Final: Cell Physiology Flashcards
A 62-year-old woman eats a high-carbohydrate meal. Her plasma glucose concentration rises, and this results in increased insulin secretion from the pancreatic islet cells. The insulin response is an example of
negative feedback
The increase in plasma insulin lowers the plasma glucose concentration back to normal and is an example of negative feedback. Negative feedback opposes change and results in stability. Positive feedback would produce a further increase in plasma glucose concentration. Chemical equilibrium indicates a condition in which the rates of reactions in forward and backward directions are equal. End-product inhibition occurs when the products of a chemical reaction slow the reaction (e.g., by inhibiting an enzyme) that produces them. Feed-forward control involves a command signal and does not directly sense the regulated variable (plasma glucose concentration).
If a region or compartment is in a steady state with respect to a particular substance, then
the amount of the substance in the compartment does not change with respect to time.
In a steady state, the amount or concentration of a substance in a compartment does not change with respect to time. Although, there may be considerable movements into and out of the compartment, there is no net gain or loss. Steady states in the body often do not represent an equilibrium condition, but they are displaced from equilibrium by the constant expenditure of metabolic energy.
The metabolic rate of a person is normally expressed in terms of the rate of heat liberation that results from the chemical reactions of the body. Metabolic rate can be estimated with reasonable accuracy from the oxygen consumption of a person. Which of the following factors tends to increase or decrease the metabolic rate of a person?
Growth hormone & Fever increase and Sleep and Malnutrition Decrease
Growth hormone can increase the metabolic rate 15% to 20% as a result of direct stimulation of cellular metabolism. Fever, regardless of its cause, increases the chemical reactions of the body by an average of about 120% for every 10°C rise in temperature. The metabolic rate decreases 10% to 15% below normal during sleep. Prolonged malnutrition can decrease the metabolic rate 20% to 30%, presumably due to the paucity of food substances in the cells.
Which of the following accounts for the largest component of daily energy expenditure in a sedentary individual?
Basal Metabolic Rate
Basal metabolic rate counts for about 50% to 70% of the daily energy expenditure in most sedentary individuals. Non-exercise activity, such as fidgeting or maintaining posture, accounts for approximately 7% of the daily energy expenditure, and the thermic effect of food accounts for about 8%. Nonshivering thermogenesis can occur in response to cold stress, but the maximal response in adults is less than 15% of the total metabolic rate.
A 54-year-old man is admitted to the emergency department after being found lying in his yard near a running lawnmower on a hot summer day. His body temperature was 106°F, blood pressure is normal, and heart rate is 106 beats/min. Which of the following sets of changes is most likely to be present in this man?
Sweating, hyperventilation and vasodilation ALL YES
This patient is suffering from heatstroke. Patients with heatstroke commonly exhibit tachypnea and hyperventilation caused by direct central nervous system stimulation, acidosis, or hypoxia. The blood vessels in the skin are vasodilated, and the skin is warm. Sweating ceases in patients with true heatstroke, most likely because the high temperature itself causes damage to anterior hypothalamic-preoptic area. The nerve impulses from this area are transmitted in the autonomic pathways to the spinal cord and then through sympathetic outflow to the skin to cause sweating.
A manual laborer moves in March from Canada to a hot, tropical country and becomes acclimatized by working outdoors for a month.
Compared with his responses on the first few days in the tropical country, for the same activity level after acclimatization, one would expect higher
SWEATING RATE
The classic changes observed in heat acclimatization are lower heart rate during exercise; an increased sweating response; and a lower core temperature during exercise, which is due to both the increased sweating response and lower thermoregulatory set point. In addition, salt is conserved by a reduced salt concentration in sweat.
A young man was found after wandering the Arizona desert for several hours under a hot sun with no source of drinking water. Which of the following solutions given intravenously would best return the individual’s intracellular volume toward normal?
ISOTONIC SALINE
The loss of hypotonic fluid due to sweating would produce a hyperosmotic osmolarity and an extracellular and intracellular contraction, given that isotonic saline would lower the individual’s osmolarity and expand the intracellular space. Whole blood (choice C), and plasma (choice D), contain a solute (albumin) that does not penetrate a capillary membrane; thus, less fluid would enter the interstitial space. This is not the case with isotonic saline.
An excessive intake of tap water over a short period of time would be expected to produce which of the following
Extracellular volume- increase
Intracellular volume- increase
Extracellular sodium conc.- decrease
Intracellular osmolarity- decrease
Ingesting excessive amounts of tap water will increase the volume of the extracellular fluid. Since tap water contains few if any electrolytes, it will dilute the extracellular sodium and reduce extracellular osmolarity. The decreased extracellular osmolarity will cause a diffusion of water into cells, increasing intracellular volume but reducing intracellular osmolarity.
Both simple and facilitated diffusion have which characteristic?
Can work in the absence of adenosine triphosphate (ATP)
In contrast to primary and secondary transport, diffusion does not require the input of additional energy and, therefore, can work in the absence of ATP. Only facilitated diffusion displays saturation kinetics and involves a carrier protein. By definition, neither simple nor facilitated diffusion can move molecules from low to high concentration. The concept of specific inhibitors is not applicable to simple diffusion that occurs through a lipid bilayer without the aid of protein.
The membrane transporter directly responsible for maintenance of low intracellular sodium concentration is the:
basolateral Na+/K+ ATPase.
The basolateral Na+/K+ ATPase, or “sodium pump” uses primary active transport to pump three sodium ions out of the cell in exchange for two K+ ions. The low intracellular Na+ concentration facilitates apical sodium entry down its concentration gradient through several different secondary active transport symporters (e.g., Na+/Ca2+, Na+/H+)
Drinking isotonic saline solution will decrease:
none of the above
Drinking an isotonic saline solution will expand the ECF volume and neither increase nor decrease the ECF or ICF osmolarity or neither increase nor decrease ICF volume. The absorption and distribution of isotonic saline from the GI tract will expand ECF volume, but will have no effect on ECF osmolarity and, therefore, will have no effect to increase or decrease the concentration of water in the ECF. Accordingly, the driving force for movement of water to and from the ECF and the ICF will be equivalent on both sides of the cell membrane and net water movement across the membrane (osmosis) will be absent.
Identify the fluid compartment that contains approximately two-thirds of the total body water.
INTRACELLULAR
Body water is approximately distributed as 2/3 intracellular fluid and 1/3 extracellular fluid
Solute movement by active transport can be distinguished from solute transport by equilibrating carrier-medicated transport because active transport
moves the solute against its electrochemical gradient.
Active transport always moves solute against its electrochemical gradient. All of the other options are shared by both active transport and equilibrating carrier-mediated transport systems.
Which of the following mechanisms causes heat loss from a normal person when the environmental temperature is 106°F and the relative humidity is less than 10%?
EVAPORATION
Evaporation is the mechanism of heat loss when the air temperature is greater than the body temperature. Each gram of water that evaporates from the surface of the body causes 0.58 kilocalorie of heat to be lost from the body. Even when a person is not sweating, water still evaporates insensibly from the skin and lungs at a rate of 450 to 600 mL/day, which amounts to about 12 to 16 kilocalories of heat loss per hour.
The diagram below shows the effects of changing the set-point of the hypothalamic temperature controller. The red line indicates the body temperature, and the blue line represents the hypothalamic set-point temperature.
shivering, sweating, vasoconstriction, vasodilation
-ALL “NO”
When the hypothalamic set-point temperature is equal to the body temperature, the body exhibits neither heat loss nor heat conservation mechanisms, even when the body temperature is far above normal. Therefore, the person does not feel hot even when the body temperature is 104°F.
Two compartments (X and Y) of water are separated by a semipermeable membrane. The concentrations of the impermeant solute at time zero are shown in the following drawing:
The “X” side has the lower solute concentration and thus the higher water concentration. Water diffuses from X to Y, thus the level of side X decreases and the level of side Y increases.
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?
K+ equilibrium potential would shift negative. (hyperpolarized)
Hypokalemia, or reduced extracellular K+ concentrations, enhances the electrochemical gradient favoring K+ efflux from cells and causes the K+equilibrium potential to shift negative. 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, but, once reached, an action potential would be initiated. K+ -channel activation always causes K+ efflux, except in rare instances
Select the mechanism most responsible for the production of the resting transmembrane potential in an axon.
A membrane that is more permeable to K+ than to Na+.
The Na+/K+ pump and differential permeability of the plasma membrane are both responsible for the resting transmembrane potential. Consistent with the pump hypothesis is the generally accepted conclusion that an active transport system extrudes 3 Na+ from the cell for every 2 K+ carried into the cell. Consistent with the permeability hypothesis is the conclusion that the plasma membrane during its resting potential is more permeable to K+ than to Na+. If a cell had an Na+/K+ pump that carried approximately equal quantities of Na+ out of and K+ into the cell, the cell would lose more positively charged particles through diffusion than it would gain. It would, therefore, tend to develop a negative transmembrane potential if there were no compensating movement of other charged particles. The Na+/K+ pump potential is only a small part of the resting potential
The resting membrane potential of a cell is measured to be -70 mV in a solution with ion concentrations resembling extracellular fluid. Which of the following manipulations would result in a hyperpolarization of the cell
Reduction in the membrane permeability to sodium ion.
Reduction in the permeability of the sodium ion will result in hyperpolarization of the cell. Ordinarily, sodium leakage into the cell contributes to the actual resting membrane potential, making it less negative than would be predicted on the basis of potassium concentration and permeability. Reduction in potassium ion permeability, on the other hand would cause depolarization of the cell, as would influx of calcium ion to increased extracellular concentration of sodium or potassium ion.
From the diagrams above, choose the letter closest to the equilibrium potential for Na+.
B
q.17 Which of the following statements is true for an uncharged solute (S) that enters a cell by facilitated diffusion?
At equilibrium, the extracellular [S] will be equal to the intracellular [S]
Facilitated diffusion is a passive process that requires a membrane carrier protein. Entry of S into the cell will dissipate the concentration gradient and drive the process toward equilibrium (no net flux of S). Since S is not charged, the chemical concentrations of S inside and outside the cell will be equal at the equilibrium. Option A is incorrect A is incorrect because the carrier will become saturated as extracellular [S] is progressively increased. Membrane potential will have no effect on the flux of an uncharged solute, so option B is incorrect. Option C is incorrect because these gases cross cell membrane by simple diffusion through the bilayer. Option E is increased because facilitated diffusion of S is not dependent of gradients of Na+ or K+ ions.
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?
Action potentials would be prolonged
The voltage-dependent Na+ channel is opened by membrane depolarization to yield the upstroke of neuronal action potential. 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 activation 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
In excitable cells, repolarization is most closely associated with which of the following events?
K efflux
The most prominent event during repolarization is potassium efflux. It is the efflux pf potassium that causes repolarization. There is still some sodium influx in the initial stage of repolarization but this quickly ceases. In repolarization the cell is moving through the relative refractory period back toward the resting phase. Thus, during this phase the cell is regaining its excitability.
In the myelinated regions of an axon:
membrane capacitance is reduced
Myelination of an axon results in decreased capacitance and increased membrane resistance. As a result, current travels through the interior of the axon but not through the membrane. Myelination produces a large increase in conduction velocity.
The molecular basis of an axonal action potential includes:
increased sodium conductance followed by increased potassium conductance.
The depolarization phase is associated with an increased sodium conductance, resulting in a sodium influx, and repolarization, by an increased potassium conductance, resulting in an accelerated potassium efflux. There are no changes in calcium conductance during the axonal action potential.
The rapid depolarization during phase 0 upstroke of the action potential in neurons is caused by opening of:
Voltage-gated Na+ channels.
Neuronal action potentials result when the threshold potential is reached voltage-gated sodium channels open. Although action potentials are evoked in some excitable cells by opening of Ca2+ channels, opening of voltage-gated Na+ channels is responsible for the phase 0 upstroke of the neuronal action potential.
The speed at which a myelinated axon conducts an action potential is directly related to
the diameter of the axon
The fastest conducting unmyelinated neurons have the largest diameter. Myelinated neurons conduct more rapidly than unmyelinated neurons with a similar axon diameter.
Which of the following characteristics of a neuron is not associated with increased conduction velocity?
Decreased fiber size
Conduction velocity is directly related to axon diameter and the amount of myelin. Myelination increases membrane resistance. Small-diameter fibers will be slow-conducting fibers.
Which of the following is a consequence of myelination in large nerve fibers?
Generation of action potentials only at the nodes of Ranvier
Myelination of the axons of large nerve fibers have several consequences. it provides insulation to the axon membrane, decreasing membrane capacitance and thereby decreasing the “leakage” of ions across the cell membrane. Action potentials in myelinated axons occur only at the periodic breaks in the myelin sheath, called nodes of Ranvier. Voltage-gated Na+ channels are concentrated at these nodes. This arrangement both increases the velocity of the nerve impulses along the axon and minimizes the number of charges that cross the membrane during an impulse, thereby minimizing the energy required by Na+ , K+ -ATPase to re-establish the relative concentration gradients for Na+ and K+.
The axon hillock of a neuron is the site of action potential initiation because of the high concentration of which of the following types of channels in the axon hillock?
Voltage-gated sodium channels
The axon hillock is the site of action potential generation due to the presence of voltage-gated sodium channels at that site. Inward rectifier-type potassium channels set the resting membrane, whereas delayed rectifier-type potassium channels, which are a type of voltage-gated potassium channel, help restore the membrane potential back to the resting value at the end of an action potential. Ligand-gated channels are not involved in propagation of the action potential along the axon.
