Brainscape's Knowledge GenomeTM


Top Flashcards (Certified)
All Flashcards

CHEM204 > Exam 3 > Flashcards

Exam 3 Flashcards

(46 cards)

Start Studying
1
Q

write the michael-menten equation

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

Assumptions for Michaelis-Menten

A

2 key assumptions: [P] is negligible early on in the rxn, reverse rxn also negligible

Michaelis-Menten kinetics are enzymes that follow the equation and have a hyperbolic dependence of V0 on [S]

These may not be just 2 step rxns - meaning of Vmax & Km can differ between enzymes.

Kd = k-1/k1 which happens with k2 ««< k1 so Km=Kd

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

Derive the michaelis menten equation given: k1[E][S] is the rate of ES formation, and k2[ES] + K-1[ES] is the rate of ES breakdown

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

Estimate Vmax or KM by visual inspection of initial rate (V0) versus [S] data.

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

how to calculate Km given vmax and [S] and graph

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

Show why the Michaelis-Menten equation simplifies to Vmax = kcat[Et] when [S]&raquo_space;> KM and use the simplified equation to calculate kcat or [Et].

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

Interpret a Lineweaver-Burk plot and the intercept data it provides to calculate Vmax or KM.

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

what do the symbols mean

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

specificity constant

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

what does the curve tell us

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

what is the rate equations

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

Describe the difference between reversible and irreversible inhibitors and provide an example of each.

A

Reversible : often structural analogs of substrates or products
They are often used as drugs to slow down a specific enzyme for a short period of time

Irreversible : either covalently bind or destroy a functional group in the active site (often toxins but maybe drugs)

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

Describe the mechanisms of competitive, uncompetitive and mixed inhibition and predict the effect each has on the observed Vmax and KM for an enzyme.

Competitive

A

Competitive inhibition: competes with the substrates for the active site of the enzymes. Often it will resemble the transition state.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Describe the mechanisms of competitive, uncompetitive and mixed inhibition and predict the effect each has on the observed Vmax and KM for an enzyme.

Uncompetitive

A

Uncompetitive inhibition: binds at the other site with the ES complex
NOTE: BINDS TO ES complex

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

Describe the mechanisms of competitive, uncompetitive and mixed inhibition and predict the effect each has on the observed Vmax and KM for an enzyme.

Mixed

A

Mixed inhibition: binds at other site distinct from active but binds with ES or E

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

Use initial rate data to calculate Vmax and KM in the absence and presence of an inhibitor and predict the type of inhibition (competitive or uncompetitive).

Competitive (graph)

A

apparent Vmax remains unchanged; apparent Km is increased – alpha Km = Km^app

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

Use initial rate data to calculate Vmax and KM in the absence and presence of an inhibitor and predict the type of inhibition (competitive or uncompetitive).

Competitive (equations)

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

Use initial rate data to calculate Vmax and KM in the absence and presence of an inhibitor and predict the type of inhibition (competitive or uncompetitive).

Uncompetitive (graph)

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Use initial rate data to calculate Vmax and KM in the absence and presence of an inhibitor and predict the type of inhibition (competitive or uncompetitive).

Uncompetitive (equations)

20
Q

Use initial rate data to calculate Vmax and KM in the absence and presence of an inhibitor and predict the type of inhibition (competitive or uncompetitive).

Mixed (graph)

21
Q

Use initial rate data to calculate Vmax and KM in the absence and presence of an inhibitor and predict the type of inhibition (competitive or uncompetitive).

Mixed (equations)

22
Q

calculate alpha’

23
Q

Show how a β-ketocarboxylic acid can undergo decarboxylation.

24
Q

Thiamine pyrophosphate (TPP)

25
Flavin adenine dinucleotide (FAD)
26
Coenzyme A (CoA-SH)
27
Nicotinamide adenine dinucleotide (NADH)
28
Provide a mechanism for the conversion of pyruvate to acetyl-CoA catalyzed the pyruvate dehydrogenase complex (PDH). Step 1: TPP mechanism with an alpha-ketoester
29
Provide a mechanism for the conversion of pyruvate to acetyl-CoA catalyzed the pyruvate dehydrogenase complex (PDH). Step 2: Lipoamide mechanism
Goal: Regenerate TPP via enzyme 1
30
Provide a mechanism for the conversion of pyruvate to acetyl-CoA catalyzed the pyruvate dehydrogenase complex (PDH). Step 3: coenzyme A - E2 mechanism
31
What are the key roles of NAD+, TPP, Lipoamide?
Key ideas: NAD+/NADH can function as a reducing or oxidizing agent; TPP functions as a way to make decarboxylation for alpha-keto esters by being a place where you can move electrons; lipoamide is used as a way to remake TPP and also to separate the decarboxylated pyruvate and to form the acetyl-CoA
32
Lipoamide
33
Apply what you have learned from the enzymatic logic of PDH to derive mechanism for other transformations catalyzed by enzymes utilizing the coenzymes listed in learning objective #1. Step 4: Reform Lipoamide
34
step 5: reform active site
35
step 6: reform FAD
36
overview of fatty acid biosynthesis
37
Explain how coenzyme A is an activator of acyl groups.
Coenzyme A adds to the donation of acyl groups to a variety of acceptor proteins as a result of the higher standard free energies of hydrolysis which is why it is called “activated”. Why it is an acceptor is done below.
38
Compare, contrast, and explain the reactivity differences between thioesters and oxygen esters.
Thioesters are more reactive compared to oxygen esters which can be shown by the hydrolysis of ester data: deltaG = -31.5 kJ/mol for thioesters vs deltaG = -21.5 for oxygen esters. There is less resonance stabilization versus oxygen esters This is the result of S being bigger than oxygen, resulting in a longer bond, as well as the size of sulfur making it a better leaving group compared to oxygen which is necessary for the reaction as shown by this condensation step: A thioester has a higher potential energy of acyl group transfer relative to that of an oxygen ester. That allows for a release of energy that pushes the reaction forward.
39
Step 1 to form fatty acid from malonyl-CoA: Condensation
40
Overview of conversion from malonyl-CoA to fatty acid 2 carbons at a time
41
Draw an arrow-formalism mechanism for the conversion of acetyl-CoA to malonyl-CoA. Step 1: Bicarbonate, Biotin, ATP
42
Step two to form malonyl CoA
43
Step 2 to form fatty acid from malonyl-CoA: Reduction
44
Step 3 to form fatty acid from malonyl-CoA: Dehydration
45
Step 4 to form fatty acid from malonyl-CoA: Reduction
46
Isomerisation mechanism

CHEM204 (3 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2025 Bold Learning Solutions. Terms and Conditions