9. Enzyme kinetics Flashcards
define exergonic reaction
a spontaneous reaction
what value of delta G does an exergonic reaction have
negative
describe the speed of exergonic reactions when catalysts aren’t used
very slow
are most chemical reactions exergonic or endergonic
exergonic
what is a catalyst
a substance that speeds up attainment of equilibrium of a chemical reaction
T or F: a catalyst changes the position of the equilibrium when it’s used
false! it only speeds up the process
what happens to a catalyst after a reaction is complete
it is recycled to participate in future reactions
T or F: a catalyst may be temporarily changed during a reaction
true; but it remains unchanged in the overall process
define catalysis
the increase in the rate of a chemical reaction due to the participation of a catalyst
what are biological catalysts
enzymes
what key properties do enzymes have that you don’t see in inorganic catalysts
high degree of substrate specificity, substantial reaction acceleration, and they function in aqueous solutions at physiological pH
why are enzymes control points for metabolism
they can accelerate chemical reactions by a factor of a million or more
what is enzyme kinetics
the study of reaction rates and how they respond to changes in reaction parameters
what letter is rate represented by
V (ie velocity)
in the reaction of S –> P, what does the rate represent
the quantity of S that disappears in a specific unit of time
what are the units of V
M/s
T or F: V (the rate) should equal the rate of appearance of P
true
for a first order reaction, what is the rate formula
V=k[S]
where V=rate
k=rate constant
and [S]=concentration of substrate
what are the units of k for a first order reaction
s^-1
describe the reaction rate (V) in regards to k
higher k value = faster reaction
for a second order reaction, what is the rate formula
V=k[S1][S2]
what are the units of k for a second order reaction
(M^-1)(s^-1)
what are pseudo-first order reactions
a second order that appears as a first when one substrate is present in far excess of the other
what is the rate formula for a pseudo-first order reaction
V=k[S1][S2], but S2 is negligible
how do we plot reaction rates?
[P] vs time
from a graph of product formed over time, what formula do we use to find V
delta P/delta t = V = k[S]
what is V0
initial velocity
how do we find V0 from a [P] vs time graph
calculate the slope at time=0 (tangent line)
describe the slope of V0 depending on different amounts of [S]
at low [S], V0 increases almost linearly with increases in [S]
at higher [S], V0 increases by smaller and smaller increments in response to increases in [S]
what is Vmax (in regards to [S] and V0)
the point where increases in [S] no longer increases V0
what is Vmax (in regards to the reaction)
Vmax is the fastest rate at which it can convert substrate to product. Adding more S doesn’t increase the rate
what are k1, k2, k-1, k-2
k1: E+S –> ES
k2: ES –> E+ P
k-1: ES –> E + S
k2: E+P –> ES
which step in the formation of a product from E and S is the slow step
ES –> E+P
what is another name for the slow step of a reaction
the rate limiting step
what is the M-M equation
V0=Vmax[S]/(Km+[S])
what are the assumptions of the M-M equation
there is an ES complex, each step has distinct rates, very early in the reaction there is no product (so k-2 is negligible), the second step is rate limiting, [Etotal]=[Efree]+[ES], and [ES] is constant (formation=breakdown)
what is Km
the Michaelis constant
what are the units for Km
M
what formula represents Km
(k-1 + k2)/k1
ES breakdown/ES formation
what does Km represent
it summarizes the rates of a reaction
it’s ES breakdown over ES formation
what happens to Km when V0=1/2 Vmax
Km=[S]
how do we plot using the M-M equation
V0 on the y axis, [S] on the x axis
the curve is hyperbolic
T or F: we can use the M-M equation for second order reactions
false; only first order
when does Km=Kd
when k2 is limiting and can be ignored, then Km=(k-1)/k1, which is equivalent to Kd
reminder: what is Kd
dissociation constant of the ES complex
what does a low Km represent
strong binding affinity between E and S
what is another name for the L-B plot
double reciprocal plot
how do we derive the L-B equation
by inverting the M-M equation
what is the L-B equation
1/V0 = (Km/Vmax)(1/[S]) + 1/Vmax
what are the axes for the L-B plot
y: 1/V0
x: 1/[S]
what are the L-B plot intercepts
x=0: 1/Vmax
y=0: -1/Km
what is the benefit of using an L-B plot vs an M-M plot
LB plot allows a better approximation of Vmax (because it’s linear, nit hyperbolic)
what is the turnover number of an enzyme
the number of substrate molecules converted into product by enzyme in a unit of time when the enzyme is fully saturated with substrate
what variable reveals the turnover number of an enzyme
Vmax
what variable is the turnover rate equal to
k2 (which is the rate limiting step)
what is the rate constant k2 often referred to as
kcat
how can you adjust V0=k2[ES] to solve for kcat?
kcat=Vmax/[Etotal]
what does kcat tell us
the amount of product we produce per second
what does 1/kcat tell us
the time each individual reaction takes
what rate constant is called the specificity constant
kcat/Km
what does the specificity constant measure
the catalytic efficiency, because it takes into account both the rate of catalysis with a particular substrate (kcat) and the nature of the enzyme-substrate interaction (Km)
why do you need both kcat and Km for the specificity constant
two enzymes catalyzing different reactions may have the same kcat (turnover number) but the rate of uncatalyzed reactions may be different and this the rate enhancements brought about by the enzymes may differ greatly
what is the equation for kcat and Km
V0=kcat/Km x [Etotal][S]
T or F: there is an upper limit to kcat/Km
true
why is there an upper limit to kcat/Km
it’s imposed by the rate at which E and S can diffuse together in an aqueous solution and form ES
what is the upper limit value for kcat/Km
10^8 to 10^9, in the units M^1s^1
define kinetic perfection
when the catalytic velocity of enzymes is only restricted by the rate at which they encounter substrate in the solution
what are the two types of reversible inhibition
competitive and uncompetitive
in competitive inhibition, where does the inhibitor bind
to the active site
how do we overcome competitive inhibition
by adding more substrate
if we add more substrate in competitive inhibition, will the same Vmax be reached as without the inhibitor?
yes
in competitive inhibition, since the [S] needs to be increased to reach Vmax, what else is increased
the Km at 1/2Vmax
by which factor will Km be increased in competitive inhibition when more S is added
alpha
how does the MM equation change to account for Km being increased by a factor of a (in competitive inhibition)
V0= Vmax[S]/aKm + [S]
on a graph, how can we tell if inhibition is competitive
using the LB plot
what are the signs of competitive inhibition on an LB plot
steeper slope=more inhibitor, more inhibitor=closer x intercept is to zero
how do you overcome uncompetitive inhibition
you can’t. Adding more S has no effect
T or F: adding more S to uncompetitive inhibition will overcome it
false; adding more S has no effect
why can’t we overcome uncompetitive inhibition by adding more S
the inhibitor binds at a site distinct from the active site, forming an ESI complex
what value is reduced in uncompetitive inhibition
Vmax
by which factor is Vmax reduced by in uncompetitive inhibition
alpha’
which way does the equilibrium shift in uncompetitive inhibition and why
shifts towards ES, because the inhibitor is removing ES from the population as ESI is formed
in uncompetitive inhibition, what is the result in the equilibrium shifting towards ES
it makes the enzyme appear as if it has a higher affinity for S than it really does
in uncompetitive inhibition, what else (other than Vmax) is reduced (and why which factor)
[S] required to reach Vmax is reduced by alpha’
in uncompetitive inhibition, what effect does the reduced [S] and Vmax have
we have a decreased Km^app value
how does uncompetitive inhibition appear on an LB plot
parallel lines
why are the lines on an LB plot parallel when there’s uncompetitive inhibition
each line has an equally reduced slope, so all lines have the same slope = parallel
what happens to the intercepts on an LB plot of uncompetitive inhibition when inhibition is increased
absolute value of the intercepts increases (ie they are more up and to the left)
