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Announcements Reading for today on glycolysis: pp. 221-234 Homework due today: Problems 8-5, 8-7 In both problems, use the favorable free energy change of ATP hydrolysis to drive the unfavorable transport of solute up its concentration gradient on a per mole basis. Reading for Wednesday, Feb. 21 on fermentation and TCA cycle: 236-241, 248-258 Reading for Friday, Feb. 23 on integration of metabolism: 258-262 Reading for Monday, Feb. 26 on respiration: 265-271 Homework due Monday, Feb. 26: Problem 9-5 Convert all concentrations to M, and your answer will be in M. Don’t worry about [H+] – use equations as given in problem. Determine [glucose] at equilibrium (ΔG’ = 0); then a concentration greater than this will favor production of G6P. Wednesday, Feb. 28 on respiration: 271-283 Friday: Exam 2 on Days 12-22, 99 points, 33 m/c questions. Note Exam 3 will be worth 81 points.

Announcements Reading for today on glycolysis: pp. 221-234 Homework due today: Problems 8-5, 8-7 –In both problems, use the favorable free energy change

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Announcements

• Reading for today on glycolysis: pp. 221-234• Homework due today: Problems 8-5, 8-7

– In both problems, use the favorable free energy change of ATP hydrolysis to drive the unfavorable transport of solute up its concentration gradient on a per mole basis.

• Reading for Wednesday, Feb. 21 on fermentation and TCA cycle: 236-241, 248-258

• Reading for Friday, Feb. 23 on integration of metabolism: 258-262• Reading for Monday, Feb. 26 on respiration: 265-271• Homework due Monday, Feb. 26: Problem 9-5

– Convert all concentrations to M, and your answer will be in M.– Don’t worry about [H+] – use equations as given in problem.– Determine [glucose] at equilibrium (ΔG’ = 0); then a concentration greater

than this will favor production of G6P.• Wednesday, Feb. 28 on respiration: 271-283• Friday: Exam 2 on Days 12-22, 99 points, 33 m/c questions.

– Note Exam 3 will be worth 81 points.

Outline/Learning Objectives

CHO metabolism

A.Metabolic reactions and ATP

B.Oxidation-reduction reactions

C.Glycolysis

After reading the text, attending lecture, and reviewing lecture notes, you should be able to:

• Describe the roles of ATP/ADP and NAD+/NADH in biological reactions.

• Summarize the purpose and place, reactants and products, critical enzymes, and net yield of glycolysis.

• Calculate free energy changes associated with glycolysis.

• Summarize the products, control steps, etc. of glycolysis.

Catabolic and Anabolic Metabolic Pathways

Catabolic (degradative)• Glycolysis, fermentation• TCA cycle, electron

transport, oxidative phosphorylation

• Glycogenolysis• Fatty acid oxidation• Amino acid degradation

Anabolic (synthetic)• Gluconeogenesis• Glycogen synthesis• Fatty acid synthesis• Amino acid synthesis

ATP is an energy intermediate

Donates Phosphate;Hydrolysis is exergonic

Accepts Phosphate;Synthesis is endergonic

Oxidation and Reduction

• Oxidation: gives up e-

– Dehydrogenation: gives up H+

• Reduction: accepts e-

– Hydrogenation: accepts H+

• Redox reactions– Usually involve 2 e-, 2 H+ in the cell– Catalyzed by dehydrogenases

NAD+ accepts, NADH donates e-

NAD+/NADH is an intermediate e- acceptor/donor• Its redox reaction is coupled to another redox

reaction in metabolic pathways:

NADH + H+ NAD+ + 2e- + 2H+ Acetaldehyde + 2e- + 2H+ ethanolAcetaldehyde + NADH + H+ ethanol + NAD+

This fermentation reaction is catalyzed by alcohol dehydrogenase.

Final e- Acceptors

• Glycolysis, fermentation (NAD+ )

– No net oxidation of glucose

• Aerobic respiration O2

– Complete oxidation of glucose using O2 as final e- acceptor: ½ O2 + 2 H+ + 2 e- H2O

• Anaerobic respiration S, H+, Fe3+

– Complete oxidation of glucose using something other than O2 as final e- acceptor.

• Glycolysis is the central catabolic pathway and first part of aerobic respiration:

Oxidation of Glucose

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Go’ = -686 kcal/mol

• In bomb calorimeter, the large EA of this rxn is overcome by heat of fire.

• In the cell, a series of reactions with small EA is overcome by body temperature and enzymes.

• Purpose: convert chemical bond energy of glucose to chemical bond energy in ATP

Glycolysis Problems

2-88

2-89

2-90

2-93

Energetics of GlycolysisSTEP REACTION ΔGo’ ΔG’

1 GLC + ATP → G6P + ADP + H+ -4.0 -8.0

2 G6P → F6P +0.4 -0.6

3 F6P + ATP → F1,6BP + ADP + H+ -3.4 -5.3

4 F1,6BP → DHAP +G3P +5.7 -0.3

5 DHAP → G3P +1.8 +0.6

6 G3P + Pi + NAD+ → 1,3BPG + NADH + H+ +1.5 -0.4

7 1,3BPG + ADP → 3PG + ATP -4.5 +0.3

8 3PG → 2PG +1.1 +0.2

9 2PG → PEP + H2O +0.4 -0.8

10 PEP + ADP +H+ → PYR + ATP -7.5 -4.0

The table above shows free energy change values in kcal/mol for the concentrations of intermediates in red blood cells. The overall ΔG’ from glucose to pyruvate is about -20 kcal/mol.

1 MConc.,pH=7

cellularConc.,pH=7

Glycolysis Summary

• Cytoplasmic• Catabolism of:

– carbohydrates G6P, F6P– fats: glycerol DHAP– some amino acids pyruvate

• Cost = 2 ATP, Gain = 4 ATP, Net Gain = 2 ATP• Reduced Coenzymes: 2 NADH + H+

• End product: pyruvate (CH3COCOO-)

• Control steps: irreversible reactions 1, 3, and 10