Exam III

Question 1

Michaelis Menton Equation

Question 2

Define V_max

Answer 2

• V_max = k_cat[E]_o
• V_max is the product of the catalytic rate constant (or turnover number) and the total enzyme concentration in the assay. Therefore, V_max depends on the conditions of the assay. The important parameter is k_cat, which is constant for a given reaction at standard conditions.

Question 3

Define K_m

Answer 3

• K_m is the concentration of substrate at which V_o equals ½V_m
• K_m = (k_-1 + k₂)/k₁
• In order to maximize the biological efficiency of an enzyme, the K_m of an enzyme is typically close the physiological substrate concentration.
• Higher K_m indicates lower affinity for substrate.

Question 4

Lineweaver-Burke Plot

Question 5

Significant points on the Lineweaver-Burke Plot

Answer 5

• The y-intercept of such a graph is equivalent to 1/V_max
• The x-intercept of the graph represents −1/K_<em>m</em>.
• The slope of the graph represents K_m/V_max.

Question 6

If 10mg of pure carbonic anhydrase (29000 g/mol) catalyzes the hydration of 0.3 g of CO₂ in 1 min at 37^oC under optimal conditions, what is the turnover number, k­_cat, of carbonic anhydrase in units of min^-1 and sec^-1.

Answer 6

You use the equation: V_max = k_cat[E]_o

To calculate V_max: -d[CO₂]/dt = 0.3g/min = 0.007 mol/min

To calculate [E]_o: 10 micrograms / 29000 g/mol = 3.4 x 10^-10 mol.

k_cat = 0.007 mol/min / 3.4 x 10^-10 mol.

Question 7

1. In a first order reaction a substrate is converted to product so that 87% of the substrate is converted in 7 min.
2. Please calculate the first order rate constant for the reaction.
3. Please calculate the half-life of the reaction-the time required for conversion of 50% of substrate to product.

Answer 7

A.) ln[A]/[A]_o = -kt so ln[0.87]/[1.0] = -k(7 minutes) = 0.02 min^-1

B) t_1/2 = 0.693/k so t_1/2 = 0.693/0.02 = 34.65 minutes.

Question 8

Given a rate constant of 10.24 s^-1 and the equilibrium constant for the dimerization process measured under the same conditions is 1.8x10⁶M^-1. What is the rate constant governing the dimer dissociation?

Answer 8

K_eq = 1.8 x 10⁶ M^-1

k_u = 10.24 s^-1

Thus, k_f = 1.8 x 10^-7 s^-1

Question 9

First Order Reactions:

• Determine rate order constant
• Determine half-life

Answer 9

• To determine the rate constant, you use the equation –kt = ln[A]/[A_o]. You plot ln[A]/[A_o] vs time, the slope is equal to –k.
• The units of the rate constant are t^-1 (time^-1).
• To find the half-life: t_1/2 = 0.693/k

Question 10

Second Order Reactions:

• Determine the rate constant
• Determine the half-life

Answer 10

• To determine the rate constant, you use the equation 1/A = kt + 1/A_o. You plot 1/[A] vs time and the slope is equal to k and the y-intercept is equal to 1/A_o.
• The units of the rate constant are M^-1t^-1
• To find the half-life: t_1/2 = 1/(kA_o)

Question 11

Reaction progress curve for an enzyme-catalyzed reaction.

Answer 11

Question 12

Enzyme Catalysis Reaction

Answer 12

Question 13

Experimental set-up: [S] >>> [E]_o, initial rate conditions

Answer 13

• Set up several tubes, each containing a different substrate concentration [S]
• Add enzyme, E, to each tube. In all tubes [S] >>> [E]_o
• Measure the product concentration versus time for each tube ([P] vs t)
• Plot [P] vs time at each [S]
• Determine the slope of each of these curves (d[P]/dt) and make sure that the initial region of the curve is being calculated so that the reverse reaction is not contributing.
• Plot initial rate, v_o, versus [S]
  
  • As [S] increases, V_o increases
    
    • At relatively low [S], the V_o increases almost linearly with increases in [S]

Question 14

Kinetic equation describing the reaction with microscopic rate constants

Answer 14

• K_m = (k_-1 + k₂)/k₁
  
  • If k₂ is the rate limiting step, then K_m approximates the equilibrium dissociation constant of the ES complex (k_-1/k₁)
  • If k_-1 is the rate limiting step, then K_m approximates k₂/k₁

Question 15

Steady-State Assumption

Answer 15

The steady state assumption is that at any single substrate concentration, the concentration of the enzyme-substrate complex initially rapidly increases to a level that depends on substrate concentration and remains constant during the time required to measure the initial rate.

Essentially, d[ES]/dt = 0.

Question 16

Significance of k_cat/k_m

Answer 16

• Different enzymes have similar efficiencies, but different K_m and k_cat values
• The best parameter for comparing activities of various enzymes or various substrates of the same enzyme is the Specificity Constant which is defined as the rate constant for conversion of E+S into E+P which is obtained from the first part of the v_o vs [S] curve.
• The larger the parameter, the more efficient the enzyme. This parameter has the units of a second order reaction: M^-1s^-1

Question 17

pH dependence of the rate with breakdown into effects on k_cat and K_m

Answer 17

• Find k_cat vs pH
  
  • The highest k_cat will be at the optimal pH
• Find 1/K_m vs pH
  
  • When K_m (binding of substrate and all intermediates), and when K_m gets larger, the affinity for substrate decreases.
• Looking at a rate vs pH profile:
  
  • Side Chain 1 is being deprotonated on the left side of the graph & Side Chain 2 is being deprotonated on the right side of the graph
  • So this shows that the max activity is when Side Chain 1 is deprotonated and when Side Chain 2 is protonated.
    
    • This shows that Side Chain 1 acts as a base and Side Chain 2 as an acid
    • Side Chain 2 is also responsible for stabilizing the intermediate

Question 18

Positive Allosteric Modulation

Answer 18

• Positive allosteric modulation (also known as allosteric activation) occurs when the binding of one ligand enhances the attraction between substrate molecules and other binding sites.
• An example is the binding of oxygen molecules to hemoglobin, where oxygen is effectively both the substrate and the effector. The allosteric, or “other”, site is the active site of an adjoining protein subunit. The binding of oxygen to one subunit induces a conformational change in that subunit that interacts with the remaining active sites to enhance their oxygen affinity.

Question 19

Negative Allosteric Modulation

Answer 19

• Negative allosteric modulation (also known as allosteric inhibition) occurs when the binding of one ligand decreases the affinity for substrate at other active sites.
• For example, when 2,3-BPG binds to an allosteric site on hemoglobin, the affinity for oxygen of all subunits decreases. This is when a regulator is absent from the binding site.

Question 20

Homotropic Allostery

Answer 20

• A homotropic allosteric modulator is a substrate for its target enzyme, as well as a regulatory molecule of the enzyme’s activity. It is typically an activator of the enzyme.
• For example, O₂ is a homotropic allosteric modulator of hemoglobin.

Question 21

Heterotropic Allostery

Answer 21

• A heterotropic allosteric modulator is a regulatory molecule that is not also the enzyme’s substrate. It may be either an activator or an inhibitor of the enzyme.
• For example, H⁺, CO₂, and 2,3-bisphosphoglycerate are heterotropic allosteric modulators of hemoglobin

Question 22

K-Type Regulation

Answer 22

Affects K_m

Question 23

V-Type Regulation

Answer 23

Affects V_max.

Question 24

Competitive Inhibition

Answer 24

This inhibition typically displays an unaffected V_max, but a higher K_m value. Affects “E + S” complex.

Question 25

Non-Competitive Inhibition

Answer 25

This inhibition typically displays an lower V_max, but an unaffected K_m value. Affects “ES” complex.

Question 26

Un-Competitive Inhibition

Answer 26

• K_m reduced
• V_max reduced

Question 27

Explain why too tight binding by an enzyme to S is not advantageous for catalysis

Answer 27

If the enzyme is bound too tightly with the substrate, the ES complex becomes too stable and the reaction is unable to proceed further.

Question 28

Kinetic studies of an enzyme at different pH’s show that an enzyme has two catalytically important residues that have pKa’s of approximately 4 and 10. Mutational experiments indicate that a glu and lys residue are essential for activity. Match the residues with the pKa’s and explain whether each is likely to act as a base or an acid. Please explain your answer.

Answer 28

The residue with the pK_a of 4 is likely to be the glutamic acid and that with a pK_a of 10 is likely to be the lysine. Because of the lower pK_a, glutamic acid is likely to be in the basic form and should act as the catalytic base. Lysine, with its higher pk_a should be in the acidic form and should act as an acid to stabilize the transition state.

Question 29

A covalent catalytic mechanism of an enzyme depends on a single active-site cysteine whose pKa is 8.0. A mutation nearby alters the environment of the active site so that the pKs of the cysteine increases to 10.0. Would the mutation cause the reaction rate to increase or decrease? Please explain your answer.

Answer 29

A covalent catalytic mechanism of an enzyme depends on a single active-site cysteine whose pK_a is 8.0. In order for the cysteine to function in the catalysis, it has to be deprotonated to act as a nucleophile. If the pK_a is increased, then the reaction rate will decrease because it will be more difficult to deprotonate cysteine.

Question 30

Predict the effect of mutating the active site Asp 102 in chymotrypsin to Asn. In your answer please indicate the effect of the mutation on both substrate binding and catalysis.

Answer 30

The Asp102 with its negative charge stabilizes the positive charge that develops on histidine after protonation from serine 195. If the Asp is replaced with an amino acid with a neutral charge, the stabilization will be lost and the reaction rate will slow down.

Question 31

What is the function of the oxyanion hole in the chymotrypsin mechanism?

Answer 31

In both tetrahedral transition states for chymotrypsin, a formal negative charge is present on the oxygen from the original carbonyl amide bond that is the hydrolysis site—an oxyanion. The role of this hole is to bind to this oxygen and stabilize it to bring down the energy of the transition state.

Question 32

Estimate the K_I for a competitive inhibitor when [I]=5mM gives a value of K_M that is three times that of the K_M for the uninhibited reaction.

Answer 32

Question 33

Dihydroxyacetone

Answer 33

Question 34

Ribose

Answer 34

Question 35

Deoxyribose

Answer 35

Question 36

Glyceraldehyde

Answer 36

Question 37

Fructose

Answer 37

Question 38

Glucose

Answer 38

Question 39

D & L Configurations

Answer 39

Normally always a “D” configuration: depends on what side the hydroxyl group (next to the CH₂OH) is on in the Fischer Projection. D = Right; L = Left.

Question 40

Cyclicization of Sugars

Answer 40

Cyclization is an intramolecular nucleophilic attack

Question 41

Acetal/Hemiacetal & Ketal/Hemiketal

Answer 41

Question 42

Pyranose vs Furanose

Answer 42

• Cyclic sugars that contain a five membered ring are called “furanoses”
• Cyclic sugars that contain a six membered ring are called “pyranoses”

Question 43

Anomeric Carbon

Answer 43

• Always directly attached to the oxygen.
• Alpha when hydroxyl group is pointing DOWN
• Beta when hydroxyl group is pointing UP

Question 44

Nomenclature of Sugars

Answer 44

1. Type of atom involved in linkage (oxygen or nitrogen)
2. Configuration of the anomeric hydroxyl of the 1st sugar (alpha or beta)
3. Name of 1st monosaccharide, root name followed by

pyranosyl (6-ring) or furanosyl (5-ring)

4. Atoms that are linked together, 1st sugar then 2nd sugar.
5. Configuration of the anomeric hydroxyl of the second sugar (alpha or beta)
6. Name of 2nd monosaccharide, root name followed by pyranose (6-ring) or furanose (5-ring)

Question 45

Name:

Answer 45

α-D-glucopyranosyl-(1→2)-β-D-fructofuranose

Question 46

Carbon Skeleton Formulas of Fatty Acids

Answer 46

• Nomenclature: X:Y^ΔN
  
  • X = # carbons
  • Y = # carbon double bonds
  • N = Position of 1^st Carbon of double bond
    
    • Example: Palmitic acid à 16:0
    • Example: 16:1⁸ à Palmitic acid with a double bond on the 8^th carbon

Question 47

Basic structures of glycerol-phospholipids

Question 48

Phosphatidic acid

Answer 48

Question 49

Phosphatidyl choline

Answer 49

Question 50

Phosphatidyl ethanolamine

Answer 50

Question 51

Phosphatidyl serine

Answer 51

Question 52

Phosphatidyl inositol

Answer 52

Question 53

Phosphatidyl glycerol

Answer 53

Question 54

Cardiolipin

Answer 54

Question 55

Gel to liquid crystalline state transition: pure lipids

Answer 55

Narrow, sharp transition.

Question 56

Gel to liquid crystalline state transition: greater unsaturation levels

Answer 56

Lower transition temperature

Question 57

Gel to liquid crystalline state transition: liquid heterogeneity

Answer 57

Broader transitions

Question 58

Gel to liquid crystalline state transition: cholesterols

Answer 58

Higher transition temperatures

Question 59

Gel to liquid crystalline state transition: divalent ions

Answer 59

Stablize the gel state relative to the liquid crystalline state.

Question 60

Serine Protease (Chymotrypsin)

Answer 60

1. The deprotonated His 57 acts as a general base to abstract a proton from Ser 195, enhancing its nucleophilicty as it attacks the electrophilic C of the amide or ester link, creating the oxyanion tetrahedral intermediate. Asp 102 acts electrostatically to stabilize the positive charge on the His.
2. The oxyanion collapses back to form a double bond between the O and the original carbonyl C, with the amine product as the leaving group. The protonated His 57 acts as a general acid donating a proton to the amine leaving group, regenerating the unprotonated His 57.
3. The mechanism repeats itself only now with water as the nucleophile, which attacks the acyl-enzyme intermediate, to form the tetrahedral intermediate.
4. The intermediate collapses again, releasing the E-SerO- as the leaving group which gets reprotonated by His 57, regenerating both His 57 and Ser 195 in the normal protonation state. The enzyme is now ready for another catalytic round of activity.

Question 61

Aspartyl Protease (HIV1)

Answer 61

Question 62

Consider the results of an initial rate measurement of the reaction of p-nitrophenyl acetate with chymotrypsin. Please explain why the product versus time curve starts out with a steep slope and then becomes more shallow.

Answer 62

The formation of the acylated enzyme is fast while the hydrolysis of the acyl intermediate is slow

Question 63

How does induced fit ensure specificity in the reaction catalyzed by hexokinase?

Answer 63

When glucose binds to the enzyme it undergoes a conformational change. Only once it undergoes the change can it catalyze the reaction. If the incorrect substrate sugar binds, it will not undergo the conformational change.