Lecture 10 Flashcards
Why is it important to measure the initial velocity for each enzyme catalysed reaction we measure?
Because as substrate is used up the reaction will slow.
What is the effect on enzyme concentration on reaction rate in excess of substrate.
Initial velocity is proportional to enzyme concentration as long as substrate is in excess (linear increase in reaction speed).
What is the effect on substrate concentration on reaction rate if enzyme concentration is kept constant? Which part is first order? Which is zero order
increases in a linear way at first but as all the active sites become occupied the rate of reaction stops increasing. The first order part is the linear increase, the zero order is the part where it doesn’t increase.
What two variable can be identified on a V vs [S] curve
Vmax (the maximum velocity when enzyme is saturated with substrate)
KM (the substrate concentration at which V = Vmax /2) this is Michaelis constant.
What is the Michaelis-Menten equation? What are the assumptions?
V = Vmax [S] / (KM + [S])
The assumptions are:
1. Enzyme substrate formation and breakdown is in rapid equilibrium, meaning the change in enzyme substrate complex concentration over time is effectively 0
2. [S] is much greater than [E] meaning that at Vmax the enzyme is saturated.
3. Initial rate means [S] does not change significantly
4. initial rate means no product is present.
What is a lineweaver-Burk plot? What are the key points?
A graph with reciprocal values for [S] and reaction rate. The slope is given by KM / Vmaxm the X intercept by -1/KM, the Y intercept is given by 1 /Vmax.
1/V is on the Y axis, 1/[S] is on the X.
It is done because the V/[S] curve is normally hyperbolic.
What are the key points on KM?
Characterises one enzyme-substrate pair (doesn’t rely on [S] but does rely on choice of S).
It is the substrate concentration needed to reach half Vmax and has units of concentration.
KM = (ES -> E+S reaction rate(k-1) + ES -> E + P reaction rate(k2)) / ( E + S -> ES reaction rate (k1))
In many cases k2 can be ignored cause its so small in many enzymes (slow step).
A low KM means a high affinity between E and S while high KM means a low affinity.
What is the physiological significance of KM?
In the cell, for a particular enzyme-substrate reaction, [S] is often below the KM, this allows for rate control to be effective, either by lowering or increasing substrate concentration.
What is kcat?
Vmax/[Et] [Et] = total enzyme concentration. It is the ‘turnover number’ or ‘catalytic rate constant’ it is the moles of substrate converted to product per mole of enzyme per second. The units are s-1.
What is the measure of enzyme efficency?
kcat/KM, the higher it is the greater the efficency (high kcat and low KM).
