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 astructural component, and the polymer glycogen acts as energy storage.
Proteins: Are responsible for virtually all cellular functions; enzymes for metabolic pathways, structuralproteins for cell structure, membrane proteins for cell regulation, cell recognition, and transport ofmolecules 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 thesame appearance. Multicellular organisms are composed of many cells. Each cell in amulticellular organism has the same DNA, but the cells can have different appearances.
iii) Are prokaryotes unicellular or multicellular? Are eukaryotes unicellular or
Prokaryotes are unicellular. Eukaryotes can be either multicellular or unicellular organisms.
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.
d) Draw the chemical structure of the following polypeptide at pH 7.
e) In the structure that you drew above, circle a peptide bond.
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
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
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.
Interactions with Minoxidil
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
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
In the model proposed here, the important features are the hydrogen bonding of Minoxidil with bothGlu51 and Asn167, and the position of the hydrophobic ring of Minoxidil in the hydrophobic pocket nearVal129 and Leu134. If you oriented Minoxidil in any other way in the pocket, only van der Waalsforces 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
In variant 2, one of the 2 hydrogen bonds remains, as does the hydrophobic pocket, and given theinformation 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
In variant 4, a hydrophobic amino acid is replaced with a charged amino acid which disrupts thehydrophobic pocket. The substitutions in variant 5 do not disrupt the hydrophobic interactions.
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
Explain the important qualities/properties of these molecules that allow them to form
Phospholipids possess hydrophilic “heads” and hydrophobic “tails” that allow them to assemble into abilayer containing a hydrophobic core when in an aqueous environment.
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 thehydrophobic 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.
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
, van der Waals
) that occurs between the side chains of the amino acids indicated.
CH2 CH2 CH2 CH Lys65
2N C CH2 CH2 Gln12
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 structurecan be closely aligned to one another.
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)
The receptors will not be able to dimerize because this substitution replaces a negatively chargedamino acid with a positively charged amino acid. The ionic bond between Asp68 and Lys65 isdisrupted, and a repulsion occurs.
The receptors will be able to dimerize because this substitution replaces a polar amino acid that canparticipate in hydrogen bonds with another such amino acid.
The receptors will be able to dimerize even though this substitution replaces a hydrophobic acid witha polar amino acid because the van der Waals forces remain. (Within the region diagrammed, theclose proximity of the charged species makes it unlikely that hydrophobic interactions are a key forcein the interaction between the two receptors.)
The receptors will be able to dimerize because this substitution replaces a hydrophobic acid withanother 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 adisulfide bond and thus covalently link the two receptors together.
The following reaction is the tenth and final step in glycolysis:
Phosphoenolpyruvate + ADP <-------------> pyruvate + ATP
∆Go = -7.5 kcal/mol
a) Calculate Keq for this reaction under standard conditions at 25oC
, and circle the correct
∆Go + RT ln ([products]/[reactants])at equilibrium,
∆G = 0, and Keq = ([products]/[reactants])
-7.5 kcal/mol = -.59 kcal/mol ln (Keq)
12.7 = ln (Keq)Keq = e12.7Keq = 3.3 x 105
At equilibrium, [phosphoenolpyruvate] > [pyruvate]
At equilibrium, [phosphoenolpyruvate] < [pyruvate]
b) The following concentrations are found in red blood cells. Calculate ∆G for the reaction at37oC.
In which direction will this reaction proceed spontaneously?
∆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/molThe reaction is spontaneous from left to right:Phosphoenolpyruvate + ADP -------------> pyruvate + ATP
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.
d) How does pyruvate kinase (the enzyme that catalyzes this reaction) change the
The enzyme lowers the activation energy only. It does not change the free energy of the reactants,products, or the overall reaction.
C + D with ∆Go as its standard free energy
the reaction velocity is given by V = Vmax
What does glutamine do? The cells of the immune system – like all the other cells in the body – need energy, and the fuels they use include glucose and glutamine. When a foreign organism is detected in the body, the cells of the immune system are sent to attack and inactivate it. The immune cells grow and divide rapidly, so the energy demand is high. Studies with isolated cells show that w
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