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 ```

(BIO) Enzyme Kinetics and Cellular Metabolism Flashcards by Alissa Leyman

Michaelis-Menten constant

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

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Vmax

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

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Competitive Inhibitors

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

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Noncompetitive Inhibitors

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

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Uncompetitive Inhibitors

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

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Kcat

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

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Enzyme Efficiency

Reflected by the quantity of Kcat/Km

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(Kcat)/(Km) > 1 when:

Turnover number is high (Large Kcat)

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

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Michaelis-Menten Curve

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

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Lineweaver- Burk Plots

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

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Glycolysis

Step 3 = committed step, rate limiting step

F6P –> (PFK + ATP) –> F16BP

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PFK

Enzyme in Step 3 for F6P –> F16BP

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

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Regulation Steps of Glycolysis

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

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Glycolysis Yields

Net 2 ATP

4 ATP + consumes 2 ATP

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Glycolysis occurs in the

Cytosol and is ANaerobic

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

Pyruvate Decarboxylation (catalyzed by pyruvate dehydrogenase complex)

Regulation of Glycolysis Conceptually

Regulation of glucose entrance into the cell
Increase in blood glucose increases insulin, leading to localization of GLUT4 at the cell surface
Regulation of the enzymes involved in the metabolic pathway (Steps 1, 3, and 10)
Irreversible steps and have large negative deltaG˚ values
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

1 Glucose molecule produces

2 Pyruvates

2 Net ATP

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

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

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

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

Per 1 Glucose:
6 NADH
2 ATP
2 FADH2

Where in the cell does the Citric Acid Cycle occur?

Mitochondrial Matrix

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

NAD and FAD

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

Activated by:
NAD+
CoA
AMP

Inhibited by plenty of fuel:
NADH
Acetyl-CoA
ATP

TCA Regulation (Main Mechanisms)

Substrate (Oxaloacetate and Acetyl-CoA) concentration
Product (NADH) concentration
Allosteric activation
Competitive feedback

TCA Regulation: Substrate concentration

Oxaloacetate and Acetyl-CoA concentration

TCA Regulation: Product (NADH) concentration

NADH inhibits PDC and other regulatory steps

TCA Regulation: Allosteric activation

ADP is an allosteric activator of TCA

TCA Regulation: Competitive feedback

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

ETC Goal

Create an H+ gradient

ETC Gradient must be present in order to

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

Reduction Potential (E˚)

The tendency to accept electrons

Electrons flow spontaneously from

Lower E˚ to Higher E˚ DeltaG = -nF (deltaE)

ETC yield of ATP

``` 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

Where does ETC occur?

Along the INNER Mito Matrix

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?

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

What is the final electron acceptor of ETC?

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

Anerobic Respiration

Glycolysis and Fermentation

Fermentation

Converts NADH to NAD+