Enzymes 2 Flashcards

1
Q

Describe a first order reaction

A

velocity of a reaction is directly proportional to reactant concentration, the reaction
V= k[A]

Units are s^-1

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2
Q

Describe a bimolecular (ssecond order reaction)

A

look like: 2A →P or A + B → P

Rate equation looks like: V = k[A]^2 or V = k[A][B]

proportionality constant for second-order reactions has the units M^-1 S^-1.

E + S → ES → E + P (E = enzyme, S = substrate, P = product)

Reaction velocity depends on the conc. of ES (complex)
Vi = k3[ES]
k3 = catalytic constant or turnover number (how much product is being made per second by the enzyme)

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3
Q

When does steady state occur in Michaelis-Menten kinetics?

A

Steady-State occurs for M-M kinetics when d[ES]/dt = 0, i.e., the rate of formation of [ES] = its rate of disappearance.
Rate during this time is defined as the Initial Velocity (vi)

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4
Q

What is the mass balance equation for enzymes?

A

E tot = [E] + [ES]

Etot = total enzyme count

E = enzyme
ES = Enzyme substrate complex

also assume rate of breakdown for ES = 0

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5
Q

What is the equation for total substrate?

A

S tot = [S] + [ES]

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6
Q

Vo (velocity) is equal to what?

A

Vo = k2[ES]

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7
Q

How can rate of product formation be found?

A

Vnet = dP/dt = k[ES]

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8
Q

What is the assumption made about [ES] at steady state?

A

We assume that the rate of formation of [ES] is equal to the rate of breakdown of [ES]. We get the equation:

[E] [S] / [ES] = k-1 (reverse reaction) + k2/k1

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9
Q

Define the Michaelis constant Km as an equation and describe what it is.

A

Km = (k-1(reverse reaction) + k2) / k1
Km is the substrate concentration at which the reaction rate is half of Vmax. It represents an enzymes affinity for a substrate. Low = high affinity, High = low affinity

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10
Q

How would you find the total amount of free enzyme in a reaction?

A

[E] = [E] tot - [ES]

subtracting the total enzyme complex count from the total enzyme count will give the concentration of free enzymes.

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11
Q

Knowing that Km = (k-1(reverse reaction) + k2) / k1, how would the [ES] steady state equation change

A

[E] [S] / [ES] = k-1 (reverse reaction) + k2/k1

becomes

[ES] = [E] [S] / Km

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12
Q

How would [E] be substituted in the steady state equation given [E] = [E] tot - [ES]?

A

[ES] = ( [E]tot - [ES] ) [S] ) / Km

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