Unit I Study Questions

Study Questions

Choose the ONE best answer.


I.1. A fluid that is composed of 120 mmol/L K+, 12 mmol/L Na+, and 15 mmol/L Cl− but is virtually Ca2+ free (<1 μmol/L) would best approximate which body fluid compartment?

A. Transcellular
B. Plasma
C. Interstitial
D. Intracellular
E. Extracellular

Best answer = D. Intracellular fluid should be recognized by its relatively high K+ concentration, which is due to the Na+-K+ ATPase found in the membrane of virtually all cells (1∙II). Extracellular fluid (ECF) has a lower K+ and higher Na+, Cl−, and Ca2+ concentration compared with ECF and can be further subdivided into plasma (fluid within the vascular space) and interstitial fluid (fluid outside the vascular space; 3∙III∙A). Because the barrier between these two ECF compartments does not prevent ion movement, their ionic composition is similar. Transcellular fluid (including cerebrospinal fluid, synovial fluid, and urine) composition is variable depending on the location and, therefore, not the best choice.


I.2. Dye indicators are important physiologic tools used to calculate unknown volumes or concentrations within the body. If a dye is membrane permeable, which of the following changes will most likely increase dye diffusion rate?

A. Lowering dye concentration
B. Increasing membrane surface area
C. Increasing membrane thickness
D. Decreasing fluid temperature
E. Lowering the dye partition coefficient

Best answer = B. Increasing membrane surface area increases the opportunity for dye to cross the membrane, which increases its diffusion rate (1∙IV∙B). It takes longer for molecules to diffuse across thick membranes than thin ones, and decreasing a fluid’s temperature increases its viscosity, which also slows diffusion rate. Concentration gradients provide the driving force for diffusion, so lowering dye concentration flattens the gradient and reduces diffusion rate. A partition coefficient is a measure of dye lipid solubility. Molecules with high lipid solubility diffuse through membranes faster than poorly soluble molecules, so lowering partition coefficient would decrease diffusion rate.


I.3. A 66-year-old male is treated with the loop diuretic furosemide (Na+-K+-2Cl− cotransport inhibitor) to reduce symptoms associated with congestive heart failure. Which of the following best describes this cotransporter’s mode of action?

A. It is a primary active transporter.
B. It is electrogenic.
C. A rise in intracellular K+ would decrease transport rate.
D. It transports Na+ and K+ into the cell and 2 Cl− out of the cell.
E. It transports Na+ against its electrochemical gradient.

Best answer = C. Cotransporters, by definition, move two or more ions in the same direction (1∙V∙C). The Na+-K+-2Cl− cotransporter simultaneously carries two anions and two cations across the plasma membrane and, thus, is not electrogenic. Primary active transporters use adenosine triphosphate to pump ions against their electrochemical gradients. Transporters that move Na+ in one direction while simultaneously bringing other ions back in the opposite direction are exchangers, not cotransporters. The Na+ gradient established by the basolateral Na+ pump (Na+-K+ ATPase) provides the electrochemical driving force for K+ and Cl− uptake, but transport rate is sensitive to transmembrane K+ and Cl− gradients. Increasing intracellular concentrations of either ion will slow net uptake.


I.4. Serum electrolytes levels are ordered on a 12-year-old boy with a gastrointestinal infection, which induced prolonged and severe vomiting episodes. Plasma K+ concentrations were found to be abnormally low (2 mmol/L). Which of the following might be expected to result from mild hypokalemia?

A. Resting potentials would shift positive.
B. K+ equilibrium potential would shift negative.
C. Neuronal action potentials would be inhibited.
D. Na+ channels would inactivate.
E. K+-channel activation would yield K+ influx.

Best answer = B. Hypokalemia, or reduced extracellular K+ concentrations, enhances the electrochemical gradient favoring K+ efflux from cells and causes the K+ equilibrium potential to shift negative (see 2∙II∙B). Because membrane potential is determined largely by the transmembrane K+ gradient, resting membrane potential (Vm) would shift negative also. A negative shift in Vm means that a stronger depolarization would be necessary to take Vm to the threshold for voltage-gated Na+ channel activation (2∙III∙B), but, once reached, an action potential would be initiated. K+-channel activation always causes K+ efflux, except in rare instances (e.g., in the inner ear; 9∙IV∙C).


I.5. A 35-year-old man carries an epilepsy gene. The gene mutation affects the neuronal voltage-dependent Na+ channel, causing it to inactivate more slowly (~50%). How might expression of this epilepsy gene affect nerve function?

A. Resting potential would settle close to 0 mV.
B. Action potentials would no longer overshoot 0 mV.
C. Action potentials would be prolonged.
D. Action potentials would rise very slowly.
E. There would be no action potentials.

Best answer = C. The voltage-dependent Na+ channel is opened by membrane depolarization to yield the upstroke of the neuronal action potential (2∙VI∙A). An inactivation gate closes shortly after activation, blocking passage of Na+ and allowing membrane potential to return to resting levels. If inactivation were slowed, membrane recovery would be delayed, and the action potential would be prolonged. Resting potential should not be affected by an inactivation defect unless it prevented the channel from closing, causing a sustained Na+ influx. Activation and inactivation are separate processes, and, therefore, the rate at which the action potential rises should be normal.


I.6. An agricultural worker is packing hot chili peppers for transport. He removes his protective mask and becomes incapacitated with a sensation of nasal burning caused by capsaicin from the peppers. What receptor type is capsaicin stimulating?

A. Transient receptor-potential channels
B. Purinergic receptors
C. Ionotropic glutamate receptors
D. Cys-loop family receptors
E. Voltage-gated Na+ channels

Best answer = A. The transient receptor potential channel (TRP) family transduces a variety of sensory stimuli, including heat, cold, and osmolality (2∙VI∙D). TRPV1 channels are stimulated by capsaicin. Purinergic, glutamate, and cys-loop receptors are activated by specific ligands (i.e., adenosine triphosphate, L-glutamate, and acetylcholine, respectively). Capsaicin is not an agonist for these receptor classes. Voltage-gated Na+ channels are activated primarily by membrane depolarization.


I.7. During a serological analysis, red blood cells (RBCs) were transferred from blood to a solution containing 100 mmol/L CaCl2 and 100 mmol/L urea and then monitored using light microscopy. How would you expect this transfer to affect RBC volume?

A. The solution is isosmotic, so no long-term effect.
B. The solution is isotonic, so no long-term effect.
C. Transient swelling would occur.
D. Swelling to the point of lysis would occur.
E. The cell would shrink by ~50%.

Best answer = B. CaCl2 dissociates into three particles (1 Ca2+ and 2 Cl−) in water. A 100-mmol/L CaCl2 solution has an osmolality of 300 mOsm/kg H2O, which approximates that of RBC intracellular fluid (ICF). 100 mmol/L urea brings total osmolality to 400 mOsm/kg H2O (the solution is hyperosmotic), but urea would rapidly enter the cell until intracellular and extracellular fluids equilibrated at ~350 mOsm/kg H2O (3∙II∙C). The solution is, thus, isotonic. Cell shrinkage would occur if urea was impermeant, but most cells are highly permeable to urea. Cell swelling in this example would only occur if ICF osmolality rose due to active accumulation of one or more of the three solutes.


I.8. Liver damage may result in decreased synthesis of plasma proteins such as albumin. What is the most significant effect of low plasma albumin on osmosis or fluid transport?

A. Interstitial fluid volume increases.
B. Vascular fluid volume increases.
C. Plasma colloid osmotic pressure increases.
D. Plasma osmolality increases.
E. Plasma osmolality decreases.

Best answer = A. Blood contains large amounts of albumin (3.5–5 g/dL) that is trapped in the vascular compartment by virtue of its large size (3∙III∙B). Its function is to help create an osmotic potential (known as plasma colloid osmotic pressure) that draws extracellular fluid (ECF) into the vasculature. A decrease in plasma albumin concentration would, therefore, allow fluid to leave the vasculature and enter the interstitium. Plasma osmolality does not change significantly with changes in protein concentration. The main determinants of ECF osmolality are ions (e.g., Na+ and Cl−) and other solutes (e.g., glucose and urea, measured as blood urea nitrogen, or “BUN”).


I.9. A 95-year-old man with widely metastatic cancer is receiving morphine to help alleviate pain. Brainstem respiratory center function has been depressed as a result, causing hypoventilation. Which of the following might be expected to result from reducing ventilation?

A. Alkalemia
B. Decreased plasma HCO3 − levels
C. Decreased renal HCO3 − reabsorption
D. Increased interstitial pH
E. Increased urinary H+ excretion

Best answer = E. Metabolism generates large amounts of volatile acid (H2CO3) that is excreted via the lungs (3∙IV∙A). Decreasing ventilation allows this acid to accumulate, producing a (respiratory) acidemia. The kidneys help compensate by increasing H+ excretion. Decreasing renal HCO3 − reabsorption would exacerbate the acidemia through loss of buffer to urine. Accumulation of volatile acid raises plasma HCO3 − levels because H2CO3 dissociates in solution to form HCO3 − and H+. H+, along with other small molecules, moves freely between the blood and interstitium. Thus, if blood is acidic, the interstitium will also have a low pH.


I.10. A researcher investigating the properties of intestinal epithelium from a patient with inflammatory bowel disease noted that the diseased areas have a low electrical resistance, whereas the healthy areas have a high resistance. What might be inferred about the properties of the healthy epithelium?

A. It forms weak transepithelial ionic gradients.
B. It is specialized for isosmotic transport.
C. It has a thick basement membrane.
D. The tight junctions are highly impermeable.
E. It lacks a basolateral Na+-K+ ATPase.

Best answer = D. High electrical resistance is characteristic of a “tight” epithelium, a property conferred in part by the impenetrability of the tight junctions between cells to ions and water (4∙II∙E). Tight epithelia are notable for their ability to establish strong osmotic and ionic concentration gradients. The areas of inflammation have a low electrical resistance, which makes them “leakier” to ions. Leaky epithelia are usually specialized for high-volume isosmotic transport. Basement membranes do not directly contribute to epithelial electrical resistance. All intestinal epithelia express a basolateral Na+-K+ ATPase.


I.11. A 52-year-old woman presents with heart palpitations and lightheadedness. An electrocardiogram shows her to be in atrial fibrillation, which has been linked to increased connexin 43 expression. Which of the following best describes connexins normally?

 A. They open during membrane depolarization.
B. They are highly ion selective.
C. They mediate Ca2+ influx from the cell exterior.
D. They allow electrical propagation through tissues.
E. They are found only in the heart.

Best answer = D. Connexins form hexameric assemblies (connexons) with a pore at their center (4∙II∙F). Connexons from two adjacent cells fuse to create a gap-junction channel that provides a pathway for electrical and chemical communication between cells. They are widely distributed. In the heart, they allow waves of contraction to spread across the myocardium (17∙III). Gap-junction channels are characterized by their wide, nonselective pores that can allow passage of small peptides. They are usually open at rest and may close upon depolarization. Ca2+ channels mediate Ca2+ influx across surface membranes, not gap-junction channels.


I.12. Hypokalemia is relatively rare in healthy individuals, but which of the following would favor increased K+ uptake by a transport epithelium for transfer to the circulation?

A. Lumen-negative potential difference
B. Increased paracellular water uptake
C. Increased Na+-K+ ATPase activity
D. High interstitial K+ concentrations
E. Apical glucose cotransport

Best answer = B. The paracellular route is a significant pathway for solute and water movement across many epithelia (4∙III∙B). High paracellular water flow rates generate solvent drag, whereby inorganic ions and other solutes are swept along with water. Because K+ is a cation, a renal or gastrointestinal lumen (for example) that is negatively charged with respect to blood decreases net uptake. The Na+-K+ ATPase increases intracellular K+ concentrations, which decreases the driving force for apical K+ uptake. High interstitial K+ concentrations also decrease the electrochemical gradient favoring net K+ uptake. Glucose cotransporters generally couple glucose movement with Na+, not K+.