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Solutions to 7.012 Problem Set 1
Question 1
a) What are the four major types of biological molecules discussed in lecture? Give
one
important function of each type of biological molecule in the cell?
Phospholipids: form the bilayers that are the cell membranes.
Carbohydrates: simple sugars supply energy for cellular processes, the polymer
cellulose acts as a
structural component, and the polymer glycogen acts as energy storage.
Proteins: Are responsible for virtually all cellular functions; enzymes for metabolic
pathways, structural
proteins for cell structure, membrane proteins for cell regulation, cell recognition,
and transport of
molecules into and out of cell.
Nucleic Acids: storage and transfer of genetic material.
b) Briefly answer the following questions.
i) What are two main differences between prokaryotic cells and eukaryotic cells?
Prokaryotic cells lack nucleus whereas eukaryotic cells have a nucleus.
Prokaryotic cells are evolutionarily more ancient than eukaryotic cells.
Eukaryotic cells have organelles and internal memebranes, prokaryotic cells do not.
ii) What is the difference between unicellular and multicellular organisms?
Unicellular organisms live and reproduce as single cells. Each cell has the same DNA
and the
same appearance. Multicellular organisms are composed of many cells. Each cell in a
multicellular organism has the same DNA, but the cells can have different
appearances.
iii) Are prokaryotes unicellular or multicellular? Are eukaryotes unicellular or
multicellular?
Prokaryotes are unicellular. Eukaryotes can be either multicellular or unicellular
organisms.
7.012 Problem Set 1 Solutions
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Question 1, continued
c) For the pairs of amino acids given below circle each side chain. Give the strongest
type of interaction that occurs between the side chain groups of each pair.
Amino Acids
GLYCINE
Interaction
GLUTAMINE
TYROSINE
GLUTAMIC ACID
ASPARAGINE
LYSINE
van der Waals
hydrogen
ionic
d) Draw the chemical structure of the following polypeptide at pH 7.
cysteine-alanine-tyrosine-phenylalanine
e) In the structure that you drew above, circle a peptide bond.
7.012 Problem Set 1 Solutions
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Question 2
The drug Minoxidil is used orally as an agent against hypertension and topically as a
stimulant for hair growth. The structure if Minoxidil is shown.
*Minoxidil is uncharged in its active state
Minoxidil
a) Below is a schematic of a Minoxidil binding site on a hypothetical protein.
i) Draw the side chains at amino acid positions 51, 129, 134, and 167.
ii) Draw Minoxidil as shown above binding in the site. Be sure to consider the
interactions between Minoxidil and the side chains when orienting Minoxidil within
the binding site.
7.012 Problem Set 1 Solutions
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Question 2, continued
b) List all the interactions that would occur between the specified amino acids and
Minoxidil in the model you have proposed by filling out the table below.
Amino acid
Glu 51
Val 129
Leu 134
Asn 167
Interactions with Minoxidil
hydrogen bond, van der waals
van der waals, hydrophobic
van der waals, hydrophobic
hydrogen bond, van der waals
c) To decide if your model is correct, you construct some altered versions of this
protein and test whether Minoxidil still binds. Assume all other amino acids remain
unchanged. The results are as follows:
Of the possible orientations for Minoxidil in the binding site, only one orientation is
consistent with the results above. Please check your model carefully, revise it if
needed and answer the following questions.
In the model proposed here, the important features are the hydrogen bonding of
Minoxidil with both Glu51 and Asn167, and the position of the hydrophobic ring of
Minoxidil in the hydrophobic pocket near Val129 and Leu134. If you oriented
Minoxidil in any other way in the pocket, only van der Waals forces exist between the
residues and Minoxidil, and the above changes would still allow these forces.
Explain in terms of your model and the likely interactions why...
i) variant 2 will bind Minoxidil but variant 3 will not bind Minoxidil.
In variant 2, one of the 2 hydrogen bonds remains, as does the hydrophobic pocket,
and given the information this is enough to allow binding. In variant 3, both hydrogen
bonds have been lost, and this disrupts the binding.
ii) variant 5 will bind Minoxidil but variant 4 will not bind Minoxidil.
In variant 4, a hydrophobic amino acid is replaced with a charged amino acid which
disrupts the hydrophobic pocket. The substitutions in variant 5 do not disrupt the
hydrophobic interactions.
7.012 Problem Set 1 Solutions
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Question 3
Growth factor receptors (shown below) are transmembrane proteins found on the cell
surface.
a) The majority of the molecules that constitute the above membrane belong to what
class of macromolecules? phospholipids
Explain the important qualities/properties of these molecules that allow them to form
membranes.
Phospholipids possess hydrophilic “heads” and hydrophobic “tails” that allow them
to assemble into abilayer containing a hydrophobic core when in an aqueous
environment.
7.012 Problem Set 1 Solutions
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Question 3, continued
A smaller schematic of the growth factor receptor is shown here.
b) Which stretch of amino acids in the above sequence is part of the transmembrane
region of the receptor? Circle these amino acids and briefly explain your reasoning.
This stretch contains amino acids that are non-polar or hydrophobic which would
readily reside in the hydrophobic environment of the membrane.
When growth factor binds to the extracellular domain of the receptor, a
conformational change occurs in the receptor. Growth factor binding causes
dimerization of two adjacent receptors in the cell membrane. Upon dimerization, the
intracellular domains of the receptors become activated. See schematic below.
7.012 Problem Set 1 Solutions
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Question 3, continued
c) Regions of the two receptors that interact upon dimerization are drawn below. In
parts (i - iv) below, name the strongest type of interaction (choose from; hydrogen
bond, ionic, covalent, van der Waals) that occurs between the side chains of the
amino acids indicated.
Interacting Side chains
Type of interaction
i) Phe50 : Val98
ii) Asp68 : Lys65
iii) Cys75 : Cys82
iv) Ser53 : Gln12
van der Waals
ionic
covalent
hydrogen
d) Explain how Gln12 and Val98, which are far apart in the primary sequence of the
protein, can be close to each other in the region of the protein diagrammed above.
When the protein folds into its final form, amino acid residues that are far apart in the
primary structure can be closely aligned to one another.
7.012 Problem Set 1 Solutions
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7.012 Problem Set 1 Solutions
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Question 3, continued
e) Molecular interactions between the two receptors are important for dimerization.
Thus, substitution of certain amino acids in the protein can affect receptor
dimerization. Predict whether the receptors will or will not be able to dimerize given
the substitutions (i - iv) below. EXPLAIN your reasoning.
i) Asp68 Arg:
The receptors will not be able to dimerize because this substitution replaces a
negatively charged amino acid with a positively charged amino acid. The ionic bond
between Asp68 and Lys65 is disrupted, and a repulsion occurs.
ii) Ser53 Thr:
The receptors will be able to dimerize because this substitution replaces a polar
amino acid that can participate in hydrogen bonds with another such amino acid.
iii) Phe50 Asn:
The receptors will be able to dimerize even though this substitution replaces a
hydrophobic acid with a polar amino acid because the van der Waals forces remain.
(Within the region diagrammed, the close proximity of the charged species makes it
unlikely that hydrophobic interactions are a key force in the interaction between the
two receptors.)
iv) Val98 Ile:
The receptors will be able to dimerize because this substitution replaces a
hydrophobic acid with another hydrophobic amino acid and the van der Waals forces
remain.
e) Substitution of one amino acid, Cys75 Gly, leads to dimerization of the
receptors with or without growth factor. Provide a brief explanation for this
observation.
This substitution positions two cysteine residues opposite each other. These two
residues can form a disulfide bond and thus covalently link the two receptors
together.
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Question 4
The following reaction is the tenth and final step in glycolysis:
pyruvate kinase
Phosphoenolpyruvate + ADP
Go = -7.5 kcal/mol
pyruvate + ATP
a) Calculate Keq for this reaction under standard conditions at 25oC, and circle the
correct statement below.
b) The following concentrations are found in red blood cells. Calculate G for the
reaction at 37oC. In which direction will this reaction proceed spontaneously?
[ADP] = 10 mM
[ATP] = 81 mM
[phosphoenolpyruvate] = 10 mM
[pyruvate] = 500 mM
G = Go + RT ln ([products]/[reactants])
G = -7.5 kcal/mol + 0.61 kcal/mol x ln (500 x 81)/(10 x 10)
G = -3.84 kcal/mol
The reaction is spontaneous from left to right:
Phosphoenolpyruvate + ADP -------------> pyruvate + ATP
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Question 4, continued
c) Draw the energy profile for this reaction under physiological conditions. On the
diagram be sure to:
1) show relative energy levels of the reactants and the products.
2) label the axes
3) label reactants and products
4) indicate the energy of activation
5) indicate G
d) How does pyruvate kinase (the enzyme that catalyzes this reaction) change the
energy profile?
The enzyme lowers the activation energy only. It does not change the free energy of
the reactants, products, or the overall reaction.
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STRUCTURES OF AMINO ACIDS at pH 7.0
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For the reaction:
A B
C D with Go as its standard free energy
Enzyme Kinetics:
For the enzyme catalyzed reaction:
where: S = substrate
E = enzyme
P = product
the reaction velocity is given by
where:
7.012 Problem Set 1 Solutions