Literature 002: Niewoehner et al., Neuron, 2014, 81, 1, p49–60

A number of disorders affect the central nervous system (CNS) such as inflammations, viral infections, Alzheimer's disease, cancers, Parkinson's disease and many, *many* more. In case of Alzheimer, which currently is not curable, people who suffer from the disease are found to have what is called Amyloid-beta plaques in the brain, aggregates of some kind of protein. These aggregates negatively affect the brain function but can be removed if targeted with special antibodies such as Aducanumab, Bapineuzumab or Gantenerumab, possibly leading to a reversal of the disease.

The treatment of diseases of the CNS is difficult, because the body protects the CNS from the outside using a special kind of barrier. In the brain, this barrier is called blood-brain-barrier. The barrier in principle consists of special cells (brain-endothelial-cells, BECs/Endothelium in the figure below) which form the blood vessels and only permit certain substances to transpass from the vessel to the brain. For example, antibodies as those mentioned above are not as such allowed to pass the barrier.

http://clincancerres.aacrjournals.org/content/13/6/1663/F3.large.jpg

In the following, out of the shown possible passways, the focus is on (C), transferrin receptor mediated transcytosis. In this pathway, a substance first binds to the transferrin receptor (Tfr) and is then transported across the cell. This is where the research of Niewoehner et al. comes in. In short, they want to find a way of transporting an active substance such as one of the above mentioned antibodies to the brain.

In "Increased Brain Penetration and Potency ofa Therapeutic Antibody Using a Monovalent Molecular Shuttle", the researchers are able to do this by using a construct which consists of the therapeutically active antibody, a linker peptide and a Tfr active antibody fragment (Tabf). They test two versions of this construct, one with only one Tfr-active fragment, one with two such fragments. The control experiments are done using just the antibody without the linker unit or the Tfr active fragment.

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

Here, dFab stands for double-Fab and sFab for single-Fab. mAb31 is the monoclonal antibody 31.

In the following, the series of experiments is addressed by the questions they answer in order to establish a causal chain of reasoning.

1 Do the active parts of the construct interfere with their respective function?

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

In B, the solid and striped bars refer to two different coating densities used in the affinity experiment. As seen, the construct still can bind both ways (Abeta referes to Amyloid beta, the target protein aggregate). Notably, sFab and dFab have very similar binding affinity.

2 Does the construct enter the BEC?

Using a laser to highlight cells where the construct entered, the number of cells at a given fluorescence intensity are counted.

http://www.cell.com/cms/attachment/2012543313/2034747624/gr2.jpg

In red the cells with two Tabf's are counted, in green with only one. The black count corresponds to no Tabf. Apparently two Tabf's lead to a slightly better internalization of the construct.

3 What happens to the constructs (one or two Tabfs) once inside the cell?

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

One hour after exposure to the construct, it is measured how much of the construct is colocalized with the Lysosome associated membrane protein 2 (Lamp2). This gives an indication of where the construct is found in the cell.

The BECs have a mechanism to identify what has entered them. It is unclear how the cell identifies the dFab and moves it to the lysosome for digestion.

4 How can one be sure that the cell does not digest the sFab construct?

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

Red and pink are two different concentrations of dFab.

After having entered the cell, the construct has three possible ways to continue: 1) to cross the cell, 2) be digested by the cell and 3) exit the cell where it has entered (see also the first figure).

In the first case, this would mean that the overall number of receptors on the surface should remain constant and this is observed (green bars). The surface concentration of receptors associated with dFab however decreases over time, meaning they do not reappear at the surface once internalized. After 72 hours, apparently new Tfr's are expressed by the cell.

5 Do the constructs cross the cell?

An in vitro experiment is designed where a layer of human BECs separates two sections of liquid (the cells in the above experiments were from mice).

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

The construct is added to the Lumen compartment (model for the blood vessel inside) and allowed to enter the cells for 1 hour. Then both layers are washed, which controls for constructs entering the abluminal layer by channels inbetween the cells, and amount of construct over time in the two compartments is measured. sFab is observed to appear in both the abluminal and luminal part, while dFab is only found reappearing in the luminal part.

This is where I get confused.

The authors note that

EC50(sFab) = 3.8 nM

EC50(dFab) = 6.4 nM

but say "This suggests that the monovalent binding mode to the Tfr, and not just the reduced affinity, is the key factor for efficient transcytosis." (p. 51) If they are referring to the measured affinities for the construct towards mouse Tfr (see question 1 where dFab has 5.3-8.9 times higher affinity than sFab), then it makes sense. Otherwise I would have said that sFab has higher affinity than dFab based on these numbers.

One more aspect I find strange:

In my opinion, in (J), the yellow bars should also appear because the way it looks, it seems to indicate that the construct only returns to the lumen, i.e. it exits the cell only in one direction. Why should there be a preference for exiting the cell in only one direction?

Edit:
After discussing this issue with fellow collegues, the following explanation arose. The cell layer is incubated with sFab/dFab for one hour. Then both the luminal and abluminal compartments are flushed. Any sFab or dFab that appears in either compartment after that has to be coming from within the cells. Since the cargo enters the cell on one side, it has to travel a distance through the cell, but because this is a diffusion process, it has no preference as to what direction it takes. But because the cell has a sorting mechanism, the dFab cargo gets filtered out while it is crossing the cell. That's why no yellow bars appear for dFab. dFab however does also appear in the luminal side, that's because this distance is too short for the cell to sort it out before dFab can exit the cell again on the side it entered. That's also the reason why the luminal bars for sFab are slightly higher than for the abluminal side. The spatial distance within the cell simply requires more time to cross.

An in vivo image of the vessels after 15 min and 8 h of exposure to the construct shows what happens.

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

In the beginning, both constructs occupy the vessel cells (A, B). After 8 hours though, only sFab has crossed the vessel cells, i.e. they are not illuminating any longer (C), dFab has remained in the cells (D). In (C), the arrows point to accumulations of the construct on Amyloid-beta plaque aggregates.

6 Does the observed difference in BEC crossing between sFab and dFab depend on the rate at which they enter the BECs?

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

No.

7 Which construct binds more to Amyloid-beta plaque?

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

The measured increased plaque binding by sFab is therefore causal attributed to the fact that sFab binds to Tfr only by one Tabf.

8 Does sFab cover plaque faster than only mAb31 even at lower concentration?

http://www.cell.com/action/showImagesData?pii=S0896-6273%2813%2901035-0

Yes. The concentration of the antibody alone (light blue is 5 times higher than green) when not having a Tabf attached does not provide any activity. Duration of exposure does not change this, in other words the blood-brain-barrier remains closed no matter how high the concentration.

9 Is there a therapeutic effect of the construct?

Yes. The baseline shows the plaque number when the mice are 4.5 months old. Then, during 14 weeks, only the linker and the Tabf (vehicle), only the therapeutic antibody (mAb31) and the sFab are injected.

The plaque concentration increases in all mice, but when treated with sFab it increases at a lower rate.

10 Remarks

I would like to better understand the experiment from question 5. I  would expect that the next thing to do is to study dependence of therapeutic effect on amount of construct administered. What are the elimination rates of the construct? Interferences with other mechanisms? How stable is the construct? What is the exact nature of the cell sorting mechanism?
Also, in case you're interested, you can find the patent for this vehicle is WO 2014/033074.

A number of questions were raised during the discussion:
- If the cell sorting mechanism could be explained, maybe also dFab could be used as a carrier. It has some benefitial characteristics over sFab. Is this possible?
- If I was to manage this project, what would I do if it turns out that the shuttle is not linked strongly enough to the cargo, meaning it lets it's cargo go and picks up something, possibly toxic, else and carries it to the brain. I would say this shuttle is too important and therefore the engineering efforts have to be directed towards making the shuttle more strongly attached to the cargo.
- One short-coming of the study, in order to prove that it is not dependent on the type of cargo, is that no other cargo was used. This would have demonstrated independence of cargo nature. Of course, this would require a major additional effort.

Some out of context notes.

I was inspired to do this summary because I wanted to see how well I can understand and communicate a study from a totally different field. I believe it's important to be open to other fields and especially to be able to understand experimental approaches. Initially, the biology specific terms were difficult to understand but after a while I think I could identify the more important aspects.

Also, it seems sometimes that because of all the pharma-bashing, one doesn't notice that these guys actually do really amazing work.

Disclaimer: I have nothing to do with this research

Coursera Certificate

My first coursera certificate. Lot's of work, but totally worth the effort.

Papers and Thesis

CalB Paper 1 PLOS ONE:
http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0049849

CalB Paper 2 PEERJ:
https://peerj.com/articles/145/

BCX Paper PEERJ:
https://peerj.com/articles/111/

BIOFET-SIM Online Paper PLOS ONE:
http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0045379

BIOFET-SIM Online Chapter Nanoscale Sensors:
http://link.springer.com/chapter/10.1007/978-3-319-02772-2_3

PhD Thesis:
http://www.scribd.com/doc/213075474/MH-Thesis

PhD Defence Slides:
http://www.scribd.com/doc/213105271/Presentation-Defence

Chem 001: Nernst Equation

Consider a galvanic cell composed of two \(\text{H}^{+}\left|\right.\text{H}_2\)-half cells where \( \left[\text{H}^+\right] \) are \(0.025 \text{M} \) and \( 5 \text{M} \) in the first and second half-cell, respectively. Calculate the electromotive force (EMF).

It's good practice to first visualize such a problem before attempting to solve it in order to gain a clear understanding of the system. The galvanic cell could look like this:

Concentration chain of two hydrogen electrodes. The blue arc indicates the salt-bridge.
The hydrogen pressures do not need to be the same (see text).

A galvanic cell with two times the same system at different electrolyte concentration is refered to as concentration chain. At the platinum electrode, the hydrogen gas is oxidized and \(\text{H}^+\) ions enter into solution. A thermodynamic equilibrium is reached when \( \left[\text{H}^+\right] \) is equal in both cells.

We can recognize that in electrode 1, \( \left[\text{H}^+\right] \) is lower and so it is in electrode 1 where hydrogen is oxidized. By convention, the electrode where oxidation occurs is referred to as anode, the one where reduction occurs as kathode. The electrons thus exit the cell at electrode 1, run through the wire to electrode 2 where they enter the cell again and reduce hydrogen ions. Electrode 1 is therefore referred to as the minus pole of the cell ("where the electrons come from"). At the anode (electrode 1), the half-reaction is

\(\text{H}_2\text{(g)} \rightarrow 2\text{H}^+\text{(aq)} + 2\text{e}^-\)

In order to calculate the EMF, it is required to calculate the potential difference between both electrodes 1 and 2 compared to the standard hydrogen electrode. The EMF for the cell is then calculated as the sum of reduction potentials

\( \Delta E = E(\text{Kath.}) + E(\text{An.}) \)

and this is where we have to be careful to choose the sign correspondingly to the reaction we are formulating.

For the reaction at the kathode (where reduction of \(\text{H}^+\text{(aq)}\) occurs), the potential (compared to the normal hydrogen electrode) is given by the Nernst equation

\( E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log{Q} \)

The \(2\) in the fraction is the number of transported electrons per reduction equivalent. The \( (-) \)-sign in front of the log-expression is used when the reaction quotient \( Q \) corresponds to a reaction formulated in direction of reduction:

\( 2\text{H}^+ \text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2 \text{(g)} \)

For the general reaction

\(  k\cdot K + l\cdot L \rightarrow x\cdot X + y\cdot Y  \)

the reaction quotient is always defined as

\( Q = \frac{a^x(X) \cdot a^y(Y)}{a^k(K) \cdot a^l(L)}  \),

where the \(a^i\)'s are activities of the species. Here, we understand the hydrogen gas has unit activity, \(a(\text{H}_2\text{(g)}) = 1\), and so 

\( E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log \frac{1}{ \left[ \text{H}^+ \right] ^2 } \)

where \( \left[\text{H}^+\right] = 5 \text{M} \) and the exponent \(2\) is the stoichiometric coefficient. We emphasize again the minus sign in front of the \( \log \)-term used when the reaction is formulated in direction of reduction. \(E^0\) is the electrode potential for the normal-hydrogen electrode which is zero by definition.

A similar equation is formulated for the anode half-reaction:

\( E\text{(An.)} = E^0 + \frac{0.06}{2}\log \frac{\left[ \text{H}^+ \right]^2}{1} \)

however here a \((+)\)-sign is used in front of the \(\log\)-term because the reaction is formulated in direction of oxidation (so the oxidized species \(\left[H^+\right]\) appears in the numerator) and \( \left[\text{H}^+\right] = 0.025 \text{M} \).

The numeric values are \(E(\text{Kath.}) = +0.041 V\) and \(E(\text{An.}) = -0.096 V\). \(E(\text{An.})\) however corresponds to an oxidation potential and so in order to calculate the EMF, we need to take its negative (such that we are adding reduction potentials). This then provides us with

\( EMF = \Delta E(\text{Kath.}) + {}^-\Delta E(\text{An.}) = 0.041 + {}^-(-0.096) = 0.137 [V] \)

The question can now be extended by asking what the EMF will be, when the hydrogen gas pressures are different from atmospheric pressure (and so the activity \(a(\text{H}_2)\) is no longer unity):

Assume \(p(\text{H}_2) = 202.6 \text{kPa}\) at the anode and \(p(\text{H}_2) = 10.13\text{kPa}\) at the kathode.

Given these pressures, the activities are \(a(\text{H}_2, \text{Kath.}) = p(\text{H}_2)/101.3 \text{kPa} = 0.1 \) and \(a(\text{H}_2, \text{An.}) = p(\text{H}_2)/101.3 \text{kPa} = 2 \). By substituting these values in the Nernst equations for the respective half-cells (using the same concentrations for \(\left[H^+\right]\)), we arrive at

\( E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log \frac{0.1}{ \left[ \text{H}^+ \right] ^2 } = 0.072 \left[V\right]\)

and

\( E\text{(An.)} = E^0 + \frac{0.06}{2}\log \frac{\left[ \text{H}^+ \right]^2}{2} = -0.105 \left[V\right] .\)

The EMF in this cell is therefore (after again switching the sign of the anode potential to obtain a reduction potential)

\( EMF = \Delta E(\text{Kath.}) + {}^-\Delta E(\text{An.}) = 0.071 + {}^-(-0.105) = 0.177 [V] .\)

Literature 001: Alexandrova et al., JACS, 2008, 130, 15907-15915

In "Catalytic Mechanism and Performance of Computationally Designed Enzymes for Kemp Elimination", Alexandrova et al. report calculated activities of enzymes that were previously prepared by the Baker group. The experimentally expressed enzymes are called KE07, KE10 (which contains the V131N mutation) and KE15.
To recall, the Kemp elimination is this reaction:

http://pubs.acs.org/doi/abs/10.1021/ja804040s

From semiempirical QM/MM calculations and transition state theory, the authors obtain \(k_{cat}\) or \(\Delta G^\ddagger\) which they compare to experiment.
For example, for KE07 the authors report a calculated \(\Delta G^\ddagger\) of 8.1 kcal/mol while the measured activation free energy is 17.1 kcal/mol. That appears as quite a discrepancy between predicted and measured activation energy and one might wonder how come this made it into JACS (see below). Also for KE15, the calculated activation energy doesn't really accurately predict the measured activation energy (calc. / exp. [kcal/mol] \(\Delta G^\ddagger\): 12.3 / 17.0). For KE10, \(k_{cat}\) wasn't measured.

However, what the authors claim is that they are not that much interested in exactly matching the measured activation energy (most likely they would use a more elaborate QM method), but much more if they can predict if the enzyme actually contributes to catalysis. I.e. they're interested in the ratio \(k_{cat}/k_{uncat}\) where \(k_{uncat}\) is the rate constant for the reaction in solvent using \(\text{OH}^-\) as base. And apparently they are able to correctly predict if this ratio is larger or smaller than 1 for all enzymes:

\(k_{cat}^{KE07}/k_{uncat} = 1.6 \cdot 10^4\)

\(k_{cat}^{KE10}/k_{uncat}\) < 3.3 (assuming \(K_M\) to be comparable or smaller than for KE07)

\(k_{cat}^{KE15}/k_{uncat} = 1.9 \cdot 10^4\)

\(k_{cat}^{KE16}/k_{uncat} = 5.2 \cdot 10^3\) (by 2-step mechanism and requiring to impose the protonation state of a catalytic D48 residue - why do they not estimate the \(pK_a\) value of the residue?)

For all enzymes, catalytic activity is observed, so qualitatively they are able to tell if an enzyme will be active or not (they calculate the uncatalyzed reaction to have an activation energy of 19.8 kcal/mol and the activation energies of all enzymes is calculated to be lower than that).
Possible explanation for discrepancies provided: no backbone sampling, accuracy of SE QM (PDDG/PM3) - even if they claim this method to be suited for elimination reactions, no polarization in MM field.

Another question can be raised: how can they be sure about their prediction if they don't consider that \(K_M\) might be different for all enzymes. So even if an enzyme might be able to catalyze the reaction, it could in principle be that it hardly binds the substrate. The way to answer this question these days is to say "The catalytic efficiency, \(k_{cat}/K_M\), has not been computed owing to the technical challenges for \(K_M\), which requires computation of the absolute free energy of binding for the substrate." The substrate does not dissociate from the active site in the equilibration and so the working assumption is to believe that all enzymes can bind the substrate and do so comparably well.

My impression: The paper provides elaborate discussions on observed mechanisms and explains them in minute (structural) detail. The weak points, which are very explicitly disclosed, are compensated by pointing out possible reasons and by the careful mechanistic analysis. A strong point is that the results provide suggestions for how to further improve the catalytic activity (i.e. increasing basicity of the catalytic base, optimization of substrate orientation and positioning of hydrogen bond donors should help increase activity). Furthermore, in the mindset that agreement with experimental activation energy is the key property to consider, the paper would not be considered good. But, the point is really to provide a method that allows to qualitatively tell if the enzyme is active or not, and it does so very well I think.

QM 006: INDO One-electron Integrals

In Frank, Computational Chemistry, Eq. 3.86 is
$$ \langle \mu_A \left| \bf{h} \right| \mu_A \rangle = \left< \mu_A \left| -\frac{1}{2}\nabla^2 - \bf{V}_a \right| \mu_A \right> - \sum_{a \neq A}^{Nuclei} \langle \mu_A \left| \bf{V}_a \right| \mu_A \rangle $$.

Is there a reason for not writing it as
$$ \langle \mu_A \left| \bf{h} \right| \mu_A \rangle = \left< \mu_A \left| -\frac{1}{2}\nabla^2 \right| \mu_A \right> - \sum_{a = A}^{Nuclei} \langle \mu_A \left| \bf{V}_a \right| \mu_A \rangle $$?

QM 005: Comparing semi-empirical methods

In a recent paper Liao et al. calculated the reaction profile for a newly suggested mechanism of acetylene hydratase. Conveniently, the atomic coordinates of all species along the reaction coordinate have been deposited in the supporting material.

Carrying out some SPE calculations of all species using MOPAC2012, we find the following profiles.

Fig. 1: Calculated Reaction profiles.

In the SPE calculations, a dielectric constant of \(\epsilon=4\) was applied. In the reference ("B3LYP" in Fig. 1), the optimization was done using the LANL2TZ(f) pseudo-potential on W, 6-311+G(d) for S and 6-31G(d,p) for the other elements. The reference SPE calculations were carried out using 6-311+G(2d,2p) on all elements but W.

Apparently, the semi-empirical methods greatly overestimate the interaction energy between W and the acetylene unit (going from 0 to 1) and underestimate the activation energy for the proton transfer in step 4.
However, without reoptimizing the transition state, PM6 is does see a very clear transition state for the hydroxylation at step 2.

Orbital diagram from supplementary material.