Final

Question 1

L20. Briefly describe the process of phagocytosis. What is the fate of materials engulfed by
phagocytosis

Answer 1

Phagocytosis can be considered to consist of 3 stages:

(1) Entrapment: the cell extends pseudopods (“false feet”) to contact and surround its
“prey;”
(2) Engulfment: the pseudopod membranes fuse, internalizing the prey in a “phagosome;” and
(3) Digestion: the phagosome fuses with a lysosome (or Golgi vesicles deliver lysosomal
enzymes to the phagosome) converting it to a “phagolysosome.”
In the phagolysosome, the “prey” is digested by hydrolytic enzymes (proteases, nucleases,
lipases, glycosidases), and the resulting products transported to the cytoplasm for recycling.

Question 2

L20. What is autophagy?

Answer 2

Autophagy literally means “self eating.” In autophagy, cells surround worn
out organelles with specialized regions of ER, enclosing them into “autophagosomes.” From there,
they are delivered to lysosomes for recycling of their components.

Question 3

L20. When starved for amino acids, cells of the Baker’s yeast Saccharomyces cerevisiae
upregulate autophagy, internalizing and digesting their own plasma membrane and plasma
membrane proteins. Why might this be beneficial to the yeast cells?

Answer 3

Autophagy of the plasma
membrane (and membrane proteins) and organelles provides a source of amino acids for the cell to
survive periods of starvation (albeit briefly)

Question 4

L20. Diagram/compare the mechanisms of pinocytosis of lucifer yellow (LY) and receptormediated
endocytosis of low density lipoprotein (LDL). Label the important intermediates and
the compartment in which LDL is uncoupled from receptor (LDL-R), and the fate of the LDL-R.

Answer 4

Diagram

Both processes use clathrin-coated vesicles to internalized molecules from outside the cell.
The formation of clathrin coated vesicles in both processes is the same:
(1) Assembly of a coat of adaptins and clathrin, driving invagination and formation of a coated pit. In
“receptor-mediated” endocytosis, receptors are recruited to the coated pits via interactions with
adaptins;

(2) Formation of a clathrin-coated vesicle, requiring GTP hydrolysis by dynamin;
(3) Uncoating, requiring ATP hydrolysis by the “clathrin-uncoating ATPase” (an HSP70 family
member) ;

(4) Delivery to the early endosome.
In pinocytosis, material enters the clathrin-coated pit/vesicle simply by diffusion. Thus the
rate of internalization is highly dependent on the extracellular concentration of the molecule being
taken up (at low concentration, very few molecules will end up in each vesicle). The presence of a cell surface receptor ( thus, “receptor-mediated” endocytosis) serves to concentrate low abundance
molecules into coated pits/vesicles for more rapid/efficient uptake.

Question 5

L20. What is the source of energy for uncoupling LDL from LDL-R?

Answer 5

Energy for uncoupling of LDL from receptor in the early endosome is provided by a pH
gradient (the endosome is acidic relative to the cytoplasm), created by an ATP-driven proton pump
in the endosomal membrane.

Question 6

L20. What stage(s) of LDL uptake might be blocked by mutations that inactivate: (1) the
cytoplasmic domain of LDL-R? (2) the adaptin complex? (3) dynamin? Explain each.

Answer 6

1. Mutations that inactivate the the cytoplasmic domain of LDL-R disrupt recruitment of LDL-R to
   coated pits.
2. Mutations that inactivate the adaptin complex would block assembly of the clathrin coat required
   for internalization.
3. Mutations that inactivate dynamin would block the final budding of the vesicle

Question 7

L20. Non-hydrolyzable analogs of GTP, such as GTPS or GMPPCP, inhibit the formation and/or
function of both COP-coated and clathrin-coated vesicles. Compare the role of nucleotide
hydrolysis in the assembly/disassembly of COP- and Clathrin-coated vesicles, indicating which
nucleotide provides the energy for assembly/disassembly of the vesicle coat, how hydrolysis is coupled to the cycle of assembly/disassembly, and what cellular protein/enzyme functions as the
“NTPase.”

Answer 7

COP-coated vesicles:
A GEF in the donor membrane activates the coat recruitment GTPase (either and
ARF or Sar1) by facilitating exchange of GTP for GDP.
The active, GTP-bound form of the coat recruitment GTPase binds to the donor
membrane and recruits COPs, which cause vesicle budding.
Hydrolysis of GTP by the coat recruitment GTPase releases the GTPase and COPs,
uncoating the vesicle. GTPS blocks UNCOATING.
Clathrin-coated vesicles:
Adaptins and clathrins assemble on the membrane to form clathrin-coated pits,
recruiting cargo receptors and their ligands…
GTP hydrolysis by the GTPase DYNAMIN is required for the final budding/scission
deeply-invaginated coated pits to form the coated vesicle. So GTPS blocks vesicle
SCISSION/FORMATION, but NOT coating or uncoating.
ATP hydrolysis by the CLATHRIN-UNCOATING ATPase is used to remove the
clathrin coats within seconds of vesicle formation.

Question 8

L21. Because of the cell-type specificity of intermediate filament subunit proteins (IFPs),
antibodies against IFPs can be used to help identify the tissue of origin of human tumors. This
can greatly improve the accuracy of diagnosis and treatment, often improving a patients
prognosis. What type of intermediate filament protein might you expect to find in:

A. a neuroblastoma (tumor of neuronal precursor cells)?

B. a basal cell carcinoma of the epidermal epithelium?

C. a rhabdomysosarcoma (muscle)?

D. a glioma (tumor of glial cells)?

Answer 8

A.Neurofilament Proteins
B. Keratins
C. Desmin
D. Glial Fibrillary acidic protein (GFAP)

Question 9

L21. Unlike actin filaments and microtubules, intermediate filaments are not “polar.” What does
this mean with regard to intermediate filament structure and function?

Answer 9

IF proteins assemble in parallel to form dimers, but dimers associate in an ant-parallel
orientation to form tetramers. In the filament cross-section, there are 8 tetramers. Because there
is an anti-parallel intermediate AND an even number of subunits in the unit filament, the two ends
of the filament look the same (thus the filament is not “polar.”). This contrasts with actin filaments,
where the actin monomers assemble head to tail, giving the filament two ends with distinct
properties (“plus- (barbed)” and “minus- (pointed)” ends), and microtubules composed of 13
protofilaments arranged in parallel.
There are no motor proteins that use IFs for a substrate to generate force… this MAY
(stress may) relate to their lack of directionality… or it may have other more mundane evolutionary
reasons.

Question 10

L21. Sketch the organization of a skeletal muscle myofibril, including thin and thick filaments,
the Z- and M-lines. Indicate the polarity of the thin filaments. What are some of the major
protein components of each structure?

Answer 10

DIAGRAM

Question 11

L21. In 10 words or less, what is the basic mechanism of muscle contraction?

Answer 11

ATP-dependent sliding of actin and myosin filaments.

Question 12

L21. Myosin and HSP70 are very different proteins with very different functions in the cell.
What common functional characteristic do they share?

Answer 12

Both proteins are using nucleotide hydrolysis to power a cycle of making and breaking highaffinity
protein-protein interactions. HSP70, by binding and releasing hydrophobic regions of
unfoled/misfolded proteins & myosin by binding and releasing actin

Question 13

L21. Briefly outline/discuss
the steps of the “crossbridge
cycle” by which myosin
uses the energy of ATP to
“walk” along the actin
filaments, generating muscle
contraction.

Answer 13

Diagram

1. No nucleotide. Myosin head is tightly bound
   to actin (“rigor”)…
2. ATP bound to myosin releases myosin from
   actin…
3. ATP hydrolysis “cocks” myosin (ADP + Pi
   remain bound). Myosin binds actin weakly…
4. Pi is released, allowing myosin to bind more
   tightly, triggering…
5. …the “power stroke” and release of ADP…
   Myosin heads “walk” towards the “plus-”
   (“barbed”) -end of actin filaments…

“Myosin is an actin-dependent ATPase that acts as a “molecular motor””

Question 14

L21.  The post-mortem
condition know as “rigor
mortis” is characterized by
stiffness and rigidity of all
the skeletal muscles. Based
upon your understanding of
the molecular interactions of
myosin and actin, propose a
molecular explanation for
rigor mortis.

Answer 14

Once ATP is depleted, all myosin heads will bind tightly to actin, locking all the muscles into
a rigid state known as “rigor mortis.” At the molecular level, cell biologists refer to the bound state
in the absence of ATP as “rigor.”

Question 15

L21. Why do you think rigor mortis subsides several hours after death?

Answer 15

As muscles and other cells die, organelle membranes lyse. Proteases released from the
lysosomes cleave myosin and actin, “tenderizing” the muscle (this is why “aged” beef is more tender).

Question 16

L21. How does Ca2+
regulate contraction
of skeletal muscle in
vertebrates?

Answer 16

Diagram

Tropomyosin dimer binds along actin filament…
Troponin complex binds to tropomyosin…
In the absence of Ca2+, tropomyosin blocks
myosin binding site…
Troponin C binds Ca2+…
Allosteric change in structure of troponins and
tropomyosin uncovers myosin binding site…
Myosin walks on actin and myofibril contracts…
Removal of Ca2+ restores inhibition…

Question 17

L21. Where does the

Ca2+ come from?

Answer 17

Ca2+ is stored
in the “sarcoplasmic
reticulum,” and
released through
voltage-gated Ca2+
channels in response
to an action potential

Question 18

L21. In the common “sliding filament” assay for myosin
motility, myosin molecules are adsorbed to a glass
coverslip. Fluorescent actin filaments are bound to the
myosin-coated coverslips. When ATP is added, the actin
filaments glide along, with a defined polarity and direction
of motion. If a single headed type I (unconventional)
myosin was adsorbed to the coverslip, which end of the gliding actin filaments (“pointed /minus”
or “barbed/plus”) would lead?

Answer 18

Myosins walk towards the plus-end of actin filaments, so the minus- end would lead

Question 19

L21. In the large internodal cells of some freshwater
algae, an inner layer of cytoplasm “streams” around the
central vacuole at rates >1 m min-1. Streaming is
powered by myosin-dependent sliding of
vesicles/organelles along actin filaments at the interface
between the stationary cortical cytoplasm and the
streaming layer (see fig inset). If the polarity of the actin
is as shown in the figure (plus “+” or minus “-“) , which
direction would the vesicle in the figure move (Circle
the arrow, and briefly explain in <10 words; 2 pts).

Answer 19

Diagram

Myosin walks towards the plus end of actin filaments

Question 20

L21. Sketch the characteristic organization of actin microfilaments in the lamellipodium of a
migrating animal cell, indicating/identifying (A) the polarity of at least two actin filaments; (B)
an ARP 2/3 complex; (C) where actin monomers are added to those filaments; and (D) the site of
actin disassembly. (E) Briefly describe the role of actin and ARP2/3 in lamellipodial extension
and cell motility

Answer 20

Arp 2/3 is activated near the
inner face of the plasma membrane
(by Rho family GTPases and other
activating proteins)...
Activated Arp 2/3 complex
nucleates actin assembly and
branching. Arp 2/3 caps minus-ends.
Assembly of the extensively
branched meshwork of actin
filaments drives extension of the
lamellipodium.
Actin filaments then
disassemble (from their minus-ends)
behind the lamellipodium...

Question 21

L21. What is the difference between an actin cross-linking protein and an actin bundling
protein? What are examples of each?

Answer 21

Most actin bundling proteins are rigid rod-like molecules with an actin binding
domain on each end, allowing them to organize actin filaments into parallel bundles.
A good example is a-actinin.

Most cross-linking proteins are longer, more flexible proteins. Again, they have
multiple actin binding domains, allowing them to cross-link actin filaments. However,
since they are long and flexible, the angle of the actin filaments is more variable.
Good examples are filamin and spectrin

Question 22

L21. Cells of the cellular slime mold Dictyostelium discodium express multiple actin bundling
proteins including -actinin and ABP. Interestingly, genetic disruption of either protein alone
has no observable effect, but disruption of BOTH proteins in the same cells resulted in marked
defects in motility and development. What would you conclude about the function of actin
bundling proteins in Dictyostelium?

Answer 22

The proteins may be REDUNDANT… meaning they perform

similar and overlapping functions. If one is defective, the other can take over its function.

Question 23

L21. Briefly compare/contrast three different ways in which the actin cytoskeleton can be linked
to the plasma membrane of the cell. Which of these can be used to integrate the actin
cytoskeleton with the extracellular matrix?

Answer 23

(1) via integrins at focal contacts and adherens junctions: plus-ends of actin filaments linked to
integrins (transmembrane ECM receptor) via linker proteins…
(2) via spectrin complexes of the membrane cytoskeleton: junctional complexes in which actin and
spectrin are linked to glycophorin via band 4.1…
(3) via unconventional myosins, as 100 kDa myosin in brush border microvilli…
(4; not mentioned in lecture) to transmembrane proteins via ERM family linkers (not discussed in
lecture).
The actin cytoskeleton is commonly linked to the ECM via integrin family receptors at focal contacts
and adherens junctions

Question 24

L21. Briefly sketch/compare/contrast the structure
and function of prokaryotic and eukaryotic flagella,
including and labeling the major structural
components of each briefly describing their function
in the generation of motility (what is the energy
source for motility, and what proteins/structures are
the motors?).

Answer 24

Diagram

Eukaryotic cilia and flagella are composed of a 9+2 arrangement of 9 outer doublet
MTs surrounding a central pair of MTs, located within the plasma membrane.

Motility is generated by ATP-dependent sliding of the outer doublet MTs powered by the “dynein”
arms. This sliding is converted to bending by nexin linking the adjacent outer doublets, and is
controlled by accessory proteins/structures of the flagella (central pair and radial spokes)

Question 25

L21. How do eukaryotic cilia differ from eukaryotic flagellae?

Answer 25

Eukaryotic cilia are generally shorter than flagellae, and have an asymmetric strokes. Flagella (euk) are often longer and have symmetric strokes.

Question 26

L21. What structural/morphological defects are apparent in pfA mutants? Briefly describe the normal mechanisms for generating flagellar motility in eukaryotes, and explain why pfA mutants have paralyzed flagellae

Answer 26

In normal flagellae, the dynein arms attached to A subfibers use ATP hydrolysis to power a crossbridge cycle similar to that of myosin, walking towards the minus-end of the adjacent Bsubfiber of the adjacent outer doublet. This MT sliding motion is converted to bending by the nexin links between outer doublets. Mutant pfA is missing its dynein arms. In their absence, the flagellae of pfA are paralyzed

Question 27

L21. Figures C and D depict flagellar cross-sections from two other mutant classes (pfB and pfC) isolated in your screen. What structural/morphological defects are apparent in flagellae from pfB and pfC mutants? Briefly relate these defects to the mechanisms of flagellar motility

Answer 27

pfB is missing the central pair/sheath. pfC is missing the radial spokes. Sliding/bending is temporally and spatially controlled by the central apparatus, including the central pair, sheath, radial spokes, etc. In their absence, the flagellae are paralyzed.

Question 28

L21. What structures contributed by the sperm during fertilization might function as centrioles to organize the centrosome that nucleates the sperm aster?

Answer 28

The basal bodies of the flagella (which are structurally identical to centrioles, and recruit other centrosome components from the egg cytoplasm).

Question 29

L21. Microtubules are highly dynamic polymers, undergoing frequent interconversions between phases of growing and shrinking. What is this behavior called?

Answer 29

“Dynamic instability

Question 30

L21. What effect would you expect GTPS, a non-hydrolyzable analog of GTP, to have on the dynamics of microtubule assembly in vitro? Explain your answer.

Answer 30

Current models of dynamic instability suggest “catastrophe” (the transition from growing to shrinking) is caused by loss of the GTP-tubulin cap at the MT end. GTPS and other nonhydrolyzable analogs of GTP stabilize MTs by slowing or preventing GTP hydrolysis and loss of the GTP-tubulin “cap,” thus keeping the MT in a growing state.

Question 31

L21. What class of proteins regulate microtubule dynamics?

Answer 31

Microtubule-associated proteins, or “MAPs

Question 32

L21. If a kinesin-like motor was assayed in such a manner, which end of the MT would lead (plusend or minus-end)?

Answer 32

Kinesin walks towards the plus-end, so the minus-ends would lead in this assay.

Question 33

L21. If dynein was adsorbed to the coverslip and ATP was added, which end of the MT would lead (plus-end or minus-end)?

Answer 33

Cytoplasmic dynein walks towards the minus-ends of MTs, so the plusends would lead in this assay

Question 34

L21. Based on your understanding of the cytoplasmic distribution/localization of the Golgi and ER relative to MT organization, which motor do you think would power: Vesicle transport from the ER to the Golgi? Vesicle transport from the Golgi to the plasma membrane?

Answer 34

Draw yourself a cell with a perinculear Golgi and ER dispersed throughout the cytoplasm. Where is the centrosome? What motor would be used for peripheral ER to Golgi (dynein... why?)? from Golgi to plasma membrane (kinesin...why?)?

Question 35

L23. The G1/S and G2/M transitions in the division cycle of eukaryotic cells are regulated by_________ protein kinases, which consist of a catalytic subunit and regulatory subunit from the________family of proteins

Answer 35

cyclin-dependent cyclin

Question 36

L23. M-phase promoting factor (MPF; also known by the acronym______) is activated during Mphase by the accumulation of its regulatory subunit

Answer 36

CDK1 Cyclin B or M-phase cyclin

Question 37

L23. Activation of the __________ by MPF targets __________ for destruction in the proteosome, resulting in inactivation of MPF and exit from M-Phase

Answer 37

anaphase-promoting complex (APC) cyclin B= M-phase cyclin

Question 38

L23. Eukaryotic cells monitor completion of important cellular processes at key stages of the division cycle referred to as_____

Answer 38

checkpoints

Question 39

L23. Programmed cell death, or______ is a process used by animals to eliminate damaged or unneeded cells

Answer 39

apoptosis

Question 40

L23. About 5% of the cells are in mitosis

Answer 40

The mitotic index, or fraction of cells in M phase is | 0.05

Question 41

L23. 1/3 of the cells label with 3 | H-thymidine in a 1 hour incubation

Answer 41

Incorporation of labeled | thymidine only occurs in S-phase. 1/3 of cells in S

Question 42

L23. The number of cells doubles every 24 hours.

Answer 42

This is a measure of the doubling time, and | thus the total cell cycle time

Question 43

L23. Cells labeled for 1 hr with 3 H-thymidine were examined at intervals by autoradiography. Initially some cells are seen to have labeled nuclei, while others are unlabeled (Why?). Labeled cells in mitosis are not seen until 6 hours after labeling.

Answer 43

Cells only incorporate labeled thymidine during S-phase... which can be detected as radioactive DNA in the nucleus. After S-phase, these cells will traverse G2 and enter M. The interval between labeling and M phase defines G2

Question 44

L23. Remember the cell cycle

Answer 44

From C, the doubling time (cell cycle time) is 24 hrs. From A, M-phase = 0.05 x 24 hrs = ~1 hr. From B, S-phase = 1/3 of 24 hrs = 8 hrs. From D, the period from the end of S phase until M-phase is ~ 6 hrs, so G2 ~ 6hrs. By subtraction, G1 ~9hrs

Question 45

L23. Sketch the MPF and CLB activity profiles (MPF activity vs. time; CLB vs. time) you would observe in an early Xenopus embryo (include 3 cycles). Label the periods corresponding to Mphase and S-phase (interphase).

Answer 45

Sketch

Question 46

L23. Treatment with cycloheximide (an inhibitor of protein synthesis) arrests the cell cycle of both adult and embryonic cells in interphase. Why? Sketch the effect adding cycloheximide would have on the CLB and MPF profiles of these cells

Answer 46

Sketch Blocking translation with cycloheximide blocks synthesis of CyclinB which is required to activate MFP

Question 47

L23. 8.c Andrew Murray deleted the first 90 amino acids from cyclin B (CLB), removing the “destruction box” that targeted CLB for destruction by the proteosome. When injected into early Xenopus embryos, this altered CLB (called cyc90) could not be degraded. Predict the effect that injecting cyc90 would have on the cell cycle. Sketch the effect of cyc90 on the profiles of MPF activity and endogenous CLB.

Answer 47

Sketch

Question 48

L23. During cycles 2-12, cells of frog embryos divide every ~30'. If treated with aphidicolin (an inhibitor of DNA synthesis), cells cultured from adult frogs will arrest in S-phase. However, embryonic cells (prior to division 13) continue dividing when treated with aphidicolin. Similarly, unlike adult frog cells, cells of early embryos treated with nocodazole continue to progress through the cell cycle. What compromises have early frog embryos made to achieve their remarkably fast cell division cycle?

Answer 48

Early frog embryos have eliminated the DNA replication and spindle assembly checkpoints in order to achieve their rapid division cycles. They have, in a sense, compromised fidelity of division for speed of division.

Question 49

L23. Explain the difference between mitosis and cytokinesis in eukaryotic cells. Which process occurs first?

Answer 49

Mitosis is the term applied to (a particular mode of) nuclear division (chromosome segregation) in eukaryotes. Cytokinesis is the term applied to cytoplasmic division in eukaryotes. In most eukaryotes, division includes MT-dependent segregation of chromosomes during mitosis and actin/myosin (animals) or MT-based (plant) mechanisms of cytokinesis.

Question 50

L23. Diagram a typical metaphase chromosome and label the following structures: chromatid, centromere, telomere, kinetochore.

Answer 50

Diagram

Question 51

L23. What is the relation between the chromatids in a chromosome? How are the chromatids held together until anaphase?

Answer 51

They are siter DNA molecules, the product of replication. The are held together by “cohesins.”

Question 52

L23. What functions do the kinetochores play?

Answer 52

Kinetochores capture MTs to attach the chromatids to the spindle.

Question 53

``` L23. Diagram a spindle in an animal cell, labeling the polar MTs, kinetochore MTs, astral MTs, and spindle poles. Label the polarity of one of each class of MT. What "organelles" function as spindle poles in most animal cells? ```

Answer 53

Diagram

Question 54

L23. Diagram a plant spindle. Briefly compare/contrast | the spindles of animal and plant cells.

Answer 54

Diagram Higher plants do not contain centrioles/centrosomes, and the spindle poles are more dispersed, giving the spindle a barrel shape

Question 55

L23. What are two models for generating the movements of anaphase A

Answer 55

(1) Dynein-related motor proteins at the kinetochore powering poleward movement along the disassembly kinetochore MTs. This model, tho’ attractive, has less experimental support. (2) Shortening of kinetochore MTs provides the energy for chromatid movement. This model requires that kinetochores remain attached to shortening MTs, which could be accomplished with a ring-shaped attachment complex (Dam1 complex in yeast) or side-binding MAPs.

Question 56

L23. What are two model mechanisms for generating the movements of anaphase B?

Answer 56

(1) Traction forces generated by cortical dynein pulling on astral MTs (occurs in some cells). (2) Anti-parallel sliding of overlapping polar MTs in the central spindle (occurs in diatoms).

Question 57

L23. Briefly compare/contrast the mechanisms of cytokinesis in animal and plant cells. Why do you think plants have evolved a mechanism of cytokinesis distinct from that used by animal cells?

Answer 57

Plants: During cytokinesis, the MT phragmoplast delivers Golgi-derived vesicles containing cell wall precursors to the growing cell plate, which divides the cell from the inside-out. Animals: A “contractile ring” of actin and myosin contracts (much like a purse string) to divide the cell. The rigid cell wall of plant cells prevents them from dividing using the actomyosin-based mechanism that animal cells use..

Question 58

L23. Mitosis can be blocked by treating cells with nocodazole, a MT inhibitor that disassembles cytoplasmic MTs and blocks spindle assembly. Somewhat paradoxically, mitosis can also be blocked with taxol, another MT poison that stabilizes MTs. How could you reconcile these observations?

Answer 58

Nocodazole blocks spindle assembly by inhibiting MT assembly, but taxol stabilizes MTs. This indicates that assembly AND DYNAMICS of MTs is required for assembly and function of the spindle.

Question 59

L25. What three features/characteristics are common to ALL cell signaling pathways?

Answer 59

A signal, a receptor, and a response.

Question 60

L25. When the small polypeptide hormone insulin is covalently coupled to 10 m diameter latex beads, it can still elicit a robust response in cells bearing the insulin receptor. In contrast, progesterone coupled to similar-sized beads does not elicit a response in its target cells. Briefly explain this observation, in light of the mechanisms of action of these two hormones.

Answer 60

Epinephrine is a hydrophilic hormone that binds to a cell surface receptor. Even when coupled to beads which prevent its entry into the cell, it can bind receptors on the cell surface and trigger a response. Progesterone, on the other hand, is a lipophilic hormone that activates an intracellular receptor. When coupled to beads, progesterone cannot enter the cell and has no access to the receptor, thus no response.

Question 61

L25. Predict the effect on glycogen breakdown of adding (or injecting into cells) the following reagents: A GMP-PNP (non-hydrolyzable analog of GTP) B 8-bromocAMP (a membrane permeable, less hydrolyzable analog of cAMP) C Excess catalytic subunit of protein kinase (PKA) D Excess regulatory subunit of PKA E excess cAMP phosphodiesterase F Caffeine (an inhibitor of cAMP phosphodiesterase)

Answer 61

A enhanced. locks Gs on B Enhanced. activates more PKA C Enhanced. Active subunit activates more phosphrylase kinase D Suppressed. Binds and inactivates catalytic subunit E Suppressd. Degrades cAMP, which inactivates PKA F Enhanced. increases [cAMP] which activates more PKA

Question 62

L25. Vibrio cholera (the bacteria that causes cholera) secretes a toxin that irreversibly modifies the -subunit of GS so that it can no longer hydrolyze GTP. What effect would cholera toxin have on cAMP-mediated signaling?

Answer 62

``` Cholera toxin irreversibly activates Gs by blocking GTP hydrolysis. Since Gs is always on and activating adenylate cyclase, intracellular cAMP levels increase, resulting in constituitive signaling by cAMP-mediated pathways. ```

Question 63

L25. During blood clotting, platelets are “activated” by thrombin, leading to activation of protein kinase C and phosphorylation of target proteins (integrins, other cytoskeletal proteins, etc) required for platelet function and clotting. Diagram the signaling pathway from the thrombin to protein kinase C.

Answer 63

Diagram Thrombin activates Receptor Activated Receptor acts as GNRP for Gq stimulating GDP release and GTP binding (alpha subunit) GTP binding activates Gq (protein) Activated Galphaq activates phospholipase C Active PLC cleaves PIP2 to IP3 and DAG IP3 opens Ca2+ chanels in ER, release Ca2+ to cyto DAG and Ca2+ activate protein kinase C Active PKC phosphorylates target proteins

Question 64

L25. Based on our discussion of cell-signaling pathways, would you expect the following drugs/treatments to: (1) stimulate/enhance; or (2) reduce/inhibit platelet activation by thrombin: A GTPgS B IP3 C EGTA (binds and sequesters Ca2+ from ER) D Phorbol myrsityl acetate (a phorbol ester that mimics diacylglycerol)

Answer 64

A. stimulate. locks Gq on B. Stimulate. Releases Ca2+ from ER C. Inhibit. Sequester Ca2+ released from ER D. Stimulates by activating PKC

Question 65

L25. Sketch the pathways (from receptor to reponse) by which transcription is regulated by a: A. Steroid Hormone

Answer 65

1. Lecture 25 slide 6

Question 66

L25. Which signaling pathway is most “direct?”

Answer 66

A. Activation/regulation of transcription by steroids (or other lipophilic signals) is more direct than the other two pathways... the receptor itself enters the nucleus and directly modulates transcription. In the other two examples, there are multiple intervening steps (G-proteins, kinases, etc).

Question 67

L25. What steps in each pathway allow for amplification in the signal/response?

Answer 67

Steroid-mediated: Amplification comes from transcription/translation... a single receptor activates a gene to synthesize many copies of an mRNA, which can direct synthesis of many copies of a protein. Some of the first genes activated (sometimes referred to as the “primary response” or “early” genes )may code for transcription factors, which in turn regulate downstream genes (“secondray” responses or “late” genes). cAMP-mediated: Each step in the pathway has the potential for amplification (Receptor to G protein; G protein to adenylyl cyclase, cAMP to PKA, PKA to CREB, etc). Ras-mediated: Each step in the pathway has the potential for amplification (SOS to Ras, Ras to Raf, Raf to Mek, Mek to ERK, ERK to transcription factor, etc)

Question 68

L25. Many signaling pathways involving receptor tyrosine kinases can be inhibited or blocked by over expressing (~3-10-fold excess) a truncated form of the receptor which contains the extracellular ligand binding and transmembrane domains, but lacks the cytoplasmic tyrosine kinase domain. Given your understanding of the mechanism of activating receptor tyrosine kinases, why might over expression of a C-terminally-truncated receptor block signaling?

Answer 68

Think about the structure of the EGF-R and how it is activated by EGF binding... most RTKs are active as dimers, and must cross-phosphorylate their partner’s cytoplasmic domain. If a receptor dimer contains one normal subunit and one truncated subunit lacking the cytoplasmic domain, it cannot be activated: there is no cytoplasmic domain to be phosphorylated, and no kinase domain to phosphorylate the normal subunit. Overexpression of the truncated subunit thus acts as a “dominant negative,” blocking the signaling pathway. This is a commonl tool used by developmental biologists to study the role of signaling pathways in organisms in which standard genetic methods are impractical

Question 69

L25. Why does the G12V mutation lead to uncontrolled cell proliferation? Discuss this result, in light of the observation that more than 30% of all human cancers have mutations in the Ras gene.

Answer 69

Ras activates a MAP kinase signaling cascade (Ras6Raf6MEK6ERK) which, among other things, activates transcription factors that regulate expression of the G1 cyclins, and thus entry into S-phase and cell division. Inappropriate or continued activation of Ras would thus lead to loss cell growth control and inappropriate proliferation (cancer)

Question 70

L25. What effect would each of the following have on the regulation of cell proliferation if expressed in normal cells: A constituitively active for of the EGF receptor?

Answer 70

Uncontrolled proliferation

Question 71

L25. What effect would each of the following have on the regulation of cell proliferation if expressed in normal cells: A dominant negative form of the EGF receptor?

Answer 71

Suppress proliferation in response to EGF

Question 72

L25. What effect would each of the following have on the regulation of cell proliferation if expressed in normal cells A mutation that inactivates SOS function?

Answer 72

Suppress signaling/proliferation dependent on | the SOS/Ras pathway.

Question 73

L25. What effect would each of the following have on the regulation of cell proliferation if expressed in normal cells A constituitively activated form of Raf?

Answer 73

Uncontrolled proliferation

Question 74

L25. What effect would each of the following have on the regulation of cell proliferation if expressed in normal cells Deletion of Raf (MAPKKK)?

Answer 74

Suppress signaling/proliferation dependent on the Ras/Raf | pathway

Question 75

L25. What effect would each of the following have on the regulation of cell proliferation if expressed in normal cells A constituitively activated form of MEK?

Answer 75

Uncontrolled proliferation.

Question 76

L25. Based on our discussion of Ras signaling pathways, would you predict that the following mutations would exacerbate or suppress cell proliferation if they were introduced into the cells expressing Ras bearing the Gly12Val mutation? A constituitively active for of the EGF receptor?

Answer 76

Exacerbate

Question 77

L25. Based on our discussion of Ras signaling pathways, would you predict that the following mutations would exacerbate or suppress cell proliferation if they were introduced into the cells expressing Ras bearing the Gly12Val mutation? A dominant negative form of the EGF receptor?

Answer 77

No effect (it’s upstream)

Question 78

L25. Based on our discussion of Ras signaling pathways, would you predict that the following mutations would exacerbate or suppress cell proliferation if they were introduced into the cells expressing Ras bearing the Gly12Val mutation? A mutation that inactivates SOS function?

Answer 78

No effect (it’s upstream)

Question 79

L25. Based on our discussion of Ras signaling pathways, would you predict that the following mutations would exacerbate or suppress cell proliferation if they were introduced into the cells expressing Ras bearing the Gly12Val mutation? A constituitively activated form of Raf?

Answer 79

Exacerbate

Question 80

L25. Based on our discussion of Ras signaling pathways, would you predict that the following mutations would exacerbate or suppress cell proliferation if they were introduced into the cells expressing Ras bearing the Gly12Val mutation? Deletion of Raf (MAPKKK)?

Answer 80

Suppress (it’s downstream of Ras, and breaks the signal | chain).

Question 81

L25. Based on our discussion of Ras signaling pathways, would you predict that the following mutations would exacerbate or suppress cell proliferation if they were introduced into the cells expressing Ras bearing the Gly12Val mutation? A constituitively activated form of MEK?

Answer 81

Exacerbate.

Question 82

L25. Where (specifically) would you find the stem cells for RBC and epidermal epithelium?

Answer 82

Hematopoietic stem cells are found in the bone marrow. Epidermal stem cells are found in the basal layer of the epidermis.

Question 83

L25. What three properties, features, or attributes do these stem cell populations share (along with other populations of stem cells?

Answer 83

1. They are determined, but undifferentiated; 2. They are capable of unlimited proliferation (within the lifetime of the organims); 3. They are self-renewing: when they divide, at least one daughter retains the properties of a stem cell.

Question 84

L25. What unique property, feature, or attribute of “embryonic stem (ES) cells” and “induced pluripotent stem cells (IPS cells)” distinguishes them from populations of “adult” stem cells?

Answer 84

They are “PLURIPOTENT” meaning they can contribute to many (in some cases all) tissues of the body.

Question 85

L25. How do mutations in APC, Ras, and p53 contribute to the formation and progression of colon cancers? Do they all have the same affect on the cells?

Answer 85

APC is a “tumor suppressor” gene, encoding the APC protein. APC “turns off” wnt signaling in the differentiated cells of the intestinal epithelium. Wnt signaling stimulates proliferation... so loss of APC function results in inappropriate wnt signaling and inappropriate cell proliferation, leading to the formation of epithelial outgrowths called “polyps.” Polyps themselves are benign, but can become cancerous through the accumulation of additional mutations. Ras is a “proto-oncogene.” Ras protein is an important component of MAP kinase signaling pathways that lead to expression of G1 cyclins and cell proliferation. Activating mutations in Ras (mutations that block its intrinsic GTPase activity) are found in >30% of human tumors. These mutations lead to constituitive activation of MAP kinase signaling pathways, and result ini uncontrolled cell proliferation. p53 is another tumor suppressor, involved in the DNA damage checkpoint and apoptosis pathway. Loss of p53 function is found in >50% of human tumors, and allows tumor cells to proliferate uncontrollably and bypass apoptosis... leading to additional genome instability and the potential for additional genetic lesions leading to loss of adhesion, increased motility, and other characteristics of metastatic cancer cells.

Question 86

L25. Sketch the pathways (from receptor to reponse) by which transcription is regulated by a: B. cAMP-mediated signaling pathy

Answer 86

2. Lecture 25 slide 18

Question 87

L25. Sketch the pathways (from receptor to reponse) by which transcription is regulated by a: C. Ras mediated signaling pathway

Answer 87

3. Lecture 25 slide 24