Lessons 1-5 – outlined answers.
Please note that these answers have been directly translated from the Swedish answers,
which correspond to the Swedish version of the lessons. These correspond to the English
ones initially, but particular towards the end of the course, I rearranged a little in the
exercises. Therefore, you will sometimes need to look at the Swedish versions of the
exercises in order to find out what question the answer belongs to  sorry about this.
CD 1:1
a) rSM14-M20 is the most structured before denaturation. Motivation: in the left column,
CD-spectra as a function of wavelength are shown. In the right column, ellipticity at a
chosen wavelength (216 nm) as a function of temperature is shown. From the
diagrams in the right-hand column we can see that the mutant with the highest
magnitude of ellipticity at 25 oC is rSM14-M20, which indicates that this mutant also
has the highest structure content. A minimum at this wavelength indicates b-structure.
That this is the native form is shown by the reduction of ellipticity at this wavelength
on denaturation.
b) The denatured state has a larger contribution of random coil (200 nm) and tendencies
towards helix formation (shoulder at 222 nm).
c) rSM14-M20 also has the highest melting point (the midpoint of the denaturation curve,
show the location in graph for complete answer)
CD 1:2
a) In this wavelength area (far-UV) the peptide group absorbs, and by studying the
ellipticity (how the molecule ‘turns’ light, the difference between absorbance of rightand left-handed circularly polarized light) it is possible by CD to estimate the
secondary structure content. At 222 nm the alpha helix shows a minimum, and the
magnitude of this is proportional to the amount of helix. The helix content can thus be
estimated from the magnitude of the signal.
b) The amount of helix is in creased because the negative signal is increased. (Note that
these spectra are difference spectra where the complex minus its parts is shown!)
c) For apo 1:1 and for holo 2:1 (peptide:CaM). Until these ratio are reached, the absolute
value of the ellipticity is almost linearly increased, but after 1:1 and 2:1 ratios have
been reached there is no more change in ellipticity.
d) In this wavelength region (near-UV) aromatic residues absorb light (Trp, Tyr, Phe),
and significant ellipticity in this area indicates ordered side chains (which thus turns
polarized light in a preferential direction).
e) In Tr1C both the ellipticity and the fine structure of the signal is more pronounced in
the presence of peptide. This indicates that tryptophanes and/or phe/tyr become more
ordered in the complex. This can occur either because Trp is present in the interaction
surface, or because the protein core in general becomes more ordered in the complex.
CD 1:3
a) The native protein, which has a high propensity of helical structure (bimodal minimum,
one of which is located at 222 nm), unfolds to a random conformation at 75 oC. At
further elevated temperatures, b-structure (minimum at 216 nm) is formed which is
then maintained as the protein is cooled down to 25 oC.
b) Perhaps the protein forms fibrils/plaques, which contain a large amount of b-structure?
Perhaps these could aggregate and accumulate in the cancer cells?
c) Only spectra for the wt are shown. It would be interesting to see how the mutations
affect the protein structure as well as its fibrillation properties.
Fluorescence
2:1
a) The longer wavelengths compared to those expected for intrinsic protein fluorescence
(Trp fluoresces at 350 nm) indicate that we are looking at the fluorescence from
DAUDA, not on the fluorescence from the protein as such. DAUDA fluorescence
increases on adding protein, indicating that DAUDA binds the protein in a way which
unlocks the fluorophore and/or protects it from solvent access. Both of these
mechanisms increase the fluorescence by increasing the quantum yield. A weak red
shift in the wavelength maximum indicates burial of DAUDA (altered chemical
environment, altered wave functions). As linoleic acid is added, the fluorescence
spectrum again resembles the initial one, indicating that the lipid acid has competed
out DAUDA from the binding site.
b) If you start by titrating in DAUDA and determine the affinity for this binding, it is
then possible to compete out DAUDA binding by increasing the concentration of lipid.
From the reduction in fluorescence, the binding constant for the lipid can be
determined. The experimental design is called competition experiment.
2:2.
a) FRET-experiment: it is possible to determine distances between fluorescing groups by
looking at singlet-singlet resonance transfer (Fluorescence Resonance Energy
Transfer). A mutation has been made to introduce W/Trp in position 188 in the protein,
and a fluorophore has been attached to DNA either upstreams (TATA22up) or
downstreams (TATA22down) from the TATA-box. Fluorophores are needed in two
positions in order to allow for the transfer of energy in the form of light between them.
Here you can explain the FRET transfer mechanism using a figure showing the energy
levels of the two systems, and discuss the conditions required to be met for FRET to
occur.
b) With only one fluorophore, the binding constant between protein and DNA could be
measured by, for example, studying quenching effects on the fluorophore in the
bound/free state, but distances would not be measurable.
c) Since the fluorescence at 440 nm, which corresponds to the FRET-signal from the
fluorophore on DNA, increases for E188W with TATA22up, the dominating
orientation of TBP on DNA is consistent with the upper picture in the right column.
2:3.
a) In figure A, circular dichroism- spectroscopy (CD) has been used, and in B
fluorescence has been measured. In figure A, the ellipticity in the wavelength region
260-290 (near-UV) has been recorded; in this region, it is possible to find out whether
the aromatics in the protein are well ordered or disordered depending on whether the
ellipticity is high or low. In figure B the fluorescence has been recorded at ~450 nm,
which indicates the use of an external probe. Based on the emission wavelength, it is
likely that ANS has been used.
b) Proteins showing dashed lines in B seem to have been partly unfolded, since they
appear to show hydrophobic surfaces that bind ANS (higher fluorescence when
bound).
c) The lower degree of folding for proteins with dashed lines agrees with the lower
ellipticity in A, which indicates a disordered core where aromatics are not well
organized. Similarly, this agrees with that the well-ordered protein – solid lines, native
a-lactalbumin – does not bind ANS since the fluorescence is low. ANS does not bind
to the surface of well folded proteins unless they have hydrophobic binding sites.
d) It would be interesting to measure the secondary structure content by CD mesurements
in the far-UV range (190-230 nm) to see whether this is affected. It would also be
interesting to determine the Tm for both proteins by measuring the ellipticity as a
function of temperature, for example at 220 nm. The more flexible protein should also
have a lower stability.
2:4
a) The approach includes the usage of a fluorescently labeled probe which binds the
active site for the protein. By analyzing which concentrations are required for
competing out the binding of other ligands, the affinity for a range of unlabeled
ligands can be determined. In this case fluorescence anisotropy – also called
polarization - is used for detection, which gives a huge effect since the molecular
weight of the probe in complex with the protein becomes incredibly much larger than
the probe itself, which results in anisotropy of the emitted light due to the slower
rotation.
b) The figure shows that the anisotropy increases when the fluorophore binds to the wild
type protein. It does not appear as if the probe binds to the mutant, it can therefore be
inactive – perhaps because the mutation is located in the active site.
c) Biotin partly competes out the fluorophore which leads to decreased anisotropy for the
sample as such – what we get out is an average of the anisotropy for bound and
unbound fluorophore.
d) The study proposes studying the anisotropy in samples in a 96-well plate or similar.
Where the anisotropy is small, the degree of inhibition will be high. This means that
the signal is an inverse function of the polarization.
Lesson three
3:1
a) The fluorophore that sits on the DNA has been used for steady-state anisotropy
measurements. When the p53 domain binds DNA the molecular weight is increased
and thereby the anisotropy. This experimental setup has been used in a titration in
order to determine the binding affinities to specific and unspecific DNA, and how this
is affected by Zn2+ binding.
b) Zn2+ binding to the core domain of p53 is essential in order for specific DNA binding
to occur. Without Zn2+, the binding of p53 to DNA appears unspecific also for the
‘specific’ sequence. This binding also appears to result in a slightly higher
stochiometry (higher than 1:1), perhaps due to binding to an unspecific site at the
‘back side’ of the DNA in the structure figure. When p53 binds specifically, perhaps
the DNA is somehow kinked to that the unspecific site on the ‘back side’ no longer
has any affinity to the p53 core domain.
3:2
On the indicated temperature transitions, the p53 protein appears to go from a helix structure
to a mixed b-sheet / random coil structure, and from there to a structure dominated by b-turns.
FT-IR has been used since the authors needed to confirm that the structural transitions also
occurred in the solid state. Using CD only solubilized material could be studied, and since
they argued that aggregates could be formed, it was necessary to also study the material using
a method suitable to study solid states.
3: 3
a) The authors would like to study the membrane bound structure, which makes FT-IR a
good method since membranes can be constructed onto a so-called ATR detector cell.
b) B and D are second derivatives of the experimental curves in A and C. The B curve
corresponds well with A (secondary structure peaks should be positioned according to
corresponding features in the second derivative) but in the D curve there is little
correspondence. In fact, the curve in D appears to be a mere duplicate of the B curve,
so this must be a misprint/mistake.
c) The peptide in A predominantly consists of b-sheet and distorted (3-10) helix. The
peptide in B primarily consists of correct alpha-helix and b-sheet. For evaluation see
course compendium page I3.
d) This is not so straight-forward… the peptide that is distorted could perhaps more
easily sneek through the membrane and thus penetrate, whereas the peptide which has
more structure could perhaps easier form correctly shaped pores and thereby punctuate
the cell. More investigations are needed.
3:4.
a) the assay that has been developed is based on fluorescence polarization. The probe is a
labeled ligand to the active site. When the probe is bound to the enzyme, the
anisotropy is high because the complex has high molecular weight. If any of the
compounds in the molecular library binds the active site better and thus releases the
probe, the anisotropy will be drastically lowered, since the probe then only ‘feels’ its
own molecular weight. The wells where the anisotropy is low thus contains ligands
that bind better than the probe, that is, so called ‘drug leads’.
b) The figure shows that the anisotropy/polarization increases when the probe is added to
the wt-enzyme. S75A must be a deficient mutant which does not bind the probe since
the anisotropy is unchanged. This mutant is therefore a control for the assay.
c) FP is the entity that binds the enzyme. Rhodamin is the fluorophore. When FP-biotin
is added it will compete out FP-Rhodamin, in equal proportions since the FP-part
binds equally well in both molecules.
d) see a) for how the assay should be used. The degree of inhibition is obtained as a 50%
value of the activity.
Lesson four
4:1
a) stuy the raw data for the ITC. The thermodynamical effect of ANAM11 mostly
resembles the effect achieved by the NAM-10 mutant. A-1 has the opposite effect (if
ANAM-11 is exoterm, A1 is endoterm). Thus, the 10 mutations outside the active site
appear to dominate ligand binding. This is also consistent with the affinities given in
the atble: the affinity is lowered most by the NAM-10 mutations. NOTE: a lower Kd
means that the affinity is higher 
b) The A-1 mutation destabilizes the protein (lowered Tm) while the NAM-10 mutations
stabilize the protein (increased Tm). The destabilizing effect of the A1 mutation can
be explained by the somewhat lowered affinity, since the ligand binding site perhaps
fluctuates too much to provide a stable ligand binding. However, the effect on the
binding constant is not at all as large as the effect on the Tm, which is quite drastic.
The NAM-10 mutations result in a somewhat elevated Tm: perhaps they result in a too
‘stiff’ protein which cannot adapt to the ligand, or the protein is stabilized into a
wrong conformation which hinders the protein to bind inhibitors (strong/stiff but
wrong). When put together in the ANAM-10 mutant, the increased stiffness of the
NAM-10 mutations and the increased mobility of the A1 mutation in the active site
results in the situation that if by chance a ligand would actually make it into the active
site despite the increased stiffness provided by the NAM10 mutations, it would be
thrown out from the site by the increased mobility provided by the A1 mutation, which
taken together lowers the affinity even more.
4:2
In the figure to the left, DSC (differential scanning calorimetry) has been used. In this
method you measure the energy required to heat a sample one degree at a time over a
larger temperature interval – this is equivalent to measuring Cp. In this way, it is possible
to determine the melting point of a protein since much more energy is required to raise the
temperature in the interval around Tm, since in order to unfold the protein, all noncovalent bonds that give the protein its structure (hydrogen bonds, VdW bonds) must be
broken for the protein to unfold.
In the figure to the right fluorescence is used, more specifically ANS fluorescence (480500 nm). A high fluorescence in this measurement indicates that ANS has bound to
exposed hydrophobic surfaces; such surfaces occur often in partly folded proteins (molten
globules etc).
The fact that we seem to have a protein with two melting points at most pHs suggests that
we have a protein with two domains that are different and have different stabilities. The
stability of one of the domains seems to be highly affected by pH (titration of His
residues?) whereas the other domain appears pretty unaffected by changes in pH.
The melting temperatures in DSC spectra agree well with temperatures that show ANS
binding. This suggests that when the protein is unfolding, hydrophobic surfaces are
exposed which bind ANS – the unfolding occurs via an intermediate, partly folded state.
At pH 5 the protein appears to be the most stable: high Tms and low ANS binding. At pH
8 the protein appears less stable, in particular in domain 2, which is evidenced by low
melting points and high ANS binding as the temperature is increased (a lot of
intermediates).
4:3
The experiment performed is a DSC experiment. The peaks correspond to melting
temperatures ™ for the two individual domains in calmodulin. In the presence of peptide,
one of the peaks move to higher temperatures, that means that the Tm is increased and one
of the domain becomes more stable. The other domain appears unaffected since its Tm is
not increased.
We know from earlier exercises that often when Calmodulin binds a peptide ligand, both
domains are affected and wrap together around the peptide. We would therefore have
expected ONE peak, not two, when the calmodulin+peptide complex denaturates, and at
higher Tms than for the individual parts.
Since we see two peaks, it seems as if the protein cannot fold around the peptide, and
since only one of the peaks move, it appears as if the peptide only binds to one of the two
subdomains.
The fact that the complex is still inactive despite the binding of peptide (this is stated in
the intro to the question) can be due to that it is required that the peptide binds BOTH
domains in order to form the biologically active structure for continued
signaling/recognition.
Lesson 5
Exercise 1
CH1-Zn2+
CD
Helix (222 nm)
CH1-Zn2+-HIF
Helix
CH1 alone
Randomcoli
fluorescence
NMR
Molten globule (ANS Molten globule
binding)
Small chemical shift
dispersion, broad
signals
Structured or
Well structured
unstructured (poor
Large chemical shift
ANS binding)
dispersion, highly
resolved spectrum
Structured or
No structure
unstructured (poor
Small chemical shift
ANS binding)
dispersion, narrow
signals
When Zn2+ binds to CH1, a molten globule structure is formed which is then further
stabilized to a well-structured protein complex together with the target protein HIF.
Exercise 2.
a) The spectrum is recorded with NMR (nuclear magnetic resonance) and is called
HSQC. The spectrum shows cross peaks between 15N and 1H in the NH group of the
peptide bond, which thus creates one cross peak for every amino acid in the protein. In
this picture the assignment of the spectrum is displayed onto the spectrum, which
means that they have already found out which peak belongs to which amino acid. This
process, called Assignment, requires running a number of other NMR experiments and
putting together the information in a jigsaw-puzzle manner. Since the peaks are well
separated and do not form a blur in the middle we can draw the conclusion that the
protein is well folded – each amino acid has a unique and special chemical
environment which renders its specific chemical shift.
b) When adding the protein TAX it seems as if some of the peaks are shifted, that is, that
their position in the spectrum changes. This indicates that the chemical environment of
the amino acid is changed, which suggests that the TAX protein binds to the KIX
protein. The new chemical environment can be due to the immediate presence of the
TAX, thus changing the chemical environment, or to that the TAX binding results in
structural changes in KIX thus changing the environment. Measurements have been
performed by adding unlabelled TAX protein to 15N-labelled KIX (we see no peaks
from Tax, only KIX-peaks that change position). By analyzing the sequence location
of the amino acids that show the largest shift changes, these can be mapped onto the
structure to see if the changes are located in a specific region of the protein. The amino
acids that show the largest shift changes are indicated in the figure to the right. It
seems as if the Tax binding results in changes in a specific surface region, which is
most likely the ligand binding site.
c) Since we now know which amino acids are located in the ligand binding site, and
where these peaks are located in the HSQC spectrum, we can use this as a basis for
SAR-by-NMR design of pharmaceutical drug candidates. We can add chemicals from
a chemical library until we find the ones that bind the ligand surface. Using organic
synthesis, we can combine these building blocks to form high-affinity ligands which
could inhibit TAX binding. These molecules could then hinder the establishment of
the viral infection.
Exercise 3.
a) Biacore /SPR. Exendin has been bound to the chip and GLP is flowed over the surface.
b) According to the affinity measurements which are accounted for in the figure legend,
the full-length protein binds best. According to the Biacore curves, the on-rate (ka)
appears highest for the N-terminal domain, but the off-rate (kd) for this construct is
very high. The reason for the higher affinity of the full length protein is most likely the
much lower off-rate, despite the slightly lower on-rate.
c) See b above
d) In this experiment, the full-length protein has been pre-incubated with extendin. Since
the full-length protein is already bound to extendin when flowing over the extendincovered chip, the complex will bind poorly to the surface. In this way, the scientists
have further confirmed that there is really a binding between these two protein
components, even in solution (sometimes proteins are destroyed on the chip causing
artefactual binding events).
Exercise 4
a) The method that has been used is AFM. A sample is attached to a surface, to the
sample a cantilever is attached. Using the cantilever it is possible to pull the sample
and feel the resistant force from the sample when stretched/extended. This method is
used to scan a surface on a molecular level as well as to investigate internal molecular
forces in proteins. It is also used to study the ligand binding strength in protein
interactions, even membrane interactions can be studied. Preparative and manipulative
actions on single molecules in single cells is possible.
B and C below gives one interpretation – a second interpretation (which we did in class)
relies on the ligand binding being the first peak. With the information given, we cannot
differentiate between these interpretations so they are equally correct.
b) In figure C, the first peak indicates that the two domains of CaM release each other
and change conformation. At the second peak the first domain starts to release the
ligand, which makes the second domain release the ligand easier (3rd peak). In figure
D, the first peak also hear indicates the release of the two domains from each other. At
the second peak, the first domain is released from the ligand, but this does not make it
easier for the second domain to release the ligand which it does at the third peak.
c) In figure C it can be observed in the diagram that a larger force is required at the first
peak than in figure D, when the two domains release each other. It can also be
observed that domain 1 is extended nearly twice as much compared to domain two in
case C when they release the ligand; this indicates high cooperativity (domain 2 easier
releases when domain 1 has let go). In case D the domains are equally extended on
ligand release, which suggests lower cooperativity.
Exercise 5
a) AFM – atomic force microscopy. In the figure, AFM has been used as a surface
microscopy technique, where we get an estimate of the surface structure by allowing
the probe to trace the surface. An alternative use for AFM is to anchor the probe in
proteins/molecules at the surface and measure the force required to pull or compress
the molecule/protein.
b) The heat-induced fibers are higher/thicker than the TFE induced.
c) Both fibres consist of b-sheet considering the highest peak in the IR spectrum. The
heat-induced also has random coil and helix structure as shown by peaks at 1650 and
1540 cm-1.
d) Figure 2 shows mainly b-sheet structure, figure 1 shows that the heat-induced
protofibrils are more than twice as thick. Taken together it is conceivable that the
TFE-induced protofibril form monolayer b-sheets and that the heat-induced form
bilayer b-sheet; alternatively, the heat-induced protofibrils could form thicker fibrils
due to the increased amounts of a-helic and random coil.