Krebs (Citric Acid) Cycle Do Now 12/10 (HW check 6.7-6.9, 6.15) OBJECTIVES: Describe the inputs and outputs of the Krebs Cycle. Describe the Krebs Cycle as the stepwise oxidation of pyruvate. TASK: 30 kJ of energy is released when 1 mol of ATP is broken down into ADP. 1 mol of glucose contains 2870 kJ of chemical energy. If aerobic respiration produces 36 mol of ATP per mol of glucose, what % of the energy stored in glucose is a cell getting from the glucose it consumes? Now, do the same calculation for Anaerobic respiration.

ATP = Energy The phosphate-phosphate bond broken when ATP  ADP + Pi releases 30 kJ/mol of energy.

Solution 36 mol ATP x 30 kJ/mol = 1080 kJ 1080 kJ / 2870 kJ = 38% In other words, aerobic respiration is about 38% efficient at getting energy from glucose into a useable form. Glycolysis?

Aerobic Respiration 4 main steps: Glycolysis splits sugar into 2 pyruvate molecules Krebs Cycle “burns” pyruvate to make energy + CO2 E- trans. Chain & Chemiosmosis in mitochondria produces MANY MANY ATP

Glycolysis Summarize…

Recap: Glycolysis & Krebs Pathway Input Output per glucose (product) Output per glucose(to next step) Glycolysis glucose 2 ATP, 2 NADH 2 pyruvate Transition Reaction (PDC) 2 CO2, 2 NADH 2 Acetyl CoA Krebs Cycle 4 CO2, 2 ATP, 6 NADH, 2 FADH2 After the Krebs Cycle: The carbon and oxygen from glucose is now in… The hydrogen from glucose is now in… The energy from glucose is now in… CO2 NADH and FADH2 ATP, NADH, and FADH2

Next Steps (tomorrow): The ELECTRON TRANSPORT CHAIN is used to create a hydrogen gradient in the mitochondria, using the energy and hydrogen from NADH and FADH2. The hydrogen gradient is the power source for the enzyme ATP synthase, which makes LOTS of ATP via chemiosmosis.