MCB110
FINAL
May 19, 2003
.
Your name and student ID
QUESTION
POINTS
1 (20 points)
2 (30 points)
3 (35 points)
4 (20 points)
5 (20 points)
6 (25 points)
7 (15 points)
8 (10 points)
9 (15 points)
10 (25 points)
11 (15 points)
12 (25 points)
13 (25 points)
14 (30 points)
15 (20 points)
TOTAL (300 points)
WARNING: Your exam will be taken apart and each question graded separately. Therefore, if
you do not put your name and ID# on every page or if you write an answer for one question on the
backside of a page for a different question you are in danger of irreversibly LOSING POINTS!
Student Name and ID #
1(a) - Which of the following organisms do you think must have the highest proportion of
unsaturated fatty acids in their membranes and why (10 p):
(a)
(b)
(c)
(d)
Antarctic fish
Cactus
Bacteria from thermal hot springs
Humans
Antarctic fish (a), as they will need the unsaturated fatty acid chains in order to lower the
critical temperature for the transition from a liquid crystal (fluid) to a gel phase (rigid) below
the temperatures in the Antarctic.
1(b) - Describe the differences in self-association of lipids in monolayers, micelles and
liposomes (10p).
In monolayers the hydrophilic lipid heads face the water at the air-water interface, with the
fatty acid chains facing the air. In micelles the lipid tails face each other in the inside of a
small sphere, with the hydrophilic heads facing the water on the outside. In liposomes the
lipids form a bilayer and give rise to a large sphere with an aqueous core.
2(a) - What are the most common arrangements of secondary structure elements in the lipidinteracting section of integral membrane proteins? Could you predict them having knowledge of
the protein sequence but lacking direct structural data? (10 p)
Transmembrane regions in integral membrane proteins are an alpha helix (single pass) or
alpha helical bundle (several passes) and beta barrels. Transmembrane helices can be
predicted from hydropathy plots by identifying hydrophobic regions in the sequence spanning
about 20-30 a.a. This method will not be able to predict beta barrels.
2(b) – During purification of membrane proteins, what different methods would you use to
separate from the membrane integral membrane proteins, peripheral proteins and lipid-anchored
proteins (10 p).
Integral membrane proteins will have to be solubilized using non-ionic detergents; peripheral
proteins can be release from the membrane with high salt or alkaline pH; phospholipases (or
enzyme that break covalent bonds between the protein and its linked hydrocarbon chain) will
be needed to release the lipid-anchored proteins.
Student Name and ID #
2(c) - You are using FRAP to study an integral membrane protein resident in the ER. What
would be your conclusion if you see a fast rate or recovery? Would you expect a faster or slower
recovery if you lower the temperature of your cells? (10 p)
The recovery after photobleaching means that the protein is free to diffuse within the ER
membrane. Lower temperature will reduce the fluidity of lipids and thus the diffusion rate of
the protein, so recovery will be slowed down.
3(a) - The distribution of K+ across and artificial membrane was measured and equal
concentrations were found on both sides. Which of the following statements is true and why (15
p):
(a)
(b)
(c)
(d)
K+ must be at equilibrium across the membrane
K+ cannot be at equilibrium across the membrane
There cannot be a membrane potential under these circumstances
More information is required to determine if K+ is at equilibrium
(d) We need to know what the electric potential is across the membrane to calculate the
electrochemical G for K+. Only if the electrochemical potential is zero will K+ be at
equilibrium ( G = 0). The fact that K+ concentration is the same on both sides does not
determine the electric potential, as we have to take into account other possible ions.
3(b) – The concentration of chloride ions in the resting neuron is 120 mM outside the cell and 12
mM inside. If there were chloride leaking channels, which way will chloride move during the
resting state? (resting potential –70 meV; 2.3RT = 1.4 kcal/mol; Faraday constant 23
kcal/mol.V). If a certain type of voltage-gated chloride channel were to open at a voltage of +50
meV which way would ions move through the channel? (20 p)
Gresting = 1.4 log [Ci]/[Co] + z 23 (-70) = -1.4 + 1.61 = 0.21 kcal/mol (moving out)
G50 = -1.4 - 1.15 = -2.65 (moving in)
Student Name and ID #
4(a) - Arrange the statements below in the proper sequence of events that occur in an action
potential (10 p):
(a)
(b)
(c)
(d)
The resting potential is reestablished
A stimulus depolarizes the membrane to a threshold
K+ gates open and K+ rushes out of the cell
Na+ gates open and Na+ rushes into the cell
(b), (d), (c) and (a) – An initial depolarization causes the opening of the Na+ channels, the
movement of Na+ into the cell, and further depolarization. This results in the opening of K+
channels. The movement of K+ out of the cell then restores the resting potential (in fact,
causing a momentary hyperpolarization).
4 (b) - A neurotransmitter that binds to a postsynaptic cell and opens K+ channels:
(a) Excites the postsynaptic cell
(b) Inhibits the postsynaptic cell
(c) Depolarizes the postsynaptic cell
(d) Must be acetylcholine
Why? (10 p)
It will inhibit the postsynaptic membrane (b) as the K+ ions will rush out of the cell
hyperpolarizing it.
5 – What are the differences between facilitative transporters, active pumps, and cotransporters?
Can you give an example of the action of such three types of transport in the brush border
epithelial cells of the intestine (20 p).
Facilitative transporters move small molecules down their concentration gradient, pumps,
using energy (most commonly ATP hydrolysis) move ions and small molecules against their
electrochemical gradient, while in cotransport the movement of an ion down an
electrochemical gradient is coupled to the movement of a small molecule against its
concentration gradient. In the brush border cell the glucose-Na+ cotransporter present in the
apical membrane facing the intestinal lumen pumps glucose against a concentration gradient
by using the Na+ gradient generated by the Na+/K+ ATPase, thus creating a high
concentration of glucose inside these cells. On the basal membrane in contact with the blood
stream, the glucose facilitative transporter moves glucose out of the cell and down its
concentration gradient.
Student Name and ID #
6 (a) – You have discovered a new t-snare that localizes to the medial Golgi cisterna. Which
additional protein should you expect to find and where will it likely be located (10 p).
A corresponding v-snare that will be present in vesicles destined to the medial Golgi. The
vesicles could originate from the Cis Golgi assuming a vesicle transport model, or from the
Trans Golgi assuming a maturation model of the Golgi.
6 (b) – By genetic screening you found a protein that results in a sec E phenotype, enlarge Golgi
and disappearance of lisosomes. When you examine your yeast cells by electron microscopy you
see that vesicles have actually started to form but are not able to pinch off. You clone the protein
and the sequence revealed a GTP binding motif to where your mutation maps. What protein
homology should you be considering to explain the function of your newly identified protein and
why? (15)
Your protein is likely to be a homologue of dynamin, which is involved in pinching off of
clathrin coated vesicles in endocytosis. Clathrin also mediates transport from the Cis Golgi to
the lisosome. In addition, dynamin is a GTP binding protein and hydrolysis is required for its
severing action. A dynamin homologue acting in the Golgi-to-lisosome vesicle movement with
a mutation in the GTP site could thus stop the final release of the vesicles resulting in a
growing Golgi, while lisosomes would fail to mature.
7 – You are trying to perform a pulse-chase experiment in your MCB 130L class to follow the
transit of secretory proteins through different cell organelles. Unfortunately you get disrupted
and go for your lunch break before you wash out the cells to get rid of radioactively labeled
amino acids. You fix the cells immediately when you come back and proceed without further
errors through the autoradiography developing process. Where do you expect to see radioactivity
and why:
1 – In secretory vesicles near the apical membrane.
2 – Nowhere, the proteins would have been secreted by now
3 – everywhere, the ER, the Golgi, and in secretory vesicles
What type of experiment conditions will give rise to the other two possibilities? (15 p)
3, everywhere. The pulse has been very large and the chase very short, so proteins containing
radioactive amino acids will be present at all stage in the secretory process. To obtained result
1 you would have needed a chase of about two hours after the pulse had ended, or even larger
for 2 (notice that the pulse time is basically irrelevant in those two cases)
Student Name and ID #
8 - If you break down liver cells by homogenization, separate broken cell membranes (pellet)
from the cytoplasm (soluble fraction) and add epinephrine to the soluble portion, the results will
be (10 p):
(a) Production of cAMP
(b) Binding of epinephrine to its receptor
(c) Activation of adenylyl cyclase
(d) None of the above
Explain your answer.
(d), none of the above. Epinephrine receptors will not be present (they will be in the pelleted
fraction) and therefore there will be no binding and no response (no activation of AC and no
production of cAMP).
9 - Which mutant form of ras is likely to cause malignancy (15 p):
(a) ras that cannot hydrolyze GTP
(b) ras that cannot bind GTP
(c) ras that cannot bind to SOS
(d) ras that cannot bind to Raf
Explain your yes or no answer for each case.
(a) YES – this will maintained ras in an activated form so that it will constantly activate
raf (MAPKKK) and thus the MAP kinase cascade that results in proliferation.
(b) NO – This will maintained ras inactive, so the MAP kinase cascade will not initiate.
(c) NO – This will also maintained ras inactive, as it will not be able to get rid of GDP and
bind GTP
(d) NO – Binding of GTP-ras to raf is required for the activation of raf and the initiation
of the MAP kinase cascade.
10 – Describe two synergistic pathways, involving two different secondary messengers, that
activate the breakdown of glucagons in muscle cells (reminder: glucagon breakdown is carried
out by the enzyme phosphorylase, which is regulated by phosphorylation) (25 p). (Do not
describe pathways that produce the opposite effect)
The enzyme phosphorylating and activating phosphorylase is phosphorylase kinase (or
glucagons phosphorylase kinase or GPK). This enzyme is made of 4 copies of 4 subunits: regulatory subunits that are phosphorylated by PKA (which is activated by cAMP), a subunit
with kinase activity, and a subunit that is calmodulin and is activated by calcium. Thus,
synergistic activation of GKP and therefore the breakdown of glucagon, is obtained by the
presence of both cAMP and Ca2+. The former is produced by adenylyl cyclase, which is
activated via G-protein couple receptor by its interaction with a GTP-bound Gs subunit. The
receptor would likely be a -adrenergic receptor that binds epinephrine (adrenaline). Ca2+
levels can be elevated because the muscle cells are contracting. In this case the ryanodine
receptor in the sarcoplasmic reticulum opens letting Ca2+ into the cytosol upon arrival of the
action potential (they can refer to the change in voltage or to the calcium binding to the
receptor). Another possibility that they could mention would be the activation of IP3 receptors
via the breakdown of PIP2 by PLC, mediated through either a G-protein coupled receptor or
an RTK.
Student Name and ID #
11 – Name and explain two reasons why eukaryotic organisms higher in the evolutionary tree
have more complex pathways with seemingly unnecessary steps (15 p).
The reasons are amplification (with each additional step the signal that originated at the
membrane gets multiplied by involving a growing number of enzymes and secondary
messengers (e.g. the MAP kinase cascade)); and adaptability/flexibility (the numerous steps
allow for the interconnection and integration of many different signals.) – This answer may
come in many different flavors, use your judgment!
12 - We have seen several examples in the course thus far of the validity of “The Goldilocks
Principle” (TGP) whereby there is an optimum solution between extremes. For example, the
fidelity of DNA synthesis of only one error per 109 nucleotides incorporated is very impressive,
but an even higher fidelity can be achieved by single mutational changes in polymerases. Why is
not higher fidelity achieved in nature? To achieve higher fidelity would require an extra input of
energy and would probably slow down replication. Also, mutations are the raw material of
evolution and thus there could be a selection for occasional mistakes in replication. On the other
hand, if the error rate was much higher, too many mutations would result and most mutations
are deleterious. Give two (2) other examples that we covered that illustrate TGP. State clearly
what are the disadvantages of the two extremes compared to the optimum value between the
extremes (25 p).
Student Name and ID #
13 - (25 Points)
Compare and contrast the processes below in terms of the following:
Role of
specific
sequences
Site-specific
recombination
Double-strand
break repair
Type-2
topoisomerase
action
Retro
transposition
Role of high
energy
cofactors
Role of
Holliday
intermediates
Role of DNA
synthesis
Presence of
ester exchange
Student Name and ID #
Question 1. (30 points)
The human zappa-gobulin gene has 6 exons and is alternatively spliced as shown below.
The line on the top of the diagram represents the genomic sequence, with the arrow
indicating the beginning of the transcription unit. Black boxes indicate the exons, which
are numbered. Below the top line are two alternatively spliced mRNA molecules, termed
A and B, that are derived from this gene.
1
2
A
1
2
B
1
3
3
4
5
6
4
5
6
4
5
6
Surprisingly, the proteins encoded by spliced transcripts A and B encode the same amino
acid sequence. For parts a-c, are the following statements is true or false? Give a reason
for every answer.
a. ( 5 points) Exons 2 and 3 must be comprised of exactly the same nucleotide sequence.
False- because multiple codons can encode the same amino acid, different nucleotide
sequences can result in the same protein.
b. ( 5 points) Exons 2 and 3 must have exactly the same length.
True- because there are no other genetic differences between transcripts A and B, exons 2 and
3 must encode the same amino acids, and therefore have the same length.
c. ( 5 points) The length of exons 2 and 3 must be divisible by three to maintain the
reading frame of the gene.
False- the reading frame will be maintained if splicing is performed accurately, and the
exon-intron boundaries can exist within codons.
Student Name and ID #
There are two human diseases associated with mutations in the zappa-gobulin gene. One
mutation causes a mild case of chapped lips. The other mutation leads to severe
immunodeficiency because the skin is fragile and no longer serves as an efficient barrier
to infection.
You clone the gene from both types of patients. You find that both patients have
deletions in the gene, as shown below. The deletions are indicated by the Greek delta
symbol.
DNA sequence analysis shown that apart from these deletions, there are no other
mutations.
1
2
3
4
5
6
1
2
3
4
5
6
You assume at first that the bottom deletion leads to the more severe disease, because it
removes an exon. You’re wrong! The upper deletion causes the severe skin defect.
Consistent with this, patients with the mild disease express a shortened version of the
zappa-gobulin protein, but patients with the severe disease have no detectable zappagobulin protein.
d. (10 points) Propose a molecular explanation for the severity of the disease and the
lack of protein caused by the upper deletion. How would you test your idea?
The deletion might contain an intronic branch point, and therefore prevent splicing between
exons 2 and 4 or 3 and 4. This would likely result in premature translational termination. In
this case, the truncated protein appears to be unstable.
This could be detected by Northern Blotting- do unspliced transcripts appear?
An alternative possibility is that an enhancer required for transcription is removed by the
deletion. This would cause there to be no detectable mRNA.
e. (5 points) Propose an explanation for the mildness of the disease caused by the lower
deletion.
Without exon 5, the protein appears to be largely, but not completely, functional.
Student Name and ID #
Question 2. (20 points)
As a member of the first manned flight to Mars, you have the first chance to analyze the a
new Martian microorganism. This single-celled microbe resembles eukaryotic cells on
earth, having a membrane-bound nucleus, which contains DNA. You isolate nuclei from
these cells and digest them with micrococcal nuclease for the indicated periods of time.
You then remove all the protein by organic extraction, and then analyze the DNA by
agarose gel electrophoresis. You also perform two control experiments: micrococcal
nuclease digestion of Martian DNA that had previously had all protein removed, and
micrococcal nuclease digestion of nuclei from cells from earth, in this case mouse
lymphocytes. You observe the following pattern of DNA migration detected by ethidium
bromide staining.
Size of the DNA molecules in base-pairs is indicated on the right-hand side of the gel.
Lane numbers are indicated on the bottom.
Student Name and ID #
A. (6 points) Why is the pattern of mouse naked DNA different from that observed with
the nuclei?.
Because the mouse DNA is packaged into nucleosomes, which protect the DNA from
digestion, leaving the linker DNA between nucleosomes more easily digested (3 points). The
nucleosomes are regularly spaced ~200 bp apart, which results in the laddered appearance of
regularly spaced bands (3 points).
B. ( 6 points) Are there proteins tightly associated with the Martian DNA in the nuclei?
Why or why not? If there are, what two things can you conclude about how they interact
with DNA?
Yes, because the DNA is protected from nuclease digestion relative to the naked DNA control
(1 point). These proteins appear to protect ~300 base-pairs of DNA because of the limit digest
size (2.5 points), but do not appear to be evenly spaced across the Martian genome, because of
the smeared rather than laddered digestion pattern (2.5 points).
C. ( 4 points) Do the Martians appear to have nucleosomes? Why or why not?
No (1 point), because there are no nuclease-protected particles of ~200 bp length (3 points).
D. (4 points) Why was it important to analyze naked DNA from these organisms at the
same time in this experiment? (lanes 1, 2, 7, 8)
Without this control, you don’t know whether you added enough nuclease activity. Any
interpretation requires that you know that DNA was protected from nuclease digestion
because of bound proteins, not because of insufficient digestion.