Lecture 14: Enzyme-substrate interactions Flashcards
Factors that affect enzyme activity.
- pH due to changing the ionisation state of side chains. Extreme pH can denature the protein- the protein unfolds.
- Denaturing reagents- detergents, for example. These also unfold the protein. Can be used to stop the reaction.
- Temperature- increase rate UNTUL thermal denaturation of the protein occurs.
- Activity is proportional to the amount of enzyme.
- Substrate concentration
- Inhibitors
The concentration of enzymes is usually ………… than the substrate concentration. Why?
less
they are catalysts, they are regenerated.
Simple models of enzyme behaviour assume certain things about their reactions. Give 4 assumptions.
- Only one molecule of one substrate binds to the enzyme- these are very rare reactions
- Enzyme and substrate form an enzyme substrate complex- rate limiting step
- The enzyme converts substrate to product and product release occurs quickly
- Product binds weakly to enzymes
The Van Slyke and Cullen model for enzyme action stated that….
The conversion of [ES] to [E] and [S] was negligible. In other words, K2 was negligible. The reaction is irreversible.
Brown and Henri and Michaelis and Menten assumed that…
E and S combine in a near equilibrium position- therefore the size of K1 and K2 are similar and K3 is much smaller
(state of equilibrium)
Briggs and Haldane said that..
the rate of formation of ES – K1- was equal to the rate of its destruction (K2 + K3). So, K1 = K2 + K3. This is the steady state model.
Draw all three models for the reaction of enzyme with substrate.
Assumption about the steady state model…
- The enzyme concentration is less than the substrate concentration
- The concentration of substrate remains constant
- The enzyme substrate concentration remains the same throughout the course of reaction- the steady state approximation.
- Product binds weakly to the enzyme
- Product does not convert back into substrate- no backwards reaction.
As substrate concentration increases, what happens to rate?
Increases UNTIL saturation is reached
Draw a Michaelis Menten plot. Label Vmax, 1/2 Vmax and Km.
Vmax
the rate when all enzyme active sites are occupied- saturation
Vmax is different for all
enzymes
V =
rate of catalysis of an enzyme
[S]
Substrate concentration
Km
[S] when V= 1/2Vmax. The Michaelis Menten constant. The substrate concentration which gives you half of Vmax for that enzyme
Describe this M-M plot.
- Enzyme A and Enzyme B both happen to have the same Vmax. But the Km of enzyme B is lower than enzyme A. This means that at ½ Vmax, there is a lower substrate concentration needed for enzyme B compared to the substrate concentration for Enzyme A to reach ½ Vmax.
- This means that enzyme A has a lower affinity for its substrate. The substrate does not bind as easily to enzyme A- you need more substrate (a higher Km) for it to reach half Vmax, in comparison to enzyme B. Enzyme B has a higher affinity for its substrate.
- Also, despite the fact they both have the same Vmax, there is a higher substrate concentration in enzyme A in order to reach Vmax.
The Michaelis - Menten equation arises from the general equation for an enzymatic reaction. Give this equation.
Give the Michaelis- Menten equation
Describe the Michaelis-Menten plot for measuring Vmax and Km. How do you plot the graph? What are some limitations? How can we overcome them?
- Direct plot of rate against substrate concentration
- Can be difficult to decide when Vmax is reached. It can also be difficult to measure rates at high substrate concentration.
- Usually requires computer programme to overcome this problem
Draw the Lineweaver-burk plot- show how you measure Vmax and Km? What does the gradient equal? What are some limitations?
- slope = Km/Vmax
- More precise- easier to determine where the straight line intercepts on the axis so it is easier to find Vmax and Km, but it is less accurate- less accurate values of Km and Vmax.
- To create this plot you have to perform graphical transformations so that all data points fall onto a straight line. However, this distorts errors on the data points to different extents. So, you are left with a rubbish estimate of what Km and Vmax is.
Describe the direct linear plot. How do you plot it? How do you find Km and Vmax? Advantages and disadvantages of this plot?
- Plot rate and substrate concentration onto aces
- Connect points
- Intersections are estimates of Km and Vmax
- Gives median values
- The advantage of this plot is that the estimates and Km and Vmax are less likely to be affected by outliers. The disadvantage is that it is difficult to detect deviations from ideal behaviour.
