Enzyme Kinetics Flashcards
E + S ⇌ ES ⇌ E + P
Based on the equation from above, formulate the first reaction rate.
k1 [E][S]
E + S ⇌ ES ⇌ E + P
Based on the equation from above, formulate the second reaction rate
k2[ES]
Describe reaction rate and equilibrium of an enzyme in terms of enzyme kinetics. How do these two relate, how do they differ?
During enzyme activity, an enzyme converts substrates into products until it reaches its equilibrium. E + S ⇌ ES ⇌ E + P
Each equilibrium sign has its own rate AKA reaction rate. Each reaction rate can be formulated and can be described based on a reactant concentration x equilibrium constant
Mathematically explain the rate of reaction. Solve for the rate of reaction for the following scenario:
60mM of fructose-6-phosphate are converted by Phosphofructokinase (PFK) into fructose-1,6,-diphosphate. If 40mM of fructose-1,6,-diphosphate was produced in 10 seconds, what is the rate of reaction?
Rate of reaction - this is the speed of reaction that occurs and can be described by The V = d[P]/dt AKA Δ[P]/Δt
40mM/10 seconds = 4mM/sec
In terms of enzyme kinetics, what factors tend to be held constant? Why?
The enzyme concentration is usually held constant because the concentration of enzymes are not usually drastically changed in the body.
What is the purpose of the michaelis-menten equation? Describe its graph?
The Michaelis-Menten describes the rate of reaction of an enzyme based on the efficiency of the enzyme and its affinity to the substrates. Graphing the michaelis menten equation on a rate reaction [velocity] vs [substrates], the equation is defined asymptotically.
A + B -> C + D
Write a rate of consumption or production for each molecule. What is its relation to others?
V = - Δ[A]/Δt = - Δ[B]/Δt = Δ[C]/Δt = Δ[D]/Δt
Rate of A and B are negative because it is consumed
Rate of C and D are positive because they are being produced
These equations are not useful. Being able to express the rate as a whole is most useful! - use the rate law and rate constant
What is the reason and use of rate law?
Rate law - uses the rate constant and the concentrations of the equation mathematically express the rate of the chemical reaction as a whole
Based on this equation, write the rate law for the forward reaction. xA ⇌ yB
In any elementary reaction, can use the coefficients of the reaction to determine the order of the reactant is in the rate law. Therefore the coefficient is the exponent in the rate law. [note: in more complex reactions, you are no longer able to use the coefficients of the chemical reaction, have to determine the rate law experimentally!]
Rate forward = k_1 [A]^x
Based on this equation, write the rate law for the forward reaction. xA ⇌ yB
In any elementary reaction, can use the coefficients of the reaction to determine the order of the reactant is in the rate law. Therefore the coefficient is the exponent in the rate law. [note: in more complex reactions, you are no longer able to use the coefficients of the chemical reaction, have to determine the rate law experimentally!]
Rate Reverse = k_2[B]^y
You’re observing a reaction and note that this reaction has reached equilibrium. Surely this reaction’s forward reaction must be equal and opposite to the reverse reaction. When is this only true?
If the reaction achieves equilibrium, then the forward reaction equals the backwards reaction. In which k[A]^x = k[B]^y
Even though this may be true, k_1 does not necessarily have to equal k_-1. This is only true when [A] = [B] when equilibrium is achieved
If you are not given the equilibrium constant and have no way to experimentally determine this constant, what can you to do find this value of a reaction?
Use the Arrhenius equation: k = Ae^(-E_a/RT) this equation describes what the equilibrium constant is dependent on
A = Frequency factor
E_a = Activation energy
R = gas constant
T = absolute temperature - temp remains constant in the body
Which portion of the arrhenius equation tends to remain constant. Why?
k = Ae^(-E_a/RT) Activation energy (E_a is the only thing that has the ability to change and affect k as everything else is constant. Enzymes decrease the activation energy, this leads to a larger e value. A larger e value leads to a larger k value, which is what increases the rate reaction rate = k[substrate]
What is the purpose of frequency factor in the arrhenius equation?
A - Frequency factor. This is the frequency of the collision seen in the reactants. According to the collision theory, the reactants have to collide with a great enough energy for the reaction to occur properly
With increase in collisions (therefore the frequency of collisions) the A value increases leading to K.
Does the enzyme affect frequency factor?
Enzymes rotate the substrates in a correct orientation and create a microenvironment that will increase the frequency of collision by decreasing the space between the substrates
Therefore Enzymes can increase A as well!!
Describe the relationship of rate of reactions and rate law.
Rate laws AKA the order of reaction show how the rate of a chemical reaction depends on reactant concentration.
Ex: For a reaction such as aA → products, the rate law generally has the form rate = k[A]^ⁿ
What order is this reaction?
Zeroth Order: if a reaction is zeroth order with respect to some reactant, then the rate is indpd of that reactant concentration
A -> B V = k[A]^0 = k
What can alter the order of reaction for a zeroth order of reaction?
No matter how we change the concentration of the reaction, the reaction rate will be the same in any scenario
A -> B V = k[A]^0 = k
Assess the reaction below:
C₉H₈O₄ (aspirin) + H2O -> C₇H₆O₃ (salicylic acid) + CH3COOH (acetic acid)
What order of reaction is this?
This is a first order of reaction in which the reaction rate is directly proportional to the reactant concentration of one of the reactants. In this manner, increase in aspirin and water, increases the products proportionally.
Rate = -Δ[A]/Δt = k[A]
What rate law does this reaction fall under: 2 NO2 → 2 NO + O2 Nitrogen dioxide decomposes into nitrogen monoxide and an oxygen molecule
Second order reactions are chemical reactions where the reaction consumes twice as much reactants to produce the one product
2A -> B
The reaction rate: V = k[A]^2
This is still a second order reaction. The result of V is still the same though. V will quadruple if A is doubled
Pseudo first order reactions are very similar to what other order reactions? What is this reaction type dependent on?
Pseudo First Order Reaction - a second order reaction that behaves like a first order reaction.
Ex: A + B -> C
The concentrations of A is very tiny but the concentrations of B is very high.
In this case, the reaction will behave like a first order. With a high concentration of B, increasing or decreasing the B, will have no affect no the rate.
However, changing the concentration of A will drastically change the rate of reaction.
Based on this chemical reaction, which step of the equation represents the steady state
Triglyceride + Lipase -> Triglyceride - lipase -> fatty acids + glycerol through lipase
Steady state - at a point of the reaction in which the ES - Enzyme Substrate Complex is constant. This assumption needs to be made when we’re looking at enzyme kinetics. We assume that the ES concentration is constant therefore the formation and loss of ES is equal to one another. Therefore the formation of ES is equal to the loss of ES. Therefore Triglyceride - lipase complex is constant
T/F - Once a substrate binds to its enzyme, it is destined to be converted into it products
False. This is true in some cases, but this is not always true. Exceptions: (1) Unbinding can occur (2) Binding can cause allosteric inhibitors (uncompetitive) to bind and lock substrate in place, but no conversion into products. (3) Binding of substrate to a an inactive enzyme due to noncompetitive inhibitor, will not cause products to be produced
Look at this generic formula for enzymes and substrates.
E + S ⇌ ES ⇌ E + P
Explain why it is generally assumed that only 3 reaction rates exist?
Overall reaction rate: Rate 1 = Rate -1 + Rate 2
k1[E][S] = K-1[ES] + K2[ES]
Rate -2 is not often indicated in a diagram as such because products tend to be more stable than the substrates and therefore are thermodynamically stable. This means that rate -2 is so small that we can see this as negligible
What molecules are reaction rates typically concerned with?
The Reactants of the reaction in progress Ex: E + S ⇌ ES ⇌ E + P k1[E][S] K-1[ES] K2[ES]
Micahelis and Menten identified their equation back in 1913. When they discovered their equation, what did they define their constant to represent? What trends does this constant have?
Vo = (Vmax[S])/(km + [S])
km - this is the inverse of affinity, meaning that as it increases in value, affinity decreases
km A. is directly proportional to the affinity of an enzyme B. measured m/s^2 C. Vmax/2 D. decreases in competitive inhibition
C. Vmax/2
Km increases with competitive inhibition, is indirectly proportional AKA inverse of the affinity of the enzyme, and has units of molar (M) or moles/L
km is the [S] where Vo = ½ Vmax
Therefore the rate is half of its max
As you study a protease collected from the B SAFE labs, you note that the enzyme’s initial activity is very slow. What is it dependent on?
Michaelis-Menten Equation. Vo = (Vmax[S])/(km + [S]) Vo = the velocity of the reaction Vmax = max velocity of the reaction Km = the michaelis constant of the enzyme S = concentration of the substrate
