Question 4(d) , how do i find the pH of the protein.

26 Jun Question 4(d) , how do i find the pH of the protein.

Question
IOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

BIOLOGY 231 Summer 2015 Exam 1 KEY

June 5, 2015

GENERAL INSTRUCTIONS:
1. Answer all questions IN INK without white-out (especially if you might want regrades).
2. NO CALCULATORS.
3. Confine each answer to the BOX provided. You are free to use any space outside the
boxes for scratch work, but ONLY what is written in the box will be graded!
4. SHOW YOUR WORK in all calculations. Be careful to use the right UNITS.
5. Make sure that your name and PU ID is PRINTED LEGIBLY on the FRONT OF
EACH PAGE. The pages may be separated for grading.
6. This exam consists of 9 questions on 12 pages and it is worth a total of 100 points,
plus 5 bonus points.

Information that may be useful:
For the reaction A + B <==> C + D
G = Go + RTln([C][D]/[A][B]) and RTln(X) = 5.7 kJ/mol*log(X)
At equilibrium, G = 0 and ([C][D]/[A][B]) = Keq therefore Go = 5.7 kJ/mol*log(Keq)
log (10X) = X; 10log(X) = X
For an enzyme that obeys Michaelis-Menten kinetics, the relationship between v and [S]
is:
v

Vmax [ S ]
K m [S ]

Or
1 Km

v Vmax

1
1

[S ] V

max

An amino acid has the following structure (amino and carboxy groups shown in their
charged form):

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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

R groups of some amino acids, shown in their charged form, if relevant:

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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

1. Size and scale of chemical and biological objects.
(a) 5 pts. In the table below, indicate the approximate size of each object.
Note that the approximations described in lecture are good enough (in fact,
theyre preferred). Be sure to indicate the size as a fraction of a meter (e.g.
something 1 nm large should be entered as 10-9 meter).
Object

Size as a fraction of a meter

A bacterial cell

10-6 m
(accept 1-5 x 10-6 m)

A eucaryotic cell (a typical small one, not
a dinosaur egg)

10-5 m
(accept 1-5 x 10-5 m)

A covalent bond between two carbon
atoms

10-10 m
(accept 1-2 x 10-10 m)
10-10 m
(accept 5 x 10-11 m to 2 x
10-10 m)

A nitrogen atom

~4 x 10-10 m
(accept 1-5 x 10-10 m)

The optimal van der Waals distance
between two atoms

2. Cellular organelles and compartments
(a) 7 pts. Please indicate which organelle/compartment has the following
functions. There may be one, more than one, or zero organelles for each
function. The same organelle may be used once, more than once, or not at all.
Function

Organelle(s)

Synthesizes cell-membrane components
and proteins to be secreted from the cell

Endoplasmic Reticulum
(ER OK, as is SER and RER)

Contains three kinds of protein filaments
(Hint, not the cytoplasm/cytosol)
Modifies proteins destined for secretion

Cytoskeleton

Digests food particles brought into the cell
The sites of protein synthesis by ribosomes
The location of reactions involving
hydrogen peroxide
Consumes carbon dioxide and releases
oxygen

3

Endoplasmic Reticulum
(ER or RER is OK), Golgi
Lysosomes
Endoplasmic Reticulum (ER or
RER is OK) and cytoplasm
Peroxisomes
Chloroplasts

BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

3. A sizes question.
9 pts. Number each of the following structures so that 1 is the largest and 9 is
the smallest by mass.
___5___ a fatty acid molecule
___2___ human chromosome #1 (about 100 million nucleotides)
___7___ an oxygen gas molecule (O2)
___8___ a water molecule
___3___ the enzyme hexokinase (about 600 amino acids)
___4___ a triglyceride molecule
___9___ a proton
___1___ a human cell nucleus
___6___ a glycerol molecule
4. Protein structure.
Consider the dipeptide with a sequence: N term-Ser-Glu-C term
(a) 5 pts. Draw the structure of this dipeptide in the box below. For each
functional group that can be charged (ionizable group), show it in the charged
form as it would be at pH = 2.5 in your drawing (Hints, you can ignore fractional
charges from the Henderson-Hasselbach equation. Be sure youve indicated all
the hydrogen atoms):

Note the A, these are electronegative atoms, but actually not good hydrogen bond
acceptors (either because they are positively charged or because their electrons are involved
in resonance). However, nothing in the course has taught this, and so these are OK answers.

(b) 2 pts. Draw an arrow pointing to each peptide bond in your drawing.
(c) 4 pts. In your drawing, mark a D next to all possible hydrogen bond donors,
and mark an A next to all possible hydrogen bond acceptors. You can use
arrows along with the Ds and As, if there is any possible confusion. (Hint, not all
hydrogen atoms are possible hydrogen bond donors. Recall, what else is
needed to make a hydrogen a possible hydrogen bond donor.)
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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

(d) 8 pts. Determine the total charge on the dipeptide at each of the following
pH values. Be sure to include the correct sign (positive or negative). You can
ignore fractional charges for pH 1, 7, 13. However, you must include fractional
charges from Henderson-Hasselbach for pH 7.8.
What is the total charge at pH = 1?

+1
What is the total charge at pH = 5?

-1
What is the total charge at pH = 10?
0

-2

What is the total charge at pH = 7.8? (Hints, remember to include HendersonHasselbach and its simple math.)
The ½ charge arises because the amino group
0
of the N terminus is half-protonated when the
-1 ½
pH equals the pK. The pK of the N terminus is
7.8 as in the table on page 2.
5. Reaction energies.
You have isolated an enzyme, creatine kinase, which catalyzes the reaction:

creatine + ATP <=> phosphocreatine + ADP
In a test tube, you combine a tiny amount of pure creatine kinase and a mixture
of creatine and ATP and find that the equilibrium constant, Keq = 100.
(a) 5 pts. Using the Keq, calculate Go for this reaction. Show your work.

Go = -5.7 kJ/mole * log (Keq)
= -5.7 kJ/mole * log (100)
= -5.7 kJ/mole * 2 = -11.4 kJ/mole

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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

(b) 5 pts. You want to see whether this reaction is spontaneous under the
conditions in a cell. You develop the appropriate assays and find that a cell has:
[creatine] = 0.1 mM
[ATP] = 10 mM

[phosphocreatine] = 5 mM
[ADP] = 2 mM

What will be the G (notGo) of the reaction starting from these conditions?

G = Go + 5.7 kJ/mole * log [(5 mM * 2 mM) / (0.1 mM * 10 mM)]
G = – 11.4 kJ/mole + 5.7 kJ/mole * log [10]
G = – 5.7 kJ/mole
(c) 2 pts. Will this reaction proceed spontaneously from left to right (as
written above) or right to left (as written above) starting from the conditions found
inside the cell? Please answer L->R or R->L.

L -> R
(d) 3 pts. Please explain part (c). There are at least 2 ways to explain part (c),
but the preferred way (for full credit) will be based on an understanding of
the ΔG from part (b). Other ways receive partial credit. (3 sentences or fewer)

Full credit: ΔG, the real ΔG starting from these conditions (from part
b), is < 0. Note the fact that ΔGo from part (a) is < 0 is irrelevant. ΔGo
starts from ideal (standard) conditions, not the conditions here.
Half credit: K (q), calculated from part b, is less than stated Keq.
(e) 5 Bonus pts. Even under conditions where the reaction proceeds
spontaneously, it is very slow. Please explain (in 3 sentences or fewer) how a
reaction can be spontaneous and slow? If the G is calculated above, what
other energy do we need to know to explain the slow reaction. (Hint, a suitable
diagram is an excellent way to answer this question.)

Spontaneous reactions can still be very slow (thermodynamics is
different from kinetics). The rate of the reaction will be
determined by the activation energy; slow reactions have larger
activation energies and spontaneous reactions can still have a
large activation energy. See the diagram on slide 6 of the
Lecture 5 notes.
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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

6. Enzyme kinetics.
You are studying the hexokinase enzyme of glycolysis from two different species,
the short-lived rodent, the pocket gopher of North America (below left) and the
related long-lived rodent, the naked mole rat of East Africa (below right).
Interestingly, the naked mole rat is one of the few mammals that is eusocial,
having a social organization like that found in ants. It is also the only mammal
that does not maintain a constant body temperature.

You measure the velocity of the two enzymes (in mmoles of product produced
per minute per nmole of enzyme) at several different substrate concentrations.
Then you plot your data using the Lineweaver-Burk equation. The best lines
through your data points are shown below.
A = naked mole rat; B = pocket gopher.

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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

(a) 8 pts. In the box below, calculate the KM (left) and vMAX (right) of the two
enzymes. Be careful to include the correct units. Show your work.
Enzyme A, KM:

Enzyme A, vMAX:

Vmax can be calculated from the
y-intercept = 1/Vmax

Km can be calculated from the xintercept = -1/Km or from the slope
= Km/Vmax
x-intercept = -1 -> Km=1 mM
or
Slope = 1 = Km/Vmax
1 = Km / 1 -> Km = 1 mM

y-intercept = 1 -> Vmax = 1 mmole
P/minute-nmole enzyme.

Enzyme B, vMAX:

Enzyme B, KM:

Km can be calculated from the xintercept = -1/Km or from the slope
= Km/Vmax
x-intercept = -4 -> Km=¼= 0.25mM
or
Slope = ½ = Km/Vmax
½ = Km / ½ -> Km = ¼ = 0.25 mM

Vmax can be calculated from the
y-intercept = 1/Vmax
y-intercept = 2 -> Vmax = ½ mmole
P/minute-nmole enzyme.

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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

(b) 4 pts. Will one of these enzymes have a smaller velocity at every substrate
concentration? If so, which one? Please explain you answers. (3 sentences or
fewer)
No, at high substrate concentrations (small 1/[S], towards the left) enzyme B has
the smaller velocity (greater 1/v). This reflects the fact that enzyme B has the
smaller Vmax.
At low substrate concentrations (large 1/[S], towards the right), enzyme A has a
smaller velocity (greater 1/v). This reflects the fact that A has a larger Km so
responds more slowly as the concentration of substrate increases.
7. Molecular structure and inhibitor types.
Examine the two molecules shown below. The top molecule is the important
vitamin and metabolite folate. Consumption of adequate folate is important for
both male and female fertility and an important nutrient for pregnant women in
preventing birth defects. The bottom molecule is an anticancer drug called
methotrexate.

Folate

Methotrexate

Inside cells folate is first reduced to dihydrofolate. Dihydrofolate is then a
substrate in the following reaction catalyzed by the enzyme dihydrofolate
reductase (DHFR)
Dihydrofolate + NADPH Tetrahydrofolate + NADP+
Tetrahydrofolate is then used in the synthesis of purines, the pyrimdine
nucleotide TMP and several of the amino acids. All of these syntheses can be
stopped by inhibiting DHFR, so theres been an intense search for inhibitors of
DHFR as anticancer, antiparasite and antibiotic drugs. One of the best reversible
inhibitors of DHFR is methotrexate.
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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

(a) 3 pts. Based on the structure of methotrexate, what kind of reversible
inhibitor is it? Please explain (in three sentences or fewer).

Methotrexate has a chemical structure that is very similar to folate,
therefore methotrexate is likely to compete with folate for binding to
the protein, making it a competitive inhibitor.

(b) 3 pts. Where on DHFR would you expect methotrexate to bind? Please
explain (in three sentences or fewer).

Since methotrexate is structurally similar to folate/a competitive
inhibitor of folate, we would expect methotrexate to bind to the active
site of DHFR.

(c) 3 pts. Please explain why irreversible inhibitors of DHFR are unlikely to be
useful as drugs (in 3 sentences or fewer).

Irreversible inhibitors of DHFR are likely to be toxic, just like
irreversible inhibitors of acetylcholinesterase (e.g. VX) are too toxic to
use as drugs. Of all the drugs in use, only a small fraction are
irreversible inhibitors and these have a special reaction mechanism
(called suicide inhibition that creates greater specificity towards the
target enzyme, although I dont expect anyone to know that).

8. Hierarchy of protein structure.
As discussed in lecture protein structure is organized into a hierarchy of primary,
secondary, tertiary and quaternary.
(a) 5 pts. Please explain how the regular secondary structures are
compatible with a large number of amino acid sequences? Hints: Which
atoms are involved in the hydrogen bonding that promotes regular secondary
structure? Where are the amino acid side chains located in the regular
secondary structures?
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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

The regular secondary structures are compatible with a large number of
sequences because the hydrogen bonding occurs only between backbone
atoms. The different side chains are not involved. Also, the side chains
point away from the main chain in the secondary structures, either away
from an alpha helix or pointing away from both sides of a beta sheet.
(b) 5 pts. What elements make up tertiary structure? Hint, there are the
secondary structure elements (name them in your answer) and something(s)
else.

Tertiary structure is made of the alpha helix and beta strand
secondary structure elements and turns and loops (sometimes called
random coil) that connect the elements and allow the protein chain
to fold back on itself to make a compact structure.

9. Protein folding.
Many kinds of interactions (bonding) play a role in protein folding.
(a) 5 pts. List the various kinds of interactions (bonding) that play some role
in protein folding.
Hydrogen bonding, electrostatic interactions, van der Waals bonding (attraction
and repulsion) and hydrophobic interactions (hydrophobic effect). Covalent bonds
doesnt have a direct role in protein folding, but peptide bonds are required to
hold the protein backbone together, so its OK (but not required) if you have
either of those as answers. Same for disulfide bonds, OK (but not required).
Many proteins dont have disulfide bonds, and their effect on folding can be small.
(b) 3 pts. Which one(s) of these interactions is responsible for providing
the free energy to make protein folding spontaneous? Please explain (3
sentences or fewer).

The free energy for protein folding comes from the hydrophobic
interaction. The folded protein excludes water molecules from
interacting with the nonpolar side chains, so this increases the
entropy of water and makes for a favorable free energy (deltaG < 0).
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BIOL 231 Summer 2015 Exam 1

Name:____ KEY _____________________
PU ID: _____________________________

(c) 3 pts. Which one(s) of these interactions plays a large role in the formation
of protein secondary structure? Please explain in 3 sentences or fewer.

Hydrogen bonding plays a large role in the formation of secondary
structures. Hydrogen bonding between the backbone atoms stabilizes
the alpha helix and beta sheets, and places both backbone (and some
side chains) in the correct final structure.
(d) 3 pts. Which one(s) of these interactions helps shape the final structure of
the protein making sure the protein atoms are well-packed (without much empty
space), but not too close to each other? Please explain (3 sentences or fewer).

Van der Waals (attraction and repulsion) helps shape the final
structure. The van der Waals attraction attracts all atoms, but the
repulsion keeps them from getting too close. Together they lead to a
well-packed interior for a protein with little empty space.

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