(BIO) Enzyme Kinetics and Cellular Metabolism

Question 1

Km

Answer 1

Michaelis-Menten constant

Represent the substrate concentration for the reaction rate to reach 1/2 Vmax

Question 2

Vmax

Answer 2

The maximum reaction rate reached by the system, when all active sites are occupied

Question 3

Competitive Inhibitors

Answer 3

Compete with the substrate to bind to the active site of the enzyme

Increase substrate concentration will reduce the chance of the inhibitor binding to the enzyme

Can be outcompeted by increase in substrate concentration

Question 4

Noncompetitive Inhibitors

Answer 4

Bind the enzyme at a site other than the active site

This can occur with or without the substrate present

Increase substrate concentration will NOT relive the inhibition, since the inhibitor binds the enzyme-substrate complex (ES) just as it binds the enzyme

Question 5

Uncompetitive Inhibitors

Answer 5

Bind only to the ES complex

This reduction in the effective ES concentration:
Increases the enzyme’s apparent affinity for the substrate
Decreases the maximum activity of the enzyme

Question 6

Kcat

Answer 6

Aka Turnover #

Denotes the number of substrate molecules converted per enzyme molecule per second

Provides a direct measure of the catalytic production of product under optimum conditions

Question 7

Enzyme Efficiency

Answer 7

Reflected by the quantity of Kcat/Km

Question 8

(Kcat)/(Km) > 1 when:

Answer 8

1. Turnover number is high (Large Kcat)

2. Enzyme has high affinity for substrate (low Km)

Question 9

Michaelis-Menten Curve

Answer 9

X axis = Substrate concentration [S]
Y axis = Reaction velocity (v)
@ 1/2 Vmax, substrate concentration = Km

Competitive Inhibition = just below normal curve but eventually reaches same Vmax, lower Km but same 1/2Vmax

Noncompetitive Inhibition = Vmax is 1/2Vmax of normal curve, starts out the same as competitive inhibition, lowest Km & 1/2Vmax

Question 10

Lineweaver- Burk Plots

Answer 10

X-axis = 1/[S]
Y-axis = 1/Vo
Slope = Km/Vmax

Competitive = same 1/Vmax, diff -1/km
Noncompetitive = diff 1/Vmax, same -1/km
Uncompetitive = diff 1/Vmax, diff -1/km

Question 11

Glycolysis

Answer 11

Step 3 = committed step, rate limiting step

F6P –> (PFK + ATP) –> F16BP

Question 12

PFK

Answer 12

Enzyme in Step 3 for F6P –> F16BP

Regulated by:
(+) AMP
(+) F26BP  --> closely related to insulin & glucagon; when insulin is active F26BP is produced
(-) ATP
(-) Citrate

Question 13

Regulation Steps of Glycolysis

Answer 13

Highly EXERgonic steps which make reverse rxn impossible!

Step 1:Glucose –> (Hexokinase + ATP) –> G6P
Step 3: F6P –> (PFK + ATP) –> F16BP
Step 10: PEP –> (Pyruvate kinase + ATP) –> Pyruvate

Question 14

Glycolysis Yields

Answer 14

Net 2 ATP

4 ATP + consumes 2 ATP

Question 15

Glycolysis occurs in the

Answer 15

Cytosol and is ANaerobic

Question 16

What reaction bridges the gap from Glycolysis to the Citric Acid Cycle?

Answer 16

Pyruvate Decarboxylation (catalyzed by pyruvate dehydrogenase complex)

Question 17

Regulation of Glycolysis Conceptually

Answer 17

1. Regulation of glucose entrance into the cell
   Increase in blood glucose increases insulin, leading to localization of GLUT4 at the cell surface
2. Regulation of the enzymes involved in the metabolic pathway (Steps 1, 3, and 10)
   Irreversible steps and have large negative deltaG˚ values
3. Regulation in response to concentrations of certain effectors
   Increase in ATP concentration signals that energy produced is greater than energy consumed, so the body can downregulate glycolysis and upregulate glycogenesis
   Increase in Citrate concentration also means that energy levels are high and intermediates are abundant

Question 18

1 Glucose molecule produces

Answer 18

2 Pyruvates

2 Net ATP

Question 19

How many rounds of the Citric Acid Cycle result from 1 round of Glycolysis?

Answer 19

1 Glucose –> 2 Acetyl CoA = 2 turns of the cycle

All C, H and O from Pyruvate end up as CO2 and H2O

Question 20

What is the net gain of ATP from the Citric Acid Cycle?

Answer 20

Per 1 Glucose:
6 NADH
2 ATP
2 FADH2

Question 21

Where in the cell does the Citric Acid Cycle occur?

Answer 21

Mitochondrial Matrix

Question 22

The electrons used in the ETC are produced by the TCA via the reduction of…

Answer 22

NAD and FAD

Question 23

Citric Acid Cycle Regulation (via PDC: pyruvate dehydrogenase complex)

Answer 23

Activated by:
NAD+
CoA
AMP

Inhibited by plenty of fuel:
NADH
Acetyl-CoA
ATP

Question 24

TCA Regulation (Main Mechanisms)

Answer 24

1. Substrate (Oxaloacetate and Acetyl-CoA) concentration
2. Product (NADH) concentration
3. Allosteric activation
4. Competitive feedback

Question 25

TCA Regulation: Substrate concentration

Answer 25

Oxaloacetate and Acetyl-CoA concentration

Question 26

TCA Regulation: Product (NADH) concentration

Answer 26

NADH inhibits PDC and other regulatory steps

Question 27

TCA Regulation: Allosteric activation

Answer 27

ADP is an allosteric activator of TCA

Question 28

TCA Regulation: Competitive feedback

Answer 28

Succinyl-CoA and intermediate, competes with acetyl-CoA to bind Citrate Synthesis

Question 29

ETC Goal

Answer 29

Create an H+ gradient

Question 30

ETC Gradient must be present in order to

Answer 30

Form ATP using the enzyme ATP Synthase

Because the binding of ADP and (PO4)3- and the subsequent release of ATP from the enzyme actually requires energy

Question 31

Reduction Potential (E˚)

Answer 31

The tendency to accept electrons

Question 32

Electrons flow spontaneously from

Answer 32

Lower E˚ to Higher E˚

DeltaG = -nF (deltaE)

Question 33

ETC yield of ATP

Answer 33

2 NADH from Glycolysis
2 NADH from PDC
6 NADH from TCA
= 10 NADH x 2.5 
=25 ATP

2 FADH2 from TCA
= 2 FADH2 x 1.5
= 3 ATP

Total ATP = 28 ATP

Question 34

Where does ETC occur?

Answer 34

Along the INNER Mito Matrix

Question 35

Infants have brown fat cells which express a protein that acts to dissipate the proton gradient, yet allows O2 consumption to continue.

What effect might this have on ATP generation?

Why might this mechanism have evolved?

Answer 35

Dissipating the proton gradient would make the ETC much LESS efficient. ATP generation would decrease, while more energy would be lost as waste

Potential evolutionary explanation: brown fat aids in thermogenesis

Question 36

What is the final electron acceptor of ETC?

Answer 36

Oxygen

Without O2, cells must rely on anaerobic respiration (glycolysis and fermentation)

Question 37

Anerobic Respiration

Answer 37

Glycolysis and Fermentation

Question 38

Fermentation

Answer 38

Converts NADH to NAD+