Download LS1a Fall 2014 Practice Problem Set 6 1. Name three ways in which

LS1a Fall 2014
Practice Problem Set 6
1. Name three ways in which enzymes can alter the rate of a chemical reaction.
2. Match the answers on the left with the definitions that best describe them on the right. (Write the
appropriate letters on the lines provided).
___ Nucleophile
___ Active Site
___ Desolvation
___ Kd
___ Scissile bond
___ ΔG‡
___ Substrate
A. Site in the enzyme where water is cleaved
B. A measure for the strength of an interaction between two
molecules
C. An electron-rich atom involved in making a new bond
D. The energetic barrier that must be overcome for a reaction to
occur
E. An electron-poor atom or molecule involved in making a new bond
F. The bond that is formed between enzyme and substrate
G. Region of the enzyme that is involved in carrying out catalysis
H. Loss of interactions of a protein or a drug with surrounding solvent
(e.g., water) molecules
I. The difference in free energy between enzyme and product
J. The bond that is cleaved by a protease
K. The molecule that an enzyme binds to and chemically modifies
II. Applied Concept Questions
3. Cyclic AMP (cAMP) is signaling molecule that can be generated in response to certain growth factors
binding to their appropriate receptors. The enzyme adenylate cyclase catalyzes the synthesis of cAMP
from ATP in the reaction shown below.
a. Draw the transition state for this reaction.
b. Shown below is an ATP molecule in an adenylate cyclase active site that catalyzes the reaction
shown above.
Indicate and briefly explain the specific role each of the following might play in the catalysis of the
reaction to form cAMP.
i. Mg2+
ii. D354
iii. D1000
iv. R1011
4. In addition to the repressor protein, high-level transcription of lac operon genes in E. coli is controlled
by a transcriptional activator protein called CAP, which helps to stabilize the interaction between RNA
polymerase and the lac promoter. In order for the CAP protein to bind DNA and help stabilize RNA
polymerase, it must be bound to a small molecule called cyclic AMP (“cAMP”). When E. coli cells run
low on glucose, the intracellular concentration of cAMP rises to allow CAP to bind to the promoter.
The cellular concentrations of cAMP are maintained at low levels in the presence of glucose.
a. The CAP protein exists in two conformationally separate states – one state that is bound to cAMP
and capable of binding DNA (CAPcAMP-represented by an oval), and another that is not bound to
cAMP and unable to bind to DNA (represented by a square). The two different states are depicted
below:
1)
2)
cAMP
+ cAMP
cAMP
+ DNA
cAMP
DNA
What effect does an increase in cellular [cAMP] have on the equilibrium shown in equation 2?
b. A mutant strain of E. coli was discovered that has a mutation in the sequence of DNA that binds to
CAP. This mutation changes the Kd of the CAPcAMP–DNA binding interaction from 10-12 M to 10-4 M.
What effect would you expect this mutation to have on CAP-dependent transcription? Briefly
explain your reasoning.
5. The emergence of multidrug-resistant bacteria has been attributed in part to membrane transporters
that enable certain bacteria to pump a broad spectrum of toxic compounds out of their cells.
Berberine (drawn in bold) is a natural antimicrobial compound that binds to a bacterial protein called
Qac R to cause enhanced transcription of a specific membrane transporter gene. The binding pocket
for berberine in the Qac R protein is partially shown below. The numbers refer to the position of the
given amino acid in the Qac R polypeptide chain.
a. Name the amino acids 55, 61, and 157 and give the strongest interaction they can form with
berberine.
b. If amino acid 157 were replaced with Asp, would this most likely increase or decrease the binding
affinity of berberine to Qac R? Justify your answer in terms of specific changes in enthalpy and/or
entropy.
c. Would you expect the binding affinity between the molecule shown below (cerine) and the
protein Qac R to be as high as the binding affinity between berberine and the protein Qac R?
Justify your answer in terms of specific changes in enthalpy and/or entropy that take place upon
binding.
d. Based on your answer to part (c), will the Kdissociation (Kd) of the binding reaction between cerine and
Qac R be greater than or less than the Kd of the binding reaction between berberine and Qac R?
Briefly explain.
6. The structure of HIV protease is stable from pH 3-7; however, the activity of the enzyme
varies within this range as shown below.
a. Based on your understanding of catalysis by HIV protease, explain this relationship between
protease activity and pH. You can assume that the tertiary and quaternary structures of the
enzyme remain relatively stable between pH 3 and 7.
b. What would be the consequence of mutating the catalytic Asp 25 into Asn? Assume
substrate binding and protease structure are not affected by the mutation.
c. What would be the consequence of mutating the catalytic Asp 25’ to Asn? Assume
substrate binding and protease structure are not affected by the mutation.
Free Energy
7. A chemist mixes equal numbers of molecules A, B, and C in a flask. Molecule B can react with
molecule A to make AB, or it can react with molecule C to make BC. The reaction energy diagram for
both reactions is shown below.
Course of the Reaction
(Reaction Coordinate)
a. After a small amount of reaction time (when only a tiny amount of products have been generated)
does the chemist have more AB or BC? Briefly Explain.
b. After one hour of reaction time the chemist would like to have more of the other product. Give
two ways the chemist could accomplish this.
c. After equilibrium has been reached does the chemist have more AB or BC? Briefly explain.
8. Lactase is an enzyme that hydrolyzes lactose into -D-galactose and D-glucose in a two-step
reaction. The overall reaction is shown below:
a. Does lactase affect the kinetics of the reaction, its thermodynamics, or both? Briefly explain in
your own words.
b. Consider the substrate, transition state, and product from the first step of the reaction. Which of
these species are stabilized by interactions with the enzyme? Which species is stabilized the
most? Briefly explain why.
c. Step one of the reaction results in a “galactosyl-enzyme” intermediate. The second step of the
reaction, which releases galactose from the active site, is shown below:
“Galactosyl-enzyme”
Intermediate
Describe two ways in which the enzyme facilitates a productive reaction between the water
molecule and the intermediate.
Answers:
1.
 Enzymes can orient substrates to position them in a maximally reactive conformation.
 Enzymes can bring substrates close together such that they can react with one another
(increasing the “proximity,” or the “effective concentration” of the reactants)
 Enzymes use acidic catalysis to make a substrate atom more electrophilic and/or base catalysis
to make a substrate atom more nucleophilic
 Enzymes help to stabilize charge accumulation and the geometry of the transition state
2. C, G, H, B, J, D, K
3a.
3b. [Note: you only need to give one reasonable explanation for full credit, more than one are
sometimes given below.]
i. Mg2+ helps neutralize the negative charges on the phosphate groups in the active site. It also
increases the electrophilicity of the phosphorous atoms (i.e., “acid catalysis”). It also helps
position the ATP in the pocket (i.e., “orientation effects”).
ii. This aspartate performs base catalysis by accepting a proton from the -OH, making the
oxygen a better nucleophile.
iii. This aspartate hydrogen bonds to the -NH2 group of ATP helping to properly position the
nucleotide in the enzyme binding pocket (i.e., “orientation effects”).
iv. The positively charged nitrogen of the arginine helps stabilize the partially negatively charge
on the oxygen of the phosphate groups. By attracting electron density towards the oxygens and
away from the phosphorous atoms, arginine also decreases the electron density on the
phosphorous atoms, making them better electrophiles (i.e., “acid catalysis”).
4a. According to Le Chatelier’s Principle, increasing cellular cAMP concentration causes the
equilibrium in both equations 1 and 2 to shift to the right, causing an increase in the amount of
CAPcAMP that is bound to DNA.
4b. This mutation is expected to result in a decreased level of CAP-dependent transcription. A
larger Kd value indicates that more CAP cAMP will dissociate from the DNA sequence to which it is
normally bound, so fewer molecules will be bound to help stabilize the interaction between
RNA polymerase and the lac promoter. This reduces the amount of RNA that is transcribed.
5a.
55: glutamate ionic interaction
61: tryptophan van der Waal’s interaction
157: asparagine hydrogen bond
5b. The Kd would increase since you would be losing a hydrogen bond and generating a repulsive
interaction.
5c. The binding affinity of cerine to Qac R should be less than that of berberine to Qac R. There will
likely not be an effect on enthalpy as the exact same interactions can be made between Qac R and
cerine as with berberine. However, the entropic loss of cerine binding Qac R will be greater than
than the entropic loss of berberin. Since it is not as rigid as berberine, cerine can adopt a greater
number of possible conformations in solution and binding to Qac R will result in a much greater
loss of entropy compared to the more rigid berberine.
5d. The Kd of cerine to Qac R will be greater than that of berberine. Since K d refers to the dissociation
of a complex to individual molecules, a larger Kd indicates that the products (unbound molecules)
are more favored relative to the reactant (bound complex), and thus indicative of weaker binding.
6a. The catalytic mechanism of HIV protease requires one protonated Asp and one deprotonated Asp.
The pKa of Asp is approximately 4. At low pH (below 4), both Asp's are protonated, so no general
base catalysis is possible. At higher pH (above 4), both Asp's are deprotonated, so no general acid
catalysis is possible.
6b. Mutation of Asp 25 into Asn would cause a loss of catalytic activity since the Asn would be unable to
mediate base catalysis.
6c. Mutation of Asp25’ into Asn would cause a loss of catalytic activity since the Asn would be unable
to mediate acid catalysis.
7a. After only a small amount of time (i.e., before the system reaches equilibrium), the
product:reactant ratio is determined by the difference in the activation energies of the reactions.
Since the activation energy for A+B+CAB+C is lower than the activation energy for
A+B+CA+BC, more AB will be formed initially.
7b. Either add heat (energy); add more B and C; and/or add an enzyme catalyst specific for the
A+B+CA+BC reaction
7c. After the reaction reaches equilibrium, the ratio of products from the two reactions is determined
by the free energies of the products. Since the free energy of A+BC is lower that the free energy of
AB+C, more BC will be formed.
8a. Lactase does not affect the thermodynamics, but it does affect the kinetics of the reaction. Like
all enzymes, lactase lowers the activation energy barrier of the reaction and allows the reaction
to reach equilibrium faster, but the enzyme does not alter the value of Keq.
8b. All three species are stabilized by interactions with the enzyme, and the transition state is
stabilized the most.
Enzymes must preferentially stabilize the transition state in order to lower the activation
energy of the reaction, which determines the reaction rate. The enzyme must bind more
favorably to the transition state than to the substrate and products so that there will be a net
decrease in activation energy.
8c.
 The enzyme holds the water molecule close to the electrophilic carbon (i.e., the “proximity
effect”).
 The enzyme increases the effective concentrations of the two substrates (another restatement of
the “proximity effect”).
 The enzyme orients the water molecule into a position that is optimal for reaction to occur (i.e.,
the “orientation effect”).
 The enzyme deprotonates water to make water a better nucleophile (base catalysis).