terms for chem exam 2 Flashcards

1
Q

the induced- fit model of enzyme actions assumes that the enzyme active site is more flexible pocket whose shape changes to accommodate the substrate molecule leading to a better fit of substrate with enzyme

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2
Q

binding energy is free energy relases n forming multiple weak bonds and interactions between an enzyme and its substrate

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3
Q

binding energy is used to lower the activation energy barrier

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4
Q

an enzyme speeds a reaction by lowering the activation energy and changing the reaction pathway, providing a lower energy route for the conversion of substrate to product

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5
Q

enzymes do decrease the activation energy

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6
Q

how does an enzyme lower activation energy?

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stablizing the transition state, puts molecules in close proximity to react(increases the local concentration of reactants)

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7
Q

enzymes put molecules in correct orientation, reactants are not only near each other on enzyme, they’re orientated in optimal position to react, making it possible to always collide in the correct orientation

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8
Q

model of enzyme action
1. initiation: reactants bind to the active site in a specific orientation
2. transition state: shape changes, using less energy
3. termination: products have a lower affinity for the active site

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9
Q

Michaelis-Menten Equation, initial rate measurements [s]&raquo_space; [E], Vo=k2, km=k2+k-1/k1

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10
Q

the theoretical maximal velocity, Vo=Vmax[s]/Km+[s],
vmax is a constant
vmax is the theoretical maximal rate of the reaction-never achieved in reality
to reach Vmax would require all enzyme molecules are bound with the substrate,

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11
Q

Vmax is the maximum rate that can be seen in the reaction
- substrate is present in excess
- enzyme can be saturated (zero order reaction)

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12
Q

Km, the Michaelis constant, a constant for any given enzyme/substrate pair (indep)
- derived from rate constants

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13
Q

Km is an inverse measure of ES binding, a relative measure of the affinity of a substrate for an enzyme (how well it binds)
-small Km means tight binding; large Km means weak binding

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14
Q

at one half of the maximal velocity, the substrate concentration at this velocity will be equal to the Km

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15
Q

higher Km=lower substrate affinity to enzyme=higher [S] required to reach 1/2 Vmax

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16
Q

Lineweaver-Burk Double Reciprocal Plot, in order to determine Km and Vmax from a graph transform the michaelis-menten equation to a linear form

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17
Q

Lineweaver-Burk Double Reciprocal Plot, 1/Vo=Km/vmax*1/[S]+1/Vmax (y=mx+b)
slope is km/vmax
y-intercept is 1/vmax
x-intercept is -1/Km

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18
Q

enthalpy- the enthalpy of a reaction is the difference in enthalpy between products and reactants; the units of (delta H rxn) are kilojoules per mole

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change of H =Hfinal-Hintial

19
Q

a catalyst is a substance that increases the reaction rate without itself being consumed in the reaction

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20
Q

catalyst provides an alternative reaction pathway that has lower total activation energy than the uncatayzed reaction

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21
Q

a catalyst will speed up both the forward and the reverse reactions, rate is faster, Ea (activation energy) is lower and the rate constant K is larger

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22
Q

for the catalyzed rxn, each hill represents an elementary step, an intermediate step is a valley (in a graph)

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23
Q

a catalyst increases the rate of the forward and the reverse reactions

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24
Q

a catalyzed rxn yields the products more quickly, but does not yield more product than the uncatalyzed rxn, and also lowers Ea by providing a different mechanism & different pathway

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25
Q

catalyst speeds up a reaction by providing a set of elementary steps with more favorable kinetics than those that exist in its absence

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26
Q

enzymes are biological catalysts: effective in small amounts, enzymes speed the reaction by lowering the activation energy making the substance more likely to react,

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27
Q

activation sites are pockets or clefts in the enzyme, make catalytic groups, uses non-covalent interactions-hydrogen bonding, ionic interactions

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