The combustion (burning) of paper to form CO2 + H2O releases heat (it is an exothermic reaction) because:

• The free energy change (ΔG) for this reaction is positive

• The products (CO2 +H2O) have more stable chemical bonds than the reactants.

• The entropy of the products (CO2 + H2O) is less than that of the reactants.

• The carbon atom of CO2 is more reduced than most of the carbon atoms in paper.

For the reaction: A + B ↔ C + D,

Keq (for the left to right direction) = 2x10^2 (i.e., Keq = 200).

Which of the following is true? (Note that ΔG°' is the free energy change as the reaction proceeds from standard conditions to equilibrium.)

• There is more A + B than C + D at equilibrium, and ΔG°' is positive

• There is more C + D than A + B at equilibrium, and ΔG°' is positive

• There is more A + B than C + D at equilibrium, and ΔG°' is negative

• There is more C + D than A + B at equilibrium, and ΔG°' is negative

The reaction F ↔ G has a free energy change (ΔG°') of – 6 kJ/mol understandard conditions (i.e., standard conditions are starting conditions of1M of F and G, and the ΔG°' is by convention for the F → G direction)

What are the relative amounts of F & G when the reaction F ↔ G reaches equilibrium?

• 10F to 1G

• 1F to 1G

• 1F to 10G

• 1F to 100G

• 6F to 1G

What relative amounts of F & G at the START of the reaction would result in a free energy change (ΔG') of + 6 kJ/mol as the reaction proceeds to equilibrium? (I am asking which starting conditions would give a ΔG' of+ 6 kJ/mol and, as noted above, ΔG' is by convention for the F → G direction.)

• 1F to 100G

• 1F to 10G

• 1F to 1G

• 10F to 1G

• 6G to 1F

The reaction A → B has a standard free energy change (DG°') of +11 kJ/mol.In the cell, the formation of B is coupled to the conversion of C → D.

To obtain a net gain in the amount of B from the coupled reaction under standardconditions, the ΔG°' for C → D must be:

• Between –1 and –11 kJ/mol

• Between +1 and +11 kJ/mol

• Greater than +11 kJ/mol (for example +20 kJ/mol)

• Less than –11 kJ/mol (for example -20 kJ/mol)

Which of the following is true about the preparatory steps of glycolysis? (The preparatorysteps are the conversion of glucose to glyceraldehyde–3–phosphate.)

• NAD+ is reduced to NADH

• A phosphate group (Pi) from fructose-6-phosphate is transferred to ADP to form ATP

• ATP is consumed (i.e., converted to ADP)

• Pyruvate is converted to lactate

For which of the following compounds is the donation of a phosphate to ADP to form ATP(a substrate-level phosphorylation) an unfavorable reaction at standard conditions?

• phosphoenolpyruvate

• 1,3-biphosphoglycerate

• glucose 6-phosphate

• substrate level phosphorylation is not favorable from any of the compounds listed

Consider the short pathway with the following standard free energy changes (ΔG°' s) foreach step:

–7.5 +7.5 –0.2 –23.

A ↔ Β ↔ C ↔ D → E

If the conversion of D → E is blocked by an enzyme inhibitor, which compound in the pathway will accumulate to the highest levels?

• A

• B

• C

• D

• E

Yeast cells can grow anaerobically (without oxygen) by fermenting glucose to ethanol and can also completely oxidize glucose to CO2 and H20 during aerobic growth.

When yeast are grown anaerobically the amount of glucose consumed per amount of growth is much higher than during aerobic growth. This is because:

• during anaerobic growth much of the glucose is used to make starch

• glycolysis is blocked in aerobic conditions

• glucose is used inefficiently in the presence of oxygen

• the yield of ATP per glucose is much higher in aerobic than in anaerobic conditions

During the fermentation of glucose to ethanol in yeast, pyruvate is converted to ethanol because:

• additional energy can be produced by the oxidation of the NADH that is formed during ethanol production

• NAD+ must be replenished for glycolysis to continue

• it is necessary to reduce pyruvate before it can be converted to CO2 and H2O in the citric acid cycle (also known as the TCA cycle)

• ethanol is more oxidized than pyruvate