2. Enzymology Flashcards
(28 cards)
In enzymology, where you are asked to comment on a kinetics graph (e.g. the Lineweaver Burk plot here), it is worth talking about what 2 things
the Vmax and the Km.
Acetylcholinesterase is found at what synapses
cholinergic synapses (synapses where acetylcholine is the neurotransmitter), in both the central and peripheral nervous systems.
what is the role of acetylcholinesterase
terminate the action of acetylcholine by hydrolysing it to choline and acetate.
in this experiment you will use acetylthiocholine and measure the production of thiocholine: why?
This allows the substituted thiol group to react with 5-thio bis-2-nitrobenzoic acid (DTNB) to produce a yellow product, the appearance of which can be measured over time in a spectrophotometer at a wavelength of 412 nm.
To create your Lineweaver Burk graph what do we plot on each axis?
x axis - 1/[Substrate]
y axis - 1/Velocity
Once you have drawn a linear best fit line through the data obtained in the absence of inhibitor, you can repeat the process using the data obtained in the presence of inhibitor, allowing you to add a second line to the same graph.
Using a Lineweaver-Burk plot, how do we determine the value for Vmax
the Y-intercept = 1/Vmax
so Vmax = 1/y intercept
Using a Lineweaver-Burk plot, how do we determine the value for Km
the X intercept = -1/Km
so Km = -1/x intercept
The Lineweaver-Burk plot shows that the Km for the enzyme is altered by the factor (1 + [I] / Ki): what is Ki and what is [I]
[I] refers to the concentration of edrophonium used and Ki is a measure of the affinity of edrophonium for the enzyme aka the inhibitor constant
Ki can be determined from the following equation:
x intercept (minus edrophonium) / x intercept (with edrophonium) = 1 +[I]/Ki
Use this equation to calculate the value of Ki example just 2 read:
To calculate the value of Ki, I worked out the x intercepts of both lines plotted. I did this by looking at the graph.
For with edrophonium, it was -5.76711.
For without edrophonium, it was -10.99176.
Then I substituted these exact values into the equation
1+ [I]/Ki = -10.99176/-5.76711 = 1.905939023
Then, I rearranged the equation, so: 1.905939023 - 1 = [I]/Ki = 0.905939023
We can use this to now work out Ki, as [I] refers to the concentration of edrophonium used, which is 0.15 uM.
0.15/0.905939023 = 0.1655740576 uM
I then converted that to nM, to give me 165.5740576 nM
165nM
Competitive Inhibition
What changes on a Lineweaver-Burk plot?
Km with inhibitor increases (substrate has to compete with inhibitor)
Vmax stays the same (high [S] outcompetes inhibitor)
Plot: Lines intersect at the Y-axis
Slope increases, X-intercept moves left (more negative)
Non-competitive Inhibition
What does the Lineweaver-Burk plot show?
Km stays the same (binding affinity of enzyme for substrate is unchanged)
Vmax decreases (enzyme efficiency is reduced)
Plot: Lines intersect at the X-axis
Y-intercept increases, slope increases
Uncompetitive Inhibition
How is it different on the Lineweaver-Burk plot?
Km decreases (inhibitor stabilizes the ES complex)
Vmax decreases (less product is formed overall)
Plot: Parallel lines (same slope, different intercepts)
Both intercepts move up and left
Edrophonium would be described as a competitive inhibitor of acetylcholinesterase based off the plot I produced. What things would indicate this
- we see that the y-intercept (which represents 1/Vmax) undergoes no change in the presence AND absence of the inhibitor. This suggests that Vmax is not affected by the presence of the inhibitor.
- The Km represents the relationship between the substrate concentration and the rate of reaction. In the presence of edrophonium, the Km is observed to increase: this indicates a reduced affinity of the enzyme for the substrate due to the presence of the inhibitor, as the reaction is slowing down as the solution does not seem to be only saturated with substrate.
Therefore, I can describe it as a competitive inhibitor as the y-intercept stays unchanged and Km increases.
On the other hand, if it were a non-competitive inhibitor, we would expect to see a different pattern: how would Km and Vmax be affected?
where km would be the same, but the Vmax would decrease because the inhibitor would bind to an allosteric site, and lead to a reduction in the number of enzyme active sites availble to bind to the substrate. As a result, the reaction rate would never reach its original maximum even at very high substrate concentrations. The y-intercept would increase in non-competitive inhibition because of the reduction in available active sites, but the x-intercept (−1/Km) stays the same.
It is not a non-competitive inhibitor as the x intercept would stay the same because the active site wouldn’t have the specific shape for the substrate to bind to the active site. However, the y intercept (1 /max) increases as fever active sites are available, this would lead to reaction velocity to decrease, so the graph would have a steeper line.
why do these changes happen in competitive inhibition
In competitive inhibition, the inhibitor competes with the substrate to binding to the enzyme’s active site. Therefore, despite the inhibitor being present, the enzyme is still able to achieve that same Vmax value once enough substrate is available, because the inhibitor can be outcompeted by higher substrate concentrations. This explains why the y-intercept stays the same, as the reaction can still reach its maximum velocity IF provided with enough substrate. Therefore, the maximum rate of reactants can form into our desired products regardless of the inhibitor being present or not.
In competitive inhibition, Km increases because the inhibitor competes with the substrate for binding to the active site, reducing the effective concentration of substrate at the active site. As the substrate concentration increases,, it can outcompete the inhibitor, allowing the reaction rate to reach Vmax, which is why the y-intercept is the same. We can see that x intercept would have a smaller value, instead shifting to the right because the gradient increases.
Edrophonium is used clinically in the diagnosis of myasthenia gravis: how
edrophonium is a reversible acetylcholinesterase inhibitor used in the diagnosis
- In the tension test, edrophonium acts the inhibitor, preventing the breakdown of acetylcholine at the neuromuscular junction. Normally, acetylcholine is broken down by the enzyme acetylcholinesterase, but by inhibiting this enzyme, edrophonium causes acetylcholine levels to rise.
The increased concentration of acetylcholine allows more acetylcholine to bind to available receptors at the neuromuscular junction, temporarily improving muscle strength. This improvement in strength helps confirm the diagnosis of MG, where muscle weakness is primarily due to reduced acetylcholine receptor function.
what is Km
Km = we’re referring to the substrate concentration at which the reaction rate is half of Vmax.
It’s essentially a measure of how much substrate is needed to get the enzyme working at a decent rate - and it’s inversely related to the enzyme’s affinity for the substrate.
what unit should we use for Ki
calculate it in whatever molar unit then convert it to nM
The ester bond in acetylcholine is essential for recognition and hydrolysis by acetylcholinesterase: what if it was replaced by something else e.g. an aromatic ring
Replacing it with an aromatic ring likely makes the molecule non-hydrolysable, meaning it would bind to the active site but not be broken down. So it just sits there, blocking ACh, acting as a competitive inhibitor, where the drug blocks acetylcholine access without being a true substrate.
Why are reversible inhibitors often preferred in clinical settings over irreversible inhibitors?
Reversible inhibitors are temporary and controllable, making them safer for clinical use. In contrast, irreversible inhibitors form covalent bonds, permanently inactivating the enzyme, which can lead to toxic accumulation of acetylcholine and prolonged side effects. Reversible drugs like neostigmine or edrophonium are used because their effects can be quickly adjusted or reversed if needed.
Michaelis-Menten equation: for V (velocity of reaction)
V = Vmax [S] / Km + [S]
Calculation Practice
a) If an enzyme reaction follows Michaelis-Menten kinetics with Km = 0.15 mM and Vmax = 0.085 absorbance units/min, what would be the reaction velocity at a substrate concentration of 0.3 mM?
V = 0.085 x 0.3 / 0.15 + 0.3 = 0.0567abs/min
Formula for new apparent Km with a competitive inhibitor:
New Km = old Km x ( 1 + ([I]/Ki))
