Enzyme kinetics - Lecture delivered by Setareh Chong. PDF

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Lecture delivered by Setareh Chong....

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Michaelis-Menten kinetics (cont.)

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The equation above describes the relationship between substrate conc. and reaction rate The graph levels off and the rate can’t increase further bc the enzyme has reached saturation

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Km can also be defined based on the rate constants of the 3 reaction constants (k1, k-1 and k2) E.g. if k-1 and k2 are larger than k1, the ESC collapses either by degrading back to the E + S, or by moving forward to E + P – this is faster than the rate at which ES is made (k1). This makes the Km large. This means you need more substrate to reach ½ of Vmax Whereas, if k1 is large and k-1 and k2 are small, you get a small Km value. This is good bc it means you don’t need as much substrate to reach a high rate As Km becomes small, the rate becomes much steeper, and reaches Vmax much quicker, at lower substrate conc.s Hence…

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*affinity is more complicated than just binding, bc it also depends on rate of breakdown of ES to either E+S or E+P

Lineweaver-Burk plot

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Its easier to transform the data so it can be plotted on a straight line Can’t get Vmax and Km, other than by mathematically manipulating the Michaelis equation by inverting it to get a linear line

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*V = rate

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The equation allows us to plot a linear graph w 1/V against 1/[S] (by plotting experimental results) From this, we can calculate what Vmax is – the y-intercept Km can be found by looking at x-intercept (x-intercept gives us -1/Km), or by calculating the gradient of the line

Energy change in terms of the enzyme substrate complex and the transition state

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The graph above shows how enzymes lower activation energy, but is simplified

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On the graph above, we have enzyme and substrate on one side, and enzyme and product on the other side. The reaction path is also more detailed Initially, there’s a slight decline in energy, followed by an energy increase (representing the transition state), then the energy in released again to form enzyme and product, first as complex (EP), then as E + P.

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The ES is an energy releasing step (shown on the graph above) - thermodynamically, the formation of this complex is favourable. However, getting from the ES to the transition state is more difficult as it requires energy input (which then collapses down back to ES, or to EP)

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There are several steps, some of which are reversible (seen at the bottom of the slide above). This is simplified bc the complex version is too difficult to be resolved mathematically – we roll all the EST and EP (reversible and irreversible) reactions into one, and call it one irreversible forward process w the k2 rate constant In term of energy changes, we have a change in the free energy (delta G). the delta G double dagger represents the activation energy. Compare this to an uncatalyzed reaction, where there wouldn’t be an ES complex forming, but instead, it would have a larger peak an require more energy for the reaction to work

There is some truth in this model – enzymes do have active sites (where the catalysis takes place) which need to fit in the substrate and interact with it with non-covalent interactions; but the problem with this model is that it is passive. It doesn’t tell us how the substrate is processed subsequent to binding to the enzyme (i.e. it explains how molecules recognise and bind proteins tightly, but doesn’t explain how then that protein might take the substrate and convert it into something else). If the S fitted the E site perfectly, catalysis would be hampered, bc the Enzyme needs to bring about a change to the S and not just bind it. This is the paradox of how enzymes work, they need to be able to bind to Substrate specifically, and they need to be able to turn Substrate into something else, which means those two things are at odds with each other.

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In this model E active site is formed in response to S binding. You have a substrate that doesn't quite fit into the enzyme but when it does bind both the substrate and the enzyme are changed so that the substrate can now fit the enzyme. Initially the active site is not perfect, but upon binding, it is able to move, which puts the Enzyme active site under strain. Both enzyme and substrate are under strain. This strain elicits the energy required for the reaction to take place; by stabilizing the transition state and not the substrate

Above is an example of a transition state using a chemical reaction. You have a tetrahedral carbon, and we want to put the OH- group onto the carbon, and knock out the bromine. The substrates and products are all stable, but the transition state isn’t. The central Carbon, is forming 4 bonds which is the state in which carbon is most stable, but then through the reaction we can react it to interact with an OH group to get the product which also has 4 bonds for the central carbon. But, in the intermediate (transition) step you have 5 bonds to the carbon (the bromine, two Hydrogens, a methyl group and OH group are all interacting with central carbon), so, the carbon is energetically unstable, but it’s a necessary intermediate for this reaction Hence, the enzyme stabilises the transition state, which increases the reaction rate.

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Here, we see a very strong binding of the substrate bc the enzyme doesn’t have to change, and the substrate fits perfectly. The formation of the ES is likely to release energy bc the E and S are such good fits. The E has a good affinity for substrate, so it binds easily w/o needing a lot of energy. However, the transition state is less stable, as the enzyme will not favour it. Binding to the substrate is so favourable, that changing this interaction to form transition state will need a lot of energy. This causes EST (enzyme substrate transition state) to increase as there’s no stabilising force – the efforts have gone into enzyme binding to substrate (a lot more energy is required to form a transition state)

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As the E and S interact w each other, there’s very little energy given out, bc binding of the 2 isn’t very tight. The ES transition will require less energy to be stabilised, bc the enzyme will favour this over the substrate itself. This lowers the energy of the transition state and makes the enzyme more likely to work In this model, the E doesn’t just rely on the E & S binding, it favours lowering the transitions state, given this is its job Hence it is the better model, bc Vmax is higher. However, this model doesn’t favour specificity and substrate binding – you will need higher conc.s of substrate to reach ½ Vmax, so Km goes up. So for ESCs that form w low conc.s of substrate, this isn’t a good model

E binds to S v tightly, so Km is low bc E & S binding is favourable, therefore less S is needed to achieve it. Vmax is also low bc energy barrier at transition state is high. There is some trade-off: the enzyme has to balance specificity vs rate

Optimal activity and specificity

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The tension between enzyme activity and binding affinity creates variation between enzymes – some are fast, but not very good at binding substrates, and others are the opposite The table above shows a lot of range e.g. threonine deaminase has a Km value of 5000 micro molars (this is how much threonine the enzyme needs for its optimal activity), compared to Arginine-tRNA synthase which has 3 different substrates and 3 different levels of Km – these are governed by the evolution of the enzyme and are an indication of how good the enzyme needs to be. Arginine tRNA synthase has a v high affinity (low Km) for tRNA itself, which is not abundant on the cell, so it has to bind it very well; it’s affinity for ATP is lower, but this isn’t bc it’s less well-adapted, it’s bc there’s more ATP readily available in the cell – the enzyme doesn’t need to get any better at binding ATP to retain activity. Hence, Km is an indication of the biological activity of the substrate

Enzymes have become adapted to be good enough for the substrate concentration available so any further improvement in their affinity doesn’t really make a difference to how the cell works On the graph above, the red line has a low Km, while the black one has a higher Km (needs more substrate to reach half the maximum rate)

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Km is a genuine constant for an enzyme. Under a given set of conditions we know how tightly a particular substrate is going to bind. Whereas, Vmax is the measure of how much activity an enzyme has – this is dependent on E conc. Kcat stands for catalytic constant. It is the measure of the activity, independent of how much enzyme is present

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Kcat is measured in seconds Rubisco is one of the slowest enzymes

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Kcat is the rate constant of the limiting step (the enzyme can’t go any faster than this) this is k2.

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Dividing Kcat by Km gives us an idea of enzyme efficiency – this is how much turnover an enzyme can manage, and how much substrate it needs to achieve it Catalytic efficiency varies between enzymes (see table above) When an enzyme evolves, a change of an amino acid in the protein could change Kcat or Km i.e. by lowering the transition state or by increasing the affinity for its substrate Note that neither of these values on their own can tell you about enzyme efficiency There are a no. of factors that have an impact on this – how difficult is a chemical reaction, how good does the enzyme have to be to carry out the reaction under biological conditions in the cell, etc.

Enzymes are adapted to have a particular function. E.g. restriction enzymes are part of the bacterial cellular protection system and can only break down invader DNA, which allows them to avoid breaking down the bacteria’s own genome. These properties of restriction enzymes can be used in the lab for e.g. genetic engineering.

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