Physiology

Question 1

What allows for the voluntary retention of urin once a full bladder has induced an increase in bladder parasympathetic tone?
A. Inhibition of S2-S4 α-motor neurons causing relaxation of the external urethral sphincter striated muscle fibers
B. Inhibition of pelvic splanchnic nerves causing a compensatory increase in bladder sympathetic activity
C. Activation of S2-S4 α-motor neurons causing contraction of the external urethral sphincter striated muscle fibers
D. Activation of neurons in the inferior mesenteric ganglion and subsequent contraction of the internal urethral sphincter
E. Activation of neurons in the inferior mesenteric ganglion and subsequent contraction of the external urethral sphincter

Answer 1

C. Activation of S2-S4 α-motor neurons causing contraction of the external urethral sphincter striated muscle fibers

The external urethral sphincter is under voluntary (somatic) control, and muscle fiber contraction causes closure of the sphincter. A full bladder sends afferent signals causing inhibition of sympathetic tone and an increase in parasympathetic activity to the bladder through the pelvic splanchnic nerves. The parasympathetic activity relaxes the internal urethral sphincter and induces bladder contraction.

Question 2

What neurotransmitter is used by preganglionic sympathetic fibers?
A. Norepinephrine
B. Epinephrine
C.Glutamate
D.Acetylcholine
E. GABA

Answer 2

D. Acetylcholine

Acetylcholine acts through nicotinic receptors at the ganglionic synapse. (A) Norepinephrine is the neurotransmitter used in most postganglionic sympathetic fibers.

Question 3

This toxin inhibits vesicle fusion in the presynaptic terminal at the neuromuscular junction.
A. Botulism toxin
B.Tetanus toxin
C. Diphtheria toxin
D.Alpha bungarotoxin
E. Tetrodotoxin

Answer 3

Botulism toxin

(B) Tetanus toxin inhibits the release of GABA and glycine in the spinal cord. (D) Alpha bungarotoxin inhibits acetylcholine from binding to its receptors at the neuromuscular junction. (E) Tetrodotoxin blocks action potentials by binding to and inhibiting fast, voltage-gated sodium channels.

Question 4

This toxin inhibits RNA translation.
A. Botulism toxin
B. Tetanus toxin
C. Diphtheria toxin
D. Alpha bungarotoxin
E. Tetrodotoxin

Answer 4

C. Diphtheria toxin

(B) Tetanus toxin inhibits the release of GABA and glycine in the spinal cord. (D) Alpha bungarotoxin inhibits acetylcholine from binding to its receptors at the neuromuscular junction. (E) Tetrodotoxin blocks action potentials by binding to and inhibiting fast, voltage-gated sodium channels.

Question 5

What decreases the rate of degradation of a passively conducted electrical signal in an axon?
A. Decreasing axonal diameter
B. Demyelinating an axon
C. Increasing axonal membrane resistance
D. Increasing extracellular resistance

Answer 5

C. Increasing axonal membrane resistance

The length constant (λ) describes the distance over which a passively conducted electrical signal decays to 37% of its initial voltage. A larger length constant means a slower rate of degradation. The equation is: λ = √(rm/(ri +ro )) where rm is the membrane resistance, ri is the internal axonal resistance, and ro is the extracellular resistance. Increasing myelination increases membrane resistance, and internal axonal resistance is decreased with increasing axonal cross-sectional area (i.e., diameter).

Question 6

What substance releases factor VIII from von Willebrand factor?
A. Fibrinogen
B. Platelets
C. Factor IX
D. Antithrombin III
E. Thrombin

Answer 6

E. Thrombin

Factor VIII is bound to von Willebrand factor (vWF) and inactive in circulation. Under the action of thrombin, factor VIII is released form vWF and thus activated. When not bound to vWF, factor VIII quickly degrades.

Question 7

In brain death, what happens to the intracranial pressure (ICP) as the mean arterial pressure (MAP) rises?
A. Cerebral autoregulation increases ICP.
B. Cerebral autoregulation decreases ICP.
C. The lack of cerebral autoregulation causes an increase in ICP.
D. The lack of cerebral autoregulation causes a decrease in ICP.
E. Cerebral autoregulation maintains a relatively constant ICP.

Answer 7

C. The lack of cerebral autoregulation causes an increase in ICP.

With brain death, there is a loss of cerebral autoregulation, so the brain’s normal response of maintaining a vascular tone sufficient to counteract increases in mean arterial pressure (MAP) is lost. The brain experiences the increases in MAP, which translate to increases in intracranial pressure.

Question 8

The sensory nerves originating in the extremities with cell bodies in the dorsal root ganglia are examples of what classification of neuron?
A. Unipolar
B. Bipolar
C. Multipolar
D. Pyramidal
E. Multiaxonic

Answer 8

A. Unipolar

Sensory neurons are classified as unipolar with an axon and dendrites on each end and a cell body branching from some point along the axon. (B) Bipolar neurons have an axon with dendrites on each end with a cell body along the axon. They are found as interneurons, like in the retina. (C) Multipolar neurons have dendrites surrounding a cell body from which a single axon originates. They are found as motor neurons. (D) Pyramidal neurons appear as bipolar neurons with numerous arboretic dendritic processes extending from the cell body and axons. Prime examples are found in the cerebellum as Purkinje cells. (E) Multiaxonic neurons do not exist.

Question 8

To what type of motion/activity will the utricle respond?
A. Injecting cold water into the ear of a person sitting upright
B. Sleeping
C. Stopping at a stop sign in a motor vehicle
D. Beginning to spin in an office chair
E. Falling at terminal velocity while skydiving

Answer 8

C. Stopping at a stop sign in a motor vehicle

Both the utricle and saccule respond to changes in linear acceleration with the utricle being oriented horizontally and the saccule vertically. Deceleration in a motor vehicle traveling straight or a head tilt downward each causes anterior displacement of the otoliths in the macula of the utricle and excites hair cells that respond to movement in that direction. In contrast, the act of jump roping mainly would be sensed by action in the saccule. (A, D) Both of these activities cause a shift of endolymph in the semicircular canals that is detected in the ampulla of each canal. The ampullae respond to angular acceleration. (B, E) The vestibular system only responds to changes in acceleration and will adapt with no or constant motion in the same direction.

Question 9

A patient has a hormone-producing pituitary microadenoma that puts him at risk (if left untreated) for peripheral neuropathies, cardiac arrhythmias, and sleep apnea. What hormone is the adenoma producing?
A. Adrenocorticotropic hormone (ACTH)
B. Growth hormone
C. Prolactin
D. Thyroid-stimulating hormone
E. Follicle-stimulating hormone

Answer 9

B. Growth hormone

Excess growth hormone in adults results in arthropathy, paresthesias, polyneuropathy, cardiomyopathy, arrhythmias, upper airway obstruction due to palatal/pharyngeal tissue overgrowth, increased risk for malignancies and colon polyps, and diabetes. (A) Excess adrenocorticotropic hormone (ACTH) secretion results in Cushing disease, characterized by weight gain, hypertension, mood changes, hypertelorism, fatigue, and “moon face.” (C) High prolactin levels are associated with infertility, spontaneous lactation, loss of libido, erectile dysfunction, and abnormal menstrual cycles. (D) Excess thyroid-stimulating hormone results in hyperthyroidism with symptoms of tremors, anxiety, weight loss, heat intolerance, brittle hair, and insomnia. (E) Excess follicle-stimulating hormone results in infertility

Question 10

What are the major proinflammatory cytokines?
A. TGF- α and VEGF
B. IL-1 and TNF-α
C. IL-6 and TGF-α
D. IL-6 and IL-13
E. IFN-γ and IL-10

Answer 10

B. IL-1 and TNF-α

The major proinflammatory cytokines are the interleukins (IL) IL-1, IL-6, IL-8, TNF-α (tumor necrosis factor-alpha), and IFN-γ (interferongamma). These produce fever, tissue destruction, and inflammation. The major anti-inflammatory cytokines include IL-4, IL-6, IL-10, IL-11, and IL-13. Of note, IL-6 can be anti- or proinflammatory depending on how it is used in a signaling cascade. (A) TGF-α and VEGF are examples of growth factors.

Question 11

What is a miniature end-plate potential?
A. Inhibitory postsynaptic potential
B. Excitatory postsynaptic potential
C. Response of the postsynaptic terminal caused by the release of a single vesicle into the synaptic cleft
D. Response of the postsynaptic terminal caused by the release of a single molecule of neurotransmitter into the synaptic cleft
E. Response of the postsynaptic terminal caused by the release of the neurotransmitters from a single neuron only

Answer 11

C. Response of the postsynaptic terminal caused by the release of a single vesicle into the synaptic cleft

Response of the postsynaptic terminal caused by the release of a single vesicle into the synaptic cleft
MEPs summate in the postsynaptic terminal to induce a response of either hyperpolarization or depolarization. (A, B) Miniature end-plate poten- tials (MEPs) can be excitatory or inhibitory.

Question 12

When do T-type calcium channels open during the action potential?
A. At the resting potential
B. Between the resting and threshold potentials
C. Between the threshold potential and maximal depolarization
D. Between maximal depolarization and the resting potential
E. Between the resting potential and “overshoot” phase (hyperpolarization)

Answer 12

B. Between the resting and threshold potentials

Between the resting and threshold potentials
Found initially in cardiac smooth muscle cells, T-type calcium channels are unique voltage-gated calcium channels that open at around –55 mV, which is slightly higher than the resting potential in cardiac cells of –60 mV. T-type calcium channels open to allow a large calcium flux into the cell to aid in the depolarization required to reach the trig- gering threshold for an action potential.

Question 13

How do class 3 cardiac antiarrhythmics (potassium channel blockers) affect action potentials and conduction velocity?
A. Shorten action potential refractory period and increase conduction velocity
B. Shorten action potential duration and maintain normal conduction velocity
C. Prolong action potential refractory period and slow conduction velocity
D. Prolong action potential duration and maintain normal conduction velocity
E. Shorten action potential refractory period and slow conduction velocity

Answer 13

D. Prolong action potential duration and maintain normal conduction velocity

Prolong action potential duration and maintain normal conduction velocity
By blocking only potassium channels (notably the inward rectifier channels), hyperpolarization (returning to a negative resting potential) is inhib- ited, and cardiac cells remain depolarized longer. This prolongs the action potential and the refrac- tory period. Conduction velocity is unaffected, as there is no prevention of the opening of subsequent sodium channels and depolarizing adjacent mem- brane segments.

Question 14

What neurotransmitters never can be used to upregulate action downstream in a neural network?
A. GABA and glycine
B. Glutamate and acetylcholine
C. Norepinephrine and epinephrine
D. Dopamine and substance P
E. All neurotransmitters can be used to upregulate downstream neural network activation

Answer 14

E. All neurotransmitters can be used to upregulate downstream neural network activation

All neurotransmitters can be used to upregulate downstream neural network activation
Although neurotransmitters are labeled as excit- atory or inhibitory, this refers only to the effect of a neurotransmitter on a specific type of receptor at a specific synapse. It is how the various receptors and neurons containing the receptors are organized that causes a downstream effect. For example, the basal ganglia circuitry is full of circuits that inhibit inhibitory circuits so that an inhibitory synapse/ receptor on an inhibitory neuron could result in the activation of a downstream neuron.

Question 15

How does caffeine exert its effects?
A. GABA receptor antagonism
B. Phosphodiesterase inhibition
C. Adenosine receptor agonism
D. Acetylcholinesterase activation
E. Ryanodine receptor antagonist

Answer 15

B. Phosphodiesterase inhibition

Phosphodiesterase inhibition
By inhibiting phosphodiesterase, cAMP degrada- tion is reduced. All of the actions of caffeine serve to upregulate the nervous system in a stimulatory manner. (A) Caffeine competitively inhibits glycine receptors. (C) Caffeine competitively inhibits ade- nosine receptors. Activation of adenosine receptors leads to the sensation of drowsiness. (D) Caffeine competitively inhibits acetylcholinesterase. (E) Caf- feine is an agonist for the ryanodine receptor.

Question 16

How does cocaine affect neurotransmission at the synaptic cleft?
A. Inhibition of the presynaptic uptake of monoamines
B. Induction of the release of monoamines
C. Prevention of the degradation of monoamines in the synaptic cleft
D. Blockade of the postsynaptic uptake of monoamines
E. Induction of the postsynaptic uptake of monoamines

Answer 16

A. Inhibition of the presynaptic uptake of monoamines

Inhibition of the presynaptic uptake of monoamines
Cocaine blocks the monoamine transporter pro- teins on the presynaptic cleft, thus preventing monoamine reuptake and vesicular storage in the presynaptic terminal. (B) Amphetamines both block the reuptake of monoamines into the presynaptic terminal and induce the release of monoamines into the synaptic cleft.

Question 17

What are the degradation products of the reaction between acetylcholine and acetylcholinesterase on the postsynaptic membrane?
A. Phosphatidylcholine and choline
B. Phosphatidylcholine and acetate
C. Acetyl CoA and acetate
D. Choline and acetate
E. Choline and acetyl CoA

Answer 17

D. Choline and acetate

Choline and acetate
Following the breakdown of acetylcholine by acetylcholinesterase in the extracellular space, acetate is transported into the intracellular space, where it is converted to acetyl CoA, which then can combine with choline to reform acetylcholine.

Question 18

What role do caspases play in necrosis?
A. Caspases signal and regulate the orderly fragmentation of DNA.
B. Caspases signal a cell to undergo necrosis.
C. Caspases typically are not part of necrosis.
D. Caspases inhibit necrosis and cell death.

Answer 18

C. Caspases typically are not part of necrosis.

aspases typically are not part of necrosis
Caspases are proteases necessary for apoptosis. They signal and regulate the controlled process by which DNA and cellular components are frag- mented and degraded. In contrast, necrosis is un- controlled cell death and results in inflammation. Caspases typically play little to no role in necrosis.

Question 18

What molecules are needed to activate the ligandgated component of NMDA receptors?
A Glutamate and glycine
B. Magnesium and glycine
C. Magnesium and serine
D. Aspartate and zinc
E. Zinc and glycine

Answer 18

A. Glutamate and glycine

Glutamate and glycine
Two molecules of either glutamate or aspartate and two molecules of either glycine or serine need to bind to an NMDA receptor in order to activate it. The receptor also has a voltage-gated component requiring depolarization of the neuron on which it is located. The voltage-gated component of the receptor is controlled by the calcium channel being blocked by either a zinc or magnesium ion when the receptor is inactive.

Question 19

What is the mechanism of action of bisphosphonates?
A. Increasing the body stores of calcium
B. Activation of osteocytes
C. Inhibition of osteocytes
D. Inhibition of osteoclasts
E. Recruitment of osteoblasts

Answer 19

D. Inhibition of osteoclasts

Inhibition of osteoclasts
Bisphosphonates bind to calcium and are taken up by osteoclasts. Bisphosphonates then induce apoptosis of these bone-reabsorbing cells. These are useful agents in osteopenia and osteoporosis. (A) Bisphosphonates have no effect on the body total stores of calcium. (B) Osteocytes are osteo- blasts that have entrapped themselves in their secretory bony matrix. They do not divide, but they do play a role in the turnover and maintenance of the bony matrix. They express TGF-β to suppress bone resorption. (C) Inhibition of osteocytes would lead to increased bone resorption. (E) Osteoblasts form new, nonmineralized bony matrix on the sur- face of mature, mineralized bone. They are regu- lated and recruited in part by osteocytes

Question 20

Where does GABA bind on the GABAA receptor?
A. On the α subunit
B. On the β subunit
C. Between the α and β subunits
D. On the γ subunit
E. Between the α and γ subunits

Answer 20

C. Between the α and β subunits

Between the α and β subunits
(E) The benzodiazepine binding site on the GABAA receptor is between the α and γ subunits

Question 21

A patient with a complete spinal cord injury has a patellar reflex in the acute period following his injury. What circuitry component must be intact?
A. Cell bodies in the midthoracic spinal cord
B. Dorsal columns between the brain and lumbar spinal cord
C. Cell bodies in the lumbar prominence of the spinal cord
D. Anterior horn cells between the brain and lumbar spinal cord

Answer 21

C. Cell bodies in the lumbar prominence of the spinal cord

Cell bodies in the lumbar prominence of the spi- nal cord
The patellar reflex occurs independently of input from the brain; however, the brain does work to suppress the reflex when all spinal circuitry is in- tact. This is why hyperreflexia and clonus can be seen with significant spinal cord compression. The actual reflex only requires the sensory nerves from the patellar tendon to be intact and synapse on cell bodies within the lumbar spinal cord. The reflex arc then stimulates motor neurons at that same level to provide the motor component of the reflex.

Question 22

What is the mechanism of action for temozolomide (Temodar)?
A. Microtubule inhibitor
B. DNA cross-linking
C. Anti-VEGF antibody
D. DNA alkylation
E. Topoisomerase inhibitor

Answer 22

D. DNA alkylation

DNA alkylation
Temozolomide and procarbazine are DNA alky- lating (methylating) agents that interfere with protein synthesis. (A) Microtubule function inhibi- tors include vincristine and vinblastine. (B) DNA cross-links by carbamylation of amino groups are formed by the nitrosoureas such as BCNU and CCNU. (C) Bevacizumab (Avastin) is an anti-VEGF anti- body. (E) Tamoxifen is a topoisomerase inhibitor.

Question 23

What are the first cleavage products of proopiomelanocortin?
A. β-lipotrophin, α-MSH, and γ-MSH
B. ACTH, α-MSH, and γ-MSH
C. α-MSH, β-MSH, β-endorphin
D. ACTH, β-lipotrophin, and γ-MSH
E. α-MSH, β-MSH, and γ-MSH

Answer 23

D. ACTH, β-lipotrophin, and γ-MSH

ACTH, β-lipotrophin, and γ-MSH
Following the initial cleavage of pro-opiomela- nocortin into ACTH, β-lipotrophin, and γ-MSH, ACTH can be processed further to α-MSH and CLIP. β-lipotrophin can be processed into β-endorphin and γ-lipotrophin with the latter eventually becoming β-MSH.

Question 24

What happens during testing of the H-reflex in electrophysiological studies when stimulation is increased to a supramaximal level?
A. The H-wave disappears
B. The M-wave disappears.
C. The H-wave increases.
D. The M-wave decreases.
E. The F-wave remains constant.

Answer 24

A. The H-wave disappears

The H-wave disappears
The H-wave is the electrophysiological equiva- lent of the stretch reflex and represents the mus- cle’s electrical response to a square wave stimulus to the skin that first propagates in the antidromic direction (away from the muscle) to the cell bodies. The reflex arc continues with an electrical signal sent in the orthodromic direction (toward the mus- cle) to elicit a response. This response is the H-wave. As stimulation amplitude increases, the H-wave diminishes and disappears with supramaximal stimulation. It is most useful for evaluating the Ia sensory afferents. (B, D) The M-wave is the ortho- dromic response recorded in the muscle to electri- cal stimulation of the skin overlying the muscle. It bypasses the reflex arc and increases with increas- ing stimulation amplitude. (E) The F-wave increases with increases in amplitude but not to the extent seen in the M-wave. The F-wave is the result of alpha-fiber stimulation and is useful to evaluate proximal (near the spinal cord) nerve conduction velocities.

Question 25

What is the substrate for nitric oxide synthetase?
A. Tyrosine
B. NADP
C. Citrulline
D. Arginine
E. Asparagine

Answer 25

D. Arginine

Arginine along with cofactors NADPH and oxy- gen react in the presence of nitric oxide synthe- tase to produce nitric oxide, NADP, and citrulline. (A, E) Tyrosine and asparagine are not involved in nitric oxide synthesis. (B, C) NADP and citrulline are products in the reaction creating nitric oxide.

Question 26

How does the isometric tension-length curve of smooth muscle compare to the curve of skeletal muscle?
A. Smooth muscle force is maximal in the tension trough.
B. The curve is wider with smooth muscle
C. Maximal tension occurs at the point of maximal contraction in smooth muscle.
D. There is only a single peak of maximal tension with smooth muscle.

Answer 26

B. The curve is wider with smooth muscle

The curve is wider with smooth muscle, as shown in this image.
(A) Tension equals the force of the muscle. (C) Tension increases to its maximum as the mus- cle is stretched beyond its ideal working length (at the point of maximal myosin and actin overlap). Tension will increase indefinitely at the far right of the tension–length curve until muscle tissues begin to tear. (D) When the single peak of active muscle tension versus length is added to the exponential relationship of tension to length in resting muscle, there is a tension peak, followed by a trough, fol- lowed by a continued rise in tension with both smooth and skeletal muscle

Question 26

What substance in excess in the extracellular space induces increased calcium ion flux through NMDA receptors on oligodendrocytes to cause excitotoxic damage?
A. Calcium
B. Glutamate
C. Acetylcholine
D. GABA
E. Epinephrine

Answer 26

B. Glutamate

Glutamate triggers the activation of NMDA
receptors.
When the magnesium ion is displaced
from its channel-blocking site, there is calcium
influx into the cytosol. Elevated calcium levels
trigger apoptotic changes in the affected cells.

Question 27

What component of the blood–brain barrier is most responsible for its integrity, creation, and maintenance?
A. Astrocytic foot processes
B. Arachnoid “cap” cells
C. Pericytes
D. Endothelial cells

Answer 27

C. Pericytes

Pericytes are the cells surrounding the endothelial
cells of capillaries and venules throughout the
body. In the brain, pericytes create and maintain
the tight junctions between endothelial cells and
regulate vesicle trafficking between endothelial
cells. It is this role that creates the selectivity of
the blood–brain barrier. Pericytes also inhibit the
expression of endothelial markers that increase
vascular permeability. (A) Astrocytic foot processes
historically were thought to be the key component
creating the blood–brain barrier; however, their
role is to provide biochemical support to the
endothelial
cells creating the barrier. (B) Arachnoid
“cap” cells are cells of the arachnoid villi from
which meningiomas originate. (D) The endothelial
cells themselves are not the key component to the
blood–brain barrier. Instead, the highly selective
tight junctions between endothelial cells allow for
the specialized function of the blood–brain barrier.

Question 28

How do organophosphates cause accumulation of acetylcholine in the synaptic cleft?
A. Irreversible binding to and inhibition of acetylcholinesterase
B. Blockade of postsynaptic acetylcholine receptors
C. Blockade of presynaptic acetylcholine receptors
D. Stimulation of release of presynaptic acetylcholine
E. Temporary reduction in binding affinity for acetylcholine and acetylcholinesterase

Answer 28

A. Irreversible binding to and inhibition of acetylcholinesterase

Organophosphates irreversibly bind to acetylcholinesterase
and inactivate this acetylcholine-degrading
enzyme. Pralidoxime can be used as an
antidote in organophosphate poisoning, as it is able
to reactivate the enzyme prior to degradation.

Question 29

How does hemoglobin’s affinity for oxygen change as blood passes through a capillary, and what shift occurs with the oxygen-hemoglobin dissociation curve?
A. The affinity of hemoglobin for oxygen decreases, and the oxygen-hemoglobin dissociation curve shifts to the left.
B. The affinity of hemoglobin for oxygen decreases, and the oxygen-hemoglobin dissociation curve shifts to the right.
C. The affinity of hemoglobin for oxygen increases, and the oxygen-hemoglobin dissociation curve shifts to the left.
D. The affinity of hemoglobin for oxygen increases, and the oxygen-hemoglobin dissociation curve shifts to the right.
E. The affinity of hemoglobin for oxygen remains the same, and the oxygen-hemoglobin dissociation curve does not shift.

Answer 29

B. The affinity of hemoglobin for oxygen decreases, and the oxygen-hemoglobin dissociation curve shifts to the right.

The affinity of hemoglobin for oxygen decreases,
and the oxygen-hemoglobin dissociation curve
shifts to the right, as shown in this image.
Increasing temperature, carbon dioxide, and
2,3-DPG along with decreasing pH all decrease the
binding affinity of hemoglobin for oxygen and thus
shift the hemoglobin-oxygen dissociation curve to
the right. Conditions opposite to these will increase
binding affinity and shift the curve to the left. As
blood passes through a capillary, metabolically
active tissues release 2,3-DPG, which induces an
allosteric change in hemoglobin and leads to oxygen
unbinding. Oxygen is taken up by tissues, and
carbon dioxide enters the blood. The partial pressure
of oxygen thus is reduced as is the pH causing
subsequent oxygen unbinding from hemoglobin. A
decrease in the partial pressure of oxygen causes a
shift along the curve to the left.

Question 30

What is the approximate resting membrane potential of a neuron, and conductance of what ion is most responsible for determining this potential?
A. –65 mV; potassium
B. –60 mV; sodium
C. +60 mV; chloride
D. –90 mV; potassium
E. +20 mV; calcium

Answer 30

A. –65 mV; potassium

All ions considered, the actual resting membrane
potential for a neuron is between –60 and –70 mV.
This is due in large part to the flux of potassium
ions from inside to outside the neuron through
open channels along the concentration gradient.
The equilibrium potential for potassium is around
–90 mV. (B) Sodium flux is from the extracellular
to the intracellular space when a membrane is at
rest, but there are far fewer sodium than potassium
channels open when a neuron is at rest.
(C) The equilibrium potential for sodium is around
+60 mV, whereas it is around –60 mV for chloride.
(D) Although
the resting membrane potential is
based largely on potassium conductance, it is more
positive than the equilibrium potential for potassium
alone. (E) The equilibrium potential for calcium
is around +20 mV, but the ion plays little role
in the membrane resting potential.

Question 31

Secretion of what proteins from the notochord establishes the ventral pole of the dorsal-ventral axis in the developing nervous system?
A. Bone morphogenic proteins
B. Wnt family proteins
C. Sonic hedgehog proteins
D. Patched proteins
E. Ras proteins

Answer 31

C. Sonic hedgehog proteins

Sonic hedgehog proteins are secreted from the
notochord and signal the development of the
floor plate and motor neurons in a concentration-dependent
fashion. (A, B) Bone morphogenic proteins
from the ectoderm overlying the notochord
and Wnt family proteins establish the dorsal axis.
(D) The patched gene (Ptc) expresses the ligand-binding
domain for the sonic hedgehog proteins.
(E) The Ras family proteins are GTPases involved
cell signal transduction and can signal aberrantly
in cancers.

Question 32

How will a patient with severe myasthenia gravis affecting a particular neuromuscular junction react to organophosphate poisoning at that same junction?
A. No effect on postsynaptic activity
B. Increased postsynaptic activity
C. Decreased postsynaptic activity
D. Increased presynaptic activity
E. Decreased presynaptic activity

Answer 32

B. Increased postsynaptic activity

Myasthenia gravis is a condition characterized by
antibodies that bind to postsynaptic acetylcholine
receptors and inhibit acetylcholine binding. This
reduces the postsynaptic activity. Medications for
myasthenia gravis act by competitively inhibiting
acetylcholinesterase, thus decreasing acetylcholine
breakdown and increasing acetylcholine in the
synaptic cleft. Organophosphates irreversibly bind
to and inhibit acetylcholinesterase and have a similar
end result on the postsynaptic activity.

Question 33

If a resting neuron with only potassium permeability was taken from its normal extracellular space and placed in endolymph, what would happen to its resting state?
A. Its resting state would become more positive.
B. Its resting state would become more negative.
C. Its resting state would remain the same.
D. Its resting state would be too variable to predict.

Answer 33

A. Its resting state would become more positive.

Due to its high potassium relative to sodium
concentration compared with perilymph and normal
neuronal extracellular fluid, endolymph has a
potential of 80 to 120 mV. The normal resting concentration
of potassium in a neuron is about 150
mmol, which is the same concentration of potassium
in endolymph. This means that there essentially
would be almost no passive flow of potassium
ions in this situation, and according to the Nernst
equation, the resting potential due to potassium
flux would be about 0 mV.

Question 34

What is the mechanism of action of strychnine?
A. Blockade of GABA receptors
B. Activation of acetylcholine receptors on neurons in the spinal cord
C. Inhibition of neurons through activation of GABA receptors
D. Inhibition of glycine receptors
E. Potentiation of neurons through activation of glycine receptors

Answer 34

D. Inhibition of glycine receptors

Strychnine is a rodenticide that works by binding
to, and thus blocking, glycine and acetylcholine
receptors on motor neurons in the spinal cord. This
leads to a hyperactivation of skeletal musculature
and death by asphyxiation due to tonic contraction
of the respiratory muscles.

Question 35

What is the hourly rate of production of cerebrospinal fluid in the adult brain?
A. 10 mL/h
B. 14 mL/h
C. 18 mL/h
D. 22 mL/h
E. 26 mL/h

Answer 35

C. 18 mL/h

Cerebrospinal fluid (CSF) is produced at an average
rate of 18 mL/h, yielding a total daily production
of 400 to 500 mL. The total CSF volume in an
adult is approximately 140 mL at any given time.
Seventy percent of CSF is produced by the choroid
plexus.

Question 36

What role do gamma motor neurons play in the muscle spindle?
A. Shorten the muscle spindles with skeletal muscle fiber contraction
B. Regulate beta motor neurons
C. Inhibit alpha motor neurons
D. Allow for the adaptation of type 2 afferents
E. Shorten the Golgi tendon organs with muscle contraction

Answer 36

A. Shorten the muscle spindles with skeletal muscle fiber contraction

The cells bodies of gamma motor neurons are
located in the anterior horns of the spinal cord.
They are activated along with their corresponding
alpha motor neurons and contract the intrafusal
muscle fibers to keep the muscle spindles taut regardless
of extrafusal muscle fiber length. (B) Beta
motor neurons activate both extrafusal and intrafusal
muscle fibers, thus shortening muscle length
and keeping muscle spindles taut. They greatly are
outnumbered by gamma motor neurons. (C) Gamma
and alpha motor neurons are activated simultaneously
in a given muscle. (D) Type 2 afferents are
highly myelinated stretch receptors in the muscle
that, unlike type 1a afferents, are non-adapting and
thus fire continuously with a stable muscle length.
(E) The Golgi tendon organs are proprioceptive
sensory receptors oriented in series with skeletal
muscle fibers. They are located where skeletal
muscles join their tendons and sense the force on
the muscles. With signaling through 1b afferents,
the Golgi tendon reflex can be initiated, which
inhibits
further contraction of the muscle involved
through inhibitory interneurons in the spinal cord.

Question 37

What enzyme promotes the conversion of 5¬hydroxy-tryptophan to serotonin?
A. Tyramine β-hydroxylase
B. Aromatic amino acid decarboxylase
C. Tryptophan hydroxylase
D. Tyrosine hydroxylase
E. Catechol-O-methyl transferase

Answer 37

B. Aromatic amino acid decarboxylase

Aromatic amino acid decarboxylase uses activated
B6 as a cofactor. (A) Tyramine β-hydroxylase
converts tyramine to octopamine (an amine that
acts in place of norepinephrine in sympathetic
neurons with chronic monoamine oxidase inhibitors).
(C) Tryptophan hydroxylase converts tryptophan
to 5-hydroxy-tryptophan during serotonin
synthesis. (D) Tyrosine hydroxylase converts
tyrosine
to levodopa during dopamine synthesis.
(E) Catechol-O-methyl transferase inactivates catecholamine
neurotransmitters including dopamine,
levodopa, norepinephrine, and epinephrine

Question 38

How is presynaptic hyperpolarization of a neuron advantageous over postsynaptic hyperpolarization?
A. It allows for inhibition of the entire presynaptic and postsynaptic neuronal chain.
B. It allows for more precise temporal summation.
C. It reduces the amount of neurotransmitter degradation occurring in the synapse.
D. It allows for more precise spatial resolution.

Answer 38

D. It allows for more precise spatial resolution.

Presynaptic inhibition has an advantage over
postsynaptic inhibition in that presynaptic inhibition
downregulates only one synapse acting on the
postsynaptic cell. The entire neuronal chain thus
does not simply become inhibited based on one
postsynaptic action. In addition, the multiple
miniature end-plate potentials (MEPPs) from several
presynaptic terminals (whether inhibitory or
excitatory)
can be grouped together at the postsynaptic
cell to induce or inhibit an action potential
(spatial resolution). (B) Temporal summation
involves the grouping of MEPPs together in a
temporal
fashion to induce an action potential
postsynaptically. (C) With presynaptic inhibition,
there are fewer quanta of neurotransmitters entering
the synaptic cleft.

Question 39

What role do L-type voltage-dependent calcium channels play in muscle contraction?
A. Release calcium from the sarcoplasmic reticulum
B. Allow for depolarization of the postsynaptic terminal
C. Reuptake calcium into the presynaptic terminal
D. Respond to depolarization and activate calcium channels on the sarcoplasmic reticulum
E. Binds to troponin to induce an allosteric change

Answer 39

D. Respond to depolarization and activate calcium channels on the sarcoplasmic reticulum

As the action potential at the neuromuscular
junction occurs, depolarization propagates through
the T-tubules in the muscle fiber. L-type voltage-dependent
calcium channels are activated, and they
open ryanodine receptors in the adjacent sarcoplasmic
reticulum. The calcium that then is released
binds to troponin C on the myofibrils to start sarcomere
contraction. (A) Once activated, calcium-release
channels (ryanodine receptors) on the
sarcoplasmic reticulum allow for stored calcium
to be released within the muscle cells’ cytoplasm.
(E) This is the role of calcium and not a channel.

Question 40

How does parathyroid hormone increase the serum calcium level?
A. Inhibits calcitonin production
B. Inhibits osteoblasts
C. Increases phosphate reabsorption in the kidneys
D. Blocks the formation of vitamin D
E. Inhibits osteoclasts

Answer 40

B. Inhibits osteoblasts

By inhibiting osteoblasts, parathyroid hormone
(PTH) is not able to inhibit osteoclasts and reduce
bone reabsorption. As bone is reabsorbed, calcium
is released into the circulation. (A) Calcitonin is
secreted by the thyroid and inhibits intestinal calcium
absorption of calcium, inhibits osteoclast activity,
stimulates osteoblast activity, and promotes
renal secretion of calcium. All of these effects work
to reduce the serum calcium concentration. Calcitonin
can be thought of as having the opposite
activity
of PTH. (C) PTH promotes the reabsorption
of calcium from the distal tubules and thick ascending
limbs in the kidneys and reduces phosphate
reabsorption. This leads to an increase in the
calcium/
phosphate ratio, meaning that more calcium
is free in the serum. Increasing renal phosphate
reabsorption
would have the opposite effect.
(D) PTH upregulates the enzyme responsible for
activating vitamin D, which in turn increases the
intestinal reabsorption of calcium. (E) Osteoclast
inhibition would reduce bone turnover and calcium
release from the bony matrix.

Question 41

A particular inhibitor is found to prevent the formation of L-dopa from tyrosine. Further studies show that the inhibitor only works once tyrosine hydroxylase has interacted with tyrosine. The inhibitor is a(n):
A. Competitive inhibitor
B. Noncompetitive inhibitor
C. Uncompetitive inhibitor
D. Mixed inhibitor

Answer 41

C. Uncompetitive inhibitor

An uncompetitive inhibitor binds to an enzyme/
substrate complex to reduce or inhibit its activity.
The enzyme/substrate must first form for this type
of inhibition to take place. (A) In a simple form, a
competitive inhibitor binds to an enzyme and,
when bound, prevents the enzyme from binding
to its substrate. In other words, the enzyme may
either
be bound to its inhibitor or substrate but not
both at the same time. (B) A noncompetitive inhibitor
binds to and acts on an enzyme to reduce its
activity and interaction with its substrate. A noncompetitive
inhibitor binds to an enzyme equally
well whether or not the enzyme is bound to its substrate.
The substrate does not have to be present
for a noncompetitive inhibitor to act. (D) A mixed
inhibitor is a type of noncompetitive inhibitor that
has a higher affinity for the enzyme in a particular
state (e.g., bound to its substrate).

Question 42

What feature of excitatory synapses forces them to transmit in only one direction?
A. Absolute refractory period across the synaptic cleft
B. Postsynaptic clefts residing on multiple postsynaptic neurons
C. Presynaptic neurotransmitter reuptake receptors
D. Vesicles and corresponding receptors each on only one side of the synaptic cleft

Answer 42

D. Vesicles and corresponding receptors each on only one side of the synaptic cleft

Excitatory synapses have vesicles and corresponding
receptors each on only one side of the
synaptic cleft; thus, there always is a “signaling”
end and a “receiving” end for vesicular traffic.
(A) The absolute refractory period forces axons to
propagate action potentials in only one direction.
(C) Presynaptic neurotransmitter reuptake receptors
exist but do not play a role in signal direction
across a synapse.

Question 43

What is the brain’s response to a rising PCO2 with regard to cerebral blow flow (CBF) and intracranial pressure (ICP)?
A. Decrease CBF and increase ICP.
B. Increase CBF and decrease ICP.
C. Decrease CBF and decrease ICP.
D. Increase CBF and keep ICP constant.
E. Increase CBF and increase ICP.

Answer 43

E. Increase CBF and increase ICP.

When the brain detects a rising carbon dioxide
level in the blood, the cerebrovasculature dilates
to increase cerebral blood flow (CBF) and thus
increase
the clearance of carbon dioxide. In turn,
this raises intracranial pressure. With the increase
in CBF, the PO2 the brain experiences also increases.
This mechanism is impaired in brain death and
severe stroke.

Question 44

What makes an action potential propagate in only one direction once it has been initiated?
A. Absolute refractory period
B. Unidirectional protein transport
C. Saltatory conduction
D. Relative refractory period
E. Temporal summation

Answer 44

A. Absolute refractory period

The absolute refractory period is determined by
the inability of voltage-gated sodium channels to
open for a period of time once they have opened
and closed during an action potential. Without a
large sodium flux from outside to inside the axon,
there can be no depolarization and thus no subsequent
action potential. The action potential thus
must propagate in an anterograde fashion to a portion
of the axon where the sodium channels have
not been activated and closed recently. (B) Protein
transport goes in both an anterograde and retrograde
fashion in axons but is not the mechanism
behind action potential direction. (C) Saltatory conduction
describes the process by which the action
potential “leaps” between nodes of Ranvier in a
design that speeds signal conduction along axons.
Action potentials are triggered only in the unmyelinated
nodes, and the voltage differential produced
at one node must be transmitted at a sufficient
level to the next node to trigger another action
potential.
(D) The relative refractory period promotes
an action potential to propagate in one
direction
but is not the absolute mechanism. The
relative refractory period refers to the point at
which a neuron is hyperpolarized by potassium flux
with voltage-gated sodium channels that are not
inactivated. The neuron is resistant to depolarization,
but a sufficient depolarization would trigger
an action potential. If the relative refractory period
was the only mechanism present, a large depolarization
would send an action potential propagating
in two directions along an axon. (E) Temporal summation
refers to the summation of multiple small
depolarizations in the postsynaptic terminal over
a given period of time with each small depolarization
alone not being sufficient to trigger an action
potential. It requires the summation of signal to
initiate the action potential.

Question 45

How would hexamethonium administration affect postsynaptic parasympathetic neurons?
A. Reduction of the depolarization threshold
B. Favoring temporal instead of spatial summation
C. Selective inhibition of the parasympathetic nervous system
D. No postsynaptic blockade
E. Augmentation of the postsynaptic response

Answer 45

D. No postsynaptic blockade

Hexamethonium is an antagonist of nicotinic
acetylcholine receptors and thus is a ganglionic
blocker in the autonomic nervous system. It binds
to the nicotinic acetylcholine receptors but not in
the binding site for acetylcholine. Hexamethonium
has no effect on the postsynaptic muscarinic
acetylcholine receptors. (C) Hexamethonium administration
will decrease signaling in both the
sympathetic and parasympathetic nervous systems.

Question 46

How do steroid hormones pass through cell membranes?
A. Through membrane steroid channels
B. By simple diffusion
C. By facilitated diffusion
D. Through G proteins
E. By active transport

Answer 46

B. By simple diffusion

Steroids are lipophilic and thus can pass through
lipophilic cell membranes without the use of channels
or receptors. (A) Membrane channels for steroids
do not exist. (C) Facilitated diffusion relies on
membrane channels to allow molecules to pass
through cell membranes along their electrochemical
and/or osmotic gradients. (D) G proteins allow
extracellular molecules to induce cell signaling
within cells, without signaling molecules passing
through the membranes. (E) Active transport
requires
the use of energy (usually in the form of
ATP) for molecules to pass through membrane
channels against a gradient.

Question 47

What is the result of light falling upon the photoreceptor cells of the retina?
A. Phosphodiesterases inhibit the phototransduction cascade.
B. Retinal is converted from the trans to the cis isomer.
C. Molecules of cAMP are hydrolyzed.
D. Hyperpolarization of the retinal cells results in phototransduction

Answer 47

D. Hyperpolarization of the retinal cells results in phototransduction

As in the auditory and vestibular systems, the
stimulus (light) induces hyperpolarization of the
retinal cells, which is due to the closure of sodium
channels and reduction of the inward sodium
flux of sodium in the case of phototransduction.
(A) Phosphodiesterases are essential for continuing
the phototransduction cascade. (B) Rhodopsin contains
retinal in the 11-cis conformation until light
evokes a configuration change to the all-trans form.
This change makes rhodopsin no longer fit into
the opsin-binding site, and upon disengagement,
opsin changes forms, continuing the phototransduction
cascade. (C) Molecules of cGMP are hydrolyzed,
which causes the closure of cGMP dependent
sodium channels.

Question 48

In which stage of the cell cycle does growth arrest by p53 occur?
A. G1 phase
B. S phase
C. G2 phase
D. M phase
E. G0 phase

Answer 48

A. G1 phase

The G1 (growth 1) phase is characterized by cell
growth and synthesis of mRNA and proteins in
preparation for DNA replication in S phase. Various
checkpoint systems like the p53 tumor-suppressor
protein prevent progression to S phase and DNA
replication if errors are present. (B) During the S
(synthetic) phase, cells replicate their DNA and are
resistant to radiation due to the numerous DNA
proofreading and repair mechanisms at play. (C) The
G2 (growth 2) phase is characterized by cell growth
and protein synthesis after cell DNA has replicated.
Some rapidly dividing cells (e.g., cancer cells)
may skip the G2 phase. There exist checkpoints
in the G2 phase that induce cell arrest if errors are
detected.
(D) The M (mitotic) phase is characterized
by division and separation of the cellular nuclear
material and division of cells into two daughter
cells. Cells are very sensitive to radiation during the
M phase. (E) The G0 (resting) phase occurs when
cells are in a quiescent state. Cells can enter the
G0 phase due to a variety of conditions including
nutrient deficiency and lack of growth factors.
Most neurons in the central nervous system are in
the G0 phase of the cell cycle.

Question 48

A motor unit is composed of:
A. A motor neuron and all of the muscle fibers innervated by that neuron
B. A muscle fiber and all of the motor neurons that innervate that fiber
C. All of the motor neurons within a peripheral nerve
D. A single motor neuron and a single muscle fiber innervated by that neuron
E. The motor and sensory neurons along with the muscle spindles associated with a single muscle fiber

Answer 48

A. A motor neuron and all of the muscle fibers innervated by that neuron

Each motor neuron innervates several muscle
fibers. There are numerous motor neurons in each
peripheral nerve.

Question 48

Where is norepinephrine synthesized?
A. Cytoplasm
B. Postsynaptic terminal
C. Neuronal membrane
D. Synaptic vesicle
E. Synaptic cleft

Answer 48

D. Synaptic vesicle

During the process of conversion of dopamine
to norepinephrine, the latter is transported into
synaptic vesicles by the vesicular monoamine transporter.
The actual synthesis of norepinephrine
mostly takes place in the synaptic vesicle. (A) Dopamine
synthesis from tyrosine takes place in the
neuronal cytoplasm. (E) The synaptic cleft is where
acetylcholine is degraded by acetylcholinesterase.

Question 48

How do cytotoxic and vasogenic edema differ?
A. Cytotoxic edema responds better to steroids than does vasogenic edema.
B. The blood–brain barrier is closed in cytotoxic edema and disrupted in vasogenic edema
C. Cells shrink in vasogenic edema but expand in cytotoxic edema.
D. Cytotoxic but not vasogenic edema occurs after ischemic injury.
E. Proteins expand the extracellular space in cytotoxic edema

Answer 48

B. The blood–brain barrier is closed in cytotoxic edema and disrupted in vasogenic edema

The blood–brain barrier (BBB) becomes disrupted
in vasogenic edema, as breakdown of the endothelial
tight junctions occurs. In addition, gliomas and
certain brain tumors can secrete vascular endothelial
growth factors that increase permeability. Proteins
and molecules leave the intravascular space
and create an oncotic force driving fluid into the
extracellular space. Fluid spreads along white matter
tracts, creating a large area of edema. In cytotoxic
edema, alteration of cellular metabolism leads
to a failure of ion pumps (notably the sodium-potassium
pumps), which in turn creates an osmotic
gradient and increased fluid in the intracellular
space. (A) Vasogenic edema responds better to
steroids, in that steroids decrease endothelial permeability
by various means. (C) Cytotoxic edema
is characterized by cell swelling, whereas cells
remain
of similar size in pure vasogenic edema.
(D) Although
ischemic injury initially is characterized
by cytotoxic edema, when cells die and the
BBB breaks down, elements of vasogenic edema
arise. (E) Proteins expand the extracellular space in
vasogenic edema due to disruption of the BBB.

Question 48

In rigor mortis, in what state are the myosin heads?
A. Myosin heads are unbound from actin with their heads uncocked.
B. Myosin heads are bound to actin following the power stroke
C. Myosin heads are bound to actin before the power stroke.
D. Myosin heads are unbound from actin with their heads cocked.

Answer 48

B. Myosin heads are bound to actin following the power stroke

The power stroke occurs as phosphate is released
from the cocked myosin heads, and they change
conformation to move relative to the actin thin filaments.
Without the addition of ATP, the myosin
heads cannot unbind actin. This accounts for the
rigid muscles and tonic sarcomere length seen in
rigor mortis. (A) The myosin heads are able to
unbind
from actin thin filaments with the addition
of ATP and return to their resting (uncocked)
positions. (C) Once cocked, myosin heads bind actin.
(D) Myosin heads cock in a high-energy, ready position
as ATP is hydrolyzed into ADP and phosphate.

Question 48

What enzyme is necessary for conversion of norepinephrine to epinephrine?
A. Monoamine oxidase
B. Tyrosine hydroxylase
C. Aromatic L-amino acid decarboxylase
D. Dopamine β-hydroxylase
E. Phenylethanolamine N-methyltransferase

Answer 48

E. Phenylethanolamine N-methyltransferase

Phenylethanolamine N-methyltransferase converts
norepinephrine to epinephrine creating homocysteine
in the process. (A) Monoamine oxidase
degrades monoamines like dopamine. (B) Tyrosine
hydroxylase converts tyrosine to L-dopa in
dopamine synthesis. It is the rate-limiting factor
in dopamine synthesis. (C) Aromatic L-amino acid
decarboxylase converts L-dopa to dopamine in
dopamine synthesis. (D) Dopamine β-hydroxylase
converts dopamine to norepinephrine.

Question 49

How would an isolated vitamin K deficiency affect the clotting cascade?
A. Upregulation of antithrombin activity
B. Reduced activity of factors XI and XII
C. Inhibition of synthesis of proteins C, S, and Z
D. Prolongation of bleeding time
E. Prolongation of the partial thromboplastin time without affecting the prothrombin time

Answer 49

C. Inhibition of synthesis of proteins C, S, and Z

Vitamin K is required for the complete synthesis
of the coagulation factors prothrombin (factor II),
factor VII, factor IX, factor X, protein C, protein S,
and protein Z. These factors are present (in the
classic coagulation model) in the intrinsic, extrinsic, and common pathways. (A) Antithrombin
inhibits
several factors in the clotting cascade, and
its effects on factors II and X are inhibited by heparin.
(B) Factor XII acts on factor XI to begin the
clotting cascade in the intrinsic pathway. (D) Bleeding
time and platelet function are unaffected by a
vitamin K deficiency. (E) With a vitamin K deficiency,
prothrombin time (a measure of the extrinsic
pathway) and partial thromboplastin time
(a measure of the intrinsic pathway) both are prolonged,
with the former being affected to a greater
extent.

Question 50

What typically prevents axonal regeneration in the central nervous system?
A. Reduced axonal growth factors following injury
B. Astrocytic scarring
C. Lack of progenitor cells
D. Lack of directional markers and/or architecture

Answer 50

B. Astrocytic scarring

Astrocytic scarring disrupts the axonal tracts
following wallerian degeneration, and astrocytes
secrete axonal growth inhibitory proteins in the
central nervous system. (A) There is a relative paucity
of axonal growth–promoting factors in the
central nervous system, but this is not due to, nor
does it change relative to, the presence of axonal
injury. (C) There is a lack of progenitor cells in the
central nervous system, but axons in the peripheral
nervous system regenerate without such progenitor
cells. (D) Directional markers and signaling
molecules do degrade following neurogenesis, but
axons still can regenerate in the peripheral nervous
system along axon tracts following the supporting
matrix (e.g., endoneurium and other such components)
after wallerian degeneration.

Question 51

During the action potential, what does closing of the sodium channels cause?
A. Membrane depolarization
B. Opening of the potassium channels
C. The beginning of the relative refractory period
D. Membrane repolarization
E. Opening of the voltage-gated calcium channels

Answer 51

D. Membrane repolarization

As the sodium channels close, the open potassium
channels are able to keep the net flux of
positive
ions out of the cell, resulting in membrane
repolarization toward the resting membrane
potential
of potassium. (B) Potassium channels are
open at this point, which allows for cellular repolarization
and eventual hyperpolarization as the
sodium channels close. (C) The relative refractory
period is the time at which the cell is hyperpolarized
so that the membrane depolarization required
to trigger a subsequent action potential is greater
than it is when the cell is at its resting membrane
potential. (E) Voltage-gated calcium channels open
with depolarization and not repolarization.

Question 52

What is the mechanism of action of hydrochlorothiazide?
A. Promotion of aquaporin channels in the distal convoluted tubules and collecting ducts
B. Inhibition of the sodium-potassium cotransporters in the distal tubules
C. Inhibition of the sodium-potassium-chloride symporters in the thick ascending limb of the loops of Henle
D. Reduction of sodium chloride and bicarbonate reabsorption in the proximal tubules
E. Inhibition of the sodium-chloride cotransporters in the distal tubules

Answer 52

E. Inhibition of the sodium-chloride cotransporters in the distal tubules

By competing for the chloride-binding site and
blocking the sodium-chloride transporters, hydrochlorothiazide
prevents the reabsorption of sodium
in the distal convoluted tubules. This maintains
the osmotic gradient in the tubules, thus promoting
natriuresis. In addition, hydrochlorothiazide increases
calcium reabsorption in the tubules through
an unrelated mechanism. (A) One mechanism by
which vasopressin acts is by promoting the insertion
of aquaporin channels in the apical membrane
of the distal convoluted tubules and collecting
ducts, allowing water to be reabsorbed into the
blood along its osmotic gradient. This creates an
antidiuretic effect. (B) Spironolactone acts on the
distal nephron on the distal convoluted tubules
where it inhibits the sodium-potassium exchangers.
This prevents the exchangers from reabsorbing
sodium and excreting potassium, resulting in a
diuretic
effect. It is considered a potassium-sparing
diuretic. (C) Furosemide is a loop diuretic that
works by inhibiting the sodium-potassium-chloride
symporters in the thick ascending limb of the loops
of Henle. Blockade of the transporter prevents
reabsorption
of sodium, chloride, and potassium.
(D) Carbonic anhydrase inhibitors reduce sodium
chloride and bicarbonate reabsorption in the proximal
tubules by blocking the luminal conversion of
bicarbonate to carbon dioxide. Without the inhibitors,
carbon dioxide diffuses passively back into the
cells of the proximal tubules. Without the conversion,
sodium remains with the bicarbonate ions in
the lumen due to their electrochemical attraction.

Question 53

What are the two major characteristics of/associated with protein molecules that prevent them from being filtered into the Bowman capsule in the renal glomerulus?
A. Large size and negative charge
B. Low oncotic pressure and positive charge
C. Small size and negative charge
D. High hydrostatic pressure and positive charge
E. High hydrostatic pressure and negative charge

Answer 53

A. Large size and negative charge

The basement membrane and podocytic epithelium
in the glomerulus have a positive charge and
thus resist passage of negatively charged molecules.
In addition, the effective pore size in the glomerular
wall is about 8 nm, preventing passage of most
proteins and other large molecules. The oncotic
pressure in the glomerular capillaries increases
as blood passes through as proteins are retained
while ions and small molecules are filtered through
the wall. This causes a gradient of flow back into
the capillaries at their distal ends as oncotic pressure
builds. Finally, the hydrostatic pressure in the
glomerular capillaries is low relative to capillaries
elsewhere in the body. This resists filtration through
the capillary walls.

Question 54

Lateral geniculate cells have what type of receptive fields?
A. Concentric
B. Rectangular
C. Motion
D. Direction
E. No receptive fields

Answer 54

A. Concentric

Lateral geniculate nucleus cells have concentric
(center-surround) receptive fields that are either
on-center or off-center, with the surrounding region
being antagonistic. Similar receptive fields are seen
in retinal ganglion cells. (B) Rectangular receptive
fields are seen in simple cells of the visual cortex.
(C, D) Motion, direction, and orientation receptive
fields are seen in complex cells of the visual cortex.
Cells are dedicated and specialized to respond to
certain aspects of visual stimuli.

Question 54

What ion entering neurons during the secondary phase of traumatic brain injury accounts for the majority of the mitochondrial malfunctioning and damage?
A. Glutamate
B. Sodium
C. Potassium
D. Calcium

Answer 54

D. Calcium

During the secondary (delayed) phase of traumatic
brain injury, neurons spill the neurotransmitter
glutamate that in turn causes a large calcium
flux into neurons. Excess intracellular calcium can
initiate apoptosis and open mitochondrial pores,
leading to metabolic dysfunction. (A) Glutamate is
the neurotransmitter involved in excitotoxic brain
injury, but the molecule only activates the receptor
that opens calcium channels. (B, C) During the primary
phase of traumatic brain injury, there is cell
damage and cellular membrane damage due to the
direct forces involved. Cellular membrane damage
can allow numerous ions and molecules to pass
freely along their concentration and electrochemical
gradients.

Question 55

What is the role of dynein in axoplasmic transport?
A. Slow anterograde transport
B. Both anterograde and retrograde transport
C. Fast anterograde transport
D. Inhibits transport
E. Retrograde transport

Answer 55

E. Retrograde transport

Retrograde transport of vesicles using dynein is
an ATP-dependent process. (A) There are two types
of slow anterograde transport. The slower mode
carries microtubules and neurofilaments, whereas
the faster slow transport mechanism carries actin
and a variety of other proteins. (B) There is no
single
carrier protein that serves bidirectional
axoplasmic
transport. (C) Both slow and fast
anterograde
transport of molecules in vesicles are
mediated by kinesins along microtubules

Question 55

What structure in the sarcomere is composed of myosin and actin filaments?
A. C-band
B. A-band
C. I-band
D. H-band
E. M-band

Answer 55

A. C-band

The C-band enlarges with sarcomere shortening,
as shown in this image. It is the region of the
A-band not including the H-band. (B) The A-band
is the region of the sarcomere containing myosin
thick filaments with actin thin filament overlaps
at the ends of the band. The A-band remains the
same size during sarcomere contraction. (C) The
I-band is the portion of the sarcomere with actin
thin filaments that are not overlapping with myosin
thick filaments. It decreases in size with sarcomere
contraction. (D) The H-band is the portion of
the sarcomere containing myosin thick filaments
that are not overlapping actin thin filaments. It
shortens with contraction. (E) The M-band is the
center anchoring point of the myosin thick filaments
and does not change in size as sarcomere
length changes.

Question 55

What enzymes allow retroviruses to produce DNA from their RNA genomes?
A. RNA polymerases
B. DNA polymerases
C. Reverse transcriptases
D. Integrases
E. Helicases

Answer 55

C. Reverse transcriptases

Reverse transcriptases create cDNA from RNA
templates in the very error-prone process of reverse
transcription. (A) RNA polymerases are involved in
RNA translation from DNA. (B) DNA polymerases
are involved in transcription and DNA replication.
(D) Integrases allow the DNA created from retroviruses
to be incorporated into host cell genomes.
(E) Helicases separate two annealed strands of
DNA, RNA, or DNA-RNA hybrids, and are involved
in the processes of transcription and translation
among other roles that they play.

Question 55

What is the rate-limiting step in dopamine synthesis?
A. Tyrosine availability
B. Tyrosine hydroxylase
C. Dopa decarboxylase
D. Dopamine β-hydroxylase
E. Tetrahydrobiopterin

Answer 55

B. Tyrosine hydroxylase

Tyrosine hydroxylase converts tyrosine to L-dopa.
(A) Tyrosine is abundant and does not limit the rate
of dopamine synthesis. (C) Dopa decarboxylase
converts L-dopa to dopamine. (D) Dopamine β-hydroxylase
converts dopamine to norepinephrine.
(E) Tetrahydrobiopterin is a cofactor for the conversion
of tyrosine to L-dopa, but being widely
available, it does not limit the reaction.

Question 56

In addition to being synthesized in the neuron and being deactivated at the site of action, what other criterion is necessary to label a substance a neurotransmitter?
A. Released nonspecifically from the presynaptic terminal
B. Exogenous administration that mimics the endogenous activity
C. Presence throughout the nervous system
D. Ubiquitous effect throughout all neural networks
E. Exerting an effect at a distant site in the body

Answer 56

B. Exogenous administration that mimics the endogenous activity

To prove that the action seen in a neuronal chain
is due to signaling by a neurotransmitter, the effect
from exogenous administration must be the same
as endogenous activity. (A) Neurotransmitters
are present in the presynaptic terminal and only
released
in a controlled amount and fashion to
exert a particular effect. (C) Some neurotransmitters
can be seen throughout the nervous system,
whereas others are found in only a few select
locations.
(D) Although a neurotransmitter may
be excitatory or inhibitory at a particular synapse,
the downstream effects can vary depending on the
arrangement of the neural network. (E) An effect
distant from the site of production/release is part
of the definition of a hormone.

Question 57

What is the main role of orexin/hypocretin in the human body?
A. Promoting arousal
B. Promoting sleep
C. Inhibiting pain
D. Appetite suppression
E. Decreasing thermogenesis

Answer 57

A. Promoting arousal

Orexin/hypocretin is a neurotransmitter that promotes
arousal and wakefulness along with appetite.
Despite decreased levels of production in the
brains of narcoleptics, narcoleptics tend to have
higher body mass indices, highlighting that the
chemical plays more of a role in arousal and thermogenesis
than in promoting that appetite. The
weight gain effect is due to the lowered metabolic
rate in orexin-deficient individuals.

Question 58

A patient receives an overdose of intravenous potassium. How would muscle action potentials be affected?
A. Resting membrane potential would increase, and the threshold to trigger an action potential would be reached with more miniature end-plate potentials (MEPPs)
B. Resting membrane potential would decrease, and the threshold to trigger an action potential would be reached with fewer MEPPs
C. Resting membrane potential would increase, and the threshold to trigger an action potential would be reached with fewer MEPPs
D. The threshold to trigger an action potential would become more negative
E. The threshold to trigger an action potential would become more positive

Answer 58

C. Resting membrane potential would increase, and the threshold to trigger an action potential would be reached with fewer MEPPs

The equilibrium potential for an ion is determined
by the relative concentration of the ion
outside compared with inside the neuron. If the
extracellular concentration of potassium increases
and the intracellular concentration remains constant,
the equilibrium potential would become
more positive. Because the resting membrane
potential
for a neuron is determined mainly by
potassium ions, it also would increase. The threshold
for trigging an action potential would remain
the same but now would be closer to the resting
potential.