C2006/F2402 ’11 -- Questions for Exam #4



C2006/F2402 ’11 -- Exam #4 – Questions & Answers

Each answer and/or explanation was worth 2 pts unless it says otherwise.

1. See Handout 22D, bottom, for the prolactin/oxytocin/lactation circuit shown on the exam.

A-1. Of the two hormones, prolactin (PL) and oxytocin, which one(s) should use cell surface receptors?

(PL only) (oxytocin only) (both) (neither) (not sure – one or both).

A-2. A receptor for oxytocin is most likely to resemble a receptor for (ADH) (GH) (TH) (TRH) (TSH).

Explain your predictions briefly.

A-1. Both PL and oxytocin are water soluble proteins/peptides. They cannot pass membranes (unlike steroids) and so each must use bind to the extracellular domain of a surface receptor. (FYI, oxytocin uses a GPCR and PL uses a TKR.)

A-2. Oxytocin and ADH are homologous peptides – encoded by very similar genes derived from a common ancestral gene. Therefore they are very similar and their receptors should be similar also. Also both are made and released in the same way, which is different from the way other hormones are handled. Both are made in the HT and released into the general circulation in the PP. The hormones are made in the neuronal cell bodies -- which are in the HT -- and released from the axon terminals of the neurons – which are in the PP.

B-1. If oxytocin’s action generates a second messenger, the enzyme responsible for generating the 2nd messenger is probably (AC) (PLC) (PKA) (PKA or PLC) (AC or PLC) (PKA or AC) (any of these).

B-2. The action of oxytocin on the breast should require (calmodulin) (phosphorylation of actin) (phosphorylation of myosin) (none of these). Circle all correct answers, and outline the intracellular steps involved here (for both parts).

B-1. Oxytocin causes contraction of myoepithelial cells, and contraction requires Ca2+. PLC generates the 2nd messenger IP3, which causes release of Ca2+ from the ER. So PLC is the obvious choice. For cAMP to cause contraction, it would have to trigger opening of Ca2+ channels in the plasma membrane. (See * below.)

B-2. Myoepithelial cells must be like smooth muscle, not skeletal, as the cells contract in response to hormonal signals, not in response to neuronal input from the somatic nervous system. So the Ca2+ must trigger contraction the way it does in smooth muscle, not the way it does in skeletal muscle. The Ca2+ must activate calmodulin (not troponin/tropomyosin), which in turn activates a kinase (MLCK) that phosphorylates and activates myosin.

*You didn’t have to remember this, but cAMP, generated by AC, does not cause contraction in smooth muscle. The cAMP activates PKA which phosphorylates and blocks activation of MLCK, preventing contraction. Therefore cAMP would probably trigger relaxation, not contraction, in myoepithelial cells.

C. PIH = PL (release) inhibiting hormone. PIH is identical to dopamine (DA), which is a catecholamine.

C-1. In cells making PIH, you would expect to find mRNA for (PIH) (enzymes for PIH synthesis) (both) (neither).

C-2. Release of PIH is most likely to increase when (baby starts nursing) (baby stops nursing) (either way) (neither – nursing shouldn’t make any difference).

C-3. Suppose synthesis of PIH is inhibited, but the stimuli for the release of PIH continue. You would expect release of PIH to stop (immediately) (after a while) (not at all). (1 pt each ans; 1 pt each for explanations of C-1 & C-3. C-2 did not need to be explained.)

C-1. PIH, aka DA, is a small molecule made by modifying tyrosine. PIH is not a peptide. You need mRNA to code for the modification enzymes to make the PIH, not for the PIH itself.

C-2. (Explanation not required.) PIH inhibits PL release, and thus blocks lactation. When the baby is nursing, release of PIH should be inhibited. When nursing stops, release of PIH should start up again.

C-3. PIH is water soluble, so it can be made in advance and stored in vesicles until it is needed. The vesicles can release their contents by exocytosis in response to the appropriate stimuli. Release will continue, after synthesis of PIH stops, until the PIH stored in the vesicles is exhausted.

D. When mothers nurse their newborns, the mothers often have very powerful uterine cramps (painful contractions) during nursing. The cramps usually stop after a day or so after birth, although mother continues nursing. The cramps probably end because of changes in (hormone levels) (receptor levels) (both) (neither). Explain the relationship between nursing and cramps.

Oxytocin levels must remain high if mother continues to nurse. Therefore the failure of the uterus to contract in response to oxytocin must be due to a change (decrease) in the level of receptors for oxytocin, not the level of the hormone.

Estrogen, a different hormone, causes synthesis of oxytocin receptors (as discussed in class much earlier in the term.) Estrogen levels fall after birth, and that presumably contributes to the down-regulation of oxytocin receptors in the uterus. It was okay if you brought this up, but you had to explain the relative roles of oxytocin and its receptors as well.

Explanation of 1-D, cont.

Note: The uterus contracts to push out the baby, but it must continue to contract &/or shrink for some time after birth in order to return to its pre-pregnancy size. (A normal uterus is about the size of a fist, not the size of a newborn baby.) The painful cramps end, but contractions continue for much longer.

E. When women nurse, their production of gonadotropins (FSH & LH) is usually low, and ovulation (egg release by the gonad) is suppressed. This is why breast feeding generally increases the interval between pregnancies. Based on the diagram, the simplest explanation of the suppression of ovulation is that suckling directly affects release of hormone(s) by the (HT) (PP) (AP) (gonads). Explain what hormone(s) is/are involved here, and how the final effect is achieved.

Gonadotropins (FSH & LH) are produced & released by the AP (2 pts) but the feedback here is on the HT, which produces releasing factors that control the output of the AP (1 pt). The diagram indicates that suckling sends a neuronal signal to the HT. That signal inhibits release of PIH, and stimulates release of PL & oxytocin. The same signal probably inhibits release of GnRH (gonadotropin releasing hormone) although it could stimulate release of a GnIH (gonadotropin inhibiting hormone). Either way, the signal from suckling would affect the HT directly, and the response of the HT would inhibit release of gonadotropins from the AP until nursing stops. Without sufficient gonadotropins, the cells in the ovary will not be stimulated, and ovulation will not occur. (The inhibition of gonadotropin release wears off, especially if frequency and intensity of suckling decline, which is why nursing is not a fool proof contraceptive. Nursing tends to delay the next pregnancy, not prevent it. )

2. A-1. For a B cell to act as an APC, it must carry out (endocytosis) (exocytosis) (both) (neither).

A-2. When a normal B cell is acting as an APC, it synthesizes (MHC I) (MHC II) (both) (neither).

A-1: 1 pt each for endo & exo; 2 pts for both. A-2: 1 pt for MHC II; 2 pts for both. No explanation required for this question.

A-1. To act as an APC, a B cell must internalize Ag, by RME, and then present the digested pieces (epitopes) on its surface by exocytosis. The Ag enters the cell by endocytosis (with the BCR). The Ag is digested, and complexes with MHC II in the EMS. The MHC II-epitope complex is transported through vesicles to the plasma membrane. Then exocytosis occurs -- the vesicle fuses with the plasma membrane, and the MHC II-epitope complex ends up in the plasma membrane, with the epitope on the outside.

A-2. All nucleated cells have MHC I on their surface (+ epitopes from the proteasome), and the B cell will also have MHC II (+ the epitopes from the internalized Ag).

B. Suppose a T cell is activating a B cell. Consider the epitope (piece of antigen) that is bound to the TCR.

B-1. This epitope should be the same as the one bound to the (MHC I of the B cell) (MHC II of the B cell) (both) (neither), AND

B-2. This epitope is probably the same as the epitope that (was bound to the BCR of the B cell)

(will be bound by the antibody secreted by the activated B cell) (both) (neither).

1 pt each answer; 2 pts for explanation of B-2.

B-1. See pictures on handout 24C. The epitope originally presented by the B cell (with MHC II) ends up in between the T cell and the B cell – it is bound on one side to the TCR of the T cell and on the other side to the MHC II of the B cell. (This did not need to be explained.)

B-2. An Ag usually has many different possible sites for binding of antibodies or B cell receptors. The BCR has one binding site. The BCR will bind to one site (part of) the Ag, trap the Ag, and internalize the entire Ag by RME. After the Ag is digested inside the B cell, all the many pieces (epitopes) derived from the Ag will be displayed with MHC II, and the TCR will bind to one of them. (The TCR, like the BCR, has only one binding site for antigen.) It is unlikely that the piece that matches the binding site of the TCR is the same as the piece that matched the binding site of the BCR. (Note: This question is not about the variable region of the BCR vs the variable region of secreted antibody – that’s question C-1.)

C. Consider the immunoglobulin made by a naïve B cell and the immunoglobulin made by the same B cell or its descendants after activation. Assume both immunoglobulins are of the same class, and there is no somatic mutation.

C-1. The two immunoglobulins -- made before and after activation – should (combine with the same epitope) (have the same number of transmembrane domains) (both) (neither) (one or the other but not both).

C-2. Consider the templates used to make the 2 immunoglobulins. Which should have same nucleotide sequence? (the DNAs used for transcription) (the mRNAs used for translation) (both) (neither).

(2 pts for DNA; 1 pt for both = right answer – 1 pt for wrong answer.)

Explanation to 2-C

C-1. The variable region of the immunoglobulin is the part that combines with an epitope. The variable region of the BCR (surface immunoglobulin) will be the same as the variable region of the antibody (secreted immunoglobulin) made after activation. Once the DNA for the variable region has been rearranged, it doesn’t rearrange again. The BCR will have a TM domain, but the secreted Ab will not.

C-2. The two immunoglobulins will have the same variable region and the same constant region (since they are the same class) except that the BCR will have a TM domain (at the carboxyl end of the H chains), and the secreted Ab will not. The DNA for both will be the same, but alternative splicing/processing of the primary transcript will give two different mRNAs (for the H chains). The mRNA made before activation will encode a hydrophobic region that will become the TM domain, and lock the BCR into the plasma membrane. The mRNA made after activation will lack the sequence coding for a hydrophobic domain, so it will encode a soluble protein that will be secreted.

3. See the info about Cyclosporin A (cycloA) at the end of the exam.

A-1. CycloA should directly block signaling in (T cells) (B cells) (both) (neither).

A-2. CycloA should block production of (effector T cells) (effector B cells) (both) (neither).

A-3. Activated calmodulin binds to and activates many different target proteins, such as NO synthase, in many different cell types. CycloA will probably block NO synthase activation by calmodulin in (T cells) (B cells) (both) (neither). Explain A-3 only. (Explain A-1 & A-2 on the back if you think there is any ambiguity.)

1 pt each correct choice (for A-2, 2 pts for both); 1 pt for explanation to 3A.

A-1 & A-2. CycloA blocks signaling and therefore prevents direct activation of T cells. So you wouldn’t get any effector T cells. However TH cells are needed for production of effector B cells (plasma cells). So cycloA blocks production of both types of effector cells. (This did not need to be explained.)

Explanation to 3A, cont.

A-3. Normally calmodulin binds to a phosphatase (calcineurin) and activates it. CycloA binds to a cytoplasmic protein to form a complex; the complex binds to the phosphatase and blocks calmodulin binding (& therefore blocks activation of the phosphatase). CycloA has no effect on calmodulin itself – it only inactivates one target of calmodulin. So any other protein target activated by binding to calmodulin (such as NO synthase) should be okay. The phosphatase should be affected by cycloA in T cells, but NO synthase should not. (Neither enzyme should be affected by cycloA in B cells.)

B. Consider 2 patients. Joe is being treated with cycloA for an autoimmune disease and Sue is not. Sue has a condition unrelated to immune function. Both get infected with hepatitis C virus, hepC. (There is no vaccine yet available for hepC.) In the cases below, what will happen? In each case, fill in the blank with one of the following: (be higher in Joe) (be higher in Sue) (be the same in both) (not occur in either patient).

B-1. Binding of a protein of the hepC virus to a BCR should _______be same in both_____________

B-2. Secretion of antibody to hepC should ___________be higher in Sue_______________________

B-3. Secretion of TCR directed against hepC should _______not occur in either patient____________

B-4. Killing of hepC-infected cells by CD8+T cells should ____be higher in Sue__________________

Provide ONE sentence of explanation for each choice.

1 pt for each choice above; 2 pts for explaining that cycloA will block both the humoral and cell-mediated branches of the immune system, because it prevents activation of both T cells (directly), and B cells (indirectly), since TH cells are needed to activate B cells.

Details for each choice above:

B-1. Production of surface antibody (BCRs), and binding between the surface immunoglobulin and a viral protein should not be affected in any way by cycloA. The consequences of binding will be altered, but not the binding.

B-2. B cells in Sue will secrete antibody, but B cells in Joe can’t be activated, so they won’t secrete antibody. (The B cells will still have surface antibody, but they won’t be able to secrete it.)

B-3. TCRs are never secreted – they remain on the surface of T cells.

B-4. Cytotoxic T cells (CD8+T cells) will be activated in Sue, but not in Joe.

4. If you inhibit the Na/K pump (the Na+/K+ ATPase) in nerve, you can still excite the nerve many times. If you inhibit the pump in skeletal muscle, the muscle becomes unexcitable after a short time. If you remove the pump inhibitor, excitability is rapidly restored to skeletal muscle.

A-1. Which of the following is the most likely to account for this difference between nerve and skeletal muscle?

(the number of ions that move across the cell membrane during an AP) (which ions move across the cell membrane during an AP) (the sign – positive or negative -- of the resting potential across the membrane)

(the level of ATP in the cell).

Problem 4, cont.

A-2. In normal muscle, the activity of the Na/K pump is affected by excitation of the muscle. The pump is most likely to be activated by (high [Na+] outside the cell) (high [Na+] inside the cell) (either way). Explain the role of the pump in maintaining muscle excitability.

A-1. In both skeletal muscle and nerve you need a Na+ (& potential) gradient across the membrane to propagate an AP in the cell membrane. Normally, in both cell types, there is low Na+ (and a negative potential) on the inside, so opening of voltage gated channels causes Na+ to rush into the cell. If enough Na+ accumulates inside, there won’t be a big gradient across the membrane, and there won’t be enough ‘push’ to drive Na+ into the cell and generate an AP. The pump is needed to set up the gradient, and to re-establish it if it runs down. You need ATP to power the pump, but both cell types normally have enough ATP to do that (and run everything else too). The only reasonable alternative here is that muscle APs involve moving a lot of Na+ into the cell per AP, so that the pump is needed almost constantly to remove the Na+ and maintain the gradient. In contrast, only a little Na+ moves per nerve AP, so the pump is not needed in nerve until many APs have gone by.

Note that one EPP will generate an AP in a muscle, but one EPSP will not generate an AP in a nerve. This is not necessarily related to the phenomenon described here. The difference in the need for the pump probably reflects the amount of Na+ that moves as the AP is propagated, not how the AP is initiated at a synapse.

A-2. Na+ is supposed to be higher outside and lower inside. If enough Na+ enters the cell, you need to turn on the pump to push it back out.

B. Suppose you have both a normal skeletal muscle fiber and a skinned muscle fiber. (See last page for how you get a ‘skinned’ fiber.) You try dripping acetyl choline (AcCh) or norepinephrine (NE) on the muscle fibers to see if you can trigger a contraction.

B-1. With normal skeletal muscle, you should get a contraction with (AcCh) (NE) (both) (neither)

(depends on the specific skeletal muscle you use).

B-2. With the skinned skeletal muscle, you should get a contraction with ____neither____ (same choices).

B-1. AcCh is the only neurotransmitter (nt) used by the somatic nervous system to cause skeletal muscle contraction. In other words, skeletal muscle has receptors for AcCh but not for any other nt.

B-2. Once the muscle is ‘skinned’ it has no receptors left for AcCh. The receptors are transmembrane proteins (& ion channels) in the plasma membrane, and once the membrane is removed, there are no receptors present. If there are no receptors, then there will be no response to AcCh – there are no ligand gated channels that can be opened by AcCh. (You can still get a response if you bypass the receptors by electrically generating an AP in the T tubule membranes, or by causing release of Ca2+ from the ER as in part C.)

C. If you soak a skinned skeletal muscle in a solution that is low in Mg2+, Ca2+ is released from the ER, and the muscle contracts. After soaking in low Mg2+,

C-1. Which of the following would have to occur before the skinned muscle could contract at all?

(an AP in the T tubule membrane) (phosphorylation of myosin) (binding of ATP to myosin)

(binding of myosin to actin) (hydrolysis of ATP) (activation of myosin kinase) (none of these are needed). Circle all correct answers.

C-2. For contraction to be maintained, which additional event(s) must occur? ___binding of ATP to myosin & hydrolysis of ATP__.

Explain briefly how this situation compares to the events that lead to contraction in a normal muscle.

1 pt for each correct term circled or written in above. 3 pts for explanation.

C-1. In the absence of Ca2+, skeletal muscle has already bound and split ATP. (1 pt) The myosin is ‘cocked’ and ready to go, but cannot bind to actin, because the binding sites on actin are covered by troponin/tropomyosin. Once Ca2+ is available, the Ca2+ will bind to troponin and move the tropomyosin out of the way. Then myosin can bind to actin, and that will lead to a power stroke = contraction (movement of actin relative to myosin). No ATP will be needed until the next step. Normally, an AP is needed to cause release of Ca2+, but the low Mg2+ treatment directly releases Ca2+, so no AP is needed. (1 pt)

Note: Phosphorylation of myosin and activation of myosin kinase are not involved here – they occur only in smooth muscle, not in skeletal.

C-2. For the muscle to maintain a contraction, the myosin and actin must separate (which requires binding of ATP) and the bridge cycle must happen all over again. (1 pt) ATP must be bound and split before myosin can contact actin and push the actin (with a power stroke as before).

Question 4, cont.

D-1. Suppose you have a skinned muscle and you inhibit the Na/K pump. Then you soak the muscle in a solution of low Mg2+. (Same conditions as in E, except for the inhibitor.) Will the muscle contract? (yes) (no) (can’t predict).

D-2. Suppose you have a normal skeletal muscle and you inhibit the Na/K pump. Then you soak the muscle in a solution containing Ca2+. You do not stimulate the muscle in any other way. Will the muscle contract? (yes) (no) (can’t predict). Explain both cases briefly.

1 pt each answer; 2 pts each for explaining each part. Short explanations:

D-1. Ca2+ is required for contraction. As long as you have Ca2+, you don’t need an AP, so you don’t need the pump. (More details below.)

D-2. Ca2+ will not enter the cell, as we assume there are no transporters or channels for it in the plasma membrane. (For contraction, the Ca2+ normally comes from the ER, not from the outside.) So a Ca2+ soak will not raise the Ca2+ in the cytosol and trigger a contraction.

Detailed explanation of D-1: Normally, AcCh opens ligand-gated channels that depolarize the cell and generate an AP in the plasma membrane. That AP spreads to the T tubule membranes, and activates proteins (voltage sensitive) that in turn open channels in the ER and release Ca2+. The Ca2+ triggers contraction. In other words, under normal conditions, Ca2+ release and contraction are dependent on the AP. Therefore, if you inhibit the pump, that inhibits the AP, and you prevent release of Ca2 +and prevent contraction. However, if you cause the cell to release Ca2+by other means, such as low Mg2+, then you don’t need the AP, and so you don’t need the pump either. In other words, you can bypass the need for the AP (& the pump) by directly triggering an event that occurs later (‘downstream’) in the pathway.

5. Cyclosporin A (cycloA) is used to suppress the immune system, but its use is limited by the fact that it causes kidney damage. One effect of cycloA is that it inhibits the Na/K pump in the ascending limb of the loop of Henle and in the ‘distal nephron’ – the remainder of the nephron, including the collecting ducts. CycloA has no effect on the pump molecules in the other parts of the nephron (or in other tissues). For this problem, ignore any other effects of cycloA.

A. How should CycloA affect the amount of glucose in the urine? In the short term, the amount should (increase) (decrease) (stay the same)* (stay the same at zero).

* one point ‘stay the same’; 2 pts for ‘stay the same at zero.’

Under normal conditions, all the glucose is reabsorbed from the filtrate, and none appears in the urine. Reabsorption of glucose occurs in the proximal tubule by secondary active transport with Na+. Secondary active transport depends on a Na+ gradient in the cells involved. CycloA does not affect the pump in the proximal tubule cells, so it should not affect glucose transport in these cells either.

B-1. If CycloA is present, and the [Na+] in the blood is in the normal range, the [Na+] in the filtrate* exiting the loop of Henle should be (higher than) (lower than) (about the same as) normal.

B-2. CycloA should decrease net Na+ transport (out of the proximal tubule) (into the proximal tubule)

(out of the distal nephron) (into the distal nephron) (out of both parts of the tubule)

(into both parts of the tubule) (out of one part of the tubule and into the other part).

* the question is intended to ask about the amount of Na+ left in the filtrate when the filtrate moves along the tubule after leaving the loop of Henle. If you took it to be about the amount Na+ that leaves the filtrate (is reabsorbed) in the loop of Henle, part credit was given.

B-1. Na+ is normally pumped out of the tubule and reabsorbed back into the body in the ascending loop of Henle. This process requires the Na/K pump, and is inhibited by cycloA. Therefore more Na+ will remain in the filtrate and less will be reabsorbed.

B-2. (No explanation required.) CycloA will not inhibit Na+ transport in the proximal tubule under any circumstances, but it will inhibit Na+ transport into the distal nephron if aldosterone is present. Normally, what happens in the distal nephron depends on the level of aldosterone. If there is no aldosterone, little if any Na+ will be removed from the distal nephron, so cycloA should have little effect. However, if there is aldosterone and cycloA, there will be less transport than if there is aldosterone alone.

C. If a normal person eats too much K+, she can correct by adjusting the K+ in the filtrate, primarily in the distal parts of the nephron. There is no active transport of K+ across the luminal surface of the tubular cells in the distal nephron.

C-1. To compensate for a K+ overload, a person should increase

(absorption of K+) (secretion of K+) (either way).

C-2. People taking CycloA are more likely to have (low blood K+) (high blood K+) (either way).

Explain C-2. (Explain C-1 on the back if you think there is any ambiguity.)

Explanation of 5-C.

C-1 (No explanation required). ‘Secretion’ in this context means release of material into the tubule from the cells lining the tubule – it means net transport from the body into the filtrate. ‘Absorption’ means the reverse.

C-2. How is K+ secreted? Since there is no active transport of K+ across the luminal surfaces of the tubular cells, K+ must be actively moved into the cells using the Na/K pump on the basolateral side. Then the K+ must diffuse passively into the tubule through ion channels on the luminal side. This normally works because the pump ensures that the [K+] in the cells is higher than the [K+] in the filtrate. If cycloA inhibits the Na/K pump, the [K+] in the cells will be too low, there will be no gradient from the cytoplasm of the cells to the inside of the tubule, and K+ will not flow into the tubule. So there will be less K+ secreted into the filtrate, and more retained in the body. Since the K+ is not pumped into the cells, it will be higher in the interstitial fluid and therefore higher in the blood.

6. Some of the kidney damage caused by CycloA is thought to be due to its stimulation of the release of vasoconstrictors, such as the paracrine endothelin, which reduces flow through the kidney.

A-1. Vasoconstriction of blood vessels by endothelin should be due to (relaxation of smooth muscle) (contraction of smooth muscle) (relaxation of striated muscle) (contraction of striated muscle).

A-2. Endothelin probably reduces the flow through the kidney by acting on

(glomerular capillaries) (the efferent arteriole) (the afferent arteriole) (either arteriole)*

(the afferent arteriole or the glomerular capillaries) (any of these).

* The question was intended to ask what happens to flow through the tubule of the kidney. ‘Either arteriole’ was accepted if you made in absolutely clear you meant flow through the blood vessels, not flow through the tubule.

A-1. Blood vessels are surrounded by smooth muscle, not skeletal muscle. Vasoconstriction occurs when the muscle around a blood vessel contracts; vasodilation occurs when the muscle relaxes.

A-2. Endothelin can’t act on capillaries, because capillaries are not surrounded by smooth muscle – only by a layer of epithelial (endothelial) cells. Endothelin must act on an arteriole. (Arterioles have a layer of smooth muscle surrounding the layer of endothelial cells.) Endothelin must restrict the flow into, not out of, the glomerulus, so it is most likely to cause constriction of the afferent arteriole, which leads into the glomerular capillaries.

B. CycloA alone causes chronic kidney damage in humans, due in part to changes in the ECM. However in rats, CycloA only causes chronic kidney damage if the rats are on a low salt diet. Researchers have recently found that a drug that is a hormone antagonist reduces chronic kidney damage by CycloA in rats on a low salt diet. The drug is probably an antagonist of (insulin) (ACTH) (ADH) (aldosterone) (beats me).

The most likely hormone causing the damage is aldosterone, as explained below. ACTH and insulin are not likely at all. ADH is not likely either, but is involved in kidney function, so half credit (2 pts) was given for ADH if well explained.

Why aldosterone? Because production of aldosterone is the usual response to low salt. So you might make too much of it, and that (along with cycloA) would cause the problem. Low salt causes secretion of aldosterone (through the angiotensin/renin system), and aldosterone normally promotes Na+ reabsorption in the distal nephron. This is a homeostatic mechanism for maintaining salt levels and blood pressure in the body. So a low salt diet would cause secretion of aldosterone, and an aldosterone antagonist could reduce any damaging effects of the hormone.

Note that in the presence of cycloA, aldosterone should not cause salt retention in the distal nephron (which is dependent on the pump). So if excess aldosterone (+ cycloA) causes kidney damage, it must be because of some other effect of aldosterone. See

Why not ADH? Because ADH is not normally produced in response to low salt, and because water loss (as opposed to salt gain) won’t fix the problem. If blood has a low osmotic pressure (is too dilute) as a result of the low salt diet, you might think you could help by inhibiting ADH release. This would lead to water loss and an increase in blood osmolarity. However, it would not increase the total salt in the body at all, although it would increase the salt concentration, and it would cause a drop in blood pressure. So this is not a reasonable homeostatic response to low salt – it doesn’t fix it. In addition, if the osmotic pressure is low to begin with, ADH will not be secreted and so adding an antagonist (or inhibiting ADH release) won’t have any effect.

You might think that a low salt diet would lead to low blood pressure, and that the damage is a consequence of the low BP, not the low osmolarity of the blood. Inhibiting ADH would cause more water loss and a lower BP, so this is not a solution.

Why not ACTH?

You might think the problem is with aldosterone, and ACTH regulates production of aldosterone. This is not the case. ACTH affects the adrenal cortex, but it primarily regulates the production of cortisol, not aldosterone. Cortisol and aldosterone are both steroids made by the adrenal cortex, but their synthesis is controlled by different inputs. The production of ACTH should not respond to low salt, and blocking ACTH shouldn’t help.

Explanation of 6B, cont.

Why not Insulin?

Inhibiting the action of insulin might affect blood pressure or blood osmolarity, but the effects of low insulin are so serious that this is not a reasonable fix. In addition, lack of insulin causes high levels of glucose in the blood (& therefore the filtrate) and this causes kidney damage in itself.

List of Abbreviations

|AC – adenyl cyclase |NE -- norepinephrine |

|AcCh – acetyl choline |NLS – nuclear localization signal |

|ACTH – adrenocorticotropin or adrenal cortex tropic hormone |NO – nitric oxide |

|ADH – antidiuretic hormone |NT -- neurotransmitter |

|AP – anterior pituitary or action potential |PIH – prolactin (release) inhibiting hormone = DA |

|APC – antigen presenting cell |PKA – protein kinase A |

|BCR – B cell receptor |PL -- prolactin |

|CycloA – cyclosporine A |PLC – phospholipase C |

|DA – dopamine |PP – posterior pituitary |

|ECM – extracellular matrix |SP – signal peptide |

|ER – endoplasmic reticulum |TCR – T cell receptor |

|FSH – follicle stimulating hormone |TF – transcription factor |

|GH – growth hormone |TH – thyroid hormone or thyroxine |

|HepC – hepatitis C (virus) |TRH – thyrotropin releasing hormone |

|HT -- hypothalamus |TSH – thyroid stimulating hormone or thyrotropin |

|LH – lutenizing hormone | |

|MHC – major histocompatibility complex | |

Info for CycloA for problem 3 (This information is NOT relevant to problems 5 & 6.)

3. When a TCR binds its ligand, the change in the receptor starts a signaling pathway that leads to activation of calmodulin. A similar pathway is triggered in the partner cell (the one with the ligand). Activated calmodulin binds to and activates a cytoplasmic phosphatase that is found in both B and T cells. The phosphatase is required to activate a TF in both types of lymphocytes. Dephosphorylation of the TF unmasks an NLS, and the TF moves to the nucleus and triggers transcription of genes needed for activation in both types of lymphocytes.

Cyclosporin A (cycloA) is used as a drug to block immune responses in transplantation and autoimmune diseases in humans. CycloA blocks immune responses by binding to a cytoplasmic protein found in T cells but not B cells. The cycloA-protein complex in T cells binds the phosphatase and blocks binding of calmodulin to the phosphatase. (I am not making this up!) CycloA has other effects, but this is the only effect on the immune system.

Info for Skinned Muscles (problem 4)

You can isolate a skeletal muscle fiber (cell), stimulate it in various ways, and measure the amount of contraction. You can also make a ‘skinned’ fiber – the plasma membrane is removed by careful dissection, and the remainder of the fiber remains intact. After you remove the membrane, the T tubules seal up, and the skinned fiber can also be stimulated to see if it contracts, etc.

If you start with a normal muscle, and stimulate the membrane electrically, you can trigger a contraction. In the case of the skinned muscle, if you stimulate the T tubule membranes electrically, you can trigger a contraction.

Why bother with a skinned muscle cell? You can manipulate the ion concentrations inside the skinned muscle cell by equilibrating the skinned cell with an appropriate solution.

Prolactin/Oxytocin/Lactation Circuit – See Handout 22D, bottom.

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