Download LS1a Problem Set #2

Name:
TF Name
LS1a Fall 2014
Problem Set #6
Due Wednesday 12/10 at 6 pm in your TF’s drop box on the 2nd floor of the Science
Center
I. Basic Concept Questions
1. (10 points) Shown below is part of a peptide that can be hydrolyzed by a protease.
a. (2 points) The P1 pocket of the protease binds to the tryptophan in the peptide,
and the P2 pocket binds to the serine. Circle the peptide bond that will be
hydrolyzed by the enzyme.
b. (4 points) The protease that cleaves this substrate shows a strong preference for
substrates that contain nonpolar, cyclic amino acid side chains such as
tryptophan or phenylalanine at position R1. The P1 pocket binds Trp more tightly
than Phe. In terms of entropy and enthalpy, why would you expect a binding
pocket that can accommodate tryptophan to bind tryptophan more favorably
than phenylalanine?
In terms of entropy, tryptophan is larger and contains more
hydrophobic surface area; as a result, more ordered water would be
released when it binds to the enzyme, leading to a more favorable (i.e.,
a larger or more positive) ΔS of binding.
In terms of enthalpy, the larger size of tryptophan provides it with
greater surface area over which van der Waals interactions can form
compared to phenylalanine. In addition, a pocket that binds tryptophan
may accept a hydrogen bond from tryptophan’s side chain –NH group.
This hydrogen bond acceptor could only make a dipole:induced dipole
interaction with phenylalanine, which would also be enthalpically
unfavorable.
c. (4 points) The P2 pocket of the protease contains a residue that forms a
hydrogen bond to the serine in the substrate. If the serine in the substrate is
mutated to alanine, would you expect Kd for binding of the protease to the
substrate to increase or decrease? Briefly explain your answer.
The P2 pocket normally forms a hydrogen bond with serine that it
cannot with alanine. The mutation therefore breaks a favorable bond
between the P2 pocket without replacing it, making the binding of the
substrate less favorable. As a result, the Kd for that binding interaction
would increase with the mutation.
2. (16 points) The diagram below shows the binding interactions between a drug and
six amino acid side chains of an enzyme that binds to the drug. The numbers
indicate the locations of the specified amino acids within the enzyme’s protein
sequence.
a. (10 points) In the table below, name the amino acid given its number in the
protein sequence and indicate the strongest type of intermolecular interaction
that can take place between the amino acid side chain and the drug (e.g., van
der Waals, ionic, etc…).
Amino acid #
Name of amino acid Strongest type of interaction possible with drug
57
Aspartate
ionic
84
Tryptophan
van der Waals (or induced-dipole:induced
dipole) [ion:induced-dipole is also
acceptable]
199
Glutamate
ion:dipole
231
Threonine
hydrogen bond
2
330
Phenylalanine
van der Waals (or induced-dipole:induced
dipole) [ion:induced-dipole is also
acceptable]
b. (6 points) The tight binding that is observed between this drug and the enzyme
is enhanced by the rigid nature of the drug as well as its ability to displace water
molecules from the space that is occupied by the drug when it binds to the
enzyme. Briefly discuss in your own words how each of these factors contributes
favorably to the free energy of binding between the enzyme and the drug.
The rigidity of the drug helps to minimize the decrease in its entropy
that occurs when it binds to the enzyme (i.e., ΔSD is made less
negative). Because this decrease in entropy is unfavorable, the free
energy of binding is made more favorable (more negative) when the
drug is pre-rigidified such that is has less entropy to lose upon binding.
The displacement of ordered water molecules that were previously
bound to the enzyme’s drug-binding pocket, releasing them into bulk
solvent upon drug binding, increases the entropy of the water
molecules (i.e., ΔSW > 0).
3. (14 points) The reaction coordinate diagram below compares the energy landscape
of an uncatalyzed reaction with that of the same reaction catalyzed by an enzyme.
a. (2 points) What is the free energy change associated with binding of the enzyme
to the substrate?
3
1 kJ/mol (either positive or negative is ok, we don’t care about the
sign)
b. (6 points) What is the difference between ΔG‡catalyzed and ΔG‡uncatalyzed? Show the
calculations you used to arrive at your answer.
1 kJ/mol
ΔG‡uncatalyzed = 10 – 6 = 4 kJ/mol
ΔG‡catalyzed = 8 – 5 = 3 kJ/mol
ΔG‡uncatalyzed - ΔG‡catalyzed = 4 – 3 = 1 kJ/mol
(again, either positive or negative value for energy is fine)
c. (6 points) A mutant version of the enzyme is discovered that generates product
at a much slower rate than the wild-type (“un-mutated”) protein. A reaction
coordinate diagram of the uncatalyzed reaction and the reaction catalyzed by the
mutant protein is shown below.
Why does the mutant protein yield product at a slower rate than the wild-type
protein?
The energy landscape of the reaction catalyzed by the mutant protein
shows much tighter binding affinity for the substrate.
The affinity of the enzyme for the substrate is so great that the enzyme
no longer lowers the energy of the transition state relative to the
energy of the E+S complex (ΔG‡ is now 5 kJ/mol for the mutantcatalyzed reaction).
4
The mutant protein generates product much more slowly because it
raises the energy of activation for the reaction, slowing down the
reaction rate.
The mutant protein raises the energy of activation of the reaction
relative the wild-type enzyme by stabilizing the E+S complex more so
than the wild-type enzyme does.
5
II. Applied Concept Questions
4. (20 points) Similar to aspartyl proteases, cysteine proteases cleave peptides using
acid/base catalysis. The enzyme first cleaves the peptide bond, releasing the aminoterminal peptide fragment in a two-step process shown below as steps (a) and (b).
The carboxy-terminal peptide fragment remains attached to the cysteine of the
enzyme, resulting in an “acyl-enzyme” intermediate as labeled below. In the final
step (not shown), the histidine positions a water molecule to attack the carbonyl of
the acyl-enzyme intermediate to release the carboxy-terminal peptide and
regenerate the cysteine of the active site.
a. (2 points) Is the sulfur of cysteine acting as a nucleophile or an electrophile in
step (a)?
Sulfur acts as a nucleophile that attacks the carbonyl carbon.
b. (6 points) What role does histidine perform in catalyzing step (a)? What role
does histidine perform in catalyzing step (b)?
In step (a), it is acting as a catalytic base by abstracting a proton off of
cysteine to make it a better nucleophile.
In step (b), it donates its proton, acting as a catalytic acid.
6
c. (6 points) Draw the transition state for step (a) of this reaction.
d. (6 points) As shown below, backbone amide hydrogens help to orient the
substrate appropriately in the active site. In addition to orienting the substrate,
what other role do the two backbone amide hydrogens perform to help catalyze
this reaction?


They stabilize growing negative charge on the carboxyl oxygen that
occurs specifically in the transition state, thereby catalyzing the
reaction by stabilizing the transition state.
The hydrogen bonds help make the carbonyl carbon more
electrophilic (acid catalysis) by placing partially-positive hydrogens
near the oxygen, stabilizing a greater negative charge on the
oxygen, pulling electron density towards the oxygen an away from
the electrophilic carbon.
7
5. (14 points) Morphine is a drug that binds to the opioid receptor and is given to
patients suffering from chronic pain.
a. (8 points) The binding pocket of the opioid receptor bound to morphine is shown
below (the morphine molecule is in bold).
For the two analogs shown below, would the Kd of binding to the same opiod
receptor be higher or lower than the Kd of morphine binding to the receptor?
Briefly explain whether the change in Kd is principally due to a change in Hbinding
or Sbinding, and why.
i.
(4 points) Analog 1:
Analog 1 would have a larger Kd than morphine because of a change
in Hbinding. When morphine is free (not bound by the pocket), its
positively-charged -NH+- group makes an ion:dipole interaction
with water. When the pocket is free, its D147 also makes an
ion:dipole interaction with water. When morphine binds the pocket,
D147 and the -NH+- form an ion:ion bond, and the released water
molecules form a hydrogen bond between them. The binding of this
region of morphine to D147 therefore replaces two ion:dipole bonds
8
with an ionic bond and a hydrogen bond. The Hbinding is roughly
zero as the energetic sum of two ion:dipole bonds is roughly
equivalent to the energetic sum an ionic bond plus a hydrogen
bond.
Analog 1 replaces the positively-charged -NH+- in morphine with a
neutral -CH- group, which would not interact with water since it is
nonpolar. The binding of analog 1 therefore replaces an ion:dipole
bond that the free pocket could form with water with an
ion:induced dipole bond when the pocket binds this region of the
drug. Replacing the positively charged nitrogen with a neutral
carbon therefore makes Hbinding for binding analog 1 positive by
replacing, making the binding of analog 1 less favorable than the
binding of morphine.
[One could also argue for full credit that making the drug more
hydrophobic by replacing the -NH+- with a -CH- group allows for
more water to be released upon analog 1 binding, such that SW is
even more positive for Analog 1 than for morphine.]
ii. (4 points) Analog 2:
Analog 2 would have a larger Kd than morphine because of a change
in Sbinding. Analog 2 lacks a bridging carbon-carbon bond that
morphine has, and is therefore less rigid than morphine. Analog 2
would therefore suffer a greater loss of entropy upon binding
compared to morphine, making the binding reaction less favorable.
b. (6 points) An opioid receptor mutant that replaces the aspartate at position 147
with a leucine was found. Which drug (morphine, analog 1, or analog 2) would
have the smallest Kd of binding with this mutant opioid receptor? Briefly explain
your answer.
Analog 1 would have a smaller Kd than morphine or analog 2 for
binding to this mutant opiod receptor.
For morphine and analog 2, the binding reaction would require that the
positively-charged nitrogen lose a ion:permanent-dipole interaction
with water when the drug is free in order to be make an ion:induceddipole interaction with leucine, such that Hbinding > 0. Analog 1 would
not have this increase in Hbinding.
Additionally, analog 1 substitutes a carbon for this positively-charged
nitrogen, making it more hydrophobic. When analog 1 is free in
solution, a solvation shell of water would form around this hydrophobic
region. When analog 1 binds the mutant opiod receptor, the
previously-ordered water molecules are released to the bulk solvent,
such that SW > 0.
9
6. (16 points) Adenosine hydrolase is an enzyme that hydrolyzes adenosine into ribose
and adenine. Shown below on the left is a diagram of the structure of an inhibitor
bound to the active site of the enzyme. The inhibitor binds much more tightly to the
enzyme than the substrate, adenosine (shown on the right), due to a critical binding
interaction between the inhibitor and glutamate 166, an amino acid in the active site
of the enzyme. The interaction between E166 and the inhibitor is indicated in the
diagram.
a. (10 points) Shown below are two proposed mechanisms for adenosine hydrolysis
(note that mechanism 1 shows only the first step of a multi-step process). Draw
the transition state associated with each mechanism. Indicate partial charges
where appropriate.
10
b. (6 points) Given the importance of the binding interaction between E166 and the
inhibitor, which of the mechanisms shown above, 1 or 2, is more likely to be the
mechanism catalyzed by this enzyme? Briefly explain your answer.
The structure of the inhibitor and the critical interaction that it forms
with the negative glutamate suggests that mechanism 1 is the
mechanism that leads to adenosine hydrolysis.
Mechanism 1 involves an accumulation of positive charge on the
oxygen of the ribose ring (the position equivalent to the position of the
positive nitrogen in the inhibitor) during both the transition state and
in the intermediate.
The interaction with glutamate lowers the energy of the transition
state in order to make the reaction proceed more quickly. Mechanism 2
does not lead to an accumulation of positive charge on this oxygen, so
if that mechanism took place, the interaction with glutamate would be
unnecessary towards stabilizing the transition state, and therefore
would be unlikely to bind an inhibitor (since inhibitors bind active sites
tightly by mimicking the transition state).
7. (10 points) The HIV Tat protein is required for high-level expression of viral genes in
infected host cells.
a. (2 points) Why is Tat necessary for HIV virulence?
Tat is required to recruit the kinase that phosphorylates RNA
polymerase. This phosphorylation is needed for efficient (i.e.,
processive) transcription of the viral genome.
b. (4 points) Shown below is a diagram of the secondary structure of the TAR
sequence. The short nucleotide sequence shown below on the right significantly
reduces Tat-mediated enhancement of viral gene expression when injected in
excess into the nuclei of infected cells. However, the sequence does not bind to
HIV Tat by itself.
11
5’-CUGGCUAACUAGGGA-3’
5’–
Propose an explanation that accounts for the observed results.
The sequence is the same as the region highlighted in red, and is
therefore complementary to the portion of the hairpin sequence
highlighted in blue. By adding an excessive amount of the 15-base RNA
polynucleotide strand shown to the right, the 15-base polynucleotide
would base pair with the blue region of TAR, preventing formation of
the hairpin structure. Because Tat recognizes and binds to the hairpin
structure, it is unable to do so in the presence of the short nucleotide
sequence.
c. (4 points) A mutant Tat protein is found that shows the same level of weak
transcription as is observed in the absence of Tat. However, the mutant Tat
protein still binds to TAR with the same affinity as the wild-type (“non-mutant”)
Tat protein. Briefly explain what could account for this.
Most likely, the mutant Tat is no longer able to recruit the protein
kinase that is required to phosphorylate RNA polymerase. Failure to
recruit the protein kinase results in transcription that is not processive.
12