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BCHM461 (Biochemistry I) > Exam III > Flashcards

Exam III Flashcards

1
Q

Michaelis Menton Equation

A
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2
Q

Define Vmax

A
  • Vmax = kcat[E]o
  • Vmax is the product of the catalytic rate constant (or turnover number) and the total enzyme concentration in the assay. Therefore, Vmax depends on the conditions of the assay. The important parameter is kcat, which is constant for a given reaction at standard conditions.
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3
Q

Define Km

A
  • Km is the concentration of substrate at which Vo equals ½Vm
  • Km = (k-1 + k2)/k1
  • In order to maximize the biological efficiency of an enzyme, the Km of an enzyme is typically close the physiological substrate concentration.
  • Higher Km indicates lower affinity for substrate.
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4
Q

Lineweaver-Burke Plot

A
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5
Q

Significant points on the Lineweaver-Burke Plot

A
  • The y-intercept of such a graph is equivalent to 1/Vmax
  • The x-intercept of the graph represents −1/K<em>m</em>.
  • The slope of the graph represents Km/Vmax.
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6
Q

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

A

You use the equation: Vmax = kcat[E]o

To calculate Vmax: -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.

kcat = 0.007 mol/min / 3.4 x 10-10 mol.

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7
Q
  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.
A

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

B) t1/2 = 0.693/k so t1/2 = 0.693/0.02 = 34.65 minutes.

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8
Q

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?

A

Keq = 1.8 x 106 M-1

ku = 10.24 s-1

Thus, kf = 1.8 x 10-7 s-1

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9
Q

First Order Reactions:

  • Determine rate order constant
  • Determine half-life
A
  • To determine the rate constant, you use the equation –kt = ln[A]/[Ao]. You plot ln[A]/[Ao] vs time, the slope is equal to –k.
  • The units of the rate constant are t-1 (time-1).
  • To find the half-life: t1/2 = 0.693/k
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10
Q

Second Order Reactions:

  • Determine the rate constant
  • Determine the half-life
A
  • To determine the rate constant, you use the equation 1/A = kt + 1/Ao. You plot 1/[A] vs time and the slope is equal to k and the y-intercept is equal to 1/Ao.
  • The units of the rate constant are M-1t-1
  • To find the half-life: t1/2 = 1/(kAo)
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11
Q

Reaction progress curve for an enzyme-catalyzed reaction.

A
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12
Q

Enzyme Catalysis Reaction

A
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13
Q

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

A
  • 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, vo, versus [S]
    • As [S] increases, Vo increases
      • At relatively low [S], the Vo increases almost linearly with increases in [S]
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14
Q

Kinetic equation describing the reaction with microscopic rate constants

A
  • Km = (k-1 + k2)/k1
    • If k2 is the rate limiting step, then Km approximates the equilibrium dissociation constant of the ES complex (k-1/k1)
    • If k-1 is the rate limiting step, then Km approximates k2/k1
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15
Q

Steady-State Assumption

A

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.

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16
Q

Significance of kcat/km

A
  • Different enzymes have similar efficiencies, but different Km and kcat 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 vo 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
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17
Q

pH dependence of the rate with breakdown into effects on kcat and Km

A
  • Find kcat vs pH
    • The highest kcat will be at the optimal pH
  • Find 1/Km vs pH
    • When Km (binding of substrate and all intermediates), and when Km 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
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18
Q

Positive Allosteric Modulation

A
  • 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.
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19
Q

Negative Allosteric Modulation

A
  • 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.
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20
Q

Homotropic Allostery

A
  • 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.
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21
Q

Heterotropic Allostery

A
  • 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
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22
Q

K-Type Regulation

A

Affects Km

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23
Q

V-Type Regulation

A

Affects Vmax.

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24
Q

Competitive Inhibition

A

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

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25
Q

Non-Competitive Inhibition

A

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

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26
Q

Un-Competitive Inhibition

A
  • Km reduced
  • Vmax reduced
27
Q

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

A

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

28
Q

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.

A

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

29
Q

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.

A

A covalent catalytic mechanism of an enzyme depends on a single active-site cysteine whose pKa 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 pKa is increased, then the reaction rate will decrease because it will be more difficult to deprotonate cysteine.

30
Q

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.

A

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.

31
Q

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

A

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.

32
Q

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

A
33
Q

Dihydroxyacetone

A
34
Q

Ribose

A
35
Q

Deoxyribose

A
36
Q

Glyceraldehyde

A
37
Q

Fructose

A
38
Q

Glucose

A
39
Q

D & L Configurations

A

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.

40
Q

Cyclicization of Sugars

A

Cyclization is an intramolecular nucleophilic attack

41
Q

Acetal/Hemiacetal & Ketal/Hemiketal

A
42
Q

Pyranose vs Furanose

A
  • Cyclic sugars that contain a five membered ring are called “furanoses”
  • Cyclic sugars that contain a six membered ring are called “pyranoses”
43
Q

Anomeric Carbon

A
  • Always directly attached to the oxygen.
  • Alpha when hydroxyl group is pointing DOWN
  • Beta when hydroxyl group is pointing UP
44
Q

Nomenclature of Sugars

A
  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)

  1. Atoms that are linked together, 1st sugar then 2nd sugar.
  2. Configuration of the anomeric hydroxyl of the second sugar (alpha or beta)
  3. Name of 2nd monosaccharide, root name followed by pyranose (6-ring) or furanose (5-ring)
45
Q

Name:

A

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

46
Q

Carbon Skeleton Formulas of Fatty Acids

A
  • Nomenclature: X:YΔN
    • X = # carbons
    • Y = # carbon double bonds
    • N = Position of 1st Carbon of double bond
      • Example: Palmitic acid à 16:0
      • Example: 16:18 à Palmitic acid with a double bond on the 8th carbon
47
Q

Basic structures of glycerol-phospholipids

A
48
Q

Phosphatidic acid

A
49
Q

Phosphatidyl choline

A
50
Q

Phosphatidyl ethanolamine

A
51
Q

Phosphatidyl serine

A
52
Q

Phosphatidyl inositol

A
53
Q

Phosphatidyl glycerol

A
54
Q

Cardiolipin

A
55
Q

Gel to liquid crystalline state transition: pure lipids

A

Narrow, sharp transition.

56
Q

Gel to liquid crystalline state transition: greater unsaturation levels

A

Lower transition temperature

57
Q

Gel to liquid crystalline state transition: liquid heterogeneity

A

Broader transitions

58
Q

Gel to liquid crystalline state transition: cholesterols

A

Higher transition temperatures

59
Q

Gel to liquid crystalline state transition: divalent ions

A

Stablize the gel state relative to the liquid crystalline state.

60
Q

Serine Protease (Chymotrypsin)

A
  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.
61
Q

Aspartyl Protease (HIV1)

A
62
Q

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.

A

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

63
Q

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

A

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.

BCHM461 (Biochemistry I) (3 decks)

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