Tutorial Practice Questions - Week 8 Flashcards

1
Q

Looking at the graph, if no activation energy was required to break down sucrose
(table sugar), would you be able to store it in a sugar bowl? Why?

A

If no Ea, sugar would break down too fast. It wouldn’t be possible to store it.

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2
Q

Which of the following sentences about endergonic and exergonic reactions is false?

a) Endergonic reactions have a positive ∆G and exergonic reactions have a negative
∆G.
b) Endergonic reactions consume energy and exergonic reactions release energy.
c) Both endergonic and exergonic reactions require energy to overcome an
activation barrier.
d) Endergonic reactions take place slowly and exergonic reactions take place
quickly

A

d)

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3
Q

Which of the following is the best way to judge the relative activation energies

between two given chemical reactions?
a) Compare the ∆G values between the two reactions.
b) Compare their reaction rates.
c) Compare their ideal environmental conditions.
d) Compare the spontaneity between the two reactions.

A

b)

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4
Q

Which of the following is not true about enzymes:

a) They increase ∆G of reactions.
b) They are usually made of amino acids.
c) They lower the activation energy of chemical reactions.
d) Each one is specific to the substrate(s) to which it binds

A

a)

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5
Q

If the ΔG’0 of the reaction A → B is 40 kJ/mol, under standard conditions the reaction

a) is at equilibrium.
b) will proceed spontaneously from left to right.
c) will proceed spontaneously from right to left.
d) will never reach equilibrium.
e) will proceed at a rapid rate

A

c)

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6
Q

Consider these reactions:

A

What is the ΔG°’ of the last reaction? ΔG°’ = -31 + 13 = -18 kJ/mol

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7
Q

7) For the reaction A → B, ΔG0 ‘ = – 60 kJ/mol. The reaction is started with 10 mmol of A; no B is initially present. After 24 hours, analysis reveals the presence of 8 mmol of A and 2 mmol of B.
Which option is the most likely explanation?

a) A and B have reached equilibrium concentrations.
b) An enzyme has shifted the equilibrium toward A.
c) The formation of B is kinetically slow; equilibrium has not been reached by 24
hours.
d) Formation of B is thermodynamically unfavourable.
e) The result described is impossible, given the fact that is ΔG0 ‘ is –60 kJ/mol.

A

c)

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8
Q

Which of the following compounds releases the most free energy on hydrolysis?

A

phosphoenolpyruvate: – 61.9 kJ/mol

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9
Q

In the following REDOX reactions:
I. Malate + NAD + —> oxaloacetate + NADH + H +
II. Acetoacetate + NADH + H+ —> Beta-Hydroxybutyrate + NAD+
III. Pyruvate + NADH + H+ —> Lactate + NAD+
IV. Succinate + FAD → Fumarate + FADH2
which molecules are the reducing agents (remember reducing agent is the molecule that
donates electrons), and which are the oxidising agents?

A

I. Malate and NADH are reducing agents, Oxaloacetate and NAD+ are oxidising
agents

II. B-hydroxybutyrate and NADH are reducing agents, Acetoacetate and NAD+ are oxidising agents

III. Lactate and NADH are reducing agents, Pyruvate and NAD+ are oxidising agents

IV. Succinate and FADH2 are reducing agents, Fumarate and FAD are oxidising agents

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10
Q

Biological oxidation-reduction reactions always involve ________________ .

a) transfer of electron(s)
b) direct participation of oxygen
c) mitochondria.
d) formation of water.
e) transfer of hydrogens

A

a)

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11
Q

Bioenergetics studies energy transformations. Muscle contractions involve the conversion of:

a) chemical energy to kinetic energy
b) chemical energy to potential energy
c) kinetic energy to chemical energy
d) potential energy to chemical energy
e) potential energy to kinetic energy

A

a) e)

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12
Q

In the citric acid cycle the FAD is reduced by succinate in the following reaction catalised by succinate dehydrogenase:

a) What are the redox pairs in this reaction?

b) What is the standard free energy change for the conversion of Succinate to
fumarate?

c) Is this reaction thermodynamically favourable in standard conditions?

A

a) FAD/FADH2
Fumarate/Succinate

b) ∆G’0 = - nF∆E’0

∆E’0 = -0.180 – (+0.030) = - 0.210 V
(E’0 from the electron acceptor minus the E’ 0 of the electron donor)

∆G’0 = -2 x 96.5 kJ/(V.mol) x -0.210 V = 36.5 kJ/mol

c) No, ∆G’0 > 0

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