C2006/F2402

C2006/F2402 ’11 Key to Review Questions for Exam #4 Last Update 05/04/2011

1. A-1. Before a person is exposed to CMV, s/he should possess some B cells that produce (surface proteins that can recognize CMV) AND

A-2. S/he should possess some T cells that produce (surface proteins that can recognize CMV) .

Both types of cells, B cells and T cells, produce surface proteins that vary from cell to cell. B cells produce surface bound antibodies (BCRs) and T cells produce TCR's. The protein repertoire is set up before the cells are exposed to antigen, because the DNA responsible for both TCRs and BCRs rearranges pre- antigen. Since there is a large repertoire of BCRs and TCRs, some of these proteins should bind to parts of (recognize) CMV. * (B cells make antibodies that are secreted, but only after exposure to antigen.)

*It was okay if you said the answer to A-2 was 'neither' because TCRs can't bind to whole or free CMV, but can bind to epitopes of CMV presented on MHC.

B-1. In the DNA from the clone that is CD8+ (CD4-), the sequence coding for CD4 should be (unaltered, but not transcribed).

B-2. The sequence of the DNA coding for MHC from the two clones will probably be (all the same) .

Only the genes for TCRs and Ab (Immunoglobulins) rearrange. All other genes are the same in all cells of an individual.
B-1. CD4 & CD8 are proteins made by cells of the adaptive immune system, but they are made in the usual way from ordinary (unaltered) genes. If a protein such as CD4 is not made, the gene is not deleted or rearranged. The gene is still there, but it is not transcribed (not expressed).

B-2. The genes coding for MHCs are polymorphic (many alleles of each gene in the population) and polygenic (multiple MHC genes per genome). Therefore people have multiple different MHC alleles per cell, and the MHC genes (& proteins) of one person are almost always different from those of another person. However, these two T cell clones come from the same person, so all the MHC genes and alleles should be the same in the two clones.

C-1. Which of the following properties do the two cell types have in common? Both (are epithelial cells) (are joined with tight junctions) (have distinct luminal and basolateral surfaces) .

C-2. In an infected person, CMV should be found in (saliva) (urine) (both) .

Explanation for C-2: Both types of cells have luminal surfaces that face a duct or tube that leads to the outside. When virus is shed from the cells, it will end up in the saliva (or filtrate) and exit the body.

No explanation required for C-1. In both cases you have epithelial cells forming a lining (of the kidney tubule or salivary lumen and/or duct). The cells must be polarized in order to carry out secretion, reabsorption, etc. In other words, the cells must have different proteins on their BL and luminal surfaces in order to transport certain materials in one direction and other materials in the other direction. To have two distinct surfaces, the cells must be tightly joined.

2. A-1. The inhibitory proteins are most likely to inhibit (exocytosis) .

A-2. This inhibitory protein is probably found (in the ER) (in the proteasome*) .

A-1. All MHC reaches the surface by exocytosis. The protein is made on the ER and inserted into the ER membrane as it is made. The MHC domain that will be extracellular binds to an epitope inside the cell, in some compartment of the EMS. Then the MHC plus epitope is carried by vesicle traffic to the plasma membrane. (Exactly where in the EMS the MHC meets the epitope depends on whether it is MHC I or MHC II.) When the vesicle fuses with the plasma membrane, the extracellular domain of the MHC (+ epitope) ends up on the outside of the cell.

Endoctyosis is needed to bring foreign proteins into the cell so that they may be degraded and the epitopes can meet up with MHC II. However, the MHC itself does not come from the outside.

A-2. The domain in question folds inside the ER, so any protein that interferes with its folding must be in the ER lumen. Details: All MHC is inserted into the ER membrane co-translationally. The domain that will end up on the outside of the cell passes through the ER membrane into the ER lumen. The luminal domain (the one that will be extracellular) folds up inside the ER lumen with the help of ER chaperones. Any protein that inhibits folding of this domain must be inside the ER, as that is where folding occurs. (See * below.)

* Misfolded protein is (retro)transported out of the ER into the cytoplasm, and degraded inside the proteasome, but the misfolding must occur first, in the ER lumen. (See explanation to A-2.) If you chose this answer and explained the role of the proteasome, you got credit for the explanation, but not the answer.

B-1. The target of the preventive immune response is probably (infected monocytes) .

B-2. The block to superinfection is most likely to be directly caused by (T_C cells) .

B-3. Will the superinfection be successful? (yes) .

B-1 & B-2. Short version of explanation: In the superinfection by the mutant, cytotoxic T cells kill the infected monocytes, so the superinfecting virus can't spread to the salivary glands etc. Since MHC I is involved, it must be the killer T cells, not the circulating Ab, that blocks the superinfection. The immune response cannot eliminate infected tissue cells, since either type of virus can set up a latent infection. Therefore it must be the monocytes that are the target of the killer T cells. Why can either virus infect the first time? Because there aren't enough cytotoxic T cells to prevent it.

Long version:

We know that both viruses can infect the first time, but only the normal can infect again, that is, can superinfect. We also know that both viruses can establish a latent infection. This indicates that the immune system cannot block an initial infection, and it cannot eliminate infected tissue cells (once they are infected with either type of virus). However the immune system can block the mutant from infecting a second time. So once a person is exposed, and the immune response has developed, the immune system can either block incoming mutant virus, or prevent monocytes from spreading it.

Since MHCI is involved, it is the differences between killing of (super)infected cells, and not differences in reacting to the virus particles themselves, that allow the immune system to respond differently to the mutant virus and to the normal. This means the block to superinfection must be cytoxic T cells (that kill infected cells) and not circulating Ab (which would stymie virus particles). We already know that the immune system cannot eliminate infected tissue cells, so it must be the monocytes with normal MHC levels that are killed when the mutant superinfects. If the monocytes are killed by cytoxic T cells, then the virus cannot spread to kidney, salivary glands, etc. If a normal virus superinfects, the infected monocytes have low levels of MHC I, and are not recognized as infected cells by killer T cells. (The TCR complex on the killer T cells can't bind to the MHC I + epitope on the monocytes.)

B-3. Why yes? If the normal virus superinfects, the monocytes have low levels of MHC I and should evade the killer T cells.

An alternative possibility -- if the first infection is with the mutant, more infected cells may be killed before the latent infection is established; this will activate more cytotoxic T cells and so the superinfection with normal virus is less likely to be successful. (This alternative was ok if fully explained.)

3. A. These TH spikes are probably due to (release of TH by degradation of thyroglobulin) .

    As pieces of the thyroid are destroyed, the thyroglobulin (TG) stored in the lumen of the follicles is released. The TG is degraded, generating large amounts of free TH each time a follicle ruptures.

    Many students chose 'increased stimulation by tropic hormone.' Even if this occurs, it does not account for the spikes (see below), and the increased stimulation (if it occurred) would lead to release of TH from TG. Therefore this answer alone was 2/4. If both answers were chosen, and explained, that was fine.

    As the thyroid is destroyed, the amount of TH that can be made gradually declines, so the amount of negative feedback to the AP should decrease, and the amount of tropic hormone (TSH) should increase. The increased stimulation should maintain the TH levels in the normal range (by increased TH production & secretion by the remaining follicles) until there isn't enough thyroid left to produce normal amounts of TH -- even if the thyroid is working at maximal levels. At that point TH levels would start to decline below the usual range. In other words, the TH levels wouldn't spike -- they would stay about constant (with small variations up and down) for a while, and then decline gradually.

    TH is never stored in vesicles. It is lipid soluble, and would diffuse through the vesicle membrane. Only water soluble signal molecules are stored in vesicles.

    Note that TH is not made separately and bound to thyroglobulin. TH is formed by modification of tyrosines that are already part of the TG peptide chain. The peptide bonds of TG must be broken to release the TH.

B-1. Once the thyroid is destroyed, you would expect people with Hashimoto to have levels of TSH that are (high) .

B-2. Galactorrhea is most likely to be due to overproduction of (prolactin) .

B-3. The direct cause of overproduction of these hormones (PL &/or oxytocin) is most likely due to unusual levels of (TRH) .

B-1. (1 pt) Low levels of TH, as in a person with untreated Hashimoto, would mean no feedback inhibition to the HT and AP. If there is no inhibitory feedback, the glands would secrete high levels of TRH and TSH respectively.

B-2 & B-3. Why TRH? If TH is very low, both TSH and TRH would be very high, so it seems reasonable that an excess of one or both could accidentally stimulate production of the 'wrong' hormone. A receptor that has a low affinity for TRH or TSH would not respond to normal levels, but could be stimulated by very high levels. The glands affected by TRH and TSH are the AP and the thyroid respectively. Prolactin (PL) is made and released by the AP, so it seems reasonable that a large excess of TRH might stimulate the cells that produce PL (see below for more details). TSH affects only the thyroid, so it should have no effect on release of either hormone.

    It seems much less likely that the low TH itself is the problem -- that lack of feedback due to the low level of TH is responsible for the production of PL and/or oxytocin. Negative feedback by TH should affect only the cells producing TSH or TRH. It is unlikely that the normal TH level stimulates the synthesis of PIH and thus controls the level of PL.

B-2. Why PL and not oxytocin? Both hormones are required in normal lactation, but it is very unlikely that oxytocin is involved here. Best argument -- because of location -- only PL is made by the AP.   TRH travels through a portal vessel from the HT to the AP. TRH goes directly to the cells of the AP. TRH does not reach the cells of the HT or PP (which release oxytocin) until it has traveled completely around the general circulation. (In the general circulation it is probably degraded; in any case it is too dilute to have any effect.) In other words, only PL producing cells (in AP) are a reasonable target for TRH -- not the oxytocin producing cells. Since the milk 'leaks' out of the breast, it is likely that PL is stimulating milk production and that there is so much milk it is dripping out of the breast without oxytocin stimulation of the contractile cells.

    Another argument -- because of function. PL is needed to produce milk, so you have to be making PL. However oxytocin is only needed to release it during normal lactation, and in this case there is so much milk it is dripping out of the breast without oxytocin stimulation of the contractile cells.

4. A-1. In the kidney, which of the following should be filtered? (TH) (oxid. TH) (tyrosine) (none of these).

A-2. In the blood, which must be bound to protein? (TH) (oxid. TH) (both) (neither).

A-3. In the kidney, which of the following should be reabsorbed? (TH) (oxid. TH) (tyrosine) (none of these).

A-1. Filtered = passes through the glomerulus, out of the blood and into the filtrate in the kidney tubule. Both oxidized TH and tyrosine (an amino acid) are filtered -- both are small water soluble molecules and pass into the kidney tubule. (This part required no explanation.)
Why isn't TH filtered or reabsorbed ? Only small soluble molecules are filtered. TH in the blood is bound to a protein (see A-2), and the complex is too large to pass into the filtrate. So TH is not filtered, and therefore there is no chance it is reabsorbed.

A-2. TH is lipid soluble, so it must be bound to a protein to travel through the blood. Oxidized TH is water soluble, so it does not need a carrier protein. (It is unlikely to bind to the normal TH binding proteins because oxidized TH does not have the same shape and chemical properties as TH.)

Note: TH is lipid soluble, but it is NOT a steroid. Steroids are a class of lipid-soluble molecules derived from cholesterol. TH has an entirely different chemical structure.

A-3. Both oxidized TH and tyrosine are filtered, but only tyrosine is reabsorbed. Amino acids are reabsorbed in the proximal tubule. They enter the filtrate, but are then transported back into the body. Transport across the luminal membrane of the cells lining the tubule occurs by secondary active transport (co-transport with Na⁺); transport across the BL membrane uses carrier mediated diffusion. TH does not appear in the urine because it is not filtered -- it never enters the tubule, so it can't be reabsorbed. There are no transporters for oxidized TH, so is it is not reabsorbed, and is excreted in the urine.

B-1. When people are treated with Rif, levels of TSH should (change whether person has Hashimoto or not) .

B-2. In a normal person taking Rif, the level of TSH should reach a steady state that is (higher than normal).

B-3. If a person with Hashimoto disease is treated with Rif, s/he can avoid hypothyroid symptoms by ( (adjusting the dosage of TH pills).

Short Version: If a person is taking Rif, some of the TH is oxidized and lost in the urine. In a normal person, the feedback system (lack of negative feedback by TH) will correct for this automatically -- the person will synthesize more TSH and therefore more TH. However the feedback system is broken in the person with Hashimoto, and s/he will not be able make more TH to compensate, even though there is more TSH. Here are the details:

    When a person takes Rif, the level of P450 enzymes rises, and some of the TH is oxidized, and excreted in the urine. Therefore the level of circulating TH falls, and there is less negative feedback on the AP (& HT). As a result, more TSH is made. This increase in TSH happens in both normal people and people with Hashimoto disease (who are taking TH pills*).

    In a normal person, the higher level of TSH stimulates the thyroid to make and release more TH until the proper level of circulating TH is restored. Since some of the TH that is made always gets oxidized and lost, the person must continue to synthesize more TH (& TSH) than normal -- without Rif -- in order to have enough circulating TH.

    In a person with Hashimoto disease (who is taking TH pills) is treated with Rif, the TSH rises, but there is no thyroid to respond to it. So the high level of TSH has no effect, the circulating level of TH remains low, and the person becomes hypothyroid. The person cannot use the usual feedback system to synthesize more TH . The only way the person can get more TH is to increase the dosage of TH pills.

*Note to B-1. If a person with fully developed Hashimoto is not taking TH pills the level of TH will be zero, the level of TSH will be high, and the increase in P450 enzymes will make no difference. However this should never happen -- a person who develops Hashimoto would have severe symptoms and consult a doctor long before the TH reaches zero.

5. Question 5 was on the review questions for exam #3..

6. A. What is the ratio of nerve endings to internal sphincter muscle cells? The intended answer was ‘about 1’, but there are good arguments for the others as well. Therefore the answer circled was ignored, and question was graded on the explanation alone.

The basic idea is that each smooth muscle cell here must be innervated individually, as the cells are not electrically coupled. The smooth muscle here is multi-unit smooth muscle – each cell contracts individually, so there are multiple units present. You need gap junctions if you want the muscle cells to contract as a group (= act as single unit smooth muscle). This is why the ratio expected is about 1.

However, since smooth muscle is usually innervated by both the PS and the S, there might be twice as many nerve endings as cells. That would make the ratio >1.

Since a single nerve ending has multiple varicosities that can contact more than one cell, at least according to the class handout, there might be multiple cells innervated by a single nerve ending, and the ratio would be <1,

B. Contraction of the bladder wall should require (Ca⁺⁺ binding to calmodulin).

For smooth muscle to contract, myosin, not actin, must be phosphorylated. In all muscle, Ca⁺⁺ is required; the binding partner for the Ca⁺⁺ is troponin in skeletal muscle and calmodulin in smooth muscle.

E-1. What should cause relaxation of the external sphincter? (neither).

E-2. For relaxation to occur, Ca⁺⁺ in the muscle cytosol must be (low).

E-3. For relaxation to occur, ATP must be (bound to myosin).

E-4. When a contracted sphincter muscle relaxes normally, ATP is (bound) (split) (neither) (both).

Explanation. Note: On the exam, only E-4 had to be explained.

E-1. Relaxation of skeletal muscle occurs when stimulation stops. No NT is used to trigger relaxation.

E-2. Ca⁺⁺ binds to troponin to trigger contraction; low Ca⁺⁺ blocks contraction. See class handout on bridge cycle for details of E-2 to E-4.

E-3. ATP must bind to myosin to break the myosin-actin connection, and allow the actin to slide ‘back’ relative to the myosin, increasing the space between the Z lines.

E-4. When skeletal muscle relaxes, after ATP is bound, the actin slides back, but the cycle doesn’t stop there. It continues until the step requiring Ca⁺⁺. Before it reaches that step, the ATPase activity of the myosin splits the ATP, so the myosin is ‘cocked’ and ready to push the actin – as soon as the actin is exposed (by Ca⁺⁺ binding to troponin etc.). The cycle stops when the actin and myosin are disengaged, and the myosin is ready for the power stroke. The splitting of ATP to bend the myosin into position for contraction occurs before the contraction cycle begins.

7. A-1. It means that all or part of the Ag acts as (part of a juxtacrine signal) .

A-2. It means that the TCR on the T_H cell binds to (a piece of Ag that was digested inside another cell) .

A-3. In order for the interaction to initiate activation, the T_H cell needs to make (CD4)* (none of these)*.

Explanation: A T helper cell must have CD4 as well as a TCR on its surface. (A cytotoxic T cell would have CD 8.) The antigen presenting cell (APC) must have an epitope digested in the APC and displayed on MHC II on its surface. The T cell and the APC stick together -- there are several juxtacrine interactions between surface proteins of the two cells. The TCR sticks to an epitope on MHC II; CD4 stabilizes the complex, by sticking specifically to MHC II (but not MHC I).

* It's okay if you said 'none of these' because the T cell has CD4 already. (The 'make or have' was intended to distinguish between components made by the T cell and those made by the APC. MHC II is needed, but it is on the APC, not on the T cell. As long as all of this was clear, 'none of these' was accepted.

Note a TCR is not the same as membrane bound antibody. TCRs and membrane bound antibodies (BCRs) are encoded by different genes. (The same genes encode BCRs -- the membrane bound form, and antibodies -- the secreted form.)

B-1. Activation of PLC is likely to lead to

a. Higher cytosolic levels of (IP3) (cAMP) (both) (neither), AND

b. Higher levels of active (PKA) (calmodulin) (both) (neither).

The adapter protein activates PLC which splits a membrane lipid generating IP3. The IP3 opens channels in the ER membrane and releases Ca²⁺, which binds to calmodulin and activates it. (Note PLC is not the same as PKC, a kinase activated by Ca²⁺.)

B-2. Activation is likely to lead to an increase in (replacement of GDP with GTP).

Explanation: That's how G proteins work. They are activated by binding GTP. The GTP is hydrolyzed to GDP to inactivate the protein. The protein is converted from ras-GDP to ras-GTP by replacement of the GDP with GTP, not by phosphorylation of the GDP.

8. A. i. These 2 hormones -- aldosterone & cortisol -- are both (steroids) (lipid soluble) .

ii. The 2 receptors -- MR & GR – are both (proteins that bind to DNA).

These hormones are both lipid soluble, so they use intracellular receptors that are transcription factors. (Note ii asks about the receptor, not about the hormone. The hormone binds the TF/receptor, and the complex binds to DNA.)

B. i. High BP is probably caused by cortisol binding to (MR). .

ii. The raised blood pressure should be caused by direct changes in the body’s handling of (Na+).

Explanations: Cortisol can bind to the MR and act as an agonist of aldosterone. It doesn't matter which hormone binds to the MR -- if either one binds, the result will be a mineralocorticoid or aldosterone effect. (It's which receptor is activated, and not which hormone does it, that matters.) If there is excess cortisol, it will bind to the MR and trigger reabsorption of Na⁺ in the distal tubule, just as aldosterone normally does. Your body won't 'know' you have too much cortisol -- it will respond as if there were a high level of aldosterone. The activated MR triggers synthesis of the components needed to reabsorb Na⁺. If more Na⁺ is retained in the blood and body fluids (ECF), then more water will be retained osmotically, and this will increase the blood pressure. The transport of water is not affected directly, but water follows the Na⁺. A greater Na⁺ gradient across a membrane will drive more water in the direction of the higher Na⁺.

C. i. Enzyme L is required to prevent (cortisol from binding to the MR).

ii. You are most likely to get you high blood pressure from (eating licorice while studying for finals).

Explanations: i. Enzyme L converts cortisol to a compound that does not bind the MR. If enzyme L is working, activation of the MR will reflect the level of aldosterone even in the presence of a large excess of cortisol. If enzyme L is not working, the MR will be activated if cortisol is high, regardless of the level of aldosterone. Therefore, if enzyme L is inactive, anything that triggers high levels of cortisol will cause Na⁺ retention and high BP.

ii. If cortisol is low, the need for enzyme L is relatively low, so effects of licorice will not be very significant. If stress and cortisol are low, it doesn't matter much if enzyme L is working or not. However, stress will raise the level of cortisol, and increase the need for enzyme L to prevent the cortisol from triggering the MR. If enzyme L is inhibited by licorice, during a stressful period, such as during exams, the high cortisol will cause high BP. Stress alone can cause high BP (although it may primarily be through vasoconstriction), but stress plus licorice will be worse. In other words, the licorice effect is dependent on high levels of cortisol.

9. A. i. A physiologist would describe this as an example of (negative feedback).

ii. When the respiratory pacemaker neurons fire, you expect the major muscles of breathing to (contract).

iii. When the H⁺ level is low in the arteries, you expect an increase in (time it takes pacemaker potential to reach threshold).

Explanations: i. Negative feedback (in physiology) means that the system makes corrections if the regulated variable deviates too much from the set point (the desired value). Neg. fb. means the system acts to negate the difference between the actual value (level of CO₂ and pH) and the set point (desired level of CO₂ and pH) so as to restore the optimal or desired value. It does NOT mean (in physio) that some process must be inhibited by an end product. In this case, you need to increase breathing rate, increase the rate of CO₂ exhalation, and increase the pH. (To increase the pH you need to decrease the level of CO₂ and acid.) This is negative feedback, because you are decreasing the deviation from the set point, and restoring the desired pH. It is not positive feedback, even though you are making controlled processes (breathing, exhalation of CO₂) go faster.

ii. It says that when the pacemaker cells fire, the motor neurons fire When neurons fire and stimulate skeletal muscles, the muscles always contract. Relaxation is the result of no stimulation. The bronchioles of the lung are surrounded by smooth muscle, but the muscles that move the diaphragm and control breathing are skeletal. Therefore, when the pacemaker cells fire, the motor neurons should fire and stimulate the major skeletal muscles.

iii. If the H⁺ level is low, that means the pH is too high and there is too little CO₂ in the body. Therefore you need to breathe more slowly to conserve CO₂ (or lose less) and increase the H⁺ level (decreasing the pH). You want to decrease the rate of firing by the pacemaker cells. Threshold and effect once it is reached are relatively invariant, and what varies is the slope of the pacemaker potential. You want it to be flatter, so the time it takes to reach threshold will be increased, and the time between APs will be longer.

B. i. If the CO₂ in the blood is too low, the rate of H⁺ secretion in the kidney should (decrease).

ii. If a person suffers kidney failure, she is most likely to have (acidosis), AND

iii. If the kidney failure is not treated, her urine is likely to be (more basic than usual).

Explanations: i. Low CO₂ means low H⁺, and so more H⁺ should be retained in the body, and less lost in the urine. Secretion refers to the transport of substances from the body to the tubule lumen, not the other way around. (Reabsorption is transport from the tubule lumen to the body.) To retain H⁺ in the body, H⁺ secretion should be decreased. (The kidney cannot reabsorb protons; see background info.)

ii. & iii. The kidney normally secretes lots of H⁺. If it falls down on the job, that H⁺ will be retained in the body, and acidosis (low pH in the blood) will result. The lack of the usual H⁺ in the urine will make the urine more basic. (If the kidney fails, the volume of urine will decrease, but any urine that is produced will be basic.)