Week 6: Enzymes

Question 1

What is enzyme rate determined by?

Answer 1

Reaction intermediate states.

Question 2

What is an enzymes main function?

Answer 2

Lower activation energy

Question 3

What is key to enzyme specificity?

Answer 3

Complementarity! Van der Waals, charge-charge + hydrogen bonding.

Question 4

Describe enzyme substrate cycle

Answer 4

Free enzyme + substrate -> ES -> EX (double dagger) - enzyme-transition state complex -> EP (enzyme-product complex) -> product + free enzyme-> starts again

Question 5

What is the lock and key model?

Answer 5

explains specificity but not catalysis

Question 6

What is induced fit model?

Answer 6

Induced fit model - enzyme active site and substrate undergo conformational changes - forces substrate to adopt conformation that destabilizes the enzyme-substrate complex and stabilizes the enzyme-transition state complex.

Question 7

How do enzymes minimize deltaGcat(double dagger) - free energy of catalysis from ES to EX

Answer 7

They stabilize EX and destabilize ES

Question 8

What are the 5 ways enzymes achieve rate acceleration?

Answer 8

Optimizing proximity and orientation of substrates, substrate and active site distortion, electrostatic catalysis, metal ion catalysis, general acid/base catalysis

Question 9

Describe Optimizing Proximity and Orientation of Substrates.

Answer 9

Facilitation of orienting molecules for easier reactions

Question 10

Describe substrate and active site distortion

Answer 10

Stabilize ES and transition state. Distorting substrate bonds for reactions.

Question 11

Describe electrostatic catalysis

Answer 11

Fixed charges (ie. on backbone) in the enzyme active site are effective at stabilizing charges in TS.

Question 12

Describe metal ion catalysis (2 ways - see images)

Answer 12

1. Metal Ions as agents of electrostatic catalysis

2. Metal ions as source of OH at neutral pH (acid/base catalysis)

Question 13

Describe GABC

Answer 13

Base catalyst - any base except hydroxide ion. Acid catalyst - any acid except H3O

GABC in enzymes use ionizable side chains ie. Glu, Asp, His

Question 14

Describe histidine GBC (see images)

Answer 14

Imidazole side chain forms bond with water which forms a bond with a resonance stabilized carbonyl group which causes the water to attach - leaving its H and donating its OH- to the leaving group

Question 15

What do proteases do?

Answer 15

hydrolyze peptide bonds

Question 16

What are 3 types of proteases?

Answer 16

Serine, acid and metalloproteases

Question 17

Describe how serine proteases work.

Answer 17

Via covalent catalysis - used by many proteases to hydrolyze peptide bonds in protein substrates. Serine proteases - use deprotonated serine R-O- as nucleophile - attacks peptide bonds - convert to ester (acyl enzyme intermediate) - then attacked by water.

Serine proteases have highly reactive serine @ active site.

Question 18

What is the catalytic triad

Answer 18

Catalytic triad - Ser 195, His 57, Asp 102

Question 19

What are three types of serine proteases

Answer 19

Chymotrypsin (cuts after Trp, Tyr, Phe), Trypsin (cuts after Lys, Arg) and Elastase (cuts after Ala and Ser)

Question 20

Describe the process of the catalytic triad generating a good nucleophile (see images)

Answer 20

The negative charge on Asp 102 orients His 57 correctly and stabilizes protonated state. His 57 acts as general base - deprotonating Ser 195 - forming alkoxide. Deprotonated Ser 195 forms covalent bond with substrate.

Question 21

Describe the catalytic cycle of chymotrypsin

Answer 21

Ser 195 - covalent catalysis

His 57 aided by Asp 102 - GABC - His 57 2x base 2x acid

Oxyanion hole - electrostatic catalysis - stabilizes two tetrahedral oxyanion intermediates

Step 1: Polypeptide substrate binds noncovalently in enzyme active site. Ser attacks the electrophilic amide C.

Step 2: The resulting tetrahedral oxyanion is stabilized by H-bonding interactions with oxyanion hole.

Step 3: Collapse of the tetrahedral intermediate and proton transfer from His lead to cleavage of C-N bond - the N-terminal peptide is bound through acyl linkage to serine. C-terminal fragment is cleaved.

Step 4: Water binds to the active site and attacks the acyl ester carbonyl.

Step 5: The resulting tetrahedral oxyanion intermediate is stabilized via enthalpic interactions with oxyanion hole

Step 6: The second peptide fragment (N-terminal fragment) is released and the enzyme returns to its initial state.

Question 22

What is two examples of acid proteases

Answer 22

Pepsin = non-specific, HIV protease - specific (HIV antiretrovirals are inhibitors of HIV protease active site).

Question 23

Describe the acid protease mechanism

Answer 23

Two asp molecules - one begins as deprotonated. The protonated one acts as an electrostatic catalyst and the deprotonated as GABC

Step 1: An oxygen on the deprotonated Asp attacks a hydrogen on water - keeping it and leaving OH-

Step 2: OH- attacks the carboxyl group which relocates its electrons to the double bonded oxygen (single bond, 3 electron pairs)

Step 3: The oxygen then adds its electrons back to make a double bond and the peptide bond is cleaved, the H on the now protonated Asp is added to the N terminus and it is now deprotonated

Question 24

Describe a zinc metalloprotease - carboxypeptidase A

Answer 24

Zn2+ can generate OH-from water in enzyme active site at physiological pH. This can be used in hydrolysis reactions, as in digestive protease carboxypeptidase A. The chemical environment in Carboxypeptidase A further inhances zincs ability to generate bound hydroxide - higher pH.

Carboxypeptidase A is a C-terminal exopeptidase - cuts peptide bonds starting at C-terminal with addition of water.

Question 25

Describe the mechanism of carboxypeptidase A

Answer 25

Has a positive Arg145 that holds the C-terminus in place. 2 charged arginine + deprotonated (charged) glutamic acid + zinc (II) Glu → acting as general base Step 1: A Zn is bound by His 69, 196 and Glu 72 - binds OH which then attacks the carboxyl which sends its electrons to the double bonded oxygen. This negative charge is stabilized by a positive Arg 127 Step 2: The electrons shift back into a double bond, and the peptide bond is cleaved - relocates its electrons to a protonated Glu 270. Step 3: water is removed and Glu 270 becomes deprotonated.. A H then joins the ZnOH and this releases the C terminus.

Question 26

Where else are acid protease mechanisms used?

Answer 26

in other hydrolases ie. esterases, phosphatases, amidases

Question 27

What are 3 strategies for regulation of enzyme activity?

Answer 27

1. Irreversible covalent activation (zymogens) 2. Irreversible covalent inhibition 3. Reversible covalent modification of the enzyme

Question 28

With what type of enzyme does irreversible covalent activation usually occur? Describe an example with chymotrypsinogen

Answer 28

with proteases - synthesized in inactive form to prevent degradation of cell that expresses them. The inactive precursor is zymogen - ie. chymotrypsinogen. The enzyme is activated by proteolysis at specific peptide bonds This is irreversible as once these peptide bonds have been cut, they cannot be reformed. In chymotrypsinogen - substrate binding site is blocked. On activation by proteolysis the substrate channel is accessible. The processing sites are not near the catalytic triad - these residues are activate but cannot attack substrates until binding site is cleared.

Question 29

Describe Irreversible covalent inhibition

Answer 29

``` Serine protease inhibitors (serpins) - class of covalent inhibitor proteins ie. antitrypsin. Serpin - binds noncovalently to active site Antitrypsin binds to trypsin to form trypsin-antitrypsin complex ```

Question 30

Describe Reversible covalent modification of the enzyme

Answer 30

Specific amino acid residues with alcohol side chains on enzymes can be reversible phosphorylated (ATP -> kinase -> ADP) or dephosphorylated (H2O -> phosphatase -> phosphate) Phosphotyrosine and phosphothreonine move towards nearby arginine residues and in doing so, they flip away from active site and make it available to substrates. Phosphatases can restore enzyme to original state ie. MAPK inactive (ATP -> MAPK kinase -> ADP) MAPK Active (H2O -> MAPK phosphatase -> phosphate) - MAPK inactive

Question 31

What is velocity? Instantaneous rate?

Answer 31

Change in speed over change in time - or change in product concentration over change in time = concentration. Instantaneous rate is the tangent

Question 32

What is the velocity for a first order reaction?

Answer 32

v0 = k[S]0 - rate is proportional to [S] - units sec-1

Question 33

What is the velocity for a zero order reaction?

Answer 33

Rate is constant: v = k - units = concentration/time

Question 34

What is saturation?

Answer 34

For a typical enzyme-catalyzed reaction, the rate “maxes out” - exhibits saturation - this dependence of rate on [substrate] is described by hyperbolic function. At low substrate - reaction is 1st order, high substrate - reaction is zero-order

Question 35

What is the Michaelis-Menten equation?

Answer 35

hyperbolic function: y = ax/(b+x) where a and b are constant. For small values of x - b+x is nearly equal to b: y = ax/b For large values of x, b+x = x equation is y = a - horizontal line - zero order reaction. Equation: v = Vmax[S]/Km+[S] Where [S] is initial substrate concentration, vmax = max rate, Km = Michaelis constant

Question 36

What are the assumptions of the Michaelis-Menten equation?

Answer 36

1. ES complex is intermediate 2. Concentration of ES remains constant - steady-state assumption 3. Initial rates of reaction are considered, the contribution of the back reaction is neglected 4. Total [enzyme] = [free enzyme] + [ES complex]

Question 37

Describe E + S -(k1)-> ES -kcat-> E + P

Answer 37

kcat and Km are properties of enzyme catalyzed reaction - depend on enzyme and substrate kcat = number of catalytic cycles per second for enzyme molecule = Vmax/[E]t Vmax depends on enzyme concentration. kcat is independent of enzyme concentration - inherent property of enzyme Km is derived from rate constants - hase units of concentration. Lower the Km the lower the concentration fo substrate needed to reach rate saturation. It is substrate concentration when v = 1/2 vmax [S] = Km when v = 1/2vmax The smaller the Km the lower the [S] required to achieve rate saturation

Question 38

How do you get Km and Vmax from a linear equation?

Answer 38

plot 1/v against 1/[S] - Lineweaver-Burk Plot (double-reciprocal plot). The x-int is -1/Km, the y-int is 1/Vmax and the slope is Km/Vmax

Question 39

Can kcat and Km tell how effective an enzyme is?

Answer 39

kcat and Km alone are not enough to tell how effective enzyme is. Together, can rate catalytic power of enzyme: enzyme efficiency = kcat/Km - second order rate constant with units /Msec - higher the better - dependent on substrate

Question 40

When are enzyme efficiencies used?

Answer 40

1. To determine substrate preference of enzyme 2. To determine how mutations affect enzyme activity