Answers to Test & Quantitative and Thought Questions
Ventricular muscle cells do not have a pacemaker
potential, and the L-type calcium channel is not
open during this phase of the action potential even
in autorhythmic cells.
Increased sympathetic nerve ﬁ ring and
norepinephrine release during exercise constricts
vascular beds in the kidneys, GI tract, and other
tissues to compensate for the large dilation of muscle
Reduced oxygen delivery to the kidneys
increases the secretion of erythropoietin, which
stimulates bone marrow to increase production of
t-PA is part of the ﬁ brinolytic system that dissolves
Quantitative and Thought Questions
No. Decreased erythrocyte volume is certainly one
possible explanation, but there is a second: The
person might have a normal erythrocyte volume but
an abnormally increased plasma volume. Convince
yourself of this by writing the hematocrit equation
as: erythrocyte volume/(erythrocyte volume +
A halving of tube radius. Resistance is directly
proportional to blood viscosity but inversely
proportional to the
of tube radius.
The plateau of the action potential and the
contraction would be absent. You might think that
contraction would persist because most calcium in
excitation-contraction coupling in the heart comes
from the sarcoplasmic reticulum. However, the
signal for the release of this calcium is the calcium
entering across the plasma membrane.
The SA node is not functioning, and the ventricles
are being driven by a pacemaker in the AV node or
the bundle of His.
The person has a narrowed aortic valve. Normally,
the resistance across the aortic valve is so small that
there is only a tiny pressure difference between
the left ventricle and the aorta during ventricular
ejection. In the example given here, the large
pressure difference indicates that resistance across
the valve must be very high.
This question is analogous to question 12-5 in that
the large pressure difference across a valve while
the valve is open indicates an abnormally narrowed
valve—in this case, the left AV valve.
Decreased heart rate and contractility. These are
effects mediated by the sympathetic nerves on beta-
adrenergic receptors in the heart.
120 mmHg. MAP = DP + 1/3 (SP – DP).
The drug must have caused the arterioles in the
kidneys to dilate enough to reduce their resistance
by 50 percent. Blood ﬂ ow to an organ is determined
by mean arterial pressure and the organ’s resistance
to ﬂ ow. Another important point can be deduced
here: If mean arterial pressure has not changed even
though renal resistance has dropped 50 percent,
then either the resistance of some other organ or
actual cardiac output has gone up.
The experiment suggests that acetylcholine causes
vasodilation by releasing nitric oxide or some other
vasodilator from endothelial cells.
A low plasma protein concentration. Capillary
pressure is, if anything, lower than normal and so
cannot be causing the edema. Another possibility
is that capillary permeability to plasma proteins has
increased, as occurs in burns.
20 mmHg/L per minute. TPR = MAP/CO.
Nothing. Cardiac output and TPR have remained
unchanged, so their product, MAP, also remains
unchanged. This question emphasizes that
MAP depends on cardiac output but not on the
combination of heart rate and stroke volume that
produces the cardiac output.
It increases. There are a certain number of
impulses traveling up the nerves from the arterial
baroreceptors. When these nerves are cut, the
number of impulses reaching the medullary
cardiovascular center goes to zero, just as it would
physiologically if the mean arterial pressure were
to decrease markedly. Accordingly, the medullary
cardiovascular center responds to the absent impulses
by reﬂ exly increasing arterial pressure.
It decreases. The hemorrhage causes no immediate
change in hematocrit because erythrocytes
and plasma are lost in the same proportion. As
interstitial ﬂ uid starts entering the capillaries,
however, it expands the plasma volume and
decreases hematocrit. (This is too soon for any new
erythrocytes to be synthesized.)
If alveolar pressure (
) is negative with respect to
atmospheric pressure (
), the driving force for
airﬂ ow is inward (from the atmosphere into the lung).
For the same change in transpulmonary pressure, a
less compliant (i.e., stiffer) lung will have a smaller
change in lung volume.
Total minute ventilation is comprised of dead
space plus alveolar ventilation. Minute ventilation
is respiratory frequency (12 breaths per minute)
multiplied by tidal volume (500 ml/breath) = 6 ml/
min. Subtract from that alveolar ventilation (4200
ml/min) and one gets 1800 ml/min.
An increase in alveolar
results from an increase
in alveolar ventilation (supply of oxygen) relative to
metabolic rate (consumption of oxygen).
The relationship between arterial
oxygen saturation is described by the oxygen-
hemoglobin dissociation curve. The greatest increase