Enzymes: Functions and Kinetics Flashcards

1
Q

Define a catalyst

A

A catalyst lowers the activation energy of a reaction and, most importantly, is regenerated at the end of the reaction. It is in the products and the reactants.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Indicate the main function of enzymes.

A

The main function of an enzyme is to lower the activation energy of a reaction. It facilitates the chemical reaction by increasing the rate. Enzymes ONLY act on the kinetics NOT the thermodynamics. They also tend to be very specific in terms of what they bind to.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Define ΔG and indicate when this term predicts a spontaneous chemical reaction with the release of energy.

A

ΔG is defined as the change in Gibbs Free Energy for a reaction. ΔG = Gproducts - Greactants - ΔG indicates a spontaneous reaction with the release of energy

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Indicate whether or not ΔG predicts the rate of the reaction.

A

NO! Spontaneity has nothing to do with the rate or kinetics. It refers to the thermodynamics of the reaction.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Draw the graph showing how G changes during a chemical reaction in the absence of a catalyst.

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Define “transition state” and draw on the graph the ΔG of activation and the ΔG of reaction for an uncatalyzed reaction.

A

Transition State: In going from substrate to product, there is a large increase in G. This corresponds to the formation of a high energy intermediate that involves unfavorable changes in the bond angles and electron redistribution. This intermediate is referred to as the transition state. The intermediate could decay back to the reactant or proceed forward to product. If the reaction was uncatalyzed, very few molecules would have the energy to make it over the hump. However, enzymes stabilize the transition state and lowers its energy making the reaction possible.

Increase rate by 10^6 – 10^14 – fold.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Describe the function of the active site.

A

The active site serves to 1) bind substrate and 2) catalyze the chemical reaction.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Describe the type of reaction that is catalyzed by carboxypeptidase A (CPA), and indicate whether or not this reaction could occur without an enzyme.

A

The type of reaction that is catalyzed by carboxypeptidase A (CPA) is a degradation reaction or hydrolysis that cleaves the C-terminal end of the protein. This protein is synthesized in the pancreas and delivered to the small intestine lumen. It plays a major role in digestion. This reaction may occur without an enzyme via hydrolysis, however, it would not occur quickly enough to be useful or make the amino acids available to the body before death would result. Thus this enzyme is essential for digestion.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Describe the role of the hydrophobic pocket and the salt bridge in recognizing and binding the C-terminal residue of a peptide substrate.

A

The role of the hydrophobic pocket and the salt bridge is to provide specificity and orient the substrate correctly.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Explain the role of the water molecule that is bound to the active site zinc ion of CPA.

A

The role of the water molecule is to carry out the hydrolysis reaction.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

In very general terms, explain how the zinc ion and certain amino acids in the active site can stabilize the transition state.

A

The amino acid Glu270 strips a hydrogen from water, creating a free hydroxyl group to attack the peptide carbonyl bond. Then, the zinc serves to bind the carbonyl’s oxygen in the peptide. Then, Glu270 transfers the proton to the NH to break the peptide bond. This shows that the whole protein or enzyme is important and that even distant amino acids in the protein help maintain a proper orientation of active site residues for stabilizing the transition state. The enzyme helps to “speed up” this reaction or in a way, force it to occur when it otherwise would not have.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Indicate whether or not enzyme amino acids located at a distance from the active site are important for substrate binding and catalysis.

A

They are

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

List the six main reaction types catalyzed by enzymes and briefly describe their functions.

A
  1. Oxidoreductase: moves electrons, hydrogens, or oxygens from one substrate to another.
  2. Transferases: Transfer a chemical group from one molecule to another.
  3. Hydrolases: Use water to cleave
  4. Lyases: Cleave bonds (-C-C-, -C-N-, or –C-O-) or sometimes make them (Synthase)
  5. Isomerases: Move a group or a double bond
  6. Ligases: Join atoms together, usually using ATP. (sythetases)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Induced Fit

A

The substrate itself can induce conformational changes. The whole protein is involved in the enzymatic reaction.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Oxidoreductase

A

moves electrons, hydrogens, or oxygens from one substrate to another.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Transferases

A

Transfer a chemical group from one molecule to another

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Hydrolases

A

Use water to cleave

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Lyases

A

Cleave bonds (-C-C-, -C-N-, or –C-O-) or sometimes make them (Synthase)

19
Q

Isomerases

A

Move a group or double bonds

20
Q

Ligases

A

Join atoms together, usually using ATP. (sythetases)

21
Q

Distinguish between two types of coenzymes, “cosubstrates” and “prosthetic groups.”

A

Cosubstrate: It is modified during the reaction and leaves the active site in this new modified form.

Prosthetic Group: It is tightly bound to the enzyme and regenerated after the reaction.

22
Q

What are many coezymes derived from?

A

vitamins

23
Q

Write the reaction rate expression for a single substrate that is converted irreversibly to one product.

A

E + S –> ES –> E + P

24
Q

Define kcat

A

kcat is the catalytic rate constant for the rate determining step. It replaces k3 because it is the rate determining step when it is relatively small. IT IS A CONSTANT!!

25
Q

Describe how the concentrations of substrate and product change over time.

A

The concentration of the substrate, since it is being used up to make product, would decrease over time. Since we are making more product, its concentration would increase over time.

26
Q

Define “initial rate.”

A

Initial rate is designated by Vo. The rate is given as the change in concentration of product over time. The initial rate is determined when there is very little substrate used up and before enough product was formed that could force the reverse reaction to occur. The rate slows down over time! Enzyme is less saturated.

27
Q

Describe how the initial rate of an enzyme-catalyzed reaction changes with increasing substrate concentration.

A

As the substrate concentration increases, the initial rate would increase since the enzyme is becoming more and more saturated. However, it will eventually reach a maximum rate or, Vmax, when the enzyme becomes completely saturated.

28
Q

Draw a graph of the initial rate versus substrate concentration.

A
29
Q

Write out the Michaelis-Menten equation with kcat and total enzyme concentration included.

A

v = kcat [Eo] [S] / Km + [S]

30
Q

Define mathematically Vmax, and rewrite the Michaelis-Menten equation with the Vmax included.

A

Vmax = kcat [Eo]

Therefore, V = Vmax [S] / Km + [S]

31
Q

Recognize that ___ and ___ are constants and that the units of ____ are moles/liter.

A

kcat and Km; Km

32
Q

Predict what the initial velocity will be when Km and [S] are equal.

A

It will be defined by the V= Vmax [S] / Km + [S] or the michaelis-menten equation

33
Q

Correlate various regions of the Michaelis-Menten plot with simplified versions of the Michaelis-Menten equation.

A
34
Q

The michaelis-menten graph is showing data with a _____ enzyme concentration. If one were to vary it, the _____ _____ would change.

A

constant; initial rate or Vo

35
Q

Explain the relationship between kcatand the “turnover number.”

A

kcat also gives the number of substrate molecules that one enzyme molecule can convert to product in one second.

kcat = Vmax / [Eo]

[kcat] = sec^-1

36
Q

Describe how Vmax and Km can be determined experimentally using either the Michaelis-Menten plot or the Lineweaver-Burk plot.

A

In michaelis-menten, Km is determined at Vmax/2. The Vmax is simply where the graph levels off.

In a lineweaver-burk plot, The y-axis intersection is 1/Vmax, the x-axis intercept is —1/Km

37
Q

Describe the basic effect of a competitive inhibitor on enzyme activity.

A

A competitive inhibitor competes for the active site and thus decreases the enzyme activity. It can be outcompeted with more substrate. The Km appears to be higher because it takes a greater [S] to reach ½ Vmax. Vmax is not affected because the inhibitor can be outcompeted with.

38
Q

Draw Michaelis-Menton plots and Lineweaver-Burk plots for enzyme alone and enzyme with a competitive inhibitor present.

A
39
Q

Describe how a noncompetitive inhibitor affects enzyme activity.

A

A noncompetitive inhibitor does not bind to the active site rather, it binds to an allosteric site. It thus distorts the active site. The binding of substrate is not affected and thus the Km value remains the same. However, Vmax is decreased because the enzyme cannot outcompete the inhibitor.

40
Q

Draw Michaelis-Menton plots and Lineweaver-Burk plots for enzyme alone and enzyme with a noncompetitive inhibitor present.

A
41
Q

Define Km (apparent) and Vmax(apparent).

A

They are the “new” Km and Vmax values for when an inhibitor is present.

42
Q

Define Ki

A

Ki is the dissociation constant for the inhibitor

43
Q

Discuss how the term (1 + [I]/Ki) leads mathematically to Km(apparent) or Vmax (apparent) for competitive and noncompetitive inhibitors, respectively.

A

This term is always going to be greater than 1 because it is “1 + x”. It leads to the apparent because it is multiplying Km by this value greater than 1 giving a higher Km for a competitive inhibitor and Vmax is divided by it giving a lower Vmax for the noncompetitive inhibitor.

44
Q
A