Practical 2:
Induced fit in maltose-binding protein
Examination: Answer the questions directly in this protocol.
Names (everyone):
1
2
Table of Contents
Getting started ............................................................................................................................................3
Exercise I: Exploring proteins in Maestro .................................................................................................3
Exercise II: Induced fit substrate binding to MBP ..................................................................................10
Exercise III: Ramachandran angles .........................................................................................................13
Exercise IV: Energetics of induced fit in MBP .......................................................................................14
Exercise V: Dynamics of induced fit in MBP .........................................................................................17
References ............................................................................................................................................................................. 19
Getting started
•
Boot the computer into Linux by restarting and holding down the F12 key.
•
Log in and open a terminal window.
•
In the terminal type:
export SCHRODINGER=/opt/schrodinger
•
Switch to the Linux temporary directory:
cd /tmp
We will create a lot of junk files and this directory is wiped when the computer is restarted.
•
Then start Maestro by typing
$SCHRODINGER/maestro &
•
While you're doing this practical look out for the Wikipedia symbol: . When it is next to a
highlighted word it means that we have found an article on Wikipedia that you may find
interesting.
Exercise I: Exploring proteins in Maestro
In this practical we will get to know the maltose-binding protein (MBP). This protein is located in the
periplasmic space between the inner and outer membrane of gram negative bacteria where it binds
to the nutrient maltose.
The binding triggers two events:
•
The uptake of maltose through the inner wall with the help of an ATP-binding cassette (ABC)
transporter .
•
A signaling cascade that makes the bacteria move towards increasing concentration of the
nutrient — an effect known as chemotaxis . [1]
3
The trigger is a large conformational change of MBP as it is binding the substrate. MBP consists of two
subunits joined together in a flexible hinge region. These subunits close around the ligand in a way that
is very similar to how Pac-Man eats pac-dots. This is the effect we are going to study.
To do this we are going to compare crystal structures of the substrate bound state (this is the holo
structure) and the ligand-free structure (this is called the apo structure).
Let's import a holo structure of MBP into our current project.
•
Open the 'Project' toolbar and click on the Get PDB button
•
In the 'PDB ID:' box type 1anf
•
Click the Download button
You're now seeing a gray wireframe model of a protein where six of the residues are colored red. This
means that Maestro found a problem with these residues.
•
Have a look in the terminal window and you'll see a bunch of warning messages.
What are they saying?
Although crystal structures are invaluable to us they are not without "error" or ambiguity. In this case
the atomic positions in some residues were not included in the PDB entry. Most likely these atoms
could not be modeled from the reconstructed electronic density.
Where are these residues located?
Since X-Rays cannot be focused the basic idea behind X-Ray crystallography is to amplify the
signal through constructive interference in a periodic crystal. For this to work the atoms have to adopt
the same conformation in the asymmetric unit throughout the crystal.
Why do you think these residues could not be resolved?
Let's have a closer look at one of them. When you hover the mouse pointer over the structure you will
see the amino acid name and number in the bottom pane of the project window.
•
Switch to selecting residues by clicking
→Residues in the Edit toolbar.
•
To center this residue on the screen click on Lys1 and then on the fit-to-screen button
in the View toolbar.
4
•
Label the residue atoms by PDB name by clicking the Label all button
Name in the Labels toolbar.
→PDB Atom
Which atoms are present in Lys1 in this structure?
We could fix this by one-residue homology modeling but we'll leave it as it is.
•
Delete all labels by clicking the Label all button
•
Deselect lysine by clicking in empty space
•
Click the fit-to-screen button
→Delete Labels
.
Hover the mouse pointer over the gray dots scattered around the structure.
What are those?
Why do we only see them close to the protein and in the binding pocket?
Now let's check out the amino acids.
•
Color the residues by property by clicking
toolbar.
→Residue Property in the Representation
Maestro has colored all residues in four colors: blue, red, cyan and green.
What do these colors signify?
Where in the protein are the green residues located mostly?
Where are the blue and red residues located mostly?
5
Why?
Protein-ligand interactions
Now we will have a look at the protein-substrate interactions at the binding site. Before we do this we
will add hydrogens to the structure.
Why do hydrogens not appear in a crystal structure?
On the menu bar, click Workflows→Protein Preparation Wizard...
•
We want to 'Assign bond orders' and 'Add hydrogens'
•
We want to 'Delete waters' that are more than 5Å away from the binding site (so keep the
tick mark and the 5Å).
•
No filling of loops or capping of termini.
•
Click 'Preprocess'
•
Maestro will warn us that we have incomplete residues but we can ignore this for now, just
click 'OK'
Hydrogens have been added to the structures and the ligand is now shown in green.
•
Let's only look at the ligand by clicking the button 'Display only'
Display Atoms toolbar. Then fit-to-screen.
The substrate is a maltose
glycosidic bonds.
→Ligands in the
which consists of two or more glucose molecules linked through α(1→4)-
•
Label the substrate atoms by PDB name.
•
The PDB atoms are numbered according to the IUPAC glycoside standard.
•
Browse the Wikipedia entry for maltose and make sure the hydrogens were added correctly.
•
If there are mistakes, use the buttons 'Decrement the bond order' or 'Increment the bond
order'
in the build toolbar followed by the button 'Add H' →Molecules in the edit
toolbar to correct them. Click on the molecule to add the hydrogens.
•
When you're done
o Delete the labels with
o Double-click the
→'Delete labels'
button to show the whole system again.
6
o
If you look at the water molecules their hydrogens all point in the same direction,
which doesn't seem very likely.
Why do you think that is?
Fortunately Maestro has an algorithm that will optimize the hydrogen bonding network.
•
In the 'Protein preparation wizard' window there is a pane called 'Refine'.
o In this pane tick 'Exhaustive sampling'
o Then click 'Optimize'
o 'Start'
When it's finished,
•
Make sure the sequence viewer is open: From the menu bar, check that the
'Maestro'→'Sequence Viewer' option is ticked. If not then tick it.
•
Find histidine 64 in the sequence viewer, click on the 'H' to select it and then fit-to-screen.
There is a label next to this histidine, what does it say?
It's hard to distinguish between heavy atoms such as N, C and O when determining a crystal structure.
•
Find the Wikipedia entry for histidine and draw the imidazole ring and its beta-carbon
below.
Imagine we couldn't tell the difference between N and C atoms.
Which two indistinguishable ways are there to draw this ring?
What happens in this calculation is that Maestro tries to determine whether the imidazole ring should
actually be flipped around or not. This is done by estimating the best possible hydrogen bonding
network.
Did histidine 64 have the correct orientation in the crystal structure?
7
•
Open the Wikipedia entry for glutamine and asparagine.
Why should we be careful with trusting the orientation of the terminal amides in gln and asn in
crystal structures?
When adding hydrogens we also have to mind whether ionizable residues should be protonated or not.
•
Go back to the Wikipedia entry for histidine.
What is the pKa of the imidazole ring in histidine?
The immediate surroundings of an ionizable residue can shift its pKa away from the value in aqueous
solution.
What happens if the pKa of a histidine is upshifted by more than one pKa-unit in the protein, at
physiological pH?
Maestro tries to determine if histidines are protonated from the hydrogen bond network. The choice is
between protonating the epsilon-nitrogen (PDB name NE, residue name HIE), the delta-nitrogen (ND,
residue name HID) or both (residue name HIP).
Where did Maestro place the hydrogen(s) and why?
Now let's look at the hydrogen bonding network of the ligand.
•
To see the H-bonds click the button 'H-bonds'
click on the ligand.
→'All H-bonds' in the labels toolbar then
With how many H-bonds does the maltose substrate interact with the protein?
8
Now let's see which residues that lie closest to interact with the ligand.
•
Display only the ligand
•
Then click the button 'Display Residues within N Å of the currently displayed atoms'
→'+3Å'
•
Color the residues by property (as we did above).
•
Let's color the maltose and water molecules by atom by
o First selecting them
•
§
Click the Select button
§
Switch to the 'Residue' pane
§
Choose 'Residue type'
§
Mark 'GLC' and 'HOH' (this is the ligand and the water molecules)
§
Click 'Add'
§
'OK'
→Select... in the Edit toolbar
Now let's color this selection
o From the menu bar open Workspace →'Atom and bond coloring...'
o Choose color scheme, e.g. 'Element (Light Carbons)'
o In the 'Apply current color scheme' make sure 'Pick:' is ticked and click 'Selection'.
•
Close the 'Atom and Bond Coloring' window.
•
Click somewhere to clear the selection.
•
To see the residue types use the 'Label all' button
toolbar.
→Residue information, in the Labels
Which kinds of general interactions do you see between the ligand and the binding site?
Display all atoms, delete the labels, center and zoom out to an overview of the protein.
•
Switch coloring to e.g. 'Element (Molecule Number Carbons)'.
9
Exercise II: Induced fit substrate binding to MBP
Now let's take a look at the conformational change that MBP goes through when it binds the substrate.
•
Load the apo structure 1omp.
•
Open the project table
•
Ctrl-Click in the 'In' column on any of the rows with the holo (ligand-bound) structure
'MALTODEXTRIN BINDING PROTEIN...' to include it to the workspace together with the
apo structure.
•
Remove water molecules with the Delete button
•
Let's display their secondary structure by clicking the button 'Show, hide or color ribbon'
in the Project toolbar.
→'Ribbons for All Residues' and then
toolbar.
•
→Waters, in the Edit toolbar.
→'Undisplay Atoms', in the Represenations
Now we will move them into a good position to compare the two.
o Click the button 'Transform' button
→Entries, in the Edit toolbar.
o Click on either structure and move them so that the N-terminal end is pointing up, the Cterminal end down and the binding site faces right.
o Click on the other structure and place it beside the first one.
o When you're done and it looks nice click the 'Fit to screen' button.
This protein is made up of two subunits interconnected bya hinge region consisting of three strands.
How do the two subunits differ between the bound and free protein?
To get a better view of this let's switch to another representation.
•
In the menu bar click Workspace→Molecular represenation...
•
In the tab that says 'Ribbons' set 'Style:' to 'CA Trace'
•
Click 'Update Existing Ribbons'
•
Close this window
We are now looking at a Cα trace structure. These are simply lines connecting the Cα carbons between
the residues. Structures with a commercial value or that the researches are not ready to disclose for
other reasons are sometimes deposited in the structure database as Cα traces. These could be structures
of biological targets in pharmaceutical research.
Now let's superposition the backbone of residues 113-150 and 190-260 in the two C-terminal domains.
10
•
From the menu bar open Tools→Superposition
•
Click on the 'ASL' tab. (This stands for Atom Specification Language)
•
Click 'Select...'
•
Click on the 'Residue' tab.
•
Click 'Residue number' and write
113-150, 190-260
in the 'Residue number:' text box.
•
Click the 'Add' button.
•
It should now read
res.num 113-150,190-260
in the 'ASL:' pane. This is the ASL for this residue range.
•
Now click 'Backbone/side chain'.
•
Tick the 'Backbone' box and click 'Intersect'
•
The ASL should now read
(res.num 113-150,190-260) AND (( backbone ) )
in the ASL pane. This is the logical expression for the backbone atoms in these residues.
•
Click 'OK'.
The two structures should now be superpositioned and you can see its RMSD
. The less the merrier.
In a general sense what does RMSD measure?
Distances are measured in Å but what is the unit of the RMSD?
How big is the RMSD for the superposition?
Judging from this value are the regions we chose to superimpose similar in structure?
11
•
Close the 'Superposition' side pane by clicking the top right 'x'.
•
Fit the complex to screen.
•
Click the button 'Save View'
name.
in the Saved views toolbar to save this view. Give it a fun
Notice how the values 1-5 right next to this button changed. Number 1 was highlighted.
•
Spin the proteins around by e.g. drag-clicking with the middle mouse on the workspace.
•
Click the highlighted #1 in the Saved views pane.
Now that we can go back to this view we are free to play around a bit.
We saw that the backbone atoms were on average less than 0.5 Å away from each other. Now have a
look at the structure.
In your opinion how well did we superposition the C-terminal domain?
If we now look at the N-terminal domain they are considerably different form each other.
How different is their secondary structure?
Where do the structures start to diverge?
This part of the protein is known as the hinge region because it acts as a joint around which the two
subunits can rotate. We will now try to measure the rotation angle between the open unbound and
closed bound protein.
•
To do this we need to see the backbone atoms.
o Click
→'Display Atoms' in the Represenation toolbar.
o Remove hydrogens with the button 'Undisplay'
atoms toolbar.
o And the side chains by
→'All Hydrogens' in the Display
→'Protein Side Chains'
o Color the atoms by 'Element (Molecule Number Carbons)'
Now we will measure the opening angle between the apo and holo structure. To a good approximation
we can take this to be the angle between the amide nitrogens of Asp30 and Tyr17 in the bound
structure and Gly32 of the unbound structure.
12
•
Open the sequence viewer from the menu bar: Maestro→'Sequence Viewer'
•
Locate Asp30 of the bound structure in the sequence viewer and click on it.
•
Fit Asp30 to screen.
•
Click the button 'Measure'
•
Click on the amide nitrogen. This is the first atom.
•
Now locate Tyr17 in the bound structure and fit it to the screen.
•
Click on its amide nitrogen. This is the second atom.
•
Locate Gly32 of the unbound structure and fit it to the screen.
•
Click on its amide nitrogen. This is the third and last atom.
•
Now restore view #1.
→ Angle, in the Labels toolbar.
Two green lines appeared that run parallel with the alpha helical axes. The angle between these lines
approximately defines the rotation angle around the hinge region.
Which value on the hinge angle did you get?
Exercise III: Ramachandran angles
Now let's measure some Ramachandran angles. The Ramachandran angles φ and ψ are the dihedral
angles Ci-1-N-Cα-C and N-Cα-C-Ni+1 in a residue.
•
From the project table include only the apo structure.
•
Open the sequence view window pane from the menu bar by ticking Maestro→'Sequence
Viewer'
•
When you hover the mouse pointer over the sequence you see the residue number.
•
Locate residues 110-112 and select them by first clicking on 110 and then shift-clicking on 112.
•
Fit them to screen.
•
Now measure the φ and ψ angles and fill in the values in Table 1.
•
Locate residues 259-261 and repeat the measurements.
•
Now switch to the holo stucture and repeat this.
•
Calculate the differences between the angles of residue 110 to make sure you calculate them
right, and fill in the rest of the values in Table 1 (it is important to click the atoms in the correct
sequence).
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Table 1. Ramachandran angles of the apo and holo structure.
φ
Residue
apo
ψ
holo
Δ
apo
holo
Δ
110
-126.5
-108.4
18.1
138.9
112.6
26.3
111
-135.8
-105.6
30.2
138.3
129.7
8.6
112
-157.9
-142.0
15.9
154.9
141.5
13.4
259
260
261
How do the angles compare between the apo and holo structure?
In what region are these residues located?
Why are the Ramachandran angles different?
Exercise IV: Energetics of induced fit in MBP
By looking at the structures we learned that the protein has a different conformation when the substrate
is bound. But does it change its conformation before or after it has bound the substrate? Or do the
conformational change and the binding event happen simultaneously? Is the substrate driving the
conformational change or does the unbound protein already exist in the closed conformation waiting
for the substrate to come along?
To understand cause and effect in biochemistry structure is not enough. The cause for conformational
change, be it during a binding event or a chemical reaction, is ultimately found in the differences in free
energy between states. These free energies can either be measured experimentally or be calculated.
The point of this course is to learn how to extract them from simulations. Calculations permit us to
make predictions. We can also decompose free energies into contributions. Knowing how each residue
contributes to a biochemical event makes it possible for us to predict mutations that will enhance or
impede this process. This could be a binding event or an enzymatic reaction.
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Defining the hinge angle to be 0o in the fully open state and 35o in the closed state (and this is what we
got when we measured it above, right?) the free energy of the unbound MBP protein increases linearly
by about 0.1 kcal mol-1 per hinge angle degree. This comes from measuring the folding and binding
free energy of hinge region mutants followed by NMR and fluorescence measurements. [2]
What hinge angle does the unbound protein prefer?
What happens with the free energy if it tries to close in the absence of substrate?
We measured the hinge angle of the fully closed protein above. What is the estimated relative
free energy of a closed protein in the absence of substrate?
The relation between the relative Gibbs free energies and the equilibrium constant is: ∆! = −!"#$%
Let's say we have prepared a solution of MBP protein at 37o C. If the concentration of open
protein is 1 mM what is the concentration of fully closed protein at equilibrium?
The binding constant Kb is defined as the equilibrium constant in the reaction E + S → E:S.
The value that is usually measured experimentally is the dissociation constant which is the equilibrium
!
constant of the reverse reaction !! = !
!
The dissociation constant Kd for a substrate to wild-type open MBP 900 nM and 6.7 nM for a mutant
with a hinge angle of 28.4o.
15
How many times stronger does the substrate bind to the partially closed mutant than the open
wild-type protein?
Similar to above the binding free energy is given as ∆!!"#$ = −!"#$!! . The Kd measurements were
done at 37o C.
What is the binding free energy of maltose to open wild-type MBP?
What is the binding free energy to the nearly closed mutant?
As with the folding free energy the binding free energy depends linearly on the hinge angle.
Given your values above how many kcal/mol per hinge degree does the binding free energy
decrease?
What is then your prediction of the binding free energy for a fully closed protein at 35o hinge
angle?
How much energy does the system have to sacrifice to drive the closure of wild type MBP to 35o
when binding maltose?
Where does this energy come from?
What is making MBP close?
16
Exercise V: Dynamics of induced fit in MBP
We've seen that the conformational change in MBP is induced by the ligand. But does maltose first
bind to the open state before this happens? Or do conformational change and binding happen
simultaneously?
In the induced fit model, the binding of the substrate drives the conformational change. An alternative
model is that the protein is found in an equilibrium between the induced and uninduced protein
conformation in solution, and that the binding of the substrate shifts this equilibrium towards the bound
conformation.
This model is called pre-existing equilibrium dynamics and can be probed with NMR or MD studies.
We are not ready to do protein MD in this practical but let's try and figure out what happens in MBP
binding by using data from simulations and measurements.
The experimentally measured association rate constant for maltose binding to MBP is 2.5·107M-1s-1. [3]
If the binding reaction only depends on the diffusion of maltose through water it is said to be a
diffusion-controlled reaction . For sugar binding in general the diffusion-controlled binding rate is
about 1·109 M-1s-1. [3]
How many binding events per microsecond per molar does that correspond to?
If the binding reaction only depends on the diffusion of maltose through water it is said to be a
diffusion-controlled reaction. For sugar binding in general the diffusion-controlled binding rate is about
1·109 M-1s-1. [3]
How many binding events per microsecond per molar does that correspond to?
How much faster is the diffusion-controlled reaction rate than the measured binding rate in
MBP?
But how long does it take for the protein to close? We can find this out with molecular dynamics!
In Figure 1 you see the result of a 30 ns simulation of the opening and closing of MBP. The closing
17
starts from the open MBP structure where the ligand has been added. The opening starts from the
closed state where the ligand has been removed.
Figure 1. Opening of MBP when removing the ligand from the holo structure and closing of MBP when adding the ligand
to the apo structure. A representative angle is shown on the ordinate and the simulation time on the abscissa. Also included
are simulations of the crystal apo and holo structures for comparison. [4]
How long does it take for ligand bound open MBP to close in this simulation?
How many times per second does this correspond to? (This is the closing rate)
If maltose binds exclusively to the open MBP in a fast diffusion-controlled reaction then the rate
constant would be about 0.1 M-1 times the simulated closing rate of the protein. [3]
What is then the association rate constant from the simulation?
18
How does this compare to the experimental association rate constant?
Given your calculations what do you think happens in this case: induced-fit or pre-existing
equilibrium dynamics?
Explain why it's hard to answer the previous question by just looking at the crystal structures?
Can you think of any way to see pre-existing equilibrium dynamics with static crystal
structures?
References
1.
Mowbray, S. & Sandgren, M. Chemotaxis receptors: A progress report on structure and function. J Struct Biol 124,
257-275 (1998).
2.
Fig 2 (inset) from: Millet, O., Hudson, R.P. & Kay, L.E. The energetic cost of domain reorientation in maltosebinding protein as studied by NMR and fluorescence spectroscopy. PNAS 100, 12700 -12705 (2003).
3.
Miller, D.M., Olson, J.S., Pflugrath, J.W. & Quiocho, F.A. Rates of ligand binding to periplasmic proteins
involved in bacterial transport and chemotaxis. J Biol Chem 258, 13665 -13672 (1983).
4.
Stockner, T., Vogel, H.J. & Tieleman, D.P. A Salt-Bridge Motif Involved in Ligand Binding and Large-Scale
Domain Motions of the Maltose-Binding Protein. Biophys J 89, 3362-3371 (2005).
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