MIT Department of Biology
7.013: Introductory Biology - Spring 2004
Instructors: Professor Hazel Sive, Professor Tyler Jacks, Dr. Claudette Gardel
7.013 Problem Set 2
Friday, February 20,
Problem sets will NOT be accepted late.
Question 1
Your friend, an expert structural biochemist, has managed to solve the crystal structure of Your
Favorite Enzyme (YFE) bound to its regulatory subunit (for explanation of x-ray crystallography,
see p. 440 of your textbook). Through in vitro biochemical approaches, you have ascertained that
the enzyme loses all activity when not bound to its regulatory subunit. YFE catalyzes the
following simple chemical reaction:
Lemons
YFE
-------------->
Lemonade
One of your new goals is to perform site directed mutagenesis and pH dependence experiments
with your protein and thus better understand the association between the enzyme and its
regulatory subunit. Site-directed mutagenesis involves selectively mutating single amino acid
residues of a protein. Your initial experiments concern the following stretch of amino acids:
a) Draw the chemical structure of the side chains of the amino acids in the below figure.
Regulatory
Subunit
Asp192
CH2
Val193
Glu194
CH
Glu195
CH3 CH3
CH2
COO
<CHARGE>
NH2 = C
NH
NH2 CH2
CH
CH
Enzyme
CH2
CH2
CH2
O=C
<VDW>
O
<CHARGE>
COO
O
<POLAR>
C
NH2
CH3
CH2
Ala206
CH2
Arg205
Gln207
CH2
NH3
CH2
CH2
CH2
Lys208
<DISULFIDE BRIDGE>
Cys283
CH2
S
Cys304
CH2
S
1
Figure by MIT OCW.
ii. For each interaction listed below, identify the strongest type of chemical bond or
force expected between each amino acid pair.
Choose from covalent, hydrogen, ionic, or van der Waals.
a. Asp192:Arg205
ionic
b. Val193:Ala206
van der Waals
c. Glu194:Gln207
hydrogen
d. Glu195:Lys208
ionic
e. Cys283:Cys304
covalent
c) Describe the expected results of the following mutations on the activity of the
enzyme. Rationalize whether and why you will see a loss of activity of the enzyme by
being specific as to how chemical bonds may be changed. [NOTE: Your explanation will
constitute the majority of points awarded.]
a. Arg205->Ala
Loss – remove salt bridge
b. Ala206->Leu
No loss – conserved substitution – hydrophobic vs. hydrophobic (VDW)
c. Gln207->Asn
No loss- conserved substitution – polar vs. polar (H-bonds)
d. Lys208->Arg
No loss – conserved substitution – charged vs. charged (Salt-bridge)
e. Cys283->Ala
Loss – Disrupt disulfide bridge in enzyme
2
Question 2
The structure of the lipid bilayer does not allow for many polar molecules to pass into the cell.
Using genetic approaches, your lab has identified pair of proteins hypothesized to be involved in
transporting two different molecules from outside of the cell into the cytoplasm.
Through common biochemical assays, Protein A has been identified as a monomer at physiological
concentrations. You decide to analyze the amino acid sequence by constructing a hydropathy plot
(which considers the relative hydrophobicity along the primary amino acid sequence—for example,
tryptophan would be scored as relatively hydrophobic and serine would be scored as relatively
hydrophilic). Your analysis yields the following result.
Relative Hydrophobic
Character
50
100
150
200
250
300
AA Residue #
Relative Hydrophilic
Character
Scale:
= 20 amino acids
Figure by MIT OCW.
a) Based upon the above hydropathy plot, and given that a transmembrane segment is
about 20 amino acids long, might you expect Protein A to be embedded in the
membrane? If so, show how it would interact with the lipid bilayer below:
Extracellular
Cytoplasm
(The protein should be associated with the membrane and should only cross once.)
Figure by MIT OCW.
3
You perform an identical analysis on Protein B and are puzzled by the results. A
hydropathy plot does not yield any indication of membrane localization. Thankfully, your
pal, an expert biochemist, has characterized Protein B as a homohexamer with strong
alpha helical character. Quaternary structural analysis predicts the following
arrangement of molecules.
Figure by MIT OCW.
b) Would you expect this protein could serve the role of a channel for the transport of
polar molecules? What is your reasoning after hearing of the above structural data.
Because of the hexamer structure of the complex, it is possible that the protein spans
the membrane.The alpha helical nature would allow for the protein to be “amphipathic”.
c) Why is the alpha helical nature of the protein crucial to your assessment of its
potential membranous or cytoplasmic localization? Draw a possible orientation of a
Protein B hexamer with respect to a lipid bilayer.
Alpha helical rotations encompass 3.6 aa residues. Therefore, polar residues could point
in toward the aqueous channel and hydrophobic residues could point into the bilayer. You
cannot rule out a membranous localization for this protein based on the hydropathy plot
cannot rule out a membranous localization for this protein based on the hydropathy plot.)
Figure by MIT OCW.
4
Question 3
H
a) Draw the tetrahedral cage-like crystalline structure of water in its solid form with
unshared electron pairs, etc. Draw at least one water molecule and its nearest neighbors.
H
O
H
H
O
H
O
H
O
H
H
O
H
Figure by MIT OCW.
b) What type of bonds was made between one water molecule and another in the figure
above? Hydrogen bonds
c) What property of molecular water allows for the structure you have drawn?
unpaired electrons attract electropositive hydrogens from other water molecules – up to 4 H-bonds can be made per
water molecule. This is the case for ice. The polar character of water allows hydrogen bonds to form.
Consider the following thermodynamic data (DG=DH -TDS.)
H20 (solid)
‡
H20 (liquid)
DH = + 5.9 kJ/mol
DG=0, at 0°C
d) Would you predict that the reaction is favorable at room temperature? __Yes____
e) Using your knowledge of the solid and liquid forms of water, explain why the DH of this reaction is
positive? Considering your explanation, why does ice melt?
Considering your explanation, why does ice melt?
Going from solid to liquid water requires the breaking of hydrogen bonds, which requires the input of
energy. Ambient heat is responsible for the melting of ice at room temperature. Also entropy is increased
in water as opposed to ice so both a higher temperature and positive entropy value is subtracted from the
DH value resulting in a spontaneous reaction at room temperature..
f) Consider the following observation about tertiary structure of proteins – hydrophobic residues tend to
be clustered and often buried deep within the protein. Explain this observation using what you know about
the favorable nature of hydrogen bonds.
(It is energetically unfavorable for hydrophobic residues to be exposed to water. Water forms ordered structures
around hydrophobic regions. It is favorable for hydrophobic residues to cluster so that water is removed from their
environment and is free to form hydrogen bonds. This also increases the entropy or disorder, which is favorable for a
free energy change.)
5
Question 4
You have recently picked up a research project that was initially undertaken by a former
student in your lab. The project involves the biochemical characterization of a protein
with a known enzymatic activity. You pull out the student’s notebook and see the
following graph:
8,000
Reaction Velocity (S-1)
7,000
Km = 30nm
Vmax = 8,000 S-1
6,000
5,000
4,000
3,000
Km = 30nm
Vmax = 4,000 S-1
2,000
1,000
0
50
100
150
200
250
[S] (nM)
Figure by MIT OCW.
a) Describe what is going on in this experiment. Estimate a vmax and Km and write these on
the graph above.
A fixed level of enzyme is initially present with varying concentrations of substrate. The
line is made by measurements from a series of experiments. Vmax is the maximal
velocity at “maximal substrate” Km is the concentration of substrate at half-maximal
reaction velocity.).
b) Why does the curve approach a maximum?
The curve levels off because the amount of enzyme is small relative to substrate and it
became saturated.
c) On the same graph, draw the corresponding curve if twice as much enzyme was added
initially in the experiment. From your new data, estimate a vmax and Km and write these
next to your new data.
6
d) On the coordinates below, draw a simple reaction profile for an energetically favorable
one-step reaction with a high activation energy. Label the reactants and products.
Relative Energy
Uncatalyzed reaction
Reactants
Enzymecatalyzed reaction
Products
Reaction Coordinate
Figure by MIT OCW.
e) Now draw the profile for the catalyzed reaction on the same graph with a dashed line.
Finally, comment on the changes that the addition of enzyme imposes on the following:
Choose from “Increases” or “Decreases“ or “No Change”.
A. Free energy of products
None
B. Overall free energy change (DG)
None
C. Activation Energy of forward reaction
Decreases
D. Activation Energy of reverse reaction
Decreases
E. Rate of forward reaction
Increases
F. Rate of reverse reaction
Increases
7
Question 5
a) Nucleotides are the building blocks for DNA polymers. Draw the below dNTPs and
indicate clearly the 5’ end and the 3’-hydroxyl.
Deoxyadenosine 5’-triphosphate (dATP)
O
O
P
O
O
<5' end>
O
O
-
P
O
O
P
H
O
O
CH
CH2
C H
H
N
O
H C
C
C
OH H
N
N
H
C
C
N
C
C
N
H
H
<3' OH>
Figure by MIT OCW.
Deoxycytosine 5’-triphosphate (dCTP)
O
O
P
O
<5' end>
O
O
-
P
O
O
O
P
O
C H
<3' OH>
C
CH
CH2
H
H
O
C
N
O
H C
C
OH H
H
H
N
C
H
N
C
O
Figure by MIT OCW.
Melting of double stranded DNA to single strands is an important technique in molecular
biology. Your favorite computer program predicts the below melting temperatures for
the corresponding oligonucleotide sequences you plan to use for a technique called PCR
(polymerase chain reaction). In this process we will learn about later in the semester,
the oligonucleotides will anneal to complementary DNA.
Oligonucleotide sequence
TGTAACCATTATCATATTCATGAC
AAGAGCGCTTGGTGCTTGCATT
Melting temperature (Tm)
50.2° C
62.1° C
b) Explain a possible reason for the disparity in melting temperature.
The GC content is the reason for the difference in melting temperature
c) What is the chemical basis for this disparity? G-C base pairs are capable of making more hydrogen bond
contacts and thus form stronger bonds. Therefore, the more GC-rich, the higher the melting temperature. To understand this
visually, consider the sites which could form hydrogen bonds in the nucleotides you have drawn above. There are three such sites in
dCTP and only two in dATP.
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STRUCTURES OF AMINO ACIDS at pH 7.0
O
O
H
C CH3
ALANINE
(ala)
H
O
O
H
N
C
H
+
C
C CH2
NH3
+
O
N
C
H
H
H
O-
O
C CH2CH2
S
CH3
H
H
C C CH3
NH OH
+ 3
THREONINE
(thr)
C CH3
NH3
+
CH3
H
H
O-
H
H
C CH2
O-
C
C
H
C CH2
OH
NH3
+
SERINE
(ser)
PROLINE
(pro)
H
H
H
H
O
O
NH3
+
TRYPTOPHAN
(trp)
O
-
O
H
N
C
NH3+
NH3
+
H C CH2
CH2
H
N
CH2
H +
H
C CH2CH2CH2CH2
LYSINE
(lys)
O
PHENYLALANINE
(phe)
O
OC H
C CH2
LEUCINE
(leu)
C CH2
O
C
H
C
NH3
+
METHIONINE
(met)
GLYCINE
(gly)
H
H
C H
NH3
+
O
C
NH3
+
O
H
H
O-
O
C
H
C C CH2CH3
ISOLEUCINE
(ile)
HISTIDINE
(his)
NH2
C
NH3 CH3
+
H
H
C
O
O
O
C
O
GLUTAMINE
(gln)
O
C H
O-
ASPARTIC ACID
(asp)
O
NH3
+
GLUTAMIC ACID
(glu)
CYSTEINE
(cys)
NH3
+
OC CH2CH2
O-
O
C CH2 C
NH2
C
H
C
H
ASPARAGINE
(asn)
O
NH3
+
NH3
+
NH3
+
O
C CH2CH2
C
O
C CH2 C
NH2
+
C
C CH2 SH
H
NH2
O-
O
C
H
C
ARGININE
(arg)
O
O
C CH2CH2CH2 N
O-
O
C
H
NH3
+
NH3
+
O
O
C
C
H
O
O
H
H
C
C
H
O
O
C CH2
OH
NH3
+
H
TYROSINE
(tyr)
H
H
C C
NH3 H
+
CH3
CH3
VALINE
(val)
9