Enzyme kinematics

Question 1

A catalyst (enzyme) does not change the reaction rate or [equilibrium] of a reaction. Only the ____ changes

Answer 1

Activation energy

Lowered like a catalyst

Question 2

The maximum velocity is ______

Answer 2

The top velocity that all variations of the experiment reached (meaning, all the different [S]

Question 3

Km occurs at _____ and represents ____

Answer 3

1/2 Vmax

The binding affinity for that reaction (inversely related)

Question 4

On the affinity graph (the normal graph that shows Vmax) the __ axis represents initial velocity and the __ axis represents [S]

Answer 4

Y
X

Question 5

Why is ES a vital stage in the reaction scheme?

Answer 5

At the formation of ES, the reaction can either proceed backwards to the E+S stage or forwards to the E+P stage. This makes the ES stage a vital part of the entire reaction

Question 6

The Briggs Haldane Assumption said that ____ reaches a steady state of rate formation where rate of formation = rate of breakdown, so that it is fixed for most of the reaction

Answer 6

[ES]

Question 7

It is the assumption that ____ is the rate limiting state of the reaction

Answer 7

ES –> E+P

Question 8

The Michaelis-Menten Equation is:

Answer 8

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

Question 9

Km is the ___ at ___ and represents the rate of speed of the reaction. It can be calculated by:

Answer 9

[S] at 1/2 Vmax

Km= (k2+k-1)/K1

Question 10

The Lineweaver-Burk equation is:

Answer 10

1/Vo= (Km/Vmax) * (1/[S]) + (1/Vmax)

Question 11

Where are 1/Vmax and -1/Km on the Lineweaver-Burk plot?

Answer 11

1/Vmax is found on the y-intercept

-1/Km is found on the x intercept

Question 12

Km is ____ proportional to enzyme affinity, meaning:

Answer 12

Inversely

The higher the Km, the lower the affinity for enzyme to substrate and therefore the higher [S] needed to reach Vmax, and vice versa

Question 13

Kcat represents the _____ and is calculated by:

Answer 13

Turnover rate (the maximum amount of molecules that are turned into products when the enzyme and substrate are bound)

Kcat= Vmax/[E]

Question 14

Kcat/Km is the _____ (2nd order rxn) which is the overall measure of catalytic affinity, meaning:

Answer 14

Specificity constant

The rate of turnover and affinity for the substrate. It can be used to compare different substrates with the same or different enzymes

Question 15

10^___ is the “perfect” diffusion rate- meaning, the fastest the enzyme can go

Answer 15

10^8 or 10^9

Question 16

Why is it important that many enzymes are “imperfect”?

Answer 16

Enzymes are responsible for a lot of cleavage, so it would be bad if they cleaved everything they came across

Question 17

The sequential method, one of the two possible methods when there are two or more reactants involved, occurs when:

Answer 17

Both of the substrates must bind to the enzyme before any product is made. Makes a complex something like: ES1S2 –> E+P1+P2

It forms a ternary complex where either substrate can come in first, but ultimately results in the formation of 2 products

Question 18

The ping-pong method, one of the two possible methods when there are two or more reactants involved, occurs when:

Answer 18

The enzyme reacts with one substrate, turns that into a product, and then reacts with another substrate. No ternary complex is formed and the substrates come in like ping-pong balls

Question 19

The Km of a reaction with a competitive inhibitor is ____ bc the inhibitor lowers the binding affinity as it outcompetes the substrate

Answer 19

Higher

(inversely proportional)

Question 20

A competitive inhibitor keeps the same ____ as the reaction without the inhibitor because

Answer 20

Vmax

At some point, you will be able to outcompete the inhibitor since it binds to the active site

Question 21

A _____ inhibitor binds at E+S and fits into the same pocket as the substrate does

Answer 21

Competitive

Question 22

An _____ inhibitor binds the ES complex and not the enzyme by itself

Answer 22

Uncompetitive

Question 23

A _____ inhibitor can bind to both the E+S and ES complex

Answer 23

Noncompetitive

Question 24

A competitive inhibitor can be identified on a Lineweaver-Burk plot by the ____

Answer 24

1/Vmax, since Vmax stays the same with or without inhibitors, all the lines will have the same y intercept regardless of slope

Question 25

An uncompetitive inhibitor can be identified on a Lineweaver-Burk plot by ____, because:

Answer 25

Parallel lines

Both the Vmax and Km are affected in the same way (decreased)

Question 26

___ can never be reached with an uncompetitive inhibitor (and Km is lowered) because ____

Answer 26

The uncompetitive inhibitor binds at the ES complex, so no matter how much substrate is present, some of it will be inhibited from reacting by the inhibitor

Question 27

A noncompetitive inhibitor can be identified a Lineweaver-Burk plot by their different ___, ___ and ____

Answer 27

Vmax, Km and slope

Question 28

Why is the slope of the Lineweaver plot different with a noncompetitive inhibitor and not with competitive or uncompetitive inhibitors?

Answer 28

Because the noncompetitive inhibitor has two ways of binding: at E+S or the ES complex

Question 29

Alpha or alpha’ calculated by:

Answer 29

1 + [I]/KI or 1 + [I]/K’I

Question 30

What does the Lineweaver-Burk plot look like for an irreversible inhibitor?

Answer 30

Just a straight line. Have no curve because every molecule that reacts is permanently inactivated

Question 31

Irreversible inhibitors can be used to find active site residues because:

Answer 31

When they bind to the active site, they deactivate the reaction. So if you bind the inhibitor to something and it stops the reaction from happening then you know you found the active site

Question 32

Every inhibitor is reversible except for the ____

Answer 32

Irreversible inhibitor

Question 33

Vmax depends on [E] because:

Answer 33

The more enzyme present, the faster the reaction will occur because there will be more active sites ready for the substrate to bind with