The Basics of Energy Metabolism Flashcards

1
Q

The metabolism of all living cells is an (open or closed) system?

A

open

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2
Q

What is the universal molecular carrier for biological energy?

A

ATP/ADP

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3
Q

Is the cell at equilibrium?

A

No, it is at a steady state and is specifically held out of equilibrium

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4
Q

What is Keq

A

the equilibrium constant of a reaction

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5
Q

What is ∆G0’?

A

The change in Gibbs free energy for a reaction under STANDARD conditions

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6
Q

What are the standard conditions?

A

naught = 1M initial products and reactants
prime = pH7

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7
Q

What’s the formula for ∆G0’?

A

∆G0’ = -R x T x ln(Keq)

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8
Q

What are the R and T in the ∆G0’ equation?

A

R = 8.32
T = 25C (298K)

RT = 2.48 kJ/mol

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9
Q

What’s the formula for Keq?

A

[products] / [reactants]

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10
Q

Keq > 1

A

products favored and exergonic (releases energy)

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11
Q

Keq < 1

A

reactants favored and endergonic (requires energy)

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12
Q

∆G0’ > 0

A

reactants favored and endergonic (requires energy)

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13
Q

∆G0’ < 0

A

products favored and exergonic (releases energy)

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14
Q

The Keq of ATP/ADP is what and means what?

A

Keq = 2.7 x 10^-6
At equilibrium there’s more ADP than ATP

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15
Q

What is the ∆G0’ of ATP/ADP and what does it mean?

A

∆G0’ = 31.8 kJ/mol

ADP (reactant) is favored and the reaction is endergonic (requires energy)

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16
Q

What is ∆G’?

A

The change in Gibbs free energy for a reaction under NON STANDARD conditions

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17
Q

What is the equation for ∆G’?

A

∆G’ = ∆G0’ + (R x T x ln ([products] / [reactants])

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18
Q

∆G’ > 0

A

reactants favored and endergonic (requires energy)

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19
Q

∆G’ < 0

A

products favored and exergonic (releases energy)

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20
Q

How do you find the actual [ATP] and [ADP]?

A

metabolics

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21
Q

What’s [ATP] for E. coli?

A

9.6 x 10^-3 M

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22
Q

What’s [ADP] for E. coli?

A

0.56 x 10^-3 M

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23
Q

What’s [PO4^3-] for E. coli?

A

1 x 10^-3 M

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24
Q

What’s the ∆G’ of E. coli for ATP synthesis and what does it mean?

A

56 kJ/mol
ADP (reactant) is favored and endergonic (requires energy, specifically 56kJ/mol to make ATP)

25
Q

Which do you expect to be larger during ATP synthesis: ∆G0’ or ∆G’?

A

∆G’
because ∆G’ = ∆G0’ + (RT x ln ([products] / [reactants])) and the [products] / [reactants] is > 1

26
Q

The electron acceptor is the (oxidant or reductant)?

A

oxidant

27
Q

The electron donor is the (oxidant or reductant)?

A

reductant

28
Q

Reduction (gains or loses) electrons?

A

Reduction Is Gain (RIG)

29
Q

Oxidation (gains or loses) electrons?

A

Oxidation Is Loss (OIL)

30
Q

NADPH is an electron (donor or acceptor)?

A

electron donor

31
Q

NADP+ is an electron (donor or acceptor)?

A

electron acceptor

32
Q

NADH is an electron (donor or acceptor)?

A

electron donor

33
Q

NAD+ is an electron (donor or acceptor)?

A

electron acceptor

34
Q

Reduction potential

A

measurement of how easily a molecule can take up electrons

35
Q

oxidation potential

A

measurement of how easily a molecule can give up electrons

36
Q

A hydride = what?

A

1 H- = 1 (H+) + 2 e-

37
Q

[H] = what?

A

1 (H+) + 1 e-

38
Q

What is the equation for ∆G0’ ?

A

∆G0’ = -n x F x ∆E0’

n = number of e-
F = 97 kJ / V
∆E0’ = standard reduction potential

39
Q

What is ∆E0’?

A

the STANDARD reduction potential
(E0’ of e- acceptor) - (E0’ of e- donor)

40
Q

∆E0’ < 0

A

more likely to give up e- (mainly oxidized)

41
Q

∆E0’ > 0

A

more likely to keep e- (mainly reduced)

42
Q

What does NAD stand for?

A

Nicotinamide Adenine Dinucleotide

43
Q

What does NAD do?

A

hydride transfer (H- = 2 e- + H+)

44
Q

What is the structure of NAD+ and NADP+?

A
45
Q

What is the structure of NADH and NADPH?

A
46
Q

What’s the structure equation for NAD(P)+ to NAD(P)H?

A
47
Q

What’s the oxidation half reaction of NAD(P)+/NAD(P)H?

A

NAD(P)H -> NAD(P)+ + (H-)

48
Q

What’s the reduction half reaction of NAD(P)+/NAD(P)H?

A

NAD(P)+ + (H-) -> NAD(P)H

49
Q

What’s the E0’ of NAD(P)+ / NAD(P)H and what does it mean?

A

-0.32 V
NAD(P)H wants to give up e-

50
Q

What is ∆G0’ of NAD(P)+ / NAD(P)H and what does it mean?

A

∆G0’ = 0

because both NAD(P)+ (e- acceptor) and NAD(P)H (e- donor) have a E0’ = -0.32 so ∆E0’ = 0 which makes the entire formula = 0

neither the product or reactant are favored

51
Q

What’s non-standard conditions?

A

pH 7 and 25 C (298 K) but [actual]

52
Q

What’s ∆E’?

A

the reduction potential under NON-STANDARD conditions (actual conditions)

53
Q

What’s the E’ of NADP+ / NADPH and what does it mean?

A

E’ = -0.373 V

NADPH wants to give up e-

54
Q

What’s the ratio of NADH / NAD+?

A

1/31
(ln this)

55
Q

What’s the ratio of NADP+ / NADPH?

A

57/1
(ln this)

56
Q

What’s the E’ of NAD+ / NADH and what does it mean?

A

E’ = -0.275 V

NADH wants to give up e-

57
Q

What is the formula for E’?

A

E’ = E0’ - ( (RT/nF) x ln( [red] / [ox]) )

58
Q

The transfer of a hydride from NADPH to NAD+ is (exergonic or endergonic)?

A

exergonic

59
Q

What’s the ∆G’ of NADPH to NAD+ and what does it mean?

A

∆G’ = -19 kJ/mol

NAD+ is favored and is exergonic (releases energy)