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Biochemistry > 6. Enzyme Kinetics > Flashcards

6. Enzyme Kinetics Flashcards

1
Q

What is enzyme kinetics? (1)

A

The study of reaction rates when enzymes are involved

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2
Q

Explain one-way first order reactions (3)

A
  1. S -> P
  2. Formula V = change in [P] (moles/L) /change in t (sec)
  3. For a very short time segment the formula is d[P]/dt
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3
Q

Explain first-order reactions (4)

A
  1. Reaction depends on the concentration of the substance (S to P)
  2. Formula V = K1[S]
    K1 = rate constant (s-1)
    [S] = concentration (mol/Ls)
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4
Q

Explain the reverse first-order reaction (5) - Net rate, equilibrium, Keq

A
  1. If the reaction can go both ways, the rate of reaction depends on the difference between the forward and backward reaction
  2. Forward rate (S to P): K1[S]
    Backward rate (P to S): K-1[P]
  3. Net rate: V = k1[S] - k-1[P]
  4. At equilibrium: net rate V= 0 so K1 = K-1
  5. Keq = K1/K-1
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5
Q

Explain one-way second order reaction (3)

A
  1. A+B = C
  2. Rate: V = k1[A][B]
  3. Units: L/mols*s
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6
Q

Explain reversible second order reaction (4)

A
  1. A + B = C and C = A + B
  2. V = k1[A][B]-k-1[C]
  3. units: L/mol*s
  4. reverse units: s-1
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7
Q

Explain how enzyme concentration affects enzyme-catalyzed reactions (3)

A
  1. Usually E < S (constant at start of a reaction)
  2. Time = 0 is a one way reaction (S-> P) so initial rate V0 = k[E][S] = k’[E]
  3. If [S] is constant, the initial rate only depends on [E].
    If [S] is the same in each assay, a graph of V0 vs [E] is a straight line
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8
Q

Explain how substrate concentration affects enzyme-catalyzed reactions (4)

A
  1. The curve is hyperbolic
  2. At low [S], most of the E is unbound.
  3. As [S] increases, more and more ES forms and V0 increases
  4. At high [S], all the E is present as ES so adding S cannot increase V0 - the E is saturated with S at Vmax
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9
Q

What is the Michaelis Constant

A

Km = (K-1 + K2)/K1

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10
Q

What is the Michaelis-Menten equation and what does it describe?

A

V = Vmax[S]/Km+[S]

  1. Describes the relationship between V0 and [S]
  2. Vmax and Km are characteristics of particular enzymes
  3. Vmax is the maximum rate of reaction at a particular [Etot]
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11
Q

What is [Etot]?

A

When the enzyme is saturated [ES] = [Etot]

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12
Q

Explain this graph (5)

A
  1. Km is the substrate concentration at which the reaction rate V0 is half of Vmax (Vmax/2)
  2. At very low substrate concentrations, the enzyme works slowly.
  3. At very high substrate concentrations, the enzyme works as fast as it can (reaches Vmax).
  4. At low [S], the reaction rate V0 increases quickly with [S]
  5. As [S] continues to increase, the rate levels off, approaching Vmax (enzyme saturation)
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13
Q

What is the equilibrium Enzyme-Substrate dissociation constant (Ks/Km) (3)

A
  1. It measures the affinity of S for E
  2. Large Ks/Km = weak binding
  3. Small Ks/Km = strong binding
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14
Q

What are the 2 ways to determine Km and Vmax

A
  1. Non-linear least-squares fitting of MM
  2. Reciprocal plot of MM
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15
Q

Explain the reciprocal plot of MM (4)

A

1/V0 = K/Vmax * 1/[S] + 1/Vmax

Slope = Km/Vmax
Y-int = 1/Vmax
X-int = -1/Km

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16
Q

Explain this graph (4)

A
  1. This describes the direct relationship between Vmax, Kcat and Etot
  2. Vmax depends on the [E] and when the E is saturated with S
  3. Vmax = K2[Etot]
  4. Vmax/Etot = K2 = Kcat
17
Q

What is kcat? Give the example mentioned in class (2)

A
  1. kcat is an E-parameter that is independent of [E] and reports how fast an E works.
  2. e.g. At pH 7, 35oC carbonic anhydrase hydrates 4x10 5 moles of CO2 yielding 4x10 5 moles of HCO3 - per mole of enzyme per second – k 2 is a 1 st-order rate constant. k cat can be used for non-M-M enzymes as well.
18
Q

What does kcat/Km represent in enzyme kinetics, and what is its significance in the reaction rate? What is the typical range for this constant? (4)

A
  1. Kcat/Km is a 2nd order rate constant that measures how fast E and S reaction
  2. Units: liters/moles*sec
  3. There is a limit on how fast the reaction can go
  4. Kcat/Km ranges from 0.36 to 2.4x10^8 M-1s-1
19
Q

Why do many enzymes exhibit an optimum pH at which maximum activity occurs? (3)

A
  1. pH may change the ionization states of AA in the active site.
  2. pH may change the ionization state of S.
  3. pH may denature the E.
20
Q

How is temperature dependence determined?

A

And by denaturation at high and low temperature

21
Q

Explain enzyme inhibitors (2)

A
  1. Some are foreign, e.g. poisons, antibiotics, pesticides,
    herbicides.
  2. Some are naturally present for regulation of enzymatic activity
22
Q

Explain competitive inhibitors (I)

A

I is similar in structure to S and binds competitively to the same site on the E as S

23
Q

Explain this example of competitive inhibitors (3)

A
  1. Malonic acid is an inhibitor of succinate dehydrogenase
  2. How it affects V0: At high [S] the I is displaced from E so Vmax is unchanged
  3. E + S + I -> ES + EI
24
Q

How does a competitive inhibitor affect Km and the enzyme’s apparent affinity for the substrate? (2)

A
  1. Km increases with a competitive inhibitor, requiring more S to reach Vmax
  2. This reflects a reduced apparent affinity of the enzyme for the substrate.
25
Q

Explain non-competitive inhibitors (3)

A
  1. I binds at a site distinct from the substrate site, usually an allosteric site
  2. it may bind to free E or to ES. Once bound, it will prevent P formation
  3. If the binding affinities of I to E and ES are identical, there will be no effect on Km
26
Q

How does a non-competitive inhibitor affect Vmax and Km?

A
  1. A non-competitive inhibitor decreases Vmax because it reduces the active enzyme concentration
  2. Km remains unchanged as substrate binding is not affected.
27
Q

Explain regulatory enzymes and this example (3)

A
  1. A metabolic pathway is one in which the product of the 1st enzyme is a substrate for the 2nd enzyme
  2. Regulation typically occurs at the beginning of the path to prevent waste
  3. E1 is Threonine dehydratase. It is a key regulatory enzyme in the pathway and is inhibited by L-Ile
28
Q

What is heteroptropic vs homotropic in end-product inhibition/feedback inhibition (2)

A

Hetero: When an enzyme is regulated by molecules that are not S or P of the enzyme being regulated
Homo: A type of regulation where S or P modulate the activity of their own enzyme

29
Q

What is directed overflow regulation (3)

A
  1. CTP is synthesized from Asp in 7 steps.
  2. CTP feedback inhibits its own synthesis by inhibiting the first step of the pathway
  3. When demand is low, CTP can still build up and it is eliminated by degradation to uracil followed by secretion
30
Q

Explain this graph (3)

A
  1. The V0 vs [S] plot is sigmoidal
  2. MM plots are hyperbolic
  3. The enzyme will be very sensitive to [S] over a narrow range and behaves like an on-off switch
31
Q

Why is S0.5 sometimes used instead of Km? (2)

A
  1. 1/V0 vs 1/[S] curves are not linear so it’s not proper to use Km
  2. S0.5 is used for [S] needed to reach 1/2 Vmax
32
Q

What are the characteristics of allosteric enzymes? (3)

A
  1. Quaternary structure: oligomeric. multi-subunit complexes
  2. Cooperativity: the binding of S to one active site of the enzyme makes the binding of subsequent S easier
  3. Each subunit can have 2 conformations: low affinity (T) or high affinity (R)
33
Q

Explain the steps of allosteric enzyme cooperativity (6)

A
  1. Without S, the equilibrium favours T and the affinity of S for E is low
  2. When small amounts of S are added, they don’t bind since T affinity is low.
  3. As S begins to bind, it reveals more high affinity sites permitting more binding
  4. Binding of S stabilizes R state, pulling equilibrium to high affinity to form Le Chateliers principle
  5. Increasing the R state reveals more high affinity sites and results in more S binding which continues until the E is saturated
  6. This can produce sigmoidal V0 vs [S] graphs
34
Q

Explain the allosteric enzyme graph with an example (4)

A
  1. Inhibitors stabilize the T-state increasing S0.5 and making the enzyme less sensitive to substrate
    Ex) ATP and citrate inhibit phosphofructokinase
  2. Activators stabilize the R-state, decreasing S0.5 and making the enzyme more sensitive to substrate
    Ex) AMP increases PFK activity
35
Q

Explain this graph about allosteric protein (4)

A
  1. Inhibitors and activators do not bind at the substrate binding site. They bind at other sites and change the conformation of the protein
  2. Ex) Homoglobin binds to O2 cooperatively which allows it to respond to changes in O2 demand by different tissues
  3. O2 binding is also inhibited allosterically by 2,3-
    bisphosphoglycerate.
  4. Because fetal hemoglobin has a lower affinity for BPG it has a higher affinity for O2. This permits the fetus to extract O2 from the mother’s blood
36
Q

Explain the covalent regulation examples (5)

A
  1. Glycogen phosphorylase is activated by phosphorylation of Ser.
  2. This is reversible by a phosphatase that removes the phosphate.
  3. Other possibilities: AMP, UMP, methyls, ADP ribose, sugars etc. can be added and removed.
  4. In cancer cells, addition of N-acetyl-Glucose to Ser-529 of
    phosphofructokinase-1 inhibits the enzyme slowing down glycolysis and speeding up the pentose phosphate pathway.
  5. This permits cells to make more biosynthetic precursors and NADPH allowing faster growth. These changes permit much faster metabolic changes than are possible by gene regulation
37
Q

Explain the irreversible inhibitor and 2 examples (3)

A
  1. Usually they form a covalent bond with an active site AA.
  2. Penicillin binds to the active site Ser in transpeptidase, an
    enzyme involved in bacterial cell wall synthesis.
  3. Diisopropylfluorophosphate binds to Ser-195 in chymotrypsin.
38
Q

What is a zymogen and give an example (3)

A
  1. Enzyme activation by proteolytic cleavage
  2. The pancreas produces inactive trypsinogen,
    chymotrypsinogen, proelastase, and procarboxypeptidase to prevent digestion of the pancreas.
  3. In the gut, a duodenal enzyme enteropeptidase activates trypsin by removing AA 1-6 of trypsinogen producing active trypsin.

Biochemistry (7 decks)

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