Answers to Test & Quantitative and Thought Questions
5-5 d
Lipid-soluble messengers cross the plasma membrane
and act primarily on cytosolic and nuclear receptors.
5-6 b
Quantitative and Thought Questions
Patient A’s drug very likely acts to block phospholipase
, whereas patient B’s drug blocks lipoxygenase (see
Figure 5–12).
The chronic loss of exposure of the heart’s receptors
to norepinephrine causes an up-regulation of this
receptor type (i.e., more receptors in the heart for
norepinephrine). The drug, being an agonist of
norepinephrine (i.e., able to bind to norepinephrine’s
receptors and activate them) is now more effective
because there are more receptors for it to combine with.
None. You are told that all six responses are mediated
by the cAMP system, thus, blockage of any of the
steps listed in the question would eliminate all six
of the responses. This is because the cascade for all
six responses is identical from the receptor through
the formation of cAMP and activation of cAMP-
dependent protein kinase. Thus, the drug must be
acting at a point beyond this kinase (e.g., at the level of
the phosphorylated protein mediating this response).
Not in most cells, because there are other
physiological mechanisms by which signals
impinging on the cell can increase cytosolic calcium
concentration. These include (1) second-messenger-
induced release of calcium from the endoplasmic
reticulum and (2) voltage-sensitive calcium channels.
Intracellular second messengers do not disappear
immediately upon removal of a fi rst messenger.
Instead, some second messengers (such as cAMP) may
linger inside the cell for seconds, minutes, or even
longer after the fi rst messenger is gone.
Chapter 6
Test Questions
6-1 b
Afferent neurons have peripheral axon terminals
associated with sensory receptors, cell bodies in the
dorsal root ganglion of the spinal cord, and central
axon terminals that project into the spinal cord.
6-2 c
Oligodendrocytes form myelin sheaths in the central
nervous system.
6-3 d
Insert the given chloride concentrations into the
Nernst equation; remember to use –1 as the valence (z).
6-4 d
A, B, and C all are correct. Using the Nernst equation
to calculate the sodium equilibrium potential gives
values of +31, +36, and +40 mV for a, b, and c. If
the membrane potential was +42 mV, the outward
electrical force on sodium would be greater than the
inward concentration gradient, so sodium would move
out of the cell in each of these cases.
6-5 e
Neither sodium nor potassium is in equilibrium
at the resting membrane potential, but the action
of the Na
-ATPase pump prevents the small
but steady leak of both ions from dissipating the
concentration gradients.
6-6 a
Because Na
is further away from its electrochemical
equilibrium than is K
, there would be more
sodium entry than potassium exit, causing local
depolarization and local current fl ow that would
decrease with distance from the site of the stimulus.
6-7 c
Due to the persistent open state of the voltage-gated
potassium channels, for a brief time at the end of an
action potential the membrane is hyperpolarized.
When the voltage-gated potassium channels
eventually close, the potassium leak channels once
again determine the resting membrane potential.
6-8 d
The IPSP caused by neuron B would summate with
(subtract from) the amplitude of the EPSP caused by
neuron A’s fi ring.
6-9 a
Dopamine, like norepinephrine and epinephrine, is
a catecholamine neurotransmitter manufactured by
enzymatic modifi cation of the amino acid tyrosine.
6-10 b
Norepinephrine is the neurotransmitter released by
postganglionic neurons onto smooth muscle cells.
Quantitative and Thought Questions
Little change in the resting membrane potential
would occur when the pump fi rst stops because the
contribution to charge separation
is very small. With time, however, the membrane
potential would depolarize progressively toward zero
because the sodium and potassium concentration
gradients, which depend on the Na
pumps and which give rise to the membrane
potential, run down.
The resting potential would decrease (i.e., become
less negative) because the concentration gradient
causing net diffusion of this positively charged
ion out of the cell would be smaller. The action
potential would fi re more easily (i.e., with smaller
stimuli) because the resting potential would be
closer to threshold. It would repolarize more slowly
because repolarization depends on net potassium
diffusion from the cell, and the concentration
gradient driving this diffusion is lower. Also, the
afterhyperpolarization would be smaller.
The hypothalamus was probably damaged. It plays a
critical role in appetite, thirst, and sexual capacity.
The drug probably blocks cholinergic muscarinic
receptors. These receptors on effector cells mediate
the actions of parasympathetic nerves. Therefore,
the drug would remove the slowing effect of these
nerves on the heart, allowing the heart to speed up.
Blocking their effect on the salivary glands would
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