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POM Exam 1 > L6: Enzymes > Flashcards

L6: Enzymes Flashcards

1
Q

competitive inhibitor

A
  • binds to substrate binding site
  • competes with substrate
  • binds only to the free enzyme not ES complex
  • affinity of substrate decreased when inhibitor is present
  • Km is increased because it takes more substrate to compete out the inhibitor
  • Vmax unchanged
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2
Q

noncompetitive inhibitor

A
  • binds to allosteric site
  • does not compete with substrate for binding to the enzyme
  • binds to either enzyme or enzymes substrate complex
  • Vmax decreases
  • Km stays the same -substrate can still bind in same manner
  • F-1,6-Bpase inhibition by AMP
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3
Q

uncompetitive inhibitor

A
  • binds to enzyme only after substrate has bound
  • binds only to ES complex
  • Km and Vmax decreased
  • AchE inhibitors for Alzheimer’s
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4
Q

irreversible inhibitor

A
  • covalently modifies and permanently inactivates the enzymes
  • Vmax decreases
  • No effect on Km
  • penicillin example
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5
Q

general characteristics of enzymes

A
  • protein catalysts
  • increase rate of reaction
  • recognize specific substrates and catalyze specific types of reactions
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6
Q

enzymes and equilibrium constant

A
  • increase rate of reaction but DO NOT CHANGE THE EQUILIBRIUM CONSTANT
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7
Q

Keq > 1

A
  • equilibrium favors products
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8
Q

Keq < 1

A
  • equilibrium favors reactants
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9
Q

kcat

A
  • the turnover number

- number of substrate molecules converted to product per second by a single enzyme active site

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10
Q

Vmax equals

A
  • kcat[E]
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11
Q

enzymes and activation energy

A
  • lower the activation energy of a reaction by stabilizing the transition state
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12
Q

transition state

A
  • the structure of a molecule as it converts from substrate to product in the enzyme-substrate complex
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13
Q

Vmax

A
  • the maximum velocity of an enzyme
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14
Q

Km

A
  • the substrate concentration that gives 1/2 the Vmax

- Michaelis-menten constant

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15
Q

Michaelis mentin equation

A

v = (Vmax[S]) / (Km+S)

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16
Q

Michaelis mentin plot

A
  • plots substrate concentration versus velocity
17
Q

Lineweaver Burke plot

A
  • inverse plot

- plots 1/substrate versus 1/velocity

18
Q

slope of line weaver burke plot

A
  • slope = Km/Vmax
19
Q

isozymes

A
  • normal gene underwent gene duplication to form two identical genes
  • these genes underwent mutation and selection to form two or more similar genes for slightly altered enzymes
20
Q

do all duplicated genes produce functional proteins?

A
  • no
21
Q

are all isozymes useful/significant?

A
  • no
22
Q

allosteric enzymes

A
  • have multiple active sites that are cooperative
  • more sensitive to [S] than other enzymes
  • give sigmoidal curve
  • bind effectors at an allosteric site that is different from active site
  • binding at one site changes the conformation of the enzyme, which affects affinity at the active site
23
Q

positive regulator

A
  • alters the active site to allow substrate to bind more effectively
  • binding of oxygen to hemoglobin
24
Q

negative regulator

A
  • alters the active site to prevent binding of substrate, which decreases reaction rate
  • PFK inhibition by ATP
25
Q

when the end product level is high?

A
  • end product acts as a negative regulator and binds to allosteric site
  • substrate cannot bind to active site, and subsequent reaction stops
26
Q

Lineweaver burke competitive inhibitor

A
  • -1/Km moves to the right because it gets bigger
  • 1/Vmax does not change
  • looks like a bow tie
27
Q

lineweaver burke noncompetitive inhibitor

A
  • -1/Km does not change
  • 1/Vmax goes further up because you are decreasing Vmax
  • looks like a fan
28
Q

line weaver burke uncompetitive

A
  • -1/Km goes further to the left
  • 1/Vmax goes further up
  • Parallel lines
29
Q

Ki

A
  • affinity for an inhibitor
30
Q

Poor inhibitors

A
  • Ki > 10 uM
31
Q

normal inhibitors

A
  • Ki = 10 uM - 100 nM
32
Q

excellent inhibitors

A
  • Ki < 1 nm (tight binding)

POM Exam 1 (16 decks)

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