Download QUESTION POINTS TOTAL (300 points)

MCB110
FINAL
May 16, 2005
.
Your name and student ID
QUESTION
POINTS
1 (15 points)
2 (10 points)
3 (15 points)
4 (15 points)
5 (15 points)
6 (25 points)
7 (20 points)
8 (20 points)
9 (15 points)
10 (20 points)
11 (15 points)
12 (15 points)
13 (50 points)
14 (25 points)
15 (25 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!
MCB 110
Page 2/14
Spring 2005 Final
Name___________________________________
SID _______________________
1. – How does unsaturation in fatty acids affect the physical properties of biological membranes?
(5 pts) When bacteria growing at 20 ˚C are warmed up to 30˚C, are they more likely to
synthesize membrane lipids with (a) saturated or unsaturated fatty acids, and (b) short-chain or
long-chain fatty acids? Explain (10 pts)
Unsaturation increases the fluidity of membrane lipids by difficult close packing of the
fatty acid chains.
(a) saturated (b) long-chain – By increasing the proportion of saturated and long-chain
fatty acids, which have higher melting points (decrease fluidity), the bacteria maintain
constant membrane fluidity at the higher temperatures
2. – Explain why lipid lateral diffusion is much faster than transverse diffusion (flipping)
across membranes. (10 pts)
Transverse diffusion is much slower due to the fact that it involves the movement
of the hydrated, polar head across the hydrophobic core of the bilayer. Lateral diffusion
does no have to overcome that energetic barrier
MCB 110
Page 3/14
Spring 2005 Final
Name___________________________________
SID _______________________
Rate of Glucose uptake
3. – The figure shows rates of glucose uptake as a function of sodium concentration for
endothelial cells and pericytes in the retina of the eye. Based on this data, what can you
say about the glucose transporter in each cell type? (15 pts)
Endothelial cells
Pericytes
[Na+]
The glucose transporter in endothelial cells
works indepently from sodium concentration
(movement of glucose into the cell, which will
be against a concentration gradient, must be
coupled to an energy source other that
cotransport of sodium. In pericytes sodium is
required, indicating a coupling mechanism of
cotransport where the movement of glucose
into the cell against a concentration gradient is
coupled to the movement of sodium down its
electrochemical gradient, that is, also into the
cell. Thus the transporter is a symporter.
4. – If the ATP supply in the cell shown suddenly vanished, would the intracellular
glucose concentration increase, decrease, or remain the same? (15 pts)
The concentration of glucose in this cell is kept higher than the outside due to the
coupling of glucose transport against a concentration gradiant by means of the Glucosesodium symporter, which relies of the generation of a concentration gradient of sodium
by the Na-K ATPase. If ATP disappears, the Na gradient will disappear and the
symporter will not be functional, resulting in a DECREASE the glucose concentration in
the cell.
MCB 110
Page 4/14
Spring 2005 Final
Name___________________________________
SID _______________________
5. – For a squid neuron the ionic concentrations for K+ and Na+ are:
ion
[intracellular]
[extracellular]
K+
Na+
410 mM
40 mM
15 mM
440 mM
If the resting potential of the plasma membrane is –70 mV, what would be the direction
of movement of each ion through an open channel permeable to that ion? (temperature 25
˚C) (15 pts)
Gint o = 1.4kcal / mol log 10
Gint o (K +) = 1.4kcal / mol log 10
[C i ]
+ z 23.06 kcal / mol.V (0.07V )
[C o ]
[410]
+ 23.06 kcal / mol.V (0.07V ) = 2.01 1.61 = 0.4kcal / mol
[15]
For K+ the free energy for movement into the cell is positive, indicating the movement is
in the opposite direction, from the inside to the outside
Gint o (Na+) = 1.4kcal / mol log 10
[40]
+ 23.06 kcal / mol.V (0.07V ) = 1.46 1.61 = 3.07kcal / mol
[440]
For Na+ the free enrgy of movement into the cell is negative, so ions will enter the cell
6.- Which compartment(s) of the cell is associated with each of the following (25 pts):
(a) Clathrin (Trans Golgi, late-endosme/lisosome, plasma membrane, and vesicles)
(b) calcium ions in a muscle cell (sarcoplasmic reticulum, partial points for SER)
(c) dolichol phosphate (RER)
(d) LDL receptors (plasma membrane)
(e) COPI proteins (Golgi compartments and vesicles between them)
(f) COPII proteins (ER and Cis Golgi and vesicles between them)
(g) Unbound SRPs (cytosol)
MCB 110
Page 5/14
Spring 2005 Final
Name___________________________________
SID _______________________
Size of labeled peptide
7. – mRNAs of different lengths encoding partial lengths of the secretory protein
prolactin were translated in the presence of microsomes and radioactively label amino
acids. The absence of a stop sequence precludes termination of co-translation
translocation. For each experiment with a given mRNA the soluble and membrane bound
components were separated and both were analysized by SDS polyacrilamide gel
electrophoresis. The prolactin peptides were visualized then by autoradiography of the
gel as shown below. Can you deduce how long the prolactin polypeptide chain has to be
in order for translating ribosomes to engage the SRP and become bound to the
microsomal membranes? How do you think this relates to the size of the polypeptide exit
channel in the ribosome and the length of the signal sequence? What event occurs inside
the microsomes that explains what is seen for mRNA of 130 or more codons? How do
you think that this length of polypedtide relates to the one addressed in the previous
section of the question (25 pts)
S M
50
S
M
70
S
M
S M
S M
90
110
130
Size of mRNA (in codons)
S M
150
The prolactin chain starts to be recruited to the membrane (is present in the membrane
fraction) when it reaches a length of 90 aa. This length of peptide has to stretch the full
length of the exit channel in the ribosome plus the length of the signal sequence, so the
full signal sequence is out of the ribosome and recognizable by the SRP.
When the chain is as large as 130 aa it becames proteolized by the signal peptidase in the
lumen of the ER (microsomes). This length includes as much as the one just described,
plus the width of the membrane bilayer.
MCB 110
Page 6/14
Spring 2005 Final
Name___________________________________
SID _______________________
8. - Describe how LDL transports cholesterol and is taken up by cells. Describe the
mechanism by which the number of LDL receptors on the cell surface is regulated. (20
pts)
LDL particles contain a core of cholesterol esters, a single layer of phospholipids and
cholesterol and a single copy of Apolipoprotein B, rendering the particle hydrophilic on
the outside. The Apolipoprotien B binds to specific LDL receptors in the plasma
membrane of the cell and gets uptaken by regulated endocytosis via clathrin vesicles.
LDL particles are delivered to the lisosome where they are broken down to their basic
constituent: amino acids, lipids and cholesterol. The concentrations of
cholesterol/cholesterol esters affects transcription of the LDL receptor so that
transcription goes down as intracellular cholesterol concentration goes up.
9. – Caffeine inhibits cAMP phosphodiesterase. How does this affect the metabolic
responses to epinephrine? What would be the effect on this response of a toxin, which
inhibits GTP hydrolysis in and a coupled Gi acting on adenylyl cyclase? (15 pts)
Epinephrine binding to its receptor results in the activation of adenylyl cyclase (via an
activated Gs) resulting in an increase in camp and the consequent activation of Protein
Kinase A. If camp is not degraded by phosphodiesterase the response to epinephrine will
persist well after the the hormone has been cleared from the blood stream.
MCB 110
Page 7/14
Spring 2005 Final
Name___________________________________
SID _______________________
10. – Type II diabetics do not respond to insulin because of a deficiency of insulin
receptors. How do you think this will affect the levels of circulating glucose immediately
after a meal? How would the rate of glycogen synthesis in muscle be affected? Explain
your answers (20 pts)
Insulin is secreated from the pancreas following an increase in blood glucose. Insulinresponsive cells such as muscle and fat cells, contain insulin receptors that when bound to
insulin trigger a cellular response that includes the fussion of vesicles containing glucose
transporters to the plasma membrane. This allow for intake of glucose that moves into the
cell down its concentration gradient. If the insulin receptor is not functional, lack of
signaling maintains the transporters in vesicles in the cytosol and glucose cannot enter the
cell. The result is that the concentration of glucose will be high in the blood .
Another effect of the activation of the insulin receptor is the activation of glucogen
synthase though the deactivation of glycogen synthase kinase by protein kinase B.
Protein kinase B, in turn, is activated by its interaction with PIP2, which is form from PI
upon activation of kinases responsive to insulin.
11. – cAMP, IP3 and Ca2+ are second messengers that need to be removed from the cell
in order to terminate signaling upon disappearance of the original extracellular signal.
Describe briefly the mechanisms by which this is carried out for each of them. (15 pts)
cAMP is converted to AMP by camp phosphodiesterase
IP3 is dephosphorylated by phosphatases
Ca2+ is removed by ATP-pumps present in the plasma membrane and the ER (or
sarcoplasmic reticulum)
MCB 110
Page 8/14
Spring 2005 Final
Name___________________________________
SID _______________________
12. Indicate which of the following mutations will have a positive or negative effect in
the activation of the MAP kinase cascade (15 pts):
(a) A mutation in the Proline-rich region of Sos
Negative. This is the region recognized by the SH3 domains in Grb2. Mutations will
potentially inhibit the recruitment of Sos to the membrane, and thus inhibit the activation
of Ras and the activation of the MAP kinase cascade
(b) A mutation in Ras that reduces its affinity for Sos
Negative. This mutation will inhibit the guanosine exchange activity of Sos on Ras, Ras
would not be activated, neither would be the MAP kinase cascade
(c) A non-funtional mutation in Ras GAP
Positive. Inactivation of Ras GAP will maintain Ras constitutively active and the MAP
kinase cascase continually activated too.
(d) A mutation in Ras that increases its affinity for GDP (versus GTP)
Negative. The active form of Ras, that activates the MAP kinase cascade is the GTPbound one
(e) A mutation that inhibits the binding of Ras to MAPKKK (Raf or MEKK)
Negative, as even active Ras will not be able to activate the first of the kinases in the
cascade
MCB 110
Page 9/14
Spring 2005 Final
Name___________________________________
SID _______________________
13. - Protein binding can alter DNA structure, and this altered structure in turn can
promote a specific replication, repair or recombination reaction. For the proteins below,
indicate (1) all important details of the preferred DNA substrate (2) the impact of protein
binding and/or activity on DNA structure (3) any cofactor requirements for effective
DNA binding and/or activity. (50 pts)
(a) RecA
(1)
(2)
(3)
(b) Gyrase
(1)
(2)
(3)
(c) DNA ligase
(1)
(2)
(3)
(d) DnaA
(1)
(2)
(3)
(e) RuvB
(1)
(2)
(3)
MCB 110
Page 10/14
Spring 2005 Final
Name___________________________________
SID _______________________
14. - One of the key steps in regulating differential gene expression in all organisms is the
control of transcription. In prokaryotes, the transcriptional machinery is relatively simple
consisting of RNA pol and Sigma factors plus a few activators and repressors. However,
in eukaryotes and especially metazoans, the transcriptional machinery has evolved into a
much more elaborate system of protein complexes that carry out multiple functions. (25
pts.)
A.
(5pts.) Name and describe three of the critical and distinct stages of the
transcription process for the synthesis of primary RNA products destined to become
mRNA.
Pre-initiation / initiation; promoter clearance/elongation; termination/ processing.
Pre-initiation: involves the formation of stable protein/DNA complexes between
promoter DNA, the RNA pol II complex and various auxiliary factors (TFIIA, B,
D,E,F,H).
Promoter clearance/ elongation: involves the ATP (energy) dependent disengagement
of the RNA pol II complex from the core promoter, the opening up or strand separation
for laying down the 1st few RNA products; the elongation step requires the processive
movement of RNA pol II aided by various elongation factors to synthesize long primary
transcripts.
Termination/processing: involves the release of the elongating RNA pol II complex
accompanied by 3’ cleavage and poly A addition of the RNA transcript.
B.
(5pts.) What are 3 different classes of large multi-subunit protein assemblies that
participate in the formation of an activated pre-initiation complex (PIC) and that
promotes gene transcription in animal cells?
•
•
•
•
Activator/enhancer complexes (enhancesomes)
RNA polymerase II and the basal transcription complex including TFIIA, B,D, E,
F, H.
Co-activator or co-repressor (co-regulatory) complexes (TBP/TAFs; SAGA;
CRSP/mediator; CBP etc.)
Chromatin remodeling and modification complexes: BAF, SWI/SNF/ Brahma,
HDAC, HAT
MCB 110
Page 11/14
Spring 2005 Final
Name___________________________________
SID _______________________
C.
(5pts.) The TATA binding protein (TBP) is part of a large transcription complex
– what is it called and what are 3 functions attributed to the TBP-containing complex?
TBP is part of the TBP/TAF complex also called TFIID. This multi-subunit complex has
the following documented functions:
• Binding to specific seq elements of the core promoter (ie TATA, INR, DPE)
• Targeted interaction with activation domains of enhancer binding activators (ie.
the TAFs can serve as co-activators linking activators to the pre-initiation
complex.)
• Some of the TAF’s have the ability to catalyze various enzymatic reactions
including kinase, acetyl transferase, UB-ligase.
• At least one TAF subunit is also involved in a chromatin recognition function via
bromodomains that bind to acetylated lysines on the N-terminal tails of H3 and
H4.
D.
(5pts.) Several of the different large multi-subunit complexes of the PIC also
contain polypeptides that carry out various enzymatic activities. Name and briefly
describe 3 such enzymes and the nature of their catalytic function.
Catalytic Activities:
RNA pol II large subunit – catalyzes RNA synthesis
TFIIH subunits – catalyze ATP hydrolysis
TFIID large subunit – catalyzes protein Kinase, acetyl transferase, Ub-ligase
E.
(5pts.) In addition to the integral components of the transcriptional pre-initiation
complex that must be assembled at eukaryotic promoters, what other class of coregulators play an important function in activating gene transcription? (HINT: recall what
the nature of the DNA template is for transcription in eukaryotes). How does this class of
co-regulators work to help activate specific genes?
A major class of co-regulators that works in concert with the basic transcriptional
apparatus include: chromatin remodeling cofactors (BAF, ISWI, etc.) and chromatin
modifying complexes: HAT’s and HDAC’s
MCB 110
Page 12/14
Spring 2005 Final
Name___________________________________
SID _______________________
15. (25 pts.)
A. (5 pts.) In addition to switching transcription up or down to regulate gene expression,
what are 5 other steps in the flow of biological information from DNA--->Protein that
can be regulated in eukaryotic organisms?
•
•
•
•
•
•
RNA processing that includes: exon splicing, 5’ capping and 3’ poly A addition
RNA transport in and out of nucleus and stability of RNA products
Control of protein synthesis (initiation, elongation, termination)
Control of protein secretion, sorting and stability
Regulating the intra cellular localization of gene products (nuclear, cytoplasmic,
membrane embedded, etc.)
Covalent modification of proteins by phosphorylation, acetylation, ubiquitination,
methylation, sumolation, etc.
B. (5 Pts.) What process during gene expression was discovered in eukaryotes that was
completely unexpected and initially made no sense? Briefly describe how this unusual
molecular “processing” event works.
One of the most unusual and unexpected findings regarding the control of gene
expression has been the discovery of RNA splicing –the highly controlled and accurate
process of removing introns and splicing multiple exons to stitch together a complete
gene coding sequence with correct uninterrupted open reading frames. RNA splicing
fundamentally involves 2 consecutive trans-esterification reactions catalyzed by RNA
itself with the help of a large complex of RNP’s (RNA/protein complexes) called the
splicesome. Another equally unusual but less dramatic finding related to splicing was the
discovery of catalytic RNA or ribozymes as well as the double stranded RNA triggered
siRNA system.
MCB 110
Page 13/14
Spring 2005 Final
Name___________________________________
SID _______________________
C. (5 pts.) In the translation of mRNA into proteins, there are several key steps that
require energy input in the form of nucleotide triphosphates. What are three of these
steps and what is the purpose of each one of these reactions?
GTP hydrolysis is used at several stages during protein synthesis including:
• formation of the initiation complex
• First step in elongation : binding of the second amino-acetyl-tRNA
• 3rd step in elongation – translocation
• the other step that requires NTP hydrolysis is charging of t-RNA’s—the
attachment of aa’s to tRNA using hydrolysis of ATP as energy.
D. (5 pts.) Briefly explain how auto catalytic RNA works and which of the two major
gene expression stages utilizes such RNA enzymes?
Auto catalytic RNA is involved in RNA splicing and in protein synthesis. These
processes generally involve the folding of RNA into very specific structures that
juxatapose various key “catalytic” residues such that multiple transesterfications
reactions can occur. Usually either a 3’ or 2’ OH participates in the esterification
reaction with a phosphate moeity at the 5’ end of the RNA.
MCB 110
Page 14/14
Spring 2005 Final
Name___________________________________
SID _______________________
E. (5 pts.) In bacteria, transcription and protein synthesis are temporally and physically
coupled. What processes in eukaryotes are “coupled” to transcription? What key
molecular components within the RNA polymerase II complex is thought to be a
molecular integrator of this coupling system?
In eukaryotes, the processes of RNA processing that include 5’ capping, splicing and 3’
end cleavage and poly A addition appears to be “coupled” to transcription. One of the
key molecular “handles” that seems to help coordinate the coupling is the CTD of RNA
pol II, a long repeating unit of 7aa rich in ser, thr etc.