Download Questions for Exam #3

C2006/F2402 ’11
Key to Exam #3
Only circle one answer unless it says otherwise. If it says ‘circle all correct answers’ there may be one, or more than one,
correct answer. Explain everything unless it says specifically not to.
Each question or answer was worth 2 pts unless it says otherwise.
1. Suppose an enzyme is normally found in the lumen of the peroxisome.
A. The mRNA for this enzyme is probably translated on ribosomes attached to (peroxisomes) (mitochondria) (the ER)
(the outer nuclear membrane) (ER or peroxisomes) (peroxisomes or mitochondria) (none of these).
B. To reach its normal location, the mRNA for this enzyme would have to pass through (a nuclear pore)
(the nucleolar membrane) (a translocon in the ER) (a coated pit) (a transport vesicle) (an endosome)
(the peroxisomal membrane) (the OMM) (the IMM). Circle all correct answers for B, and explain both parts A & B briefly.
Peroxisomal proteins are made on free ribosomes in the cytoplasm; the mRNA (not the protein) has to exit the nucleus to be
translated in the cytoplasm. It is the protein, not the mRNA, that must cross the peroxisomal membrane.
2. Hypoglycemia (low blood sugar) can be a serious problem in newborns. The most common cause of hypoglycemia in
infants is having a diabetic mother. This type of hypoglycemia resolves after a while, but it persists for a few days after
birth.
A-1. The hypoglycemic infant is probably producing (too much insulin) (not enough insulin)
(too much glucagon) (either too little insulin or too much glucagon) (too much of either hormone) (beats me).
A-2. In the pregnant mother, the levels of blood sugar were probably (normal) (low) (elevated) (can’t predict).
The baby was exposed to high (maternal) blood sugar in utero, and made high levels of insulin in response. When the child
is born, there is too much insulin for a normal level of sugar intake and/or blood sugar, and the excess insulin causes too
much glucose uptake and therefore hypoglycemia. It takes a short while for the infant to readjust the levels of insulin
production to match the levels of blood sugar.
B-1. Suppose hypoglycemia occurs shortly after feeding in a newborn. In that case, the hypoglycemia probably results
primarily from abnormally (low glucose release) (high glucose uptake) (high glucose release)
(low glucose uptake) (high uptake & low release) (low uptake & high release).
B-2. Of the organs listed, which is/are the major effector(s) causing the problem? (pancreas) (muscle) (liver) (brain).
Circle all correct answers.
The problem occurs right after feeding, when the infant is in the absorptive state, not later in the post-absorptive. Therefore
the hypoglycemia must be caused by too much uptake (due to the high insulin) right after feeding, not too little release
between feedings (due to low glucagon or persistent insulin).
3. Hypoglycemia (low blood sugar) activates the sympathetic branch of the ANS. This triggers sweating, shakiness, etc.,
which usually causes adults to eat before the condition gets so bad that the brain can’t function. Sweat glands are innervated
by neurons of the sympathetic system. These neurons have their bodies in the CNS, and their axons synapse on the sweat
glands. (These are the only neurons that innervate the sweat glands.) The neurons release AcCh, and receptors on the
sweat glands are muscarinic receptors.
A. These neurons to the sweat glands most closely resemble ordinary (preganglionic sympathetic neurons)
(postganglionic sympathetic neurons) (postganglionic PS neurons) (beats me).
These neurons have their cell bodies in the CNS, and secrete AcCh, like normal preganglionic ANS neurons. (Note
preganglionic PS neurons is not given as a choice.) The nature of the receptor on the next cell, the effector, is not a
property of the neuron, and doesn’t matter here.
B. The receptors on the sweat glands are the ones to be expected on (the effectors of the PS)
(the effectors of the sympathetic) (the effectors in the somatic system) (ganglia of the PS)
(ganglia of sympathetic) (ganglia of the somatic). Circle all correct answers.
Of the choices given, only the effectors of the PS have muscarinic AcCh receptors. One of the following specifics had to be
mentioned for full credit: AcCh is the NT in ganglia of the ANS, and in the effectors of the somatic system, but the
receptors are all nicotinic. The effectors of the sympathetic branch of the ANS usually use adrenergic (NE) receptors. There
are no ganglia in the somatic system.
4. Vasopressin (VP) is a peptide hormone that has two different GPCRs. VP receptor A (VPRA) is found in peripheral
blood vessels; VP receptor B (VPRB) is found in kidney cells. VP causes peripheral blood vessels to constrict; VP
stimulates water transport in kidney cells.
A. From the info given above (ignore any additional info below or on other pages), what can you conclude about the
VP receptors?
A-1. The 2 GPCRs could bind to (the same G protein) (different G proteins) (either way) (neither – GPCRs do not bind
directly to G proteins).
A-2. Which VPR should be part of a channel? (VPRA) (VPRB) (both) (neither) (one or the other, but not both) (beats
me).
Different GPCRs can bind to different G proteins and have different effects. However dif. GPCRs can bind to the same G
protein, and trigger the same pathway, yet still have different effects in dif. cells. You’ll get different effects if the target
proteins present in the cells – the ones that get modified at the end of the signaling cascade -- are different.
B. VP (a peptide) and epinephrine have the same effect on the smooth muscle of the arterioles. (Epi has no effect on
water transport in kidney cells.) In arteriole smooth muscle, the two hormones, VP and epi, could both activate or bind to
(the same receptor) (the same G protein) (both) (neither).
Receptors are specific for their ligands, and VP and epi are very different in chemical shape and properties. So two
different receptors are required. However, both can bind to the same G protein. This G protein then triggers the same
pathway (say IP3 or cAMP) and produces the same effect, regardless of which ligand-receptor combination was used. .
C. You have an agonist of VP that binds to VPRA but not to VPRB. Suppose a student who has just been meditating
takes the VP agonist.
C-1-a. You measure the amount of vasoconstriction and/or water transport in kidney. You would expect water
transport in the kidney to (increase) (decrease) (stay the same), AND
C-1-b. You would expect vasoconstriction to (increase) (decrease) (stay the same). No explanation needed for C-1; use
back if you think it’s necessary.
C-2. Suppose the student who took the agonist gets an unexpected phone call from the dean demanding that s/he come
in immediately for a meeting. Given all the information so far, will the phone call have any effect, in addition to, the
effect(s) of the agonist? You would expect an additional effect on (water transport in the kidney) (vasoconstriction) (both)*
(neither). Explain your reasoning – it’s the reasoning here that counts, not the answer!
For C, 1 pt each answer; 3 pts for explanation. (*‘both’ was okay if properly explained as at * below.)
A call from the dean should cause stress, and subsequent release of epinephrine. (2 pts) Epi causes vasoconstriction,
and so does the VP agonist. If the system is already saturated, and max. constriction is already reached, or the G protein is
already completely activated, then adding epi won’t make any difference. But since the agonist and epi use different
receptors, there is a good chance that the effects will be additive, and having epi will activate more G protein and result in
more vasoconstriction.
*Stress will activate the sympathetic system, and the activation will ultimately cause release of VP. This will cause a
change in water transport, and may cause a change in vasoconstriction. However, the agonist has already stimulated the
VP receptors, so it’s not clear that the VP will produce more vasoconstriction. The VP effect on water transport may take a
while to show up – the VP causes more water retention, and reduces production of urine. However it doesn’t affect what’s
already in the bladder – it’s the nervous system that causes the bladder muscle to relax and hold more, not the VP.
D. VP causes a rise in IP3 in smooth muscle of arterioles and a rise in cAMP in kidney cells
D-1. Suppose you inject VP and an inhibitor of PDE into a suitable experimental organism. Compared to the response
with VP, alone, you would expect (more contraction) (less contraction) (more water transport) (less water transport)
(no difference). Circle all correct answers and explain.
PDE converts cyclic AMP to plain AMP. If you block PDE, the cAMP lasts longer, and so does the activation of PKA and
any other response to cAMP. PDE has no effect on the IP3 pathway.
D-2. Suppose you inject VP and an inhibitor of GTP hydrolysis. Compared to the response with VP, alone, you would
expect (more contraction) (less contraction) (more water transport) (less water transport)
(no difference). Circle all correct answers (2 pts each) and explain.
VP works in both tissues by activating a G protein. Activation of any G protein requires GTP/GDP exchange; deactivation
requires hydrolysis of the GTP to GDP. If you block hydrolysis, the activation of the G protein lasts longer, and so do the
processes stimulated (directly or indirectly) by the G protein.
C2006/F2402 ’11
Exam #3
ID #:______________________________
5. There are many different glycogen storage diseases (GSD). In each condition, a different step in glycogen metabolism is
affected. To summarize, people with GSD type 1 have no glucose phosphatase. People with type 5 have no muscle
phosphorylase. People with type 6 have no liver phosphorylase. (More details on GSD and the enzymes on last page.) Some
people with GSD develop hypoglycemia (low blood sugar).
A. A normal person can compensate for low blood sugar by (glycolysis) (glycogen breakdown)
(glycogen synthesis) (gluconeogenesis) (none of these). Circle all correct answers. (2 pts each answer.)
Liver is the primary effector for releasing glucose to correct for low blood sugar. The glucose can come from
glycogenolysis (depolymerization of glycogen), or from gluconeogenesis -- manufacture of glucose from smaller molecules,
such as lactate. Both occur in the liver. Muscles also breakdown glycogen, but they don’t release glucose – they release
lactate, which the liver either breaks down further for energy generation, or converts to glucose for release or storage (as
glycogen).
B. Which of the following should be able to compensate for high blood sugar? People who are
(GSD type 1) (GSD type 5) (GSD type 6) (normal) (none of these). Circle all correct answers. (2 pts for circling them all; 2
pts for explaining why.)
To compensate for high blood sugar, a person has to be able to take sugar up from the blood; if there is extra, beyond
immediate energy needs, the person must be able to store the glucose by synthesizing glycogen (or fat). All these people can
do that. The GSD types have problems with breaking down glycogen, not with making or storing it.
C. Consider the homeostatic mechanism for regulation of blood glucose. In a person with GSD 1,
C-1. The primary problem is with the (effector(s)) (sensor(s)) (information flow from sensor to effector(s))
AND
C-2. There has been a change in the (set point) (regulated variable) (controlled process(es))
(critical values at which corrections occur). Circle all correct answers.
(1 pt each answer; 2 for explanation). The problem here is glucose release (the controlled process) by the liver (the
effector). The regulated variable (glucose) and the desired value (the set point) remain unchanged. Glucose release is
controlled by adjusting the rate of Glucose-phosphate hydrolysis, but the physiologically important process is glucose
release, not hydrolysis.
D. Which of the following is most likely to develop hypoglycemia? A person with GSD type (1) (5) (6)
(1 or 5) (1 or 6) (5 or 6) (any of these are equally likely).
E. Who is the LEAST likely to develop hypoglycemia? A person with GSD type (1) (5) (6) (1 or 5) (1 or 6)
(5 or 6) (any of these are equally unlikely).
2 points each answer; 2 points (each) for explaining the situation in people with each of the three types of GSD. (10 pts
total for D & E). Note that phosphorylase breaks down glycogen (a glucose polymer) to glucose phosphate, and
phosphatase removes the phosphate from the glucose.
People with GSD1 can not release glucose from the liver at all – both glycogen breakdown and gluconeogenesis produce
glucose phosphate, and the glucose is trapped in side the cell if the phosphate cannot be removed. Therefore these people
are very likely to develop hypoglycemia.
People with GSD6 can generate (& release) some glucose from gluconeogenesis in liver. They can’t break down liver
glycogen, but they can breakdown muscle glycogen to lactate, convert the lactate to glucose phosphate in liver, remove the
phosphate, and release the glucose. So they can develop hypoglycemia, but it’s not as severe a problem as with type 1.
People with GSD5 have no problems with release of glucose from the liver. They may generate less lactate (from muscle)
for the liver to use for gluconeogenesis, but they have normal liver function. Since the muscle is not directly involved in
release of glucose, and their problem is with muscle, they are least likely to develop hypoglycemia.
C2006/F2402 ’11
Exam #3
ID #:______________________________
6. Channels of the TRP family are required for sensing stimuli in many organisms. In some cases, the receptor for the
stimulus is ionotropic and in some case it is metabotropic.
One of the TRP channels, call it TRPQ, opens in response to heat. TRPQ is a nonspecific cation channel. TRPQ is
found in the sensory neurons that detect heat; these neurons can fire APs. Exposure to the compound capsaicin, the active
ingredient in hot peppers, also opens TRPQ channels in sensory neurons.
A. When the worms are exposed to heat, you expect
A-1. Ca++ to (go out of the cell) (go in to the cell) (stay put – no significant movement across the membrane).
A-2. Na+ to (go out of the cell) (go in to the cell) (stay put – no significant movement across the membrane).
A-3. Cl- to (go out of the cell) (go in to the cell) (stay put – no significant movement across the membrane).
A-4. When the cells are exposed to heat, you expect the cells to become (hyperpolarized) (depolarized)
(either way). No explanation required; use back if you think it’s necessary.
FYI only: All three ions, Ca++, Na+, and Cl- are higher on the outside of the cell. In addition, the inside of the cell is
negative, relative to the outside. So clearly the positive ions (cations) will move in if channels are opened for them, and the
cell will become less negative, or depolarized. What Cl- will do if channels open depends on the electrical gradient.
However in this case, there are no channels for Cl-, so it won’t move anyway.
B. If you expose worms to heat, above 40oC, they avoid the heat – they crawl to the other side of the Petri dish. If you
add a little capsaicin to the medium in the Petri dish, the worms do not respond at room temperature. However, when you
heat up the dish with capsaicin in it, the worms are more sensitive to heat. They migrate away at a lower temperature, say
32oC. For each part of this question, circle all correct answers.
B-1. In the presence of the low amount of capsaicin at room temp., the sensory neurons that detect heat should
(fire action potentials) (have a depolarizing receptor potential) (have an EPSP) (have an IPSP)
(have a hyperpolarizing receptor potential) (none of these).
B-2. At 35oC plus the low amount of capsaicin, the sensory neurons that detect heat should
(fire action potentials) (have a depolarizing receptor potential) (have a hyperpolarizing receptor potential) (have an EPSP)
(have an IPSP) (none of these).
B-3. Suppose the worms are at 35oC plus the low amount of capsaicin, and you increase the concentration of capsaicin.
Then which of the following should increase in the sensory neurons? (the number of spikes or APs)
(the size of the APs) (the width of the APs) (the absolute magnitude of the receptor potential) (none of these).
B-4. Capsaicin changes (the threshold for firing an AP) (the distance to threshold)
(the Na+ equilibrium potential) (none of these).
Explain briefly how capsaicin increases sensitivity to heat.
2 pts each part of B, 8 total. 2 pts for the explanation.
There is no EPSP in receptor cells; they are presynaptic only. There is a graded response in the receptor cells that is
proportional to the stimulus – this is called the receptor potential, not an EPSP or IPSP. In this case, the rec. pot. is
depolarizing (see A). The more channels are opened, the bigger the receptor potential. If it is big enough, it reaches
threshold and the cells fire action potentials. Both capsaicin and heat open channels and depolarize the cells. The receptor
potential caused by capsaicin and the rec. pot. caused by heat are summed, and if the total is over threshold, the cell fires
an AP. The bigger the rec. pot., due to either heat or capsaicin, the more APs. If capsaicin is present, the cell is already
partially depolarized and the amount of heat needed to depolarize to threshold is decreased. (Note this is a one cell system
– the receptor cells themselves fire APs.)
C. By genetic engineering, you add the gene for the TRPQ channels to epithelial cells that do not normally respond to
stimuli. Then you use a patch clamp to record from the membrane of the recombinant cells. If you record from an attached
patch, you can detect a current in response to heat (over 40oC) or capsaicin. If you record from a detached patch, you also
detect a current in response to heat. (For more complete description, see last page.)
C-1. From these results, it is likely that the heat receptor proteins of the worm are
(ionotropic) (metabotropic) (either way).
C-2. Hot peppers probably taste ‘hot’ because they (bind to TRPQ channel proteins)
(bind to separate receptors that open TRPQ channel proteins)* (generate heat in the mouth)
(open TRPQ channels, but can’t tell if it’s direct or indirect).
*This answer was accepted if properly explained, as at * on next page.
Answer to 6 C, cont.
C-1. The only new protein made in the recombinant epithelial cells, one that is not made in normal epithelial cells, is the
TRPQ channel. This is sufficient to give a response to both heat and capsaicin. In the detached patches, there is no
possibility for generation of a second messenger, and there is still a response to heat. So the obvious explanation is that the
receptor is ionotropic – that is, the receptor is part of the channel. There is no separate receptor. (The only other possibility
is that normal epithelial cells already have unused receptors for heat, and corresponding G proteins, and that the G protein
or a membrane bound component can open the added channels without a soluble second messenger.)
C-2. It does not say in the problem whether or not there was a response to capsaicin in the detached patches. Most students
took the wording to mean capsaicin wasn’t tested, or the results weren’t reported, which was the intended meaning. In this
case, given that the receptor is ionotropic, it makes sense that capsaicin binds to the receptor/channel itself, and opens the
channels, as heat does. There is no reason to invoke a separate receptor, because epithelial cells do not normally respond to
capsaicin or heat, and should not have any part of the signaling system except the added TRPQ channels. (Note: In the
actual experiments, the response was measured, and the detached patches do respond to capsaicin.)
*Some students took the wording of the question to mean that there was no response to capsaicin in the detached patches,
so capsaicin and heat must use different signaling pathways. In this case, capsaicin must act through a separate receptor and
a second messenger generated in the cytoplasm. But where will the separate receptor come from? The only new proteins
made in the recombinant cells are the TRPQ channels. So this solution will only work if the proteins needed to bind
capsaicin and generate the second messenger (to open the channels) are found in normal epithelial cells even though they
are not used to detect heat or capsaicin or to open TRPQ channels. This is not a likely situation, so this answer was only
accepted if thoroughly explained.
Abbreviations and Acronyms
AcCh = acetyl choline
ANS = autonomic nervous system
AP = action potential
cAMP = cyclic AMP
Epi = epinephrine
EPSP = excitatory postsynaptic potential
ER = endoplasmic reticulum
GPCR = G protein Coupled Receptor
GSD = glycogen storage disease
GTP = guanosine triphosphate
IMM = inner mitochondrial membrane
IP3 = inositol tri phosphate
IPSP = inhibitory postsynaptic potential
NE = norepinephrine
OMM = outer mito. membrane
PDE = phosphodiesterase
PKA = protein kinase A
PS = parasympathetic
S = sympathetic
TRPQ = a type of channel
VLCFAs = very long chain fatty acids
VP = vasopressin
VPR = vasopressin receptor
Details on GSD for Problem 5.
People with GSD type 1 has no glucose phosphatase.
People with type 5 have no glycogen phosphorylase in muscle.
People with type 6 have no glycogen phosphorylase in liver.
Glycogen phosphorylase catalyzes the breakdown of glycogen according to the following reaction:
(glucose)n+1 + Pi → (glucose)n + glucose-phosphate
Phosphatase catalyzes the reaction:
glucose-phosphate → glucose + Pi
Complete description of experiment in Problem 6 C.
You have ordinary epithelial cells from a different organism (not a worm). The cells do not respond to any external stimuli
(heat, light, pressure, etc.) By genetic engineering, you add the gene for the TRPQ channels to the cells. No other genes are
added. Then you use a patch clamp to record from the membrane of the recombinant cells that make the TRPQ channels.
You attach a fine glass electrode (pipette tip) to the surface of the cell membrane, put a solution in the pipette and record
any current changes across the patch of membrane covering the tip.(This is an attached patch.) If you do this, you can
detect a current (ion flow) across the patch of membrane on the cell surface in response to heat (over 40 oC). You can also
detach the patch of membrane from the cell surface, and record from the detached patch. You stick the tip of the electrode
(with detached membrane patch) into a solution of ions (no proteins) and record. You detect a current in the detached patch
in response to heat.