Week 9 Session Notes - sorry for late upload

Point 9.1

Using a set number of degrees of freedom – we don’t overanalyse, we save ourselves a huge amount of mathematics etc

Elasticity comes from entropy. Entropy determines the elasticity of an object

Rubber band force pulling it back becomes less at high temperature

Elastic energy cost increases with temperature

Shape of curve determines elastic energy (equation 9.4 assumes circular curve)

Approximate a curve as a bunch of little circular segments

Figure 9.4

C) still a coil

d) unwinds and becomes straight

slope of A = entropic elasticity

figure 9.5

gives 4 different models to describe 9.4’s regime A and part of B

talked about reasons why we’d stretch out dna and other things

1D chain model – chain containing segments that pointed either left or right. Only two parameters occurring are temperature (which is fixed), and the persistence length (length of segment)

3D chain – segments can point in any direction. Adds two degrees of freedom (the two extra dimensions)

Elastic rod model – 3dfjc model plus an energy cost for every bend. Each segment cares about what’s going on with its neighbours

Figure 9.7

Find it weird that the structure become more ordered at higher temperatures, as opposed to denaturing. Does show however that at low temperatures it is denatured, and for all we know, it may denature again at a higher temperature.

Cold denaturing – loss of electrostatics, lowering of disulfide bonds

Long protein can be approximated as an infinitely long coil

Short protein – ends can’t be stabilised as easily, due to fraction of residues that can’t become part of the helix

Figure 9.8

Rotation determines fraction of helices

Double conc = double rotation

One helical rotation turns light to the left, the handedness of rotation turns light to the right.

CD

Figure 9.9

Effects of changing degree of cooperativity

In long chains, hard to tell if cooperativity matter

In short chains though, makes a hell of a difference

Cooperativity allows for sharp transitions

Figure 9.10

HbO2 -><- Hb + O2

Keq = [Hb][O2]/[HbO2]

Y = [HbO2]/([Hb] + [HbO2] = fraction bound

Bad to have myoglobin.....20% oxygenation difference in tissue

Haemoglobin has 4 binding sites....effective value of 3 due to its binding nature

Small fraction have 1, 2 or 3 bound

Response to Mitch's Post on Enzyme Inhibition

So when inhibitors are involved with our simple enzyme reaction system, there are two main ways in which this inhibitor can incorporate itself.

The first way in which it can be involved is to bind to the enzyme while not bound to the substrate. By forming the EI complex, the inhibitor can prevent the substrate from binding and forming the ES complex, thereby prohibiting the formation of a product.

The second way in which an inhibitor can be involved is by binding to the ES complex to form the ESI complex. Again, by binding to this, the production of the product in prevented. In this case, it is because the binding of the inhibitor usually distorts the shape of the enzyme thereby making it lose its fitting bind to the substrate disabling the catalytic effects that come from a fitted binding of the ES.

I have heard that there can be some situations where there is the step EI + S <-> ESI, but I cannot name any examples, and I personally find it hard to imagine that happening, as surely the EI complex would straight away prevent it from binding the substrate due to either being bound in the same pocket that the substrate would be binding it, or it would distort the conformation of the enzyme, and as such the binding pocket of the substrate.

Michaelis-Menten and simple kinetics

Let's start with the general procession of an enzymatic reaction (which is unaffected by inhibitors). We have:

E + S <-> ES<-> EP <-> E + P

In the case where the is none, or next to no product present initially, the chemical potential is a large negative number, resulting in a steep downhill slope for the third step, allowing us to write EP -> E + P

We now have:

E + S <-> ES <-> EP -> E + P

We shall also assume that E+S, ES, and E+P are separated by large barriers, allowing each transition to be treated independently. The transition involving binding of substrates from solution is characterised by a first order rate, proportional to the substrate concentration.

Throwing a single enzyme into a vat of substrate at initial concentration cs, and negligible product, will spend a fraction of time Pe unoccupied, and the rest (1-Pe) bound to substrate. These two times can be said to be nearly constant in time, so the enzyme converts the substrate at a constant rate.

We are now left with the final reaction of:

E + S <-> ES -> E + P

Forward rate of the first step = csk1

Reverse rate of the first step = k-1

Forward rate of the second step = k2

Michaelis constant = (k-1 + k2)/k1

vmax = k2Ce

Michaelis-Menten rule

v = vmax(cs / (Km + cs))

Ratchet?

I have to say Nelsons use of analogies in this chapter seems overly complicated, such as his introduction of the G and S-ratchets.



So in order to understand his analogy I propose to consider a different analogy. Instead of a rod moving through a membrane driven by thermal fluctuations; take a simple chemical reaction.

A + B <-> C + D

The reaction can move either left or right. There is only a small difference in energies which can easily be supplied by thermal fluctuations.
The equilibrium can become biased when the activation energy for the products becomes large. In this way the amount of product will increase and reactant decrease (like the sliding of the rod).
By simply removing amounts of the product (Le Chatelier’s Principle) the reaction will keep forming products and use up the available reactants (like applying a force on the rod).

Molecular Machine Example

One amazing molecular machine is ATP synthase.



It operates in a rotational catalysis mechanism. The β subunits perform the catalysis of ADP + Pi -> ATP. The γ subunit is responsible for rotating the β subunits and is driven by the influx of Hydrogen ions.

As the β subunits are rotated their conformation changes depending on their contacts with the γ subunit. Firstly ADP and Pi are bound loosely within one β subunit. The rotational conformation change then produces ATP by binding the reactants tightly and stabilizing the ATP form. In the next rotation, the ATP molecule is only loosely bound by the β subunit and the molecule can dissociate. There are many videos of this process available on Youtube.

This synthase is a great example of both a cyclic machine (10.1.1) and an enzyme catalyses a reaction by binding to the transition state (10.3.3). Can you think of any other examples?

(Information taken from textbook: Lehninger- Principles of Biochemistry
Image from article: Molecular Motors: Turning the ATP motor, Richard L. Cross, Nature 427, 407-408(29 January 2004), doi:10.1038/427407b)

Enzyme Inhibition

The Lineweaver-Burk equation

 

As my first post this week was quite long, I’ll make my second a short one and save you some reading time. I’m pretty sure we’ve all done CHEM2002, and have learnt about enzyme competition before. This chapter mentions competitive inhibition and noncompetitive inhibition. To refresh our memories, competitive inhibition is where the inhibitor molecule binds to the same site as the enzyme’s substrate. Thus they are in direct competition with each other at the binding site.

Noncompetitive inhibition is where the enzyme has two binding sites, which are effectively mutually exclusive. When the inhibitor binds to its binding site, the enzyme cannot bind the substrate any more, and thus the reaction again is inhibited. We learnt a third kind of inhibition, uncompetitive inhibition. This is where the inhibitor binds to the enzyme substrate complex.

These three kinds of inhibition have different effects on the Lineweaver-Burk plot as the inhibitor concentration is increased. If the inhibition is competitive, the y-intercept of the Lineweaver-Burk plot is constant, but the gradient changes. If the inhibition is noncompetitive, the x-intercept of the plot is constant, but the slope changes. If the inhibition is uncompetitive, the gradient is constant, but the y-intercepts changes (it increases). This means that competitive inhibition does not affect the vmax of the reaction and noncompetitive inhibition does not affect Km. Uncompetitive inhibition lowers both the Km and vmax.

Kinesin Motion

I thought a key topic in this chapter was the motion of kinesin. I was confused at first about how this molecule works, but I think I understand now. Please correct me if I am wrong anywhere. Each head of the kinesin molecule has two sites, one which binds one ADP molecule and one which binds to the microtubule. Both these sites bind strongly to their respective substrates. However, the head cannot bind strongly to both substrates at once. This is an example of non-competitive inhibition.

Because these sites bind strongly, and there is an abundance of ADP and microtubule binding sites in the cell compared to kinesin molecules, at least one of these sites is bound at any time. So if the molecule is not bound to the microtubule, then both these heads of the dimer have ADP bound. When the molecule is near a microtubule, eventually one of the heads will lose its ADP and bind to the microtubule. I suppose that a backward step could be taken at this point, and the head bound to the microtubule releases the microtubule and rebinds ADP. However, as the concentration of ATP in the cell is much higher than the concentration of APT, an ATP molecule is more likely to bind to the vacant site on the kinesin head bound to the microtubule.

Unlike ADP, kinesin is able to bind ATP and the microtubule strongly. When the ATP molecule binds, the head does not let go of the microtubule. The neck linker of the bound head then attaches to the head. This state is a local energy minimum. The other head is free to diffuse around; however, the position of the neck linker biases this diffusion. This is the asymmetry which is necessary for directed molecular motion.

Eventually the other head diffuses close enough to the next microtubule binding site for it to bind. However, this head still has ADP bound, so it can only bind weakly to the next site. It is likely that it binds and unbinds several times. However, eventually the head will spontaneously release the ADP, and will be free to bind the microtubule.

The protein chain connecting the two heads is only just long enough to reach the next microtubule binding site, so when both heads are bound, the molecule is strained due to stretching. Since the protein chain is only just long enough to reach the next microtubule binding site, even with the bias in the neck linker, it is nearly impossible for the unbound head to accidentally diffuse close enough to the previous site ad take a step backwards, because the neck linker will always point in the forward direction.

Perhaps this added strain on the molecule deforms the ATP binding site on the initially bound head, which causes it to hydrolyse the ATP bound to it. The head only weakly binds the resulting ADP and the microtubule, so the molecule will likely release one of these substrates. Due to the extra strain the molecule is under in its stretched state, the head preferentially releases the microtubule, to cancel this stress.

The energy released by ATP hydrolysis detaches the neck linker from the head, and the molecule is left with one head bound to the microtubule and one head bound to an ADP molecule. This is the state it began in, thus the cycle is available to repeat, providing the substrate concentrations remain the same. I presume this process will repeat until an ADP binds to the head bound to the microtubule instead of an ATP molecule, and the kinesin molecule detaches from the microtubule altogether.