Enzyme assay and detection Flashcards
What is rate of reaction
change in [substrate] or [product] per unit time (Mmin^1 or molL^-1min^-1)
- conc of substrate used OR produce formed/time
- negative sslope of the curve ([S] vs time)
- initial reaction rate given by v0 (directly propeortional to concentration of enzyme
enzyme activity
- moles of substrate converted per unit time
- rate x volume
- represents the quantity of enzyme present
- usually measured in 1 enzyme unit (1 umol/min ie 10^-6 mol/min
** should not change as you purify the protein**
- realistically, it can slgihtly go down, might lose some enzyme during purification
what is specific activity
- activity / mass of enzyme (activity per unit mass of TOTAL PROTEIN)
- moles of substrate converted per unit time per unit mass,
- measured in Katal (1/kg) or umol/min*mg or umol/min*ug
- when 2 different pure enzymes are compared, specific activity is a measure of the enzyme purity
ex: pure enzyme has SA of 10umol/min*mg, sample iwth a SA of 2 umol/min*mg is 20% pure, less pure sample includes impurities making up 80% of the mass)
What is molar activity?
- moles of substrate converted per unit time per mole of enzyme (1/min)
- molar activity= specific activity x molar mass enzyme (note unit conversions)
- molar activity is equal to the turnover number the number of catalytic reaction cycles per molecule of enzyme/sec (ratio of moles to mol will be same)
An enzyme is purified from a complex mixture of proteins, what usually happens to the total amount of peotein and the specific activity during purification
-total amount of peotein decreases and specific activity increases
* note once enzyme is pure SA should stay same
**try lec slides 3 pg 27
What is kinetics
- mathematical analysis of how reaction rate varies as a function of reactant concentration
- plot substrate or product concentration over period of time (this is a progressive curve
- enzyme reaction does not follor a simple rate law
How is analysis of enzyme reaction performed
- 2 steps (binding and catalysis)
- enzyme is recylced in rxn [E] does not change as rxn proceeds
Who are michaelis and menten
- performed systematic experimental verification by Micharlis and Menten of what was discovered by Henri in 1905
- decided to perform calcualtion at time 0 meansing [product] =0, therefore reaction at step 2 can be ignored
- analysis of enzyme reactions
What is the turnover number
- number of times per seconrd that the enzyme completes a reaction cycle
- Specific activity x molar mass
- also equal to molar activity
- represented by K2
What is the michaelis-menten equation
derived from: E + S ⇌ ES → E + P
- graph for enzymes is different bc happenes in two steps
V0= Vmax [s]/(Km + [S]
- must be derived from values we know: [Etotal], [S], k’s,
*must solve for ES, assume rxn at steady state rate formation =rate consumption
*dervied at time 0
What is the equation for the initial rate of reaction
V0= K2[ES]
- note concentration of ES cannot be determined, unstable intermediate
What does Vmax tell about the enzyme
- every substrate needs to bind to enzyme to be converted to product, will come to maxiumum rate when every enzyme bound to substrate
- at high [S] all enzymes are bound to substrate
Vmax is a pseudo constant, only constant if amount of enzyme is fixed
Vmax= K2[Etot] : Etot= ES (all eznyme bound to substrate)
- K2 is true constant, turnover number of enzyme
- iF Vmax a > Vmax b (a has faster catalytic rate
*note: vmax on graph is an estimation
What is Km
- concentration os substrate at which V0 is equal to 50% of the maxiumum rate
- low Km indicates that eenzyme bonds to substrate well; less [S] is needed to occupy enzyme (enzyme able to reach half Vmax at half substrate concentration)
- high Km indicates that enzyme binds and utilizes substrate poorly; more [S] is needed to occupy the volume
- at V0=0.5 Vman, [S]=Km
*tells about ability to bind and recognize substrate molecules
enzyme mutated lys replaced with ala, Vmax is same as original non mutated but Km is 50 fold higher what does this mean
- Vmax tells about caltyltic activity
- Km tells about binding ability
- lys must have been important for binding (higher Km means worse at binding to substrate)
** Km getting higher is bas, less efficient
What is percentage of Vmax is an enzyme functioning when the substrate concentration is 25% of Km
V0/Vmax =[S]/Km + [S]
- 20%
What is linear transformation
- convert set of data to obtain straight line graph (give more percise values from slopes and intercepts to find Vmax and Ka)
- take reciporcal of both sides of michaelis menten equation
- 1/V0 = y and 1/[S] =x
- slope= Km/Vmax
- y-int= 1/Vmax
- X= -1/Km (-yintercept/slope)
- obtain Km by extending the graph onto negative axis
What is competitive inhibition
- affects ability to bind substrate
- arises when inhibitor can only bind to unoccupied enzyme
- inhibitor binding is governed by equilibirum constant K
Ki= ( [E] [I] )/ [EI]
- formation of EI complex means less E availbale to bind to substrate
- inhibitor and substrate compete for availble enzyme - hight [S] can OVERCOME competitive inhibitor
*is enzyme already bound to substrate, inhibitor cannot bind*
*EI is an unproductive complex*
What is non competitive inhibition
affects catalytic rate (mixed inhibiiton)
- arises when inhibitor can bind to BOTH E and ES
- governed by equilibrium constant Ki
- formation on ESI and EI means less ES to unergo catalysis, but substrate can still bind to EI without yielding product
- inhibitor binding site is different from substrate binding site (hence they can both bind)
- bound inhibitor may disorganize the catalytic component of enzyme
- if EI and EIS steps each have a different Ki, get mixed inhibiton
*No rxn like inactivators just binding stopping rxn
How does competitive inhibitor effect rate rxn graph
- no effect on Vmax, but apparent Km is increased
Km’ = Km_( 1+ [I] / Ki)_ (inhibition factor)
- inhibition factor = 1 + [I]/Ki (increases as [I] increases)
- if [I] is set equal to Ki : Km’ = Km (1+1) = 2Km
- Ki =concentration of inhibitor that causes Km to double
*need higher substrate concentration to reach Vmax
- competitive inhibition
- no efect on Vmax (all have same Y-intercept)
- x-intercept changes: 1/Km, Km’ increases as [I] increases, so x-intercept gets smaller
- slope increases as [I] increases
*look at magnitude of x-axis not sign*
SLOPE Km’/Vmax (Km increases and vmax same)
How does non competitive inhibition effect graph
- Vmax decreases as [I] increases: V’ max = Vmax / (1 + [I]/Ki0
- y intercept INCREASES : (1/v’max)
- Km unchanged (mixed inhibition may show a small effect
- if [I] is set equal to Ki, inhibition factor = (1+1)=2, Ki is concentration of inhibitor that causes Vmax to halve
- slope INCREASES as [I] increases , LB slope = Km/V’max
Summarize inhibition effects
- Km is increased by multiplying by ( 1 + [I] / Ki )
- Vmax is decreased by dividing by ( 1 + [I]/Ki )
- for MIXED INHIBITION both Vmax and Km change
