 43

Signal can be any optical signal e.g. Abs, fl 0.5-2 msec mixing time Solutions forces from syringes into mixing cylinder Flow stops when opposing piston fills chamber linked to a measuring device

Question 44

Nucleoside diphosphate kinase- measurement by stop flow

Answer 44

Shuttles PI between ATP and NDPs Ping pong mechanism Steady state assays may be difficult because of product inhibition So need to analyse before too much product is made kcat = 1000 s-1 (so 1x10-3 to complete reaction) Trp fluorescence decreases when ATP added

Question 45

One step ligand binding

Answer 45

Linear Rate = L X k+1 + k-1 Slope = k+1 Intercept = k-1

Question 46

What is the enzymes lifetime

Answer 46

kobs = 1/tau | So if kobs is 1000, then it takes the enzyme 10ms to bind S

Question 47

What info does the relative amplitude give?

Answer 47

Extent of binding Keq Keq = k-1/k+1

Question 48

Two step ligand binding

Answer 48

Has a fast and slow step Intercept is k-2 Substrate for half Vmax = 1/K1 Vmax is k+2 + k-2 Rate is K1 k+2 [L] + k-2

Question 49

3 different type step reactions

Answer 49

Single irreversible- linear, zero intercept Single reversible- linear, non zero intercept Two step- hyperbola, non zero intercept

Question 50

The proteasome

Answer 50

Peptide is ubiquitinsted by enzyme The filtered to proteasome 2 X 19S caps, one 20S subunit with 2a, 2b subunits The beta units are catalytic, alpha structural ATP needed for the two 19 caps to bind and unfold target 3 activities: trypsin, chymotrypsin, caspase

Question 51

GroEL chaperone

Answer 51

Hsp70 binds to hydrophobic patch Enters the GroEL complex Initially the peptide binds to a hydrophobic pocket The lid then binds groES The hydrophobic residues turn to polar which releases the protein and promotes folding The lid is then released

Question 52

6 reasons why multi enzyme complexes exist

Answer 52

Catalytic enhancement- reduces diffusion time between enzymes Substrate channel img Servicing- one subunit passes reagent to others Sequestration of reactive intermediates Coordinate regulation Coordinate expression - enzymes expressed together e.g. Operon

Question 53

Tryptophan synthase

Answer 53

A2B2 tetramer indole-3-glycerol phosphate -> indole + G3P Beta subunit then indole + serine -> Trp Beta subunit needs PLP, active Vit B6 Alpha and beta connected by a substrate channel No side products so efficient

Question 54

3 functions of DNA polymerase

Answer 54

Exonuclease Proofreading Functional polymerase

Question 55

E1

Answer 55

Pyruvate dehydrogenase TPP removes Co2 from pyruvate This makes hydroxyethylTPP TPP stabilises intermediates when it's ring acts as an electrophile

Question 56

E2

Answer 56

Dihydrolipoyl acetyltransferase Transfers acetylation group to form a thioester bond One bond is reduced and the other binds to acetyl The acetyl reacts with CoASH to form acetyl CoA The Lip-SH SH is recycled

Question 57

E3

Answer 57

Dihydrolipoyl dehydrogenase Both SH of the Lip are reduced and need to be oxidised again E3 internal disulphide bond reduces as the lipoate is oxidised The E3 and FAD are non covalently bound- they alternate between red and ox forms by a ping pong mechanism E3 when red catalyses NAD -> NADH + H+ So regenerates oxidised E3

Question 58

Lipoid acid as a cofactor

Answer 58

Attaches to E3 by a lysine Very reactive Shuttles acyl

Question 59

What reasons for enzyme enhancement does the PDH complex have?

Answer 59

Catalytic enhancement- E2 Lionel swinging arm and catalytic core Channelling- swinging arm and CoA Servicing- E1/2 and E3 Sequestration- E1 (TPP) intermediate passed to E2 Coordinate regulation- inhibition by products and phosphorylation of E1

Question 60

Signals which report binding events

Answer 60

``` Size- sedimentation, gel filtration Shape- SAXS Optics Enthalpy Conductance NMR Kinetic ```

Question 61

Dialysis

Answer 61

Two buffer chambers separated by membrane Ligand diffuses across to reach eqm When protein is added to one side, the ligand evens out Difficult to measure free and bound ligand though Radioactive ligand

Question 62

Fractional saturation

Answer 62

Rearrange the Kd equation Kd/L = E free/ E bound Theta = EL/Etot, or = L/(L+Kd)

Question 63

Fluorescence titration

Answer 63

EL/Etot = L/(L+Kd) Plot the E/Etot (fractional sat) against L Kd is then the gradient The intercept gives the affinity info When the enzyme is 50% saturated, then free L = Kd But the total L will be = Lfree + Kd (amount on enzyme)

Question 64

Multiple identical binding sites

Answer 64

Identical sites looks same as single site Kd = E L/ EL X EL L/EL2 And fractional sat becomes = nL /L + Kd These two equations can be rearranged to give the scatchard and hill plots

Question 65

Scatchard plot

Answer 65

Theta/L against theta Slope 1/Kd Intercept is n

Question 66

Hill plot

Answer 66

Log (theta/n-theta) against log L Slope is n Intercept log Kd Only a linear plot if there is no Cooperativity Therefore helps to define its nature If not there is an S shape where the intercept of lines is Kd Gives an average of all Kds in a linear system

Question 67

Multiple non identical sites

Answer 67

Ligand fills one before the other | Theta = nL/L+ Kd X then same equation for site 2

Question 68

Using hill plots for multiple cooperative sites

Answer 68

Linear line each time a ligand binds The slope between the 2 lines is the hill coefficient This is always less than the number of binding sites, because of the energy lost in the reaction

Question 69

Fluorescence for measuring ligands

Answer 69

Fluorescent E | Or fluorescent L, where background needs to be subtracted

Question 70

Fluorescence Polarisation

Answer 70

Fluorophores with absorption transition moments aligned with the polarised light are excited If moves then has a different polarisation Small diffuses quickly, large don't So formation of a complex changes the polarisation

Question 71

FP applied to DNA unwinding by helicase

Answer 71

When the fluorescent DNA binds to a helicase, the molecular tumbles slower and there is loss of polarisation dsDNA- fast With helicase- slow Addition of ATP activates helicase-> fast ssDNA

Question 72

Total internal reflection fluorescence

Answer 72

Excitation beam penetrates 100 nm into solution Protein bound to surface Will only excite fluorophores ligand that are bound to the protein These then remit the light

Question 73

Single molecule fluorescence

Answer 73

There is only a single molecule on the slide | Can define the Kd for a given ligand conc

Question 74

Limitations of fluorescence studies

Answer 74

Not good for weak affinities Fl must change on binding Fl labels can alter protein behaviour

Question 75

Isothermal titration calorimetry

Answer 75

``` Two chambers Ligand and ligand with protein Ligand acts a temp control Measure temp change with binding Measure heat/cooling to keep both sides equal temp ```

Question 76

Competition binding

Answer 76

How a ligand which gives signal competes with a silent ligand Fl ATP analogues Use competition to define how normal ATP binds So get a Kd for each ATP type Then because E is the same you can merge the equations Kd1[L]/Kd2[L] = [EL]/[EL second ligand]

Question 77

Solid phase/ high throughput binding assays

Answer 77

Monitor binding to immobilised target Ligand can be washed off and new added Used for drug screening Can use kinetic and thermodynamic techniques Defines Kd, stoichiometry and enthalpy in one measurement Delta G = -RT lnKd

Question 78

Surface plasmon resonance

Answer 78

Protein is covalently bound to foil Binding of ligand changes the resonance angle of the light reflecting off the surface This is because photons are absorbed and the wavelength lowers Only valuable for large assay numbers Surface may alter properties False indication of ligand conc Direct coupling via residue or indirect via antibody/ His or GST tags

Question 79

Reconstitution of TM proteins

Answer 79

Tether vesicles to SPR Increases fluidity by assisting ligand access Tethered membranes have lower fluidity but will be homogenous

Question 80

Examples of proteins that change conformations

Answer 80

CCT- has a lid that closes upon unfolded substrates Prions, G proteins, channels and pumps Binding of substrate is composed of multiple energy wells

Question 81

Proteins sensors with no ligand

Answer 81

Temp, field strength, mechanical distortion | Kconf = E'/E

Question 82

Building a temperature sensor

Answer 82

Alpha helix with a FRET pair at each end Helix collapse gives FRET Done around active site

Question 83

Examples of mechanical gates Channel Kinase

Answer 83

Ryanodine receptor is a Ca channel in skeletal, smooth and heart muscle. Locks receptor in open or closed. Release of Ca from SE. Ryanodine poison causes severe muscle contractions Enzymes are Titin kinase, also talin vinculin assembly

Question 84

Resting calcium level

Answer 84

10-100 nM

Question 85

Structure of myosin head

Answer 85

Lower and upper 50 KDa domains have a cardiomyopathy loop ATP site near upper Then have a converter, essential light chain, and regulatory LC which make up the level arm

Question 86

Optical sensor of myosin cycle

Answer 86

Removes all Trp W512 is on the end of the helix and detects SW2 movement The Trp changes from buried to exposed and there is less quenching Engineer Trp on SW1 to elect actin binding pocket Pyrene Fl on Cys near C terminus of actin detects myosin binding and release Mant label on ATP to detect binding as ribose tag doesn't affect binding

Question 87

The myosin cycle

Answer 87

ATP binding, ATP tightly bound and actin released - SW1 SW2 closes and recovery stroke ATP hydrolysis Actin rebinds Cleft closure, PI release and power stroke ADP release

Question 88

Mutations in myosin

Answer 88

R238 E459 -> E238 R459 Part of the switch like in G proteins Needs a positive and negative salt bridge to form between Sw1 and sw2

Question 89

Extending the lever arm to test hypothesis

Answer 89

Spectrin like repeats replace and extend arm Immobilised myosin and added actin Longer neck increased velocity

Question 90

Role of myosin 1c

Answer 90

Maintains tension in stereocillia Strain dependent ADP release mechanism Shape change in cross bridge opposed by strain Less energy released as need to stretch

Question 91

What does Keq need to be to Do work Act as sensor Act as a gate (won't release ADP unless pulled)

Answer 91

Large =1 Small

Question 92

Troponin as Ca sensor

Answer 92

Troponin binds Ca Change in conformation to move tropomyosin and reveal the actin binding sites Ca opens the cleft Allows binding of the switch peptide This moves the inhibitory peptide away from the tropomyosin

Question 93

Calcium pumps

Answer 93

``` Lower Ca levels to 0.1 uM P type ATPases Phosphorylation of P domain by ATP Causes the external cleft to open Ca leaves Hydrolysis of Pi returns to normal ```

Question 94

ABC family of transporters

Answer 94

ATP linked Transport of ions, peptides, amino acids Linked to diseases such as macular a degeneration and CF

Question 95

ABC importers

Answer 95

Substrate bind to binding protein ATP binding to cytoplasmic domain of transporter Binding protein binds to extra cellular transporter Substrate taken into channel The ATP is hydrolysed when the channel flips to inner ADP release

Question 96

ABC exporters e.g. Flipase

Answer 96

Ligand from extra cellular binds in cleft ATP binding opens top Hydrolysis of one ATP flips the ligand over Hydrolysis of both returns to original conformation

Question 97

ATP synthase

Answer 97

Protons enter the rotor domain which turns the stator 3 ab domains where ADP and PI and converted to ATP All cycles out of phase Turning of rotor observed by avidin labelled actin filament The Keq is 1, energy needed to release product

Question 98

Bacterial flagella motor

Answer 98

C ring- fligG, fliM and fliN Receptors lead to AspPi of CheY Binds to base of flagella motor and turn clockwise -> tumble Chemoattractant leads to CheY releasing PI helped by CheZ Stops binding and causes anti-clockwise rotation for swimming Swimming up conc gradient stops tumbling