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CARDIOVASCULAR PHYSIOLOGY CONCEPTS
8.
Choice “d” is correct because inspiration
reduces intrapleural pressure, which
expands the right atrium, lowers its
pressure, and thereby enhances venous
return. Choice “a” is incorrect because
an increase in cardiac output must
increase venous return because the cir-
culatory system is closed. Choice “b” is
incorrect because decreased sympathetic
activation of the veins causes them to
relax, which increases their compli-
ance. This reduces preload on the heart,
which leads to a reduction in cardiac
output and venous return. Choice “c”
is incorrect because a Valsalva maneu-
ver increases intrapleural pressure,
compresses the vena cava, and reduces
venous return.
9.
Choice “a” is correct because decreased
venous compliance shifts the systemic
function curve to the right, which
increases the mean circulatory filling
pressure (value of the x-intercept).
Choice “b” is incorrect because changes
in systemic vascular resistance alter the
slope of the systemic function curve,
but not its x-intercept. Choice “c” is
incorrect because a decrease in blood
volume causes a parallel shift in the sys-
temic function curve to the left, which
decreases mean circulatory filling pres-
sure. Choice “d” is incorrect because
mean circulatory filling pressure, by
definition, is the intravascular pressure
when cardiac output is zero, and there-
fore it is independent of cardiac output.
10. The correct answer is “b” because a
decrease in systemic vascular resistance
increases the slope of the systemic func-
tion curve, which increases cardiac
output and right atrial pressure. Choices
“a” and “d” are incorrect because
decreased blood volume and increased
venous compliance decrease right atrial
pressure and cardiac output by causing
a leftward parallel shift in the systemic
function curve. Choice “c” is incorrect
because increased heart rate increases
cardiac output a small amount and
decreases right atrial pressure.
ANSWERS TO PROBLEMS AND CASES
PROBLEM 5-1
Under constant flow conditions, AP °c AR
(from Equation 5-1). Furthermore, R «= 1/r4
(from Equation 5-6). Therefore, AP «= 1/r4.
Using this relationship, we find that decreas-
ing diameter (or radius, which is proportion-
al to diameter) by 50% (to 1/2 its original
radius) increases AP by a factor of 16 (recip-
rocal of 1/2 to the fourth power). Therefore,
the new pressure gradient along the length
of vessel will be 32 mm Hg (2 mm Hg
X
16).
PROBLEM 5-2
In this problem, the two smaller daughter
arterioles (Rd) are parallel with each other
and in series with the parent arteriole (Rp).
Therefore, the total resistance (R^ can be
found by the following equation:
Substituting the relative resistances given in
this problem, we obtain
RT =1 +
1
1
+
1
4
4
= 3
PROBLEM 5-3
From Equation 5-10, we know that
SVR = (MAP ~ cvp)
CO
Because CVP is zero, this equation simplifies
to
SVR =
MAP
RT = RP
+
CO
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