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Biochemistry II > Jim Reid > Flashcards

Jim Reid Flashcards

1
Q

what is the formula to find t in a unimolecular irreversible reaction with the [A] at the start and a given time
eg. time till 95% completion

A

ln([A]t / [A]0) / -k
eg. ln(0.05) / -k

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2
Q

what order are unimolecular reactions always

A

first, there is only on concentration term

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3
Q

how do you work out the ratio of [A]eq : [B]eq in unimolecular reversible reactions

A

1 : k
at eq. forwards and reverse k is the same

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4
Q

what is the formula to work out [B] at a given time in an unimolecular irreversible reaction

A

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

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5
Q

headings of the table of the numerical approach unimolecular irreversible

A

t, [A], rate of change of [A] in uMs-1, d[A] in whatever time interval you want

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6
Q

what is the formula for [A] at a given time in an unimolecular irreversible reaction

A

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

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7
Q

in a unimolecular reversible reaction what is the average time A
aka lifetime

A

A = 1/k1

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8
Q

in a unimolecular reversible reaction what is the average time B

A

B = 1/k-1

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9
Q

what is the proportions of [A] and [B] at equilibrium

A

[A] = 1/1+K, [B] = K/1+K

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10
Q

what does kobs =

A

k+1 + k-1
observed rate

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11
Q

how to work out [B] from [A]0 and [A]T in unimolecular reversible

A

[B] = [A]0 - [A]T

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12
Q

what is kobs in unimolecular reversible (theory)

A

This is the only term with time dependence it will be a term between 1 and 0 determining how much of the reaction has taken place

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13
Q

for a question such as How long will it take for the reaction to reach 95% completion? for a unimolecular reversible reaction what equation would you use

A

t = ln(x)/-kobs
derived from x = e^-kobs

in this case x = 0.05 because that’s the amount of reaction which would be remaining aka [A]t/[A]0

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14
Q

equation for the time depenance of bimolecular irreversible at equilibrium

A

d[A]/dt = -k+1[A][B] + k-1[C] = 0

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15
Q

fraction of protein bound to ligand equation

A

[PL]/[P]T = [L]/[L]+Kd

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16
Q

what does [A]t = in bimolecular irreversible

A

[A]0 x e^kobst
where kobs = k[B]
when [B]»[A]

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17
Q

rate of change

A

Δ[A] / Δt = -k[A]

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18
Q

definition of rate

A

change in concentration over time

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19
Q

Rate constant

A

coefficient of proportionality relating the rate of a chemical reaction at a given temperature to the concentration of reactant (in a unimolecular reaction) or to the product of the concentrations of reactants.

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20
Q

what is 1/k1 (explanation in words)

A

average lifetime in state A

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21
Q

what is the general rule for the left side of the rate of change equation

A

sum of the rate terms eg.
-k[A] in irreversible
-k1[A] + k-2[B] in reversible
-k[A][B] in bimol. irreversible
-k1[A][B] + k-1[C] bimol. reversible

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22
Q

what is the average time in B in a reversible reaction

A

1/k-1 (backwards rate)

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23
Q

what is a sense check for ratio/proportion calculations(reversible)

A

B should be higher because forward rate should be higher (check the question for this)
The numbers should add up the the original concentration

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24
Q

what needs to assumed (or done to experimental conditions) to use in analytical approach in bimolecular irreversible

A

that [A]»[B] so it is pseudo first order
so we can treat [B] as if it doesn’t change in the reaction

25
Q

how to work out table in numerical approach

A

to get the next conc in the table:
times starting conc by -k, this gives change over time in moles per second, then get rid of the s by multiplying by the time interval

26
Q

how can a simple enzyme (E+S <–> ES –> E + P) be written if k-1&raquo_space; k+2

A

E+S <–> ES (pseudo first order?)

27
Q

simple enzyme - product formation is driven by k+2 - what is this equation?

A

d[P]/dt = k+2[ES]

28
Q
A
29
Q

Km relation to simple enzymes with k-1&raquo_space; k+2 assumption

A

[E]eq[s]eq/[ES]eq = k-1/k+1 = Km

30
Q

dissociation constant from protein bound to ligand (bimolecular reversible?) -in terms of concentrations

A

Kd = [P][L]/[PL]

31
Q

equation for what fraction of protein is bound to ligand

A

[PL]/[P]total = [L]/[L]+Kd

32
Q

what assumption/experimental set up can we make to know [L]

A

total ligand = free ligand + bound ligand
[P]total &laquo_space;[L]total so [PL] «[L]total
therefore [L] total is roughly the same as [L] free
you can then rewrite the equation replacing [L] with [L]total
[PL]/[P]total = [L]t/[L]t+Kd

33
Q

when does the [P]total &laquo_space;[L]total assumption fail

A

when you need comparable amounts of protein and ligand
for the assumption to work ligand should be 10 fold

34
Q

how do we get the constants from a plot (bimolecular reversible)
what is along the x and y axis

A

y intercept = k-1
gradient = k+1

x = [B] y= kobs

35
Q

what does time dependence mean

A

rate (constant)
d[x]/dt = …

36
Q

what is the difference between Kd and Km

A

Kd is the concentration of ligand or substrate at which half of the receptors or enzymes are bound (thermodynamics). Km is the concentration of substrate at which the reaction rate is half of the maximum rate (kinetics)

37
Q

how to work out fraction of protein with ligand bound when [P]total &laquo_space;[L]total

A

[PL]/[P]total = [L]t/[L]t+Kd

38
Q

how to work out concentration of enzyme substrate complex when k-1&raquo_space; k+2
what can you do with this to work out the overall initial rate (Vi) of the enzyme

A

[ES] = [E]total x [S] / [S]+Km

times [ES] by k+2/kcat

39
Q

what is kcat of an enzyme

A

kcat=k+2

40
Q

Vmax

A

Vmax = kcat[E]total

41
Q

equation for the initial rate, simple enzyme

A

initial rate Vi = Kcat[E]t x [S] / [S] + Km

or Kcat[E] can be replaced with Vmax

42
Q

what is Vmax/2 on an enzymatic plot (initial rate against [S])

A

where [S] is equal to the Km

43
Q

what is initial rate Vi of an enzyme (concept)

A

d[p]/dt

44
Q

what is the rate order of Michalis Menton

A

it is undefined the rate is not a simple product of substrate concentration
but for high substrate conc it is first order
low [s] is second order

45
Q

what is the equation for initial rate Vi of an enzyme at [S]«Km

A

Vi = Kcat/Km * [E]t[S]

46
Q

what is the steady state

A

the [ES] complex is made pretty much as rapidly as it falls apart
d[ES]/dt = ~0

47
Q

equation for Km in the steady state

A

km = k-1 + k+2 / k+1

48
Q

what is the presteady state

A

when ES is being made, there are non-linear changes in E and P concentration

49
Q

which approaches work best in the presteady/steady states

A

presteady - exponential approaches
steady - Michalis Menton

50
Q

how to do numerical method

A

multiply the [A] by the rate and make it negative, then times this by the time interval to get the amount of A lost over that time period, repeat

51
Q

what does kobs = in bimolecular irreversible (assuming pseudo first order)
why?

A

k1[B]
[B] no longer changes - become part of the constant

52
Q

equilibrium constants for different conditions

A

unimolecular reversible - Keq
bimolecular reversible - Kd
simple enzymes - Km

53
Q

what is the equation for initial rate for simple enzyme when [S] = Km

A

Vmax/2

54
Q

what is the equation for initial rate Vi of an enzyme at [S]»Km

A

Vi = Vmax
which also equals kcat[E]t

55
Q

what is the rate of formation of [ES] in the steady state

A

k+1[E][S]

56
Q

what is the rate of loss of [ES] in the steady state

A

k-1[ES] + k+2[ES]

57
Q

Km in the steady state (k-1 and k+2 are comparable)

A

[E][S]/[ES] or k-1+k2/k+1

58
Q

steady state d=

A

d[ES]/dt ~=~ 0

Biochemistry II (8 decks)

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