Module 6 Flashcards

(16 cards)

1
Q

What is the Enzyme-Substrate Complex equation?

A

E + S <-> ES -> P+E

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2
Q

What are the assumptions of the ES Complex?

A

ES has to be constant
Substrate is in great excess over enzyme
K-1 is much greater than k2

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3
Q

What is Ks and why it is important?

A

Ks= k-1/k1
Dissociation constant

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4
Q

What does a low and high Ks mean?

A

Low Ks means enzyme binds to substrate tightly, high binding affinity
High Ks means enzyme binds to substrate poorly, weak binding

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5
Q

What is the Michaelis-Menten Equation?

A

V0=Vmax[S]/ Km + [S]

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6
Q

What is Km equal to and what does it represent?

A

Km= (k-1/k2)/k1
Represents the substrate concentration when velocity is half maximal

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7
Q

What does low and high values of Km mean?

A

Low Km means that the enzyme reaches half maximal velocity with less substrate, affinity is high
High Km means that enzyme reaches half maximal velocity with more substrate, affinity is low

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8
Q

How to measure catalytic efficiency?

A

k2/Km = measure of enzyme catalytic efficiency

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9
Q

What is another equation for k2?

A

Vmax/[E total]

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10
Q

What is the Lineweaver-Burk Equation?

A

1/v0 = (Km/Vmax) 1/[S] + 1/Vmax

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11
Q

What are the axes on the Lineweaver-Burk Graph?

A

x axis: 1/ [S]
y axis: 1/ v0

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12
Q

What does the slope and x & y intercepts represent on the Lineweaver-Burk Graph?

A

Slope= Km/Vmax
x intercept: -1/Km
y intercept: 1/Vmax

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13
Q

What is the upper limit of catalytic efficiency?

A

can’t be bigger than k1

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14
Q

In competitive inhibition, what changes on a Lineweaver-Burk graph?

A

All intersects at y intercept
The x-intercept is -1/(a * Km)

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15
Q

In uncompetitive inhibition, what changes on a Lineweaver-Burk graph?

A

All planer, no intersect
The x-intercept is -a’/Km
The Y intercept is a’/Vmax

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16
Q

In mixed inhibition, what changes on a Lineweaver-Burk graph?