2008 Exam 1 Answer Key

General Human Physiology Exam 1 February 6, 2008 Liard 1. The baseline value of variable X is 50 and increases to 100 in response to a disturbance in the presence of a control system. In the absence of the control system, the baseline value of X is 50 and increases to 250 in response to the same disturbance. What is the gain of the control system? A. B. C. D. E.

-0.67 -1.0 -2.0 -3.0 -4.0

Explanation: The change in variable X is +50 in the presence of the control system and + 200 in its absence. The correction is 50-200 = -150. The incomplete prevention of change by the control system (error) is 50. The gain of the system is then calculated to be -3 by the following formula:

This is analogous to the example given in Guyton & Hall on page 8. 2. Which of the following is an example of positive feedback in the body? A. B. C. D.

Clotting of blood. Return of blood pressure toward normal after a hemorrhage. Increased respiration rate caused by accumulation of carbon dioxide in the blood. Decreased sympathetic nervous system activity in response to increased blood pressure.

Explanation: In a positive feedback, the initiating stimulus causes more of the same. As discussed in Guyton & Hall, page 8, when a blood vessel is ruptured and a clot begins to form, multiple enzymes called clotting factors are activated within the clot itself. Some of these enzymes act on other unactivated enzymes of the immediately adjacent blood, thus causing more blood clotting. This process continues until the hole in the vessel is plugged and bleeding no longer occurs. 3. Which of the following substances has the highest extracellular fluid to intracellular fluid concentration ratio for most mammalian cells? A. B. C. D. E.

Bicarbonate ions. Potassium ions. Sodium ions. Proteins. Chloride ions.

Explanation: see values given in the table of Lecture 4, page 36 of the handouts.

2008 Exam 1 Answer Key

4. In a human subject, the cardiac output is 8000 ml/min, the arterial content of oxygen is 0.2 ml/ml blood and the central venous content of oxygen is 0.05 ml/ml blood. What is the oxygen consumption? A. B. C. D. E.

200 ml/min. 400 ml/min. 600 ml/min. 800 ml/min. 1200 ml/min.

Explanation: mass balance indicates that the amount of oxygen leaving the lungs is equal to that entering the lung. What enters the lungs is the cardiac output multiplied by the oxygen central venous content: 8000 ml/min X 0.05 ml/ml blood, or 400 ml/min, plus the oxygen consumption. What leaves the lungs is the cardiac output multiplied by the oxygen arterial content: 8000 ml/min X 0.20 ml/ml blood, or 1600 ml/min. The difference between the two is the oxygen consumption (1600-400 = 1200 ml/min). 5. A mass Y of a substance is added to container A; its measured concentration is C. A mass Y/2 of the same substance is added to container B: its measured concentration is 3C. Which of the following statements is correct? A. B. C. D. E.

The volumes of containers A and B are equal. The volume of container A is 3 times that of container B. The volume of container A is 6 times that of container B. The volume of container B is 3 times that of container A. The volume of container B is 6 times that of container A.

Explanation: Half of the amount is added to B and the concentration is 3 times higher than in A. This means that the volume of B is 1/6 of the volume of A.

6. In the vascular system represented above, what statement concerning the resistances RA and RB is correct? A. B. C. D. E.

There is not enough information to compare the two resistances. Resistance RA is equal to 4 RB. Resistance RA is equal to 3 RB. Resistance RA is equal to 1/4 RB. Resistance RA is equal to 1/3 RB.

Explanation: ΔP across RA is 20, ΔP across RB is 60. Since F is the same, RB is equal to 3 RA (using equation R = ΔP/F). 2008 Exam 1 Answer Key

Stekiel 7. Which of the following statements for a biological cell under normal in vivo conditions is FALSE? A.

∆F (free energy) = 0 across the cell plasmalemmal membrane (i.e. membrane separating the cytoplasm from the intracellular fluid). B. The cell is in a steady state. C. The intracellular and extracellular H2O concentrations are equal. D. There must be a constant input of metabolic free energy. E. Intracellular and extracellular osmotic pressures are equal. Explanation: Answer A is correct because the statement in answer A is not true. There must be a continual utilization of free energy by a biological cell to maintain its normal in vivo structure and function, hence, ∆F (free energy) ≠ 0. 8. Which of the following statements concerning hyperkalemia is FALSE? A. B. C. D. E.

A patient with a plasma K+ concentration of 8.0 mM would be classified as hyperkalemic. Hyperkalemia can be alleviated by blockade of the plasmalemmal Na+/K+ pump activity. Hyperkalemia can result from intravascular hemolysis. Hyperkalemia can produce a cell membrane depolarization. A classic pathophysiological consequence of hyperkalemia is a cardiac arrhythmia.

Explanation: Answer B is correct because it is false that blockade of the plasmalemmal Na+/K+ pump activity would alleviate hyperkalemia. Rather, such blockade would exacerbate hyperkalemia. All of the other answers are true. 9. Which of the following statements for the potassium equilibrium potential (EK) across a cell membrane is TRUE? At EK, the electrochemical energy difference (∆μ) for K+ is zero across the cell plasmalemmal membrane. B. The absolute magnitude of EK would be infinite if the intracellular and extracellular K+ concentrations were made equal. C. The absolute magnitude of EK is directly proportional to the extracellular [K+]o concentration (assuming no change in the intracellular [K+]i concentration). D. The absolute magnitude of EK is directly proportional to the cell membrane permeability for K+ (PK). E. EK is inversely proportional to the cell membrane permeability for K+ (PK). A.

Explanation: At EK, the chemical potential energy gradient is equal and opposite to the electrochemical energy gradient for K+. Hence, answer A is correct. Relative to the other answers, EK would be zero if the intracellular and extracellular K+ concentrations were made equal. Also, the absolute magnitude of EK is inversely proportional to [K+]o and is independent of (PK).

2008 Exam 1 Answer Key

10. Passive transmembrane flux of a solute is directly proportional to all of the following EXCEPT: A.

its transmembrane chemical concentration gradient if the solute is uncharged (i.e. a neutral molecule). B. its transmembrane chemical concentration gradient if the solute is charged (i.e. a cation or anion). C. its transmembrane electrical gradient if the solute is uncharged. D. its transmembrane electrical gradient if the solute is charged. E. its transmembrane electrochemical gradient if the solute is charged. Explanation: Exclusive of H2O, the sole passive driving force for transmembrane flux of an uncharged solute is the chemical concentration energy gradient. The passive driving force for transmembrane flux of a charged solute is the algebraic sum of the chemical concentration energy gradient and the electrical energy gradient. Thus the passive transmembrane flux of an uncharged solute is not directly proportional to the electrical energy gradient. So the statement in answer C is false and C is the correct answer. 11. As predicted by Fick’s 1st Law of Diffusion, the rate of diffusion of oxygen: A. would be directly proportional to altitude when measured across the lung alveolar cell wall. B. is directly proportional to alveolar cell thickness when measured across the alveolar cell wall. C. is directly proportional to the oxygen concentration of venous blood entering the lung when measured across the alveolar cell wall. D. is directly proportional to the oxygen concentration of arterial blood when measured across the blood capillary wall in the peripheral circulation. E. is directly proportional to the oxygen concentration of the extracellular fluid across the blood capillary wall in the peripheral circulation. Explanation: According to Fick’s 1st Law of Diffusion, the rate of transfer of solute across a cell membrane is directly proportional to the chemical potential energy residing in the transmembrane concentration difference of the solute. Also, it is inversely proportional to the length of pathway (approximated most closely by membrane thickness). Thus, answer D is the only correct answer. 12. According to the Goldman, Hodgkin, Katz equation for transmembrane potential (Em), which of the following specific changes (i.e. with no other changes) can produce a significant (i.e. more than a few millivolts) hyperpolarization of a resting electrically excitable cell membrane? A. B. C. D. E.

A decrease in K+ permeability (PK). A decrease in Na+ permeability (PNa). A decrease in extracellular [Na+]o concentration. An increase in extracellular [Na+]o concentration. A decrease in extracellular [K+]o concentration.

Explanation: Answer E is correct because a specific decrease in extracellular [K+]o concentration alone will increase the chemical potential gradient for K+. This will cause more separation of positive from negative charge across the membrane and thus increase the absolute magnitude of the Em (i.e. cause hyperpolarization). Answer A will cause a decrease in absolute magnitude of the Em (i.e. cause depolarization). Relative to answers B, C, and D, a decrease in PNa+ or changes in [Na+]o will have no significant effect on Em for a resting electrically excitable membrane.

2008 Exam 1 Answer Key

13. The first ion to rapidly increase its transmembrane permeability across an electrically excitable nerve or muscle cell membrane following generation of a local depolarization to the threshold potential is: A. B. C. D. E.

Ca2+ Mg2+ K+ Na+ Cl-

Explanation: Generation of a local depolarization to the threshold potential in an electrically excitable nerve or muscle cell membrane first causes a rapid increase in the membrane permeability for Na+. This results from rapid activation (i.e. opening) of voltage-dependent Na+ channels followed by a slower activation of voltage-dependent K+ channels. 14. Which of the following is NOT a property of an aquaporin protein? A.

An aquaporin is a tetramer that forms four transmembrane pores for transmembrane diffusion of H2O. B. The transmembrane diffusion of H2O through aquaporins is an example of facilitated diffusion. C. Aquaporins provide the only pathway for the active transport of H2O up its concentration gradient. D. Aquaporins are selective for water molecules. E. H2O dipoles are separated from bulk water and are “reorientated” for single file passage through the aquaporin protein pore.

Explanation: The statement in answer C is incorrect because H2O cannot be moved up its concentration gradient across a cell membrane by any form of active transport. The other statements in the other answers are true. 15. Which of the following statements concerning the transmembrane movement of H2O is FALSE? A.

Transmembrane movement of H2O by osmosis can be up or down its concentration gradient depending upon the direction of the gradient (i.e. from cytoplasm to extracellular fluid or vice versa). B. Movement of H2O by osmosis is always accompanied by a volume change. C. Movement of water by diffusion is not accompanied by a volume change. D. Osmotic pressure is measured as the counter hydrostatic pressure that must be applied to prevent H2O from moving down its concentration gradient. E. Osmotic pressure is directly proportional to the number of solute particles per unit volume in a solution. Explanation: The statement in answer A is false because H2O can only move down its concentration gradient by osmosis regardless of the direction of the concentration gradient. Answers B and C describe a fundamental difference between the diffusion versus osmosis of H2O down its concentration gradient. Answer E is the definition of osmotic pressure as a colligative property of a solution.

2008 Exam 1 Answer Key

16. The transmembrane transport of a solute by facilitated diffusion: A.

can occur up or down its concentration gradient depending upon the source of chemical bond energy. B. uses ATP high energy bonds as an energy source for transport. C. will be zero if the extracellular and intracellular solute concentrations are made equal. D. is not rate limited by a high solute concentration. E. requires movement of a coupled ion down its concentration gradient. Explanation: Answer C is correct because the energy source for facilitated diffusion of a solute is its concentration gradient. Thus facilitated diffusion always occurs down the concentration gradient of the solute. 17. The activity of the Na+/K+ pump in an electrically excitable cell membrane can be increased by all of the following EXCEPT: A. B. C. D. E.

an increase in extracellular [K+]o concentration. an increase in intracellular [Na+]i concentration. addition of a cardiac glycoside such as digitalis to the extracellular fluid. an increase in the cell's electrical activity (i.e. action potential generation). an increase in sodium permeability (PNa+).

Explanation: The activity of the Na+/K+ pump in an electrically excitable cell membrane is a "coupled" active transport system. Its activity is directly proportional to the extracellular [K+]o concentration and the intracellular [Na+]i concentration. Its activity is blocked (not enhanced) by cardiac glycosides acting on the extracellular side of the pump. 18. Which of the following is NOT essential for secondary active transport of an uncharged solute? A. B. C. D. E.

A membrane-bound carrier. Direct utilization of chemical bond energy. Utilization of energy derived from the electrochemical gradient of the “driving” solute. Movement of the "driving" solute down its electrochemical gradient Movement of the transported ("driven") solute up its chemical potential energy gradient.

Explanation: The defining properties of a secondary active transport carrier system include a "driving" solute that is coupled to a "driven" solute. The potential energy contained in the electrochemical gradient of the "driving" solute (upon its movement down this gradient) is utilized to move the "driven" uncharged solute across the membrane up its chemical potential energy gradient. No direct utilization of chemical bond energy is involved. 19. Which of the following types of muscle contraction were NOT measured in the skeletal muscle lab exercise? A. B. C. D. E.

Twitch. Incomplete tetanic. Isometric. Isotonic. Temporally summated.

Explanation: An isometric force transducer was used to measure both passive and active force. This provided the means to measure the A, B, C, and E forces.

2008 Exam 1 Answer Key

20. Which of the following statements concerning the generation of force by an isolated, normal, single, skeletal muscle fiber is TRUE? A.

A single skeletal muscle fiber performs external work when generating an isometric contractile force. B. A single skeletal muscle fiber cannot generate active contractile force when placed in a physiological salt solution without Ca2+. C. A single skeletal muscle fiber cannot exhibit temporal summation. D. A single skeletal muscle fiber cannot exhibit rigor. E. The contractile force generated by a tetanic contraction of a single skeletal muscle fiber is greater than that generated by a twitch contraction. Explanation: Answer E is true because in a skeletal muscle fiber, the free "activator" Ca2+ concentration in the sarcoplasm (required for actin-myosin bridge formation) is always greater during a tetanic stimulus. Relative to answer A, a single muscle fiber does not perform any external work during an isometric contraction (since the external length of the muscle doesn't change) although it does perform internal work (since at least some sarcomeres shorten). Relative to answer B, a single muscle fiber can generate active contractile force as long as Ca2+ is available for release from the sarcoplamic reticulum. 21. From the plot of muscle tension (or force) vs. muscle length that was generated in the skeletal muscle lab, all of the following statements are true EXCEPT: A. B. C. D. E.

total tension was zero when active tension was equal to passive tension. total tension was equal to active tension when passive tension was zero. total tension was equal to passive tension when active tension was zero. active tension was maximal at muscle resting (or "optimal") length. passive tension was zero at muscle equilibrium length.

Explanation: Answer A is correct because the statement in it is false. By definition, total tension equals the sum of active tension plus passive tension. Thus total tension equals active tension when passive tension is zero (i.e. muscle at equilibrium length) and total tension equals passive tension when active tension is zero. Also, resting (or optimal) length is defined as the length at which active tension is maximal because overlap of actin and myosin filaments is "optimal" for bridge formation. Lombard 22. In the smooth muscle laboratory exercise, which mechanism is most likely responsible for the reduced amplitude of rhythmic contraction that occurs in the jejunum in response to an increase in Ca2+ concentration in the physiological salt solution? A. B. C. D. E.

Liberation of acetylcholine from neural elements in the intestinal wall. An increase in membrane Ca2+ permeability in the smooth muscle cells. Release of nitric oxide from the smooth muscle cells. Activation of Ca2+-activated K+ channels in the cell membrane after Ca2+ reaches a critical threshold level in the cell cytoplasm. Increased activity of cell membrane sodium/potassium ATPase.

Explanation: The greater electrochemical gradient favoring Ca2+ entry causes the threshold for opening of Ca2+ activated K+ channels to be reached earlier. When the K+ channels open, the membrane hyperpolarizes, and voltage gated Ca2+ channels close. This terminates Ca2+ entry (and contraction), until K+ permeability decreases again and the membrane depolarizes. 2008 Exam 1 Answer Key

23. In the smooth muscle laboratory exercise, which of the following would give the greatest increase in contractile force in the aorta? A. B. C. D. E. F.

Epinephrine in the presence of phentolamine. Increased Ca2+ concentration in the physiological salt solution. Removal of the endothelium. Addition of propranolol to the tissue bath. Acetylcholine. Increasing bath K+ concentration to 50 mM.

Explanation: Increasing K+ concentration to 50 mM depolarizes the membrane and opens voltage gated Ca2+ channels, leading to a large influx of Ca2+, which causes a large, sustained contraction. 24. In the smooth muscle laboratory exercise, the rhythmic contractions of intestinal smooth muscle could be eliminated by which of the following? A. B. C. D. E.

Stretching the intestine. Adding epinephrine to the tissue bath. Adding propranolol to the tissue bath. Adding phentolamine to the tissue bath. Adding atropine to the tissue bath.

Explanation: Epinephrine inhibits rhythmic contractions of intestinal smooth muscle by activating alpha and beta adrenergic receptors, similar to the action of the sympathetic nervous system in the flight or fight response. Stretching the intestine would stimulate contraction, and the other drugs are receptor blockers that would not affect the intrinsic contractile activity of the intestine. 25. In the smooth muscle laboratory exercise, acetylcholine: A. B. C. D. E. F. G.

contracts intestinal smooth muscle by activating muscarinic cholinergic receptors. relaxes intestinal smooth muscle by activating muscarinic cholinergic receptors. inhibits rhythmic contractions of aortic smooth muscle by activating muscarinic receptors. contracts intestinal smooth muscle by activating alpha adrenergic receptors. relaxes intestinal smooth muscle by activating alpha-adrenergic receptors. contracts intestinal smooth muscle by activating beta-adrenergic receptors. relaxes intestinal smooth muscle by activating beta adrenergic receptors.

Explanation: Similar to the parasympathetic nervous system activation where ACh is the neurotransmitter, addition of ACh to the muscle causes contraction. This is mediated through a muscarinic, cholinergic receptor. Adrenergic receptors inhibit contraction of intestinal smooth muscle.

2008 Exam 1 Answer Key

26. In the smooth muscle laboratory exercise, elevated K+ concentration in the tissue bath: A. B. C. D. E. F. G. H. I.

contracts intestinal smooth muscle and relaxes aortic smooth muscle. relaxes intestinal smooth muscle contracts aortic smooth muscle. relaxes intestinal smooth muscle and relaxes aortic smooth muscle. contracts intestinal smooth muscle and contracts aortic smooth muscle. contracts intestinal smooth muscle and has no effect on aortic smooth muscle. relaxes intestinal smooth muscle and has no effect on aortic smooth muscle. contracts aortic smooth muscle and has no effect on intestinal smooth muscle. relaxes aortic smooth muscle and has no effect on intestinal smooth muscle. has no effect on aortic smooth muscle and no effect on intestinal smooth muscle.

Explanation: Elevated K+ concentration in the tissue bath would depolarize smooth muscle cells in both tissues, opening voltage-gated Ca2+ channels, and allowing Ca2+ influx and smooth muscle contraction. 27. In skeletal muscle, which of the following does NOT contribute to the regulation of intracellular Ca2+ levels? A. B. C. D. E.

Entry of extracellular Ca2+ through the sarcolemma. Pumping rate of the sarcoplasmic reticulum Ca2+ ATPase. Calsequestrin. Action potentials in the transverse tubule system. Frequency of nerve firing in the motor neurons.

Explanation: Unlike either smooth or cardiac muscle, entry of extracellular Ca2+ does not contribute to changes in intracellular Ca2+ either directly or indirectly (via Ca2+ -induced Ca2+ release). All the others either affect Ca2+ release or reduce Ca2+ levels in the cytoplasm via pumping into the SR or binding in the SR (calsequestrin) which enables the SERCA to pump more Ca2+ into the SR. 28. Which of the following statements about individual sarcomeres of skeletal muscle is FALSE? A.

Sarcomere length, as judged by the distance between the Z lines, would decrease when the muscle contracts. B. The width of the I band would increase when the muscle is stretched and decrease if the muscle contracts. C. The width of the A band would be similar to the length of the thick filament. D. The width of the A band would increase by muscle contraction. Explanation: All these responses arise directly from the sliding filament mechanism of contraction, because the width of the A band is determined by the length of the thick filament (which doesn’t change). The thin filaments slide between the thick filaments during contraction and relaxation, affecting the amount of overlap between the filaments, and thus the appearance of the other components of the sarcomere.

2008 Exam 1 Answer Key

29. Which of the following statements regarding isometric and isotonic contractions of skeletal muscle is TRUE? A. B.

No ATP is hydrolyzed during an isometric contraction, because the muscle does not shorten. An isotonic contraction can become an isometric contraction if the load on the muscle is increased sufficiently. C. The maximum velocity of an isotonic contraction is similar to that of an isometric contraction. D. The maximum velocity of an isotonic contraction of a skeletal muscle is similar in all muscles, regardless of the source of the muscle. Explanation: As seen in the force-velocity relationship, very heavy loads on the muscle result in a contraction velocity of zero, which is an isometric contraction (i.e., the muscle doesn’t shorten).

30. In skeletal muscle, which of the following will NOT increase contractile force? A. B.

Increased frequency of action potential firing in the motor nerve. Shortening the muscle from rest length (which corresponds to the length in the body) to 75% of rest length. C. Administration of an anti-cholinesterase drugs such as neostigmine. D. Increasing stimulus strength from threshold. E. Increasing the weight the muscle has to lift prior to an isotonic contraction. Explanation: According to the length force relationship, maximum overlap of thick and thin filaments occurs at rest length, so maximum contractile force that the muscle could generate would occur at that length, and would decrease at shorter lengths. In isotonic contractions, the muscle develops force until it can lift the weight (load) and shorten at a constant force. When the weight on the muscle is increased, the muscle needs to develop more force to lift the load when the muscle is activated.

31. In skeletal muscle, myosin: A. binds Ca2+ during muscle contraction. B. interacts with the active sites on the troponin molecule. C. hydrolyzes ATP at different rates, depending on the degree of phosphorylation of the light chain. D. binds ATP immediately prior to cross bridge release. E. all of the above statements are true. Explanation: ATP binding is essential for the cross bridge to release from the active site. After cross bridge release, the ATP is hydrolyzed, cocking the myosin head for the next power stroke, which occurs spontaneously after the cross bridge binds to the active site on the thin filament. ATP hydrolysis rate (and therefore cross bridge cycling rate) for a given myosin isoform in skeletal muscle is constant for that muscle type, in contrast to smooth muscle, where the amount of phosphorylation of the light chain determines the rate of ATP hydrolysis.

2008 Exam 1 Answer Key

Force

Length 32. The above diagram shows the length-force relationship in a skeletal muscle, as depicted in your textbook or constructed using the same approach employed in the skeletal muscle laboratory exercise. As the length of the muscle is increased, which of the following statements regarding this relationship is TRUE for the point indicated by the arrow? A. B. C. D. E.

The muscle has no active force, but passive force is increasing. The muscle has no passive force. Active force is decreasing while passive force is increasing. The muscle is at its rest length (Lo), i.e. the optimal length for contraction. All of the above statements are true.

Explanation: At the point indicated by the arrow, the force depicted by the upper curve (total force) is decreasing while the lower curve (passive force) is increasing. Subtracting the point on the lower curve from the point on the upper curve at any given muscle length gives active force for that muscle length, which is greater than zero, but less than active force at the rest length of the muscle. 33. Which of the following statements about smooth muscle contraction is FALSE? A. A smooth muscle cell can shorten more than a skeletal muscle cell of the same length. B. Smooth muscle exhibits a length-tension relationship. C. Unlike skeletal muscle, smooth muscle does not have a force-velocity relationship because smooth muscle contracts tonically while skeletal muscle exhibits twitch contractions. D. Contractile force in a smooth muscle can be similar to or greater than contractile force in a skeletal muscle of similar cross-sectional area. E. Increases in intracellular Ca2+ levels play an essential role in initiating both smooth muscle contraction and skeletal muscle contraction. Explanation: Smooth muscle has a force-velocity relationship that is similar in shape to that of skeletal muscle, but has a much lower velocity of contraction for any given load on the muscle. This is because of the slow rate of cross bridge cycling in smooth muscle, as compared to skeletal muscle (or cardiac muscle).

2008 Exam 1 Answer Key

34. Which of the following stimuli can modulate active contractile force in smooth muscle? A. B. C. D. E.

Neurotransmitters. Changes in oxygen levels. Hormones. Pressure, stretch, or distention. All of the above.

Explanation: Smooth responds to a wide variety of excitatory and inhibitory inputs to produce an integrated level of contraction. Excitatory and inhibitory inputs can include neurotransmitters such as norepinephrine and acetylcholine and even gases such as NO; circulating hormones such as epinephrine and angiotensin; physical stimuli such as stretch (myogenic activation), or metabolic stimuli such as reduced oxygen levels or increased levels of vasodilator metabolites. 35. Myosin light chain kinase (MLCK): A.

removes phosphates from the regulatory light chain of the myosin molecules, terminating contraction in smooth muscle. B. binds to tropomyosin in order to regulate smooth muscle contractile force. C. is activated by a Ca2+-calmodulin complex, and initiates smooth muscle contraction by phosphorylating the regulatory light chain on the myosin molecule. D. phosphorylates calponin and caldesmon, allowing the myosin cross bridges to reach the active sites on the thin filament. E. is located on the alpha helical portion of the thick filament, near the hinges for the myosin cross bridges. Explanation: When the muscle is excited, Ca2+ has to bind to calmodulin for MLCK to be activated. This allows the MLCK to phosphorylate the light chain on the myosin molecule in order to initiate ATP hydrolysis and cross bridge cycling. 36. Which of the following are found in skeletal muscle, but not in smooth muscle? A. B. C. D. E. F.

Tropomyosin. Troponin Sarcoplasmic reticulum (SR). Dense bodies. Cell-cell junctions for electrical transmission. Intermediate filaments.

Explanation: Smooth muscle has no troponin. All of the others are found in smooth muscle, but not in skeletal muscle, with the exception of the SR and tropomyosin, which are found in both.

2008 Exam 1 Answer Key

37. Under normal conditions, which of the following statements regarding the membrane potential of smooth muscle vs. skeletal muscle is FALSE? Action potentials in skeletal muscle are carried by Na+ ions while action potentials in smooth muscle are carried by Ca2+ ions. B. Spontaneous oscillations in membrane potential can occur in smooth muscle, but not in skeletal muscle. C. Membrane permeability to K+ is higher in smooth muscle than in skeletal muscle. D. Extracellular Ca2+ entry can contribute to changes in membrane potential in smooth muscle, but not in skeletal muscle. E. K+ is the predominant ion determining resting membrane potential in both skeletal muscle and smooth muscle.

A.

Explanation: Membrane potential in smooth muscle is less negative than that in skeletal muscle and less negative than that predicted by the Nernst equation for the K+ equilibrium potential. This is because K+ permeability in smooth muscle is lower than that in skeletal muscle, not higher. 38. Cerebral vasospasm (a severe constriction of cerebral arteries that can reduce blood flow to the brain tissue) can be alleviated by which of the following? A. B. C. D.

Application of a drug that opens K+ channels in the smooth muscle cells. Treatment with a drug that inhibits the sarcoplasmic reticulum Ca2+ ATPase. Treatment with a sodium/potassium pump inhibitor such as digitalis or ouabain. Inhibiting the sodium-calcium exchanger in the cell membrane.

Explanation: Opening of membrane K+ channels would hyperpolarize the smooth muscle membrane, closing voltage-gated Ca2+ channels and leading to vascular relaxation. All the others would aggravate the vasospasm by favoring vascular smooth muscle contraction.

2008 Exam 1 Answer Key

Greene

A

B

C

39. Match letter A in the diagram with the associated event of the cardiac cycle. A. B. C. D. E. F. G. H. I. J. K. L. M.

Left ventricular end systolic pressure Left ventricular end systolic volume Left ventricular end diastolic pressure Left ventricular end diastolic volume Third heart sound P wave Q wave T wave R wave End of ejection phase Closure of pulmonic valve End of isovolumic contraction Closing of aortic valve

2008 Exam 1 Answer Key

A

B

C

40. Match letter B in the diagram with the associated event of the cardiac cycle. A. B. C. D. E. F. G. H. I. J. K. L.

Left ventricular end systolic pressure Left ventricular end systolic volume Left ventricular end diastolic pressure Left ventricular end diastolic volume Third heart sound P wave Q wave T wave R wave End of ejection phase Closure of pulmonic valve End of isovolumic contraction

2008 Exam 1 Answer Key

A

B

C

41. Match letter C in the diagram with the associated event of the cardiac cycle. A. B. C. D. E. F. G. H. I. J. K. L. M.

Left ventricular end systolic pressure Left ventricular end systolic volume Left ventricular end diastolic pressure Left ventricular end diastolic volume Third heart sound P wave Q wave T wave R wave End of ejection phase Closure of pulmonic valve End of isovolumic contraction Closing of aortic valve

42. Ventricular output is 5 liters per minute in a normal individual at rest. Arterial blood pressure is 120/80 mmHg. Pulmonary blood pressure is 25/8 mmHg. Heart rate is 50 beats per minute. Which of the following statements is CORRECT? A. B. C. D. E.

Pulmonary blood flow is slightly greater than systemic blood flow. Left atrial pressure is greater than right atrial pressure. Stroke volume is 50 mls. Stroke work of the left ventricle is less than the right ventricle. None of the above is true in this individual.

Explanation: The correct answer is B. For any given ventricular output in the physiological range, preload of the left heart is slightly higher than that of the right heart. The other answers are incorrect because a) pulmonary and systemic blood flow must be the same, c) stroke volume is ventricular output divided by heart rate 5000 ml per min /50 beats per min = 100 ml, d) stroke work is approximately equal to stroke volume x afterload which is much greater in the left ventricle.

2008 Exam 1 Answer Key

43. Which ionic conductance is the greatest contributor to the rapid depolarization phase of cardiac muscle? A. B. C. D. E.

Potassium. Calcium. Hydrogen. Sodium. Chloride.

Explanation: A large increase in conductance of sodium is responsible for the rapid depolarization of cardiac muscle. Potassium and calcium contribute to the plateau phase. Hydrogen and chloride ions are not involved in depolarization. 44. In cardiac muscle cells: A. B. C. D. E.

calcium is released from the T tubules. ATP is not required for reuptake of calcium into the SR. beta receptor stimulation decreases intracellular calcium. resting membrane potential is approximately -50 mV. none of the above are true.

Explanation: In contrast to skeletal muscle, calcium is released in large amounts from the T tubules. Calcium is then taken back up into the SR, a process requiring ATP. Stimulation of beta receptors increases intracellular calcium, augmenting contraction. The resting membrane potential of a cardiac muscle cell is approximately -90 mV. 45. An increase in heart rate from 60 to 90 beats per minute results in which of the following? A. B. C. D. E.

A prolongation of the R-R interval. A shortening of the P-Q interval. S-T segment elevation. Prolongation of the R-T interval. Shortening of the T-P interval.

Explanation: a moderate increase in heart rate reduces the time spent in diastole, this would result in a shorter interval between ventricular repolarization (T) and the next atrial depolarization (P). 46. Which part of the left ventricle is the first to repolarize? A. B. C. D. E.

Outer apical surface. Inner apical surface. The endocardial surface near the base. A site of injury. The region adjacent to the AV node.

Explanation: The outer apical surface repolarizes first resulting in a positive T wave.

2008 Exam 1 Answer Key

47. A left axis deviation in the EKG could be caused by: A. B. C. D. E.

right bundle branch block. pulmonary hypertension. ventricular tachycardia. slowed conduction in AV node . none of the above.

Explanation: none of the above is the correct answer. Right bundle branch block would result in a rightward shift. Pulmonary hypertension would cause right ventricular hypertrophy and a rightward shift. Ventricular tachycardia and slowed AV nodal conduction would not cause a deviation in the net vector. 48. The right arm potential is -0.1 mV, left arm potential +0.4 mV, left leg potential is +1.5 mV. What is the deviation measured in lead I? A. B. C. D. E.

-0.5 mV -0.3 mV 0 mV + 0.3 mV + 0.5 mV

Explanation: Lead 1 measures the potential between the left arm and the right arm with positive reference in the left arm. Thus the deviation would be 0.4 – (-0.1)= +0.5 mV. 49. What is the explanation for the slowed conduction in the AV nodal fibers compared to fibers of the left or right bundle branches? A. B. C. D. E.

A reduced number of cells in the AV node. An increased potassium conductance in the membrane of the AV nodal cells. The parallel architecture of the penetrating AV bundle slows conduction. Increased numbers of fibroblasts in the AV nodal structure increases resistance and thus slows conduction. Fewer gap junctions in successive AV nodal cells increases resistance and slows conduction.

Explanation: E is the correct answer. Gap junctions between the cells provide for ionic conductance thereby coupling cells. AV nodal fibers have fewer gap junctions connecting them. This higher resistance slows the propagation of excitation between the cells. 50. If two vessels with diameters D1 and D2 and resistances R1 and R2 are in series with one another, the total resistance of the pair is given by: A. B. C. D. E.

R1D1 x R2D2 R1 + R2 R1 x R2 (R1 + R2)/(R1xR2) 1/(R1+R2)

Explanation: B is correct. The total resistance in a network of two resistances in series is the sum of the resistances.

2008 Exam 1 Answer Key

51. Which of the following are in order of decreasing cross sectional area? A. B. C. D. E.

Aorta, small arteries, arterioles, capillaries, venules, small veins. Capillaries, venules, aorta, vena cava, small arteries. Capillaries, small veins, arterioles, vena cava, aorta. Aorta, vena cava, capillaries, venules, arterioles. Vena cava, venules, capillaries, small veins.

Explanation: C is correct. According to the table on page 162 of Guyton and Hall the order is: Capillaries, venules, small veins, arterioles, small arteries, vena cava, and aorta. 52. Mean arterial blood pressure is 150 mmHg. Pulse pressure is 60 mmHg. Right atrial pressure is 0 mmHg. Heart rate is 80 beats per minute. Cardiac output is 6000 ml/min. Calculate TPR. A. B. C. D. E.

0.5 PRU 1.0 PRU 1.5 PRU 2.0 PRU 2.5 PRU

Explanation: Resistance is given by pressure gradient divided by flow. In this case the pressure gradient is 150mmHg - 0 mmHg and the flow is 6000 ml/min or 100 ml/sec (the proper unit for calculating peripheral resistance units PRU). 150 mmHg/100 ml/sec = 1.5 PRU. 53. A vascular bed that normally holds 500 ml of blood suddenly experiences an increase of volume of 100 ml, causing pressure to rise by 10 mmHg. The compliance of that vascular bed is: A. B. C. D. E.

50 ml/mmHg. 1/50 ml/mmHg. 0.1 ml/mmHg. 10 ml/mmHg. 5 ml/mmHg.

Explanation: Compliance is given by the ratio of increase in volume to increase in pressure (dV/dP) in this case that is 100 ml/10mmHg = 10 ml/mmHg. 54. When a person who is about 6 feet tall stands completely still, the pressure in the veins of his/her feet commonly reaches: A. B. C. D. E.

10 mmHg. 30 mmHg. 60 mmHg. 90 mmHg. 120 mmHg.

Explanation: The correct answer is 90mmHg. The elevated pressure in the feet is because of the hydrostatic pressure generated by the weight of the column of blood in the veins.

2008 Exam 1 Answer Key

55. Which of the following is most likely to occur in a patient with arteriosclerosis? A. B. C. D. E.

An increase in systolic aortic blood pressure. A decrease in aortic stiffness. An elevated blood flow in the descending aorta. An increase in aortic backflow. All of the above would occur.

Explanation: Arteriosclerosis is associated with a increased stiffness of the aorta. This reduction in compliance causes an augmented pulse pressure and an increase in systolic blood pressure. Blood flow would not change appreciably. 56. Which of the following statements regarding vasomotion is FALSE? A. B. C. D.

Vasomotion reduces the resistance to blood flow. Vasomotion is an intermittent contraction of precapillary sphincters. Vasomotion alters the delivery pattern of oxygen. Vasomotion causes capillary blood flow to be pulsatile.

Explanation: Vasomotion is the intermittent contraction of precapillary sphincters and metarterioles. The contraction of these vessels causes capillary blood flow to be intermittent or pulsatile causing periods of low flow and resulting decreases in local delivery of oxygen. Vasomotion can result in elevated peripheral resistance. 57. Blood flow in an individual vessel segment is determined by: A. B. C. D. E.

total peripheral resistance. inflow pressure. outflow pressure. inflow-outflow pressure. cardiac output.

Explanation: The pressure gradient is one of the main determinants of blood flow in an individual vessel segment. The others are blood viscosity, vessel diameter, and vessel length.

2008 Exam 1 Answer Key

58. Interstitial fluid colloid osmotic pressure is 9 mmHg, mean capillary pressure is 20 mmHg, plasma colloid osmotic pressure is 23mmHg, interstitial free fluid pressure is -5 mmHg. In which direction will the net flux of fluid be? A. B. C. D.

Out of the capillary, Into the capillary, There will be no net flux of water, There is not sufficient information given to calculate the direction of net flux.

Explanation: The direction of flux can be determined by calculating the Starling equilibrium. In this case the sum of the outward forces is 20+9 = 29 mmHg. The sum of the inward forces is 23+ (-5) = 18 mmHg. The sum of the mean forces is 29 out – 18 inward = 11 mmHg net outward force, so fluid would move out of the capillary (answer A).

Cowley 59. The “myogenic response”: A. B. C. D. E.

is the most important mechanism for the regional regulation of blood flow. is most active in large arteries. is a negative feedback mechanism that results in an increase in vessel wall tension. is initiated by an increase blood flow and wall shear stress. represents a positive feedback control system.

Explanation: The myogenic response represents a positive feedback mechanism since it is initiated by an increase in luminal pressure and stretch of the vessel. The response reduced vessel diameter and this increases the pressure in the lumen of the vessel even more resulting in a greater myogenic constriction, which leads in turn to a greater luminal pressure with greater constriction (a positive feedback vicious cycle). The myogenic mechanism is not of great importance in the regulation of blood flow since the arterioles which are up-stream of the tissue do not have a way of sensing the metabolic needs of the tissue. The myogenic response is most active in small arterioles and not in large arteries. 60. Which of the following would NOT occur with daily supplements of L-arginine? A. B. C. D. E.

Reduced vascular tone. Greater conversion of arginine to citrulline. Increased cellular levels of cGMP. Increased vascular tone. Greater NO production.

Explanation: Increased intake of L-arginine provides excess substrate for the enzyme nitric oxide synthase (NOS) to convert into citrulline and nitric oxide (NO) within the vascular endothelium and vascular smooth muscle. The increased NO results in increased cellular levels of cGMP and reduced vascular tone (not increased vascular tone).

2008 Exam 1 Answer Key

61. The oxygen demand theory of blood flow autoregulation would NOT predict which of the following? A. B. C. D. E.

Cyanide poisoning would vasodilate arterioles. High altitude climbing would vasodilate arterioles. An increased supply of blood flow to the tissue would vasoconstrict arterioles. Excess delivery of blood flow would reduce CO2 levels. Arterioles would vasodilate in face of an increase of arterial pressure.

Explanation: An increased oxygen demand according to the oxygen demand theory would result in arteriolar vasodilation. Cyanide poisoning makes it more difficult for O2 to be released from the hemoglobin of red cells so local tissue oxygen levels decrease resulting in vasodilation. The same occurs with high altitude as the partial pressure of O2 is reduced. Conversely, an increase in blood flow to the tissue will deliver excess oxygen to the tissues and result in a faster washout of tissue CO2, both of which would vasoconstrict the arterioles. An increase in arterial perfusion pressure to a tissue would also initially increase blood flow and overperfuse the tissue with an excess delivery of oxygen resulting in vasoconstriction – not vasodilation. 62. If you were to design a new drug that could be used to enhance vasodilation and reduce blood pressure in hypertensive patients, which of the following would be an appropriate target? A. B. C. D. E.

A drug that inhibited catechol-O-methytransferase. A drug that stimulated angiotensin AT1 receptors. A drug that reduced levels of neutral endopeptidase. A drug that increased or mimicked phenylethanolamine-N-methyltransferase (PNMTP). A drug that stimulated production of angiotensin converting enzyme (ACE).

Explanation: A drug that inhibited reduced catechol-O-methytransferase would increase NE at the post-synaptic tissue and increase vasoconstriction. The same would true for a drug that stimulated angiotensin AT1 receptors or stimulated the production of ACE. A drug that increased or mimicked phenylethanolamine-N-methyltransferase would increase production of epinephrine and result in vasoconstriction and increased cardiac work. However, a drug that reduced levels of neutral endopeptidase would probably increases the levels of ANP and result in enhanced vasodilation and sodium excretion. Such compounds have indeed now been developed and are in various phases of drug testing.

2008 Exam 1 Answer Key

63. In the heart, brain, skeletal muscle, liver, kidney, and intestine, one minute of vascular occlusion is followed by which of the following changes when the occlusion is removed? A.

a period of 15 to 30 seconds during which flow progressively increases to the preocclusion value B. a rapid (less than 2 seconds) return to the preocclusion flow. C. a period of 15 seconds or more during which flow is markedly higher than it was prior to occlusion. D. a change in flow produced primarily by adrenergic sympathetic neurons. E. a change in flow produced primarily by cholinergic sympathetic neurons. Explanation: C is the correct answer. This is called reactive hyperemia. In the heart, for example, 10 seconds of coronary artery occlusion causes, on release of occlusion, 15 seconds during which the flow is two to four times that prior to occlusion. 64. Viagra is used for the treatment of impotence and is known to act primarily by: A. B. C. D. E.

increasing renin secretion. inhibiting sympathetic nerve activity. stimulating nitric oxide synthase (NOS) activity. inhibiting phosphodiesterase activity. increasing cGMP activity.

Explanation: There are nonadrenergic, noncholinergic fibers in the nervi erigentes and these contain large amounts of NO synthase, the enzyme that catalyzes the formation of NO. Viagra inhibits the breakdown of cGMP by phosphodiesterase. 65. Which of the following is the most important mechanism for initiating an increase in blood flow to active skeletal muscle during exercise? A. B.

An increased cardiac output. An increased concentration of epinephrine in the vicinity of the blood vessels of active skeletal muscle. C. An increased concentration of norepinephrine in the vicinity of blood vessels of active skeletal muscle. D. An increased concentration of acetylcholine in the vicinity of the blood vessels of active skeletal muscle. E. The action of metabolites on skeletal muscle blood vessels. Explanation: The major factors regulating skeletal muscle blood flow during exercise are changes in the distribution of cardiac output initiated by vasodilation in the active skeletal muscle related to the metabolic autoregulatory processes and an associated vasoconstriction in most other areas due to sympathetic stimulation and elevations of circulating epinephrine which elevates arterial pressure helping to maintain the increased blood flow to the working muscles. The increased cardiac output is clearly important to sustain the increased muscle and is sustained by the increased venous return to the heart coming from the enhanced skeletal muscle blood flow, but the increase of cardiac output is a consequence and not the initiating factor in the increased blood flow to the working muscle. The increased epinephrine and norepinephrine levels in the vicinity of the skeletal muscle blood vessels fail to vasoconstrict the vessels due to the overriding effects of local autoregulation. Acetylcholine is not believed to participate in any measurable way in humans.

2008 Exam 1 Answer Key

66. The increased vascular endothelial shear stress and perhaps the reduced tissue oxygen levels associated with strenuous exercise of skeletal muscle result in: A. B. C. D. E.

reduced NO release from the endothelial cells. increased release of VEGF. reduced sprouting of endothelial cords. reduced angiogenesis. reduced release of FGF.

Explanation: Stimulation of vascular endothelial growth factor (VEGF) and fibroblast growth factor (FGF) occurs with increased blood flow in skeletal muscle vessels resulting in the disruption of the basement membrane and sprouting of endothelial cords from the underlying endothelial cells resulting in the budding and growth of new blood vessels. Accordingly, B is the correct answer.

67. In the above figure, an abnormality has caused the renal output curve to shift 50 mmHg toward higher blood pressure. This shift occurred rapidly (i.e. a step change). The following question addresses the consequences of such a shift of the pressure-diuresis relationship. One of the immediate (within 3 minutes) consequences of this abnormality would be: A. B. C. D. E.

an increase of cardiac output. an increase of drinking. an increase of heart rate. a reduction of urine excretion. a large increase of total peripheral resistance.

Explanation: An immediate upward shift of the renal function curve would mean that the kidneys would immediately require an increased arterial perfusion pressure to sustain a normal excretion of sodium and water. As occurs with a partial clamping of the renal artery which results in such an immediate upward shift of the renal function curve, the arterial pressure does not immediately rise to offset this occlusion. Therefore, the immediate change that occurs is that urine excretion is greatly reduced and you begin retaining fluids.

2008 Exam 1 Answer Key

68. In the above figure, an abnormality has caused the renal output curve to shift 50 mmHg toward higher blood pressure. This shift occurred rapidly (i.e. a step change). The following question addresses the consequences of such a shift of the pressure-diuresis relationship. Twenty-four hours following the initiation of this abnormality, one would expect: A. B. C. D. E.

a reduction of cardiac output. edema of the extremities. an increase of heart rate. a reduction of blood volume. an increase of blood volume.

Explanation: Following the reduction in urine excretion there is a retention of fluids, and an expansion of blood volume. Cardiac output would therefore increase and this would drive an increase of arterial pressure. The rise of arterial pressure would slow the heart rate initially via the baroreceptor reflexes.

2008 Exam 1 Answer Key

69. In the above figure, an abnormality has caused the renal output curve to shift 50 mmHg toward higher blood pressure. This shift occurred rapidly (i.e. a step change). The following question addresses the consequences of such a shift of the pressure-diuresis relationship. One month following the initiation of this abnormality one would expect: A. B. C. D. E.

a large increase of blood volume. a large increase of total peripheral resistance. a reduction of heart rate. a reduction of cardiac output. none of the above.

Explanation: After several days of an elevated cardiac output and an associated overperfusion of many of the tissues, tissue autoregulation of blood flow would result in vasoconstriction of the vessels and increase the total peripheral resistance and the resistance to venous return. This would be amplified in several weeks by hypertrophy of the arteriolar walls and reduction in the capillary density in certain organs (rarefaction) such as the skeletal muscle. All of this leads to a steady-state elevation of total peripheral resistance which sustains the elevated blood pressure. As this occurs, only a very small (~5%) elevation of blood volume and cardiac output is required in the long run to sustain this elevation of total peripheral resistance and sustained hypertension. 70. Partial constriction of both common carotid arteries would result in: A. B.

a reduction of systemic arterial blood pressure within 30 seconds. a reduction in baroreceptor afferent nerve activity going to the nucleus tractus solitarius of the medullary area in the brain stem. C. an increase in cardiac vagal nerve activity. D. a reduction in renin secretion by the kidney. E. a reduction in afferent nerve impulses from the aortic baroreceptors. Explanation: With constriction of the common carotid arteries there is a reduction in the stretch of vessels at the bifurcation of the common carotids which is the site of the carotid baroreceptors (stretch receptors). This results in reduction in the baroreceptor afferent nerve activity going to the medullary centers. This results in stimulation of the efferent sympathetic nerves which increases arterial pressure in the rest of the systemic circulation; it results in a reduction in the cardiac vagal nerve activity to the heart which increases heart rate; it results in stimulation of renin secretion from the juxtaglomerular cells of the kidney which are innervated by sympathetic fibers. The final 2008 Exam 1 Answer Key

possibility E require some greater thought. The afferent nerve activity from the aortic baroreceptors actually would increase (the opposite of the carotid sinus nerve activities) due to the increased aortic blood pressure resulting from the occlusion of the carotid arteries and the carotid sinus reflex responses. 71. The carotid sinus nerve activity of a normotensive individual (Norm) with a mean arterial pressure of 100 mmHg compared to an individual with chronic hypertension and a mean arterial pressure of 150 mmHg (Hazel). A. B. C. D. E.

Nerve activity of Norm is greater than Hazel. Nerve activity of Hazel is greater than Norm. Nerve activity of Norm is equal to Hazel. Nerve activity of Norm is normal, that of Hazel is reduced. None of the above.

Explanation: Since the baroreceptors adapt over the period of several days after a chronic change of arterial pressure, the afferent nerve activity of Hazel (the hypertensive subject) would be the same as that of Norm (the normotensive subject). 72. A large acute expansion of the blood volume would reduce the circulating hormone levels of all but the following: A. B. C. D. E.

Vasopressin. Aldosterone. Atrial natriuretic peptide. Renin. Epinephrine.

Explanation: An acute expansion of blood volume results in stimulation of low pressure cardiac atrial reflexes. Also, cardiac output increases and raises arterial pressure, which stimulates the carotid and aortic baroreceptors. Together increased nerve activity from these stretch receptors will reduce release of vasopressin from the posterior pituitary, epinephrine from the adrenal medulla, and renin from the kidney which in turn will reduce adrenal aldosterone secretion. All of these will enhance renal sodium and water excretion retention and reduce systemic vascular arteriolar tone. However, the atrial stretch of the heart will stimulate an increased release of atrial natriuretic peptide, a response that will contribute to the natriuresis and vasorelaxation.

2008 Exam 1 Answer Key

73. This figure represents the approximate potency of various arterial pressure control mechanisms at different time intervals after the onset of a disturbance to the arterial pressure. Using your knowledge of the various pressure control mechanisms answer the following question. Which of the response curves in the above figure represents that of the renin-angiotensin system? A. B. C. D. E.

Curve #1 Curve #2 Curve #3 Curve #5 Curve #8

Explanation: Answer D (Curve #5 on the graph) is the correct answer. Following a decrease of arterial pressure, renin is released from the kidney within 1-2 minutes with the immediate production of angiotensin I which circulates to the lungs producing angiotensin II. This results in immediate arterial vasoconstriction and a correction of arterial pressure over a period of about 20 minutes. Twenty minutes is required since renin has a circulating half-life of about 20 minutes, so it requires about this amount of time for circulating levels to reach a steady-state and a similar time to achieve the maximum feedback gain for the control of arterial pressure with this system. The system has a maximum negative feedback gain of about 2.0 reaching a peak in about 20 minutes as represented in the above figure.

2008 Exam 1 Answer Key

74. This figure represents the approximate potency of various arterial pressure control mechanisms at different time intervals after the onset of a disturbance to the arterial pressure. Using your knowledge of the various pressure control mechanisms answer the following question. Which mechanism would predominate during the first hour following a severe hemorrhage in which arterial pressure fell to levels below the autoregulatory capacity of the heart, brain, and kidneys? A. B. C. D. E.

Curve #8 Curve #5 Curve #1 Curve #4 Curve #7

Explanation: Answer D (Line 4 on the graph, CNS ischemic response) is the correct answer. If arterial pressure falls below the autoregulatory limit of the brain, cerebral ischemia would result and stimulate the cerebral ischemic reflex. This reflex results in a rapid (within a minute or so) and great stimulation of the sympathetic outflow as represented by Line 4.

2008 Exam 1 Answer Key

75. This figure represents the approximate potency of various arterial pressure control mechanisms at different time intervals after the onset of a disturbance to the arterial pressure. Using your knowledge of the various pressure control mechanisms answer the following question. What mechanism would predominate in the weeks following the surgical repair and closure of a large arterial-venous shunt (A-V shunt) that initially resulted in a significant increase of total peripheral resistance? A. B. C. D. E.

Curve #1 Curve #3 Curve #5 Curve #7 Curve #8

Explanation: Answer E (line 8 on the graph, renal-blood volume pressure control) is the correct answer. The immediate closure of a large A-V shunt results in an increase of total peripheral resistance with an accompanying rise of arterial blood pressure. The arterial pressure control system represented in this figure that would predominate in the weeks following this rise of pressure would be the renal pressure-natriuresis-diuresis control system which would gradually reduce the blood volume and cardiac output thereby returning arterial pressure to normal. 76. An increase in arterial pressure from 100 to 130 mmHg produces all the following effects, EXCEPT: A. B. C. D. E.

an increased nerve traffic from the arterial baroreceptors. an increase in vagal tone. a decrease in sympathetic tone. an increased diameter of the carotid sinus. a stimulation of vasopressin release.

2008 Exam 1 Answer Key