Cardiovascular Physiology Flashcards
A 53-year-old woman is found, by arteriography, to have 50% narrowing of her left renal artery. What is the expected change in blood flow through the stenotic artery?
(A) Decrease to ½ (B) Decrease to ¼ (C) Decrease to 1⁄8 (D) Decrease to 1⁄16 (E) No change
The answer is D [II C, D].
If the radius of the artery decreased by 50% (1/2), then resistance would increase by 24, or 16 (R = 8ηl/πr4). Because blood flow is inversely proportional to resistance (Q = ΔP/R), flow will decrease to 1/16 of the original value.
When a person moves from a supine position to a standing position, which of the following compensatory changes occurs?
(A) Decreased heart rate (B) Increased contractility (C) Decreased total peripheral resistance (TPR) (D) Decreased cardiac output (E) Increased PR intervals
The answer is B [IX A; Table 3.4].
When a person moves to a standing position, blood pools in the leg veins, causing decreased venous return to the heart, decreased cardiac output, and decreased arterial pressure. The baroreceptors detect the decrease in arterial pressure, and the vasomotor center is activated to increase sympathetic outflow and decrease parasympathetic outflow. There is an increase in heart rate (resulting in a decreased PR interval), contractility, and total peripheral resistance (TPR). Because both heart rate and contractility are increased, cardiac output will increase toward normal.
At which site is systolic blood pressure the highest?
(A) Aorta (B) Central vein (C) Pulmonary artery (D) Right atrium (E) Renal artery (F) Renal vein
The answer is E [II G, H, I].
Pressures on the venous side of the circulation (e.g., central vein, right atrium, renal vein) are lower than pressures on the arterial side. Pressure in the pulmonary artery (and all pressures on the right side of the heart) are much lower than their counterparts on the left side of the heart. In the systemic circulation, systolic pressure is actually slightly higher in the downstream arteries (e.g., renal artery) than in the aorta because of the reflection of pressure waves at branch points.
A person’s electrocardiogram (ECG) has no P wave, but has a normal QRS complex and a normal T wave. Therefore, his pacemaker is located in the
(A) sinoatrial (SA) node (B) atrioventricular (AV) node (C) bundle of His (D) Purkinje system (E) ventricular muscle
The answer is B [III A].
The absent P wave indicates that the atrium is not depolarizing and, therefore, the pacemaker cannot be in the sinoatrial (SA) node. Because the QRS and T waves are normal, depolarization and repolarization of the ventricle must be proceeding in the normal sequence. This situation can occur if the pacemaker is located in the atrioventricular (AV) node. If the pacemaker were located in the bundle of His or in the Purkinje system, the ventricles would activate in an abnormal sequence (depending on the exact location of the pacemaker) and the QRS wave would have an abnormal configuration. Ventricular muscle does not have pacemaker properties.
If the ejection fraction increases, there will be a decrease in
(A) cardiac output (B) end-systolic volume (C) heart rate (D) pulse pressure (E) stroke volume (F) systolic pressure
The answer is B [IV G 3].
An increase in ejection fraction means that a higher fraction of the end-diastolic volume is ejected in the stroke volume (e.g., because of the administration of a positive inotropic agent). When this situation occurs, the volume remaining in the ventricle after systole, the end-systolic volume, will be reduced. Cardiac output, pulse pressure, stroke volume, and systolic pressure will be increased.
An electrocardiogram (ECG) on a person shows ventricular extrasystoles.
The extrasystolic beat would produce
(A) increased pulse pressure because contractility is increased
(B) increased pulse pressure because heart rate is increased
(C) decreased pulse pressure because ventricular filling time is increased
(D) decreased pulse pressure because stroke volume is decreased
(E) decreased pulse pressure because the PR interval is increased
The answer is D [V G].
On the extrasystolic beat, pulse pressure decreases because there is inadequate ventricular filling time—the ventricle beats “too soon.” As a result, stroke volume decreases.
An electrocardiogram (ECG) on a person shows ventricular extrasystoles.
After an extrasystole, the next “normal” ventricular contraction produces
(A) increased pulse pressure because the contractility of the ventricle is increased
(B) increased pulse pressure because total peripheral resistance (TPR) is decreased
(C) increased pulse pressure because compliance of the veins is decreased
(D) decreased pulse pressure because the contractility of the ventricle is increased
(E) decreased pulse pressure because TPR is decreased
The answer is A [IV C I a (2)].
The postextrasystolic contraction produces increased pulse pressure because contractility is increased. Extra Ca2+ enters the cell during the extrasystolic beat. Contractility is directly related to the amount of intracellular Ca2+ available for binding to troponin C.
An increase in contractility is demonstrated on a Frank-Starling diagram by
(A) increased cardiac output for a given end-diastolic volume
(B) increased cardiac output for a given end-systolic volume
(C) decreased cardiac output for a given end-diastolic volume
(D) decreased cardiac output for a given end-systolic volume
The answer is A [IV D 5 a].
An increase in contractility produces an increase in cardiac output for a given end-diastolic volume, or pressure. The Frank-Starling relationship demonstrates the matching of cardiac output (what leaves the heart) with venous return (what returns to the heart). An increase in contractility (positive inotropic effect) will shift the curve upward.
In a capillary, Pc is 30 mm Hg, Pi is −2 mm Hg, πc is 25 mm Hg, and πi is 2 mm Hg.
What is the direction of fluid movement and the net driving force?
(A) Absorption; 6 mm Hg (B) Absorption; 9 mm Hg (C) Filtration; 6 mm Hg (D) Filtration; 9 mm Hg (E) There is no net fluid movement
The answer is D [VII C 1].
The net driving force can be calculated with the Starling equation
Net pressure = (Pc - Pi) - (πc-πi)
= [(30-(-2)) - (25-2)] mmHg
= 32 mmHg - 23 mmHg
= + 9 mmHg
Because the net pressure is positive, filtration out of the capillary will occur.
In a capillary, Pc is 30 mm Hg, Pi is −2 mm Hg, πc is 25 mm Hg, and πi is 2 mm Hg.
If Kf is 0.5 mL/min/mm Hg, what is the rate of water flow across the capillary wall?
(A) 0.06 mL/min (B) 0.45 mL/min (C) 4.50 mL/min (D) 9.00 mL/min (E) 18.00 mL/min
The answer is C [VII C 1].
Kf is the filtration coefficient for the capillary and describes the intrinsic water permeability.
Water flow = K X Net pressure
= 0.5 mL/min/mmHg X 9 mmHg
= 4.5 mL/min
The tendency for blood flow to be turbulent is increased by
(A) increased viscosity
(B) increased hematocrit
(C) partial occlusion of a blood vessel
(D) decreased velocity of blood flow
The answer is C [II D 2 a, b].
Turbulent flow is predicted when the Reynolds number is increased. Factors that increase the Reynolds number and produce turbulent flow are decreased viscosity (hematocrit) and increased velocity. Partial occlusion of a blood vessel increases the Reynolds number (and turbulence) because the decrease in cross-sectional area results in increased blood velocity (v = Q/A).
A 66-year-old man, who has had a sympathectomy, experiences a greater-than-normal fall in arterial pressure upon standing up. The explanation for this occurrence is
(A) an exaggerated response of the renin– angiotensin–aldosterone system
(B) a suppressed response of the renin– angiotensin–aldosterone system
(C) an exaggerated response of the baroreceptor mechanism
(D) a suppressed response of the baroreceptor mechanism
The answer is D [IX A].
Orthostatic hypotension is a decrease in arterial pressure that occurs when a person moves from a supine to a standing position. A person with a normal baroreceptor mechanism responds to a decrease in arterial pressure through the vasomotor center by increasing sympathetic outflow and decreasing parasympathetic outflow. The sympathetic component helps to restore blood pressure by increasing heart rate, contractility, total peripheral resistance (TPR), and mean systemic pressure. In a patient who has undergone a sympathectomy, the sympathetic component of the baroreceptor mechanism is absent.
The ventricles are completely depolarized during which isoelectric portion of the electrocardiogram (ECG)?
(A) PR interval (B) QRS complex (C) QT interval (D) ST segment (E) T wave
The answer is D [III A].
The PR segment (part of the PR interval) and the ST segment are the only portions of the electrocardiogram (ECG) that are isoelectric. The PR interval includes the P wave (atrial depolarization) and the PR segment, which represents conduction through the atrioventricular (AV) node; during this phase, the ventricles are not yet depolarized. The ST segment is the only isoelectric period when the entire ventricle is depolarized.
In which of the following situations is pulmonary blood flow greater than aortic blood flow?
(A) Normal adult (B) Fetus (C) Left-to-right ventricular shunt (D) Right-to-left ventricular shunt (E) Right ventricular failure (F) Administration of a positive inotropic agent
The answer is C [I A].
In a left-to-right ventricular shunt, a defect in the ventricular septum allows blood to flow from the left ventricle to the right ventricle instead of being ejected into the aorta. The “shunted” fraction of the left ventricular output is therefore added to the output of the right ventricle, making pulmonary blood flow (the cardiac output of the right ventricle) higher than systemic blood flow (the cardiac output of the left ventricle). In normal adults, the outputs of both ventricles are equal in the steady state. In the fetus, pulmonary blood flow is near zero. Right ventricular failure results in decreased pulmonary blood flow. Administration of a positive inotropic agent should have the same effect on contractility and cardiac output in both ventricles.
A 30-year-old female patient’s electrocardiogram (ECG) shows two P waves preceding each QRS complex. The interpretation of this pattern is
(A) decreased firing rate of the pacemaker in the sinoatrial (SA) node
(B) decreased firing rate of the pacemaker in the atrioventricular (AV) node
(C) increased firing rate of the pacemaker in the SA node
(D) decreased conduction through the AV node
(E) increased conduction through the His-Purkinje system
The answer is D [III E 1 b].
A pattern of two P waves preceding each QRS complex indicates that only every other P wave is conducted through the atrioventricular (AV) node to the ventricle. Thus, conduction velocity through the AV node must be decreased.
An acute decrease in arterial blood pressure elicits which of the following compensatory changes?
(A) Decreased firing rate of the carotid sinus nerve
(B) Increased parasympathetic outflow to the heart
(C) Decreased heart rate
(D) Decreased contractility
(E) Decreased mean systemic pressure
The answer is A [VI A 1 a-d].
A decrease in blood pressure causes decreased stretch of the carotid sinus baroreceptors and decreased firing of the carotid sinus nerve. In an attempt to restore blood pressure, the parasympathetic outflow to the heart is decreased and sympathetic outflow is increased. As a result, heart rate and contractility will be increased. Mean systemic pressure will increase because of increased sympathetic tone of the veins (and a shift of blood to the arteries).
The tendency for edema to occur will be increased by
(A) arteriolar constriction
(B) increased venous pressure
(C) increased plasma protein concentration
(D) muscular activity
The answer is B [VII C 4 c; Table 3.2].
Edema occurs when more fluid is filtered out of the capillaries than can be returned to the circulation by the lymphatics. Filtration is increased by changes that increase Pc or decrease πc. Arteriolar constriction would decrease Pc and decrease filtration. Dehydration would increase plasma protein concentration (by hemoconcentration) and thereby increase πc and decrease filtration. Increased venous pressure would increase Pc and filtration.
Inspiration “splits” the second heart sound because
(A) the aortic valve closes before the pulmonic valve
(B) the pulmonic valve closes before the aortic valve
(C) the mitral valve closes before the tricuspid valve
(D) the tricuspid valve closes before the mitral valve
(E) filling of the ventricles has fast and slow components
The answer is a [V E].
The second heart sound is associated with closure of the aortic and pulmonic valves. Because the aortic valve closes before the pulmonic valve, the sound can be split by inspiration.
