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MBB261 > Craven > Flashcards

Craven Flashcards

1
Q

What is the equation for a unimolecular irreversible reaction?

A

k

A —–> B

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2
Q

In a unimolecular irreversible system, why do molecules last in state A for diff amounts of time?

A
  • undergo diff collisions

- stochastic process

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3
Q

In a unimolecular irreversible system, what is the average amount of time spent in A?

A
  • DIAG*

- 1 / k

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4
Q

In a unimolecular irreversible system what is the rate of change of no. molecules in state A?

A
  • -kNA

- where NA = no. molecules in state A

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5
Q

In a unimolecular irreversible system what is the rate of change of [A]?

A
  • -k[A]
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6
Q

What order is the rate constant in a unimolecular irreversible system?

A
  • 1st order
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7
Q

What is the numerical solution of rate equation, for unimolecular irreversible systems?

A
  • set of values of [A] at set of times for particular values (discrete)
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8
Q

Why is the numerical solution always slightly approx compared to analytical solution, and how can it be made more accurate?

A
  • assumes ROC of [A] constant during whole timestep

- use smaller timestep

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9
Q

What is the analytical solution of the rate equation for unimolecular irreversible systems, and how is it calc?

A
  • works out equation for [A] as function of time

- [A] = [A]0e^-kt

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10
Q

What are the advantages of the analytical solution of the rate equation, for unimolecular irreversible systems?

A
  • works for any [A]0, k and t

- exact

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11
Q

What are the disadvantages of the analytical solution of the rate equation, for unimolecular irreversible systems?

A
  • need to know lots of maths

- only poss to find in simple cases

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12
Q

What are the advantages of the numerical solution of the rate equation, for unimolecular irreversible systems?

A
  • maths easy

- totally general, can apply to v complex biological models

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13
Q

What is the disadvantage of the numerical solution, for unimolecular irreversible systems?

A
  • slightly approx
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14
Q

What is the [A] at t=0, and why?

A
  • [A] = [A]0e^kx0

- ∴ [A] = [A]0

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15
Q

What is the [A] if let t become v big?

A
  • kt v big
  • e^kt v big
  • so e^-kt = 0
  • [A] = 0 (all A used up)
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16
Q

How is half life calc?

A
  • 0.693/k
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17
Q

How can [B] be calc in terms of [A]?

A
  • [B] = [A]0 - [A]0e^-kt

- [B] = [A]0 (1-e^-kt)

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18
Q

What is the equation for a biomolecular irreversible reaction?

A

k

- A + B —–> C

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19
Q

What order is the rate constant in a biomolecular irreversible system?

A
  • 2nd order
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20
Q

In a biomolecular irreversible system, what is the average time spent in A?

A
  • 1 / k[B]
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21
Q

In a biomolecular irreversible system, how does [A] relate to [B]?

A
  • from perspective of A, rate of making collisions w/ B is dep on [B]
  • in ideal solution rate directly proportional to to [B]
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22
Q

In a biomolecular irreversible system, how is the rate of change of [C] calc?

A
  • k[A][B]
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23
Q

What is the diffusion controlled limit in biomolecular irreversible systems?

A
  • typically most collisions unsuccessful, but collision can occur straight away
  • if [B] = 1mM, time for collision ≈1μs
  • av time = 1 / k[B]
  • so 1μs = 1 / k x 1m
    10^-6s = 1 / k x 10^3M
    k = 10^9M^-1s^-1
  • approx largest value of 2nd order rate constant, so reaction can’t go faster than this
  • but k generally a lot smaller
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24
Q

Why are calcs of reaction time courses much more complex for biomolecular irreversible system than unimolecular?

A
  • likelihood of A reacting (in next moment of time) changes as B used up
  • or opp true
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25
Q

When and why can biomolecular irreversible systems have a pseudo 1st order rate constant?

A
  • when [B]0&raquo_space; [A]0
  • eg. [A]0 = 1μm and [B]0 = 1000μm
  • after long time A –> 0 and B –> 999μm
  • conc of B only drops small fraction, so can be treat as constant
  • k[B] = k’
  • so rate of change of A = -k’[A]
  • [A] = [A]0e^-k’t
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26
Q

What is the equation for a unimolecular reversible reaction?

A

k1
- A ⇌ B
k-1

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27
Q

In a unimolecular reversible system, what is the average time a molecule spends in state A and B?

A
  • A = 1 / k1

- B = 1 / k-1

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28
Q

What is a single molecule time course for a unimolecular reversible system?

A
  • DIAG*

- stochastic process

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29
Q

How is the eq constant, K calc for a unimolecular reversible system?

A 
- K = time spent in B / time spent in A 
      = k1 / k-1
- at any 1 moment in time
  K = no. molecules in B / no. molecules in A 
     = k1 / k-1
     = [B]eq / [A]eq
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30
Q

In a unimolecular reversible system, how are the rates of A–>B and B–>A calc?

A
  • A –> B = k1 x NA

- B –> A = k-1 x NB

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31
Q

How does K allow us to calc [A] and [B] at eq as a ratio and a fraction?

A
  • [A]eq : [B]eq
    1 : K
    1 / 1+K : K / 1+K
  • [A]eq = (1 / 1+K) x total conc A
  • [B]eq = (K / 1+K) x total conc B
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32
Q

Which way up is K defined in A ⇌ B?

A
  • conventionally K = [RHS]eq / [LHS]eq = [B]eq / [A]eq
  • but poss to define K as [A]eq / [B]eq and get reciprocal value
  • so always define which way up you calc K as
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33
Q

What is the equation for a biomolecular reversible reaction?

A

k1
- A + B ⇌ AB
k-1

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34
Q

How are biomolecular reversible systems v important to mol bio?

A
  • drug based medicine
  • receptors
  • TFs
  • enz + substrate/inhibitor
  • protein 1 + protein 2 ⇌ complex
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35
Q

How can biomolecular reversible systems be quantified?

A
  • assess how strong interaction is –> mutate residue and see how strength of interaction changes or mod drug molecules
  • predict whether 2 molecules will bind significantly under conditions of known total concs of basic constituents
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36
Q

What order are the rate constants in a biomolecular reversible system?

A
  • k1 (kon) is 2nd order

- k-1 (koff) is 1st order

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37
Q

What is KD and what kind of units does it have?

A
  • dissoc constant

- units of conc

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38
Q

What can be defined as an alt to KD, and how is it calc?

A
  • KA (assoc constant) = reciprocal of KD
  • KA = [AB]eq / [A]eq[B]eq
    = k1 / k-1
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39
Q

How is rate of AB formation in a biomolecular reversible system calc?

A
  • k1[A][B]
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40
Q

How rate of loss of AB in a biomolecular reversible system calc?

A
  • k-1[AB]
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41
Q

How is KD calc?

A
  • at eq k1[A]eq[B]eq = k-1[AB]eq
  • ∴ KD = [A]eq[B]eq / [AB]eq
    = k-1 / k1
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42
Q

How does [B] affect behaviour of a single molecule in a biomolecular reversible system?

A
  • DIAG*
  • low [B] means spends more time in A than AB
  • higher [B] means time in A shorter, but time in AB same on av
  • at saturated level of B, time in A only visible as vertical lines
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43
Q

For higher affinity (favour binding), what value should KD and its components have?

A
  • small KD

- small k-1 (koff) / big k1 (kon)

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44
Q

How can KD be measured simply?

A
  • measure when all A in free state, fluorescence = 250
  • measure when all A bound as AB, fluorescence = 100 (A bound to B doesn’t fluoresce as well)
  • then measure amount of B need to add to get to fluorescence of 175 (50 + 125) –> when half A free and half bound as AB
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45
Q

How can it be useful to rearrange [AB]eq / [A]eq[B]eq = 1 / KD?

A
  • [AB]eq / [A]eq = [B]eq / KD
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46
Q

What is the rate equation for P + L ⇌ PL?

A
  • [PL]eq / [P]eq = [L[eq / KD
47
Q

How can it be shown that if conc of free ligand is equal to KD, then half receptor molecules will be bound?

A
  • if [L]eq = KD, then [PL]eq / [P]eq = 1
    ∴ [PL]eq = [P]eq
    ∴ 50% bound
48
Q

In a biomolecular reversible system, if [P]tot &laquo_space;KD, then how can the fraction of ligand bound be calc?

A
  • then [L] ≈ [L]tot

- fraction bound = [L]tot / [L]tot + KD

49
Q

In a biomolecular reversible system, if we assume [P]tot &laquo_space;KD and know [P] ≤ [P]tot, what can we calc about KD?

A
  • ∴ [P] &laquo_space;KD

- ∴ [PL] / [L] = [P] / KD = <

50
Q

In a biomolecular reversible system, how can you generalise formula for fraction of P bound, given values of [L]tot and KD - as ratios and fractions?

A
  • as ratios –> [PL] : [P]
    [L] : KD
  • as fractions –> [L] / [L] + KD
    –> KD / [L] + KD
51
Q

What is the equation for a simple enzyme reaction?

A

k1 k2
- E + S ⇌ ES —-> E + P
k-1

52
Q

What order are the rate constants in a simple enzyme system?

A
  • k1 is 2nd order

- k-1 and k2 are 1st order

53
Q

Why can’t you talk about 1st and 2nd order rate constants together?

A
  • completely diff units
54
Q

For a simple enzyme system what would the time course of a single molecule look like if k-1&raquo_space; k2, and what would be the times spent in E and ES?

A
  • DIAG*
  • av time in E = 1 / k1 [S]
  • av sum of time in ES = 1 / k2
55
Q

For a simple enzyme system, how does the time course of a single molecule change if [S] is increased?

A
  • DIAG*
  • av time in E less
  • so product released after shorter time periods
56
Q

For a simple enzyme system, how does the time course for a single molecule change if [S] is v v high, so never waiting for S?

A
  • DIAG*
  • product released after v short periods of time
  • ≈ 1 / k2
57
Q

What does the Michaelis-Menten equation tell us?

A
  • how rapidly product will be formed for given substrate conc
58
Q

In a simple enzyme system, if Stot&raquo_space; Etot and k2 = 0, then how can [ES] be calc?

A
  • no ES –> E + P occurring
  • so at eq KD = k-1/k1 = [E]eq[S]eq / [ES]eq
  • ∴ [ES] = Etot x [S] / [S] + KD
59
Q

How will [ES] change if k2 &laquo_space;k-1 (instead of k2 = 0)?

A
  • barely alters [ES}
  • ES –> E + P is 1st order unimolecular so ROF of product = k2[ES]
    = k2 (Etot[S] / [S] + KD)
60
Q

What is the Michaelis-Menten equation, and how is this derived from the case when k2 &laquo_space;k-1?

A
  • rate of P formation = kcat x Etot x [S] / [S] + Km

- if k2 &laquo_space;k-1, then k2 = kcat and KD = Km

61
Q

What is kcat?

A
  • max rate enzyme can form product if never waiting for substrate
62
Q

What is Vmax?

A
  • in a particular case (cell, experiment) the total rate of reaction = kcat x Etot
  • this is often called Vmax, ie. fastest rate of production of P
63
Q

How do kcat and Vmax differ?

A
  • kcat is a fundamental property of particular enzyme

- Vmax is specific to particular case or experiment

64
Q

What does Km tell us?

A
  • parameter that tells us how high [S] needs to be to get supply of substrate to each enzyme molecule sufficiently high
65
Q

What does it mean if [S] = Km?

A
  • enzyme molecules “idle” half the time
66
Q

What does it mean if [S]&raquo_space; Km?

A
  • enzyme molecules able to be loaded w/ substrate all the time
67
Q

What is the effect on [ES] if k2 NOT &laquo_space;k-1?

A
  • S in principle changing v slowly = steady state

- ie. if pu in more S to maintain its value, then stays constant, so ES stays constant

68
Q

At steady state what is the rate of ES formation?

A
  • k1[E][S]
69
Q

At steady state what is the rate of ES loss?

A
  • k-1[ES] + k2[ES]

= (k-1 + k2) [ES]

70
Q

At steady state how do rate of ES formation and loss relate, and what does this mean for KD?

A
  • they are equal
  • ie. k1[E][S] = (k-1 + k2) [ES]
  • ∴ [E][S] / [ES] = k-1 + k2 / k1 = Km
  • [E][S] / [ES] also = k-1 / k1 = KD
    ∴ Km equivalent to KD
71
Q

How does [S], [E], [ES] and [P] change over a reaction?

A
  • DIAG*
  • when rof of P = 0, is pre steady state
  • initial drop in [S] as ES formed, then decreases at same rate as [P] increases
  • after steady state, as S used up, ES will drop and E will rise
72
Q

In a metabolic pathway, how are ideas of flux and steady state transfer important?

A
  • allow calc of formation/loss of compounds
  • ie. values coming in must equal those going out
  • eg. 10 5
    W ⇌ X ⇌ Y
    7 ?
    ? = (7+5) - 10 = 2
73
Q

What is the problem w/ representing a simple enzyme as E + S ⇌ ES –> E + P, and how can it be imporved?

A
  • implies P falls off immediately after reaction occurs, but P likely to be close chemical derivative of S, so likely to bind E fairly well
    ∴ E + S ⇌ ES –> EP –> E + P
  • but P –> S also cat by active site, so EP –> ES may be signif
    ∴ E + S ⇌ ES ⇌ EP –> E + P
  • if [P] becomes too high have to inc rebinding of P
    ∴ k1 k2 k3
    E + S ⇌ ES ⇌ EP ⇌ E + P
    k-1 k-2 k-3
74
Q

Is it easy to look at steady state kinetics?

A
  • at steady state all systems obey Michaelis-Menten

- but kcat and Km will be complicated combo of rate constants

75
Q

What is pre steady state kinetics?

A
  • looking at early stages after mixing to try to unpick diff systems
76
Q

In what instances can binding alter behaviour of molecules?

A
  • how strongly can bind another molecule
  • how effective they are as enzymes
  • how open they are as enzymes
  • how open they are as ion channels
  • how strong signal they relay is as receptors
77
Q

When might enzymes bind more than 1 molecule?

A
  • inhibition
  • enz that binds 2 substates (eg. kinase binds ATP and protein substrate)
  • Hb (binds 4xO)
  • ligand gated ion channels can be, eg. pentameric and bind 5 ligands
78
Q

What concepts does binding of more than 1 diff ligand bring in?

A
  • independence
  • cooperativity (+ve or -ve)
  • competition
  • mutually exclusive binding (eg. comp inhibitor)
  • allostery
  • orthostery
79
Q

What concepts does binding of more than 1 of same ligand bring in?

A
  • cooperativity
  • hill coefficient (Hb)
  • “sharp switching” (eg. ion channels)
  • allostery (as mechanism for cooperativity
80
Q

What kind of direct ligand interaction takes place if 2 sites distant and binding 1 ligand doesn’t affect other site?

A
  • DIAG*

- independent

81
Q

What kind of direct ligand interactions can take place if 2 sites close and binding 1 ligand doesn’t affect other site?

A
  • DIAG*
  • if no particular clash between ligands = independent
  • if favourable interaction, eg. charge-charge interaction = +vely cooperative
  • if unfavourable interaction, eg. charge-charge repulsion = -vely cooperative
82
Q

What kind of direct ligand interaction takes place if both ligands bind site and no poss of simultaneous occupation of site?

A
  • DIAG*

- mutually exclusive (=competitive)

83
Q

What is the problem w/ representing proteins as solid when looking at binding of ligands?

A
  • they are dynamic and flex

- binding 1 site may affect other site even w/o direct interaction between ligands –> allosteric conformational change

84
Q

What are orthosteric and allosteric sites?

A
  • orthosteric = where substrate or main receptor ligand binds
  • allosteric = where some 2nd ligand binds
  • not fixed terms, like in P + L, ligand could be protein (not always clear distinction)
85
Q

How does +ve cooperativity happen via allosteric conformational change of protein?

A
  • DIAG*
  • when one ligand bound, changes in both binding sites occur and that ligands binding site it drawn “closer to optimal” and other site “looks better”
  • when both bind, further improvements in binding sites occur
86
Q

How does value of KD change if 2 ligands are +vely cooperative?

A
  • KD lower when binding site of other ligand occupied, as can bind to protein better
87
Q

How does the value of KD change if 2 ligands bind protein independently?

A
  • same as if only 1 ligand present
88
Q

How does value of KD change if 2 ligands are -vely cooperative?

A
  • KD higher when binding site of other ligand occupied, as can’t bind protein as well
89
Q

How does value of KD change if 2 ligands are competitive, and how could this be overcome?

A
  • v v high KD

- if put enough of other ligand in

90
Q

If binding of 1 ligand improves binding of other, is the opp true?

A
  • yes, symmetrical
  • must alt KD by same factor
  • also true of -ve cooperativity
91
Q

What is the binding curve for Hb and Mb?

A
  • DIAG*
  • Mb like normal binding curve
  • Hb curve is a sigmoid
92
Q

What is the role of Hb and Mb?

A
  • Hb is oxygen carrier in blood

- Mb is oxgen carrier in muscles

93
Q

What is the mol reason for the behaviour of Hb (sigmoidal shape)?

A
  • “unit” of Hb can bind 4 molecules of O
  • binding not indep
  • binding +vely cooperative
94
Q

How can binding of molecules affect each other in a system where protein can bind 2 molecules of same ligand, and what are the effects on KD?

A
  • DIAG*
  • independent = binding as good whether or not site occupied, same KD applies whether or not other site occupied
  • +ve cooperativity = binding stronger if other site occupied (allosteric effect), KD smaller for binding when 2nd site full
95
Q

Why does +ve cooperativity make the binding curve sigmoidal?

A
  • DIAG*
  • at low conc of 1 ligand, other site likely empty
  • at high conc of 1 ligand, other site likely full
  • curve between those for if 2nd site always empty and if 2nd site always full
96
Q

In Hb how does binding of molecules affect strength of other molecules binding, and what is the effect on KD?

A
  • binds 4 molecules much better than 1/2/3

- KD only decreased when all 4 bind

97
Q

What is the Hill equation, and when does it apply?

A
  • for systems capable of binding Nsites molecules of ligand

- av no. sites bound = Nsites x ( [L] / [L]^nh x [KD]^nh

98
Q

What is the value of nh when there is indep binding and for v extreme +ve cooperativity?

A
  • indep = 1

- +ve cooperativity = Nsites (data exactly follows Hill equation)

99
Q

What curve do real systems w/ +ve cooperativity typically have for the Hill equation?

A
  • approx Hill form, w/ nh somewhere between 1 and Nsites
100
Q

What is the Hill equation useful for?

A
  • good way to approx characterise binding curve showing sign s of +ve cooperativity
  • as full analysis would need consideration of all indiv KDs, but most data wouldn’t be precise enough to trust this detailed analysis
101
Q

How does binding curve vary for diff values of Hill coefficient?

A
  • higher = sharper switching

* DIAG*

102
Q

How can binding curve for Hb and Mb be explained in muscle capillaries?

A
  • Hb has low affinity for O at low concs of external O
  • so O dissociates easily at low ambient O
  • does not happen in case of Mb, so Mb will bind O released from Hb
103
Q

How can binding curve for Hb and Mb be explained in the lungs?

A
  • Hb has high affinity for O at high concs of external O

- so O will bind easily at high ambient O

104
Q

How can competitive inhibition be overcome?

A
  • if make conc of substrate high enough, can have virtually all sites full at any 1 moment
  • so regain full maximal rate
105
Q

What is Ki (inhibition constant), and when does it apply?

A
  • dissoc constant for binding of inhibitor in absence of substrate
  • applies in competitive an allosteric case
106
Q

How does competitive inhibition affect Km and kcat?

A
  • Km increases

- kcat unchanged

107
Q

In an enzyme catalysed reaction how does response (rate of observed reaction) relate to the fraction of E sites which are filled at any 1 time?

A
  • directly prop (linear)
108
Q

How can allosteric inhibition cause an inhibitory response in enzyme?

A
  • DIAG*
  • S binding site diff when I site occupied
  • enz may hold onto sub worse when I occupied
  • enz may take longer to achieve chem reaction when I site occupied
109
Q

How can allosteric inhibition cause an activatory response in enzyme?

A
  • DIAG*
  • debatable whether this should be labelled as inhibitor
  • S binding site diff when I site occupied
  • enz may hold onto sub better when I occupied
  • enz may take less time to achieve chem reaction when I site occupied
110
Q

What ligands can a receptor bind?

A
  • binds ligand that nature intended to set off signal (orthosteric/endogenous agonist)
  • another ligand can compete for same binding site or bind at diff site (enhance/inhibit activity)
111
Q

In a receptor what is ligand conc related to?

A
  • signalling related to ligand conc

- fractional occupancy at receptor related to ligand conc

112
Q

In what way is the response in a cell complex for receptor-ligand interactions?

A
  • cellular response may or may not be directly prop to fraction of receptor sites filled
  • for some responses, as long as few filled then signalling will occur, and after certain point increasing no. won’t make any diff to signalling strength
  • alt if binding has to stop a signal firing, then need virtually all filled
113
Q

What parallels can be drawn between receptor binding and enzyme cat reactions?

A
  • strongest parallel is for concepts of occupancy
  • for concept of response, receptors in cellular contexts lead to more complex ideas –> still somewhat loosely relate response to occupancy
114
Q

How can antagonists affect agonists in receptor binding?

A
  • may alt binding of endogenous agonist
  • may alt signalling ability of endogenous agonist
  • might compete for same site as endogenous agonist

MBB261 (6 decks)

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