Reading...
Play button
Play button
Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
64 Cards in this Set
- Front
- Back
|
A 20-year-old patient with a large ventricular septal defect (VSD) underwent PA
banding in childhood and was lost to follow-up. A recent echocardiogram revealed the following: peak systolic velocity across the VSD 3 m/s,TRvelocity 5 m/s, estimated RA pressure 10mmHg, cuff blood pressure in the right arm 146/70mmHg, peak flow velocity across the pulmonary band 4.7 m/s. The following statement is true: A. This patient has normal PA pressure B. The patient has severe pulmonary hypertension C. The patient has features of left ventricular (LV) failure D. PA pressure cannot be determined |
Answer: A.
The RV systolic pressure is 110mmHg based on TR velocity (554þ10¼ 110 mmHg). VSD peak velocity is 3 m/s corresponding to an LV–RV pressure gradient of 36 mmHg. Given the systemic systolic pressure of 146mmHg (hence an LV systolic pressure of 146 mmHg), the VSD gradient is again concordant with an RV systolic pressure of 110 mmHg. In the absence of PS, this is the pressure in the proximal PA. The pressure gradient across the band is 4.74.74¼88 mmHg. Hence the PA systolic pressure distal to the band is 110 – 88¼22 mmHg. Though technically this patient has severe elevation of proximal PA pressure, the PA vascular perfusion pressure is normal, indicating the absence of pulmonary artery disease, making this patient a candidate for surgical closure of VSD. |
|
|
The patient has an LVOT velocity of 1 m/s, TVI of 25 cm, LVOT diameter 2 cm,
aortic transvalvular velocity of 1.5 m/s, heart rate 70 beats/min and the cardiac output in this patient is: A. 5.5 L B. 4.5 L C. 6.3 L D. Cannot be determined based on the given data |
Answer: A.
The stroke volume equals the cross-sectional area x TVI of LVOT, which is 3.14x1x1x25=78 cc. Stroke volume multiplied by heart rate, i.e. 78 x 70= 5.5 L/min, equals cardiac output. |
|
|
A patient with aortic stenosis has an LVOT diameter of 2 cm, LVOT velocity (V1)
2.5 m/s, transaortic valve velocity (V2) 5 m/s and two-dimensional examination showed moderate systolic anterior motion of the mitral leaflet. Valvular aortic stenosis in this patient is: A. Mild B. Moderate C. Severe D. Cannot be calculated based on given data |
Answer: D.
In a patient with serial stenosis in close proximity, the continuity equation cannot be applied because of difficulty in obtaining precise subvalvular velocity and crosssectional area of the flow in the LVOT. In a person without systolic anterior motion the cross-sectional area of subvalvular flow is roughly equal to the cross-sectional area of the LV outflow tract. Subvalvular obstruction will result in flow streams such that the cross-sectional area of flow is less than the anatomic LVOT area. |
|
|
In a patient with valvular PS with right PA branch stenosis, the following measurements
were obtained: tricuspid regurgitation (TR) velocity 4 m/s, right atrial (RA) pressure 6 mmHg, systolic velocity across the pulmonary valve 2.5 m/s, velocity across the discrete branch stenosis 2.5 m/s. The systolic pressure in the right pulmonary branch distal to the stenosis is likely to be: A. 20mmHg B. 5mmHg C. 70mmHg D. Cannot be estimated |
Answer: A.
In this patient, the estimated right ventricular systolic pressure (RVSP) is 64+6 mmHg=70 mmHg. Pressure drop across the pulmonary valve is equal to 25 mmHg, resulting in a systolic pressure of 45mmHg in the main PA. As there is 3–4 cm between the pulmonary valve and the right PA, flow streams would have normalized and would allow us to estimate the pressure drop at the branch stenosis without the limitations of stenosis in series unless there is a substantial pressure recovery. As the pressure drop across the branch stenosis is 25 mmHg, estimated systolic pressure distal to the branch stenosis is 45-25 or 20 mmHg. |
|
|
Bicuspid aortic valve may be associated with:
A. Coronary anomalies B. Coarctation of the aorta C. Atrial septal defect D. None of the above |
Answer: B.
Bicuspid valve is associated with coarctation of the aorta. Biscuspid aortic valve occurs in 1–2% of the population. In these people aortic coarctation is rare, but 25% of patients with coarctation have a bicuspid aortic valve. |
|
|
A dilated coronary sinus could be seen in all of the following conditions except:
A. Right atrial hypertension B. Persistent left superior vena cava C. Coronary A–V fistula D. Unroofed coronary sinus E. Azygos continuity of inferior vena cava |
Answer: E.
Coronary sinus can be dilated due to increased pressure or flow. There is increased flow in the coronary sinus in the left superior vena cava, which drains into the coronary sinus, coronary A–V fistula due to increased shunt, and unroofed coronary sinus due to increased flow from left atrium (LA) to coronary sinus. Right atrial hypertension causes increased pressure, which will lead to dilated coronary sinus. |
|
|
Atrial septal defect (ASD) of sinus venosus type is most commonly associated with:
A. Anomalous drainage of right upper pulmonary vein into right atrium B. Anomalous drainage of left upper pulmonary vein into right atrium C. Persistent left upper superior vena cava D. Coronary artery anomalies |
Answer: A.
ASD of sinus venosus type is most commonly associated with anomalous drainage of the right upper pulmonary vein into the right atrium. |
|
|
Ostium primum ASD is most commonly associated with:
A. Cleft anterior mitral leaflet B. Cleft in septal leaflet of tricuspid valve C. Patent ductus arteriosus D. Aortic stenosis |
Answer: A.
Ostium primum ASD is most commonly associated with a cleft anterior mitral leaflet. This is a form of endocardial cushion defect. |
|
|
Dilatation of the pulmonary artery is seen in all of the following conditions except:
A. Atrial septal defect B. Valvular pulmonary stenosis C. Infundibular pulmonary stenosis D. Pulmonary hypertension |
Answer: C.
Infundibular pulmonary stenosis is not associated with dilatation of the pulmonary artery. Poststenotic dilatation is seen only in valvular pulmonary stenosis and not in subvalvular pulmonary stenosis. In ASD the pulmonary artery dilates due to increased flow and in pulmonary hypertension dilatation is due to increased pressure. Idiopathic dilatation of the pulmonary artery can also occur. Marfan syndrome is a cause of pulmonary artery dilatation as well. |
|
|
52-year-old patient with a 31mm St. Jude mitral valve has severe shortness of breath.
Left ventricular function and aortic valve are normal. The disc motion of the prosthetic valve is normal. Analysis of transmitral flow with continuous wave Doppler revealed an E-wave velocity of 2.6 m/s, A-wave velocity of 0.6 m/s, E-wave pressure half-time 40 ms, diastolic mean gradient of 6mmHg at a heart rate of 60/min, isovolumic relaxation time (IVRT) 30 ms. This patient is likely to have: A. Mitral regurgitation B. Pannus growth into the prosthetic valve C. Prosthetic valve thrombosis D. Normal prosthetic valve function |
Answer: A.
Normal IVRT is 70–100 ms, and pressure half-time is 65–80 ms for a prosthetic mitral valve. With a normal cardiac output the mean gradient would be 3–4mmHg at a heart rate of 60/min. Shortened IVRT, short pressure half-time and high E/A ratio indicate high LA pressure. A stenotic prosthetic valve would have caused increase in pressure half-time and an increase in mean gradient far more than 6mmHg at a heart rate of 60/min. A mildly increased gradient despite a shortened pressure half-time indicates increased transvalvular flow suggestive of mitral regurgitation, which may be difficult to visualize from a transthoracic echo. Hence a transesophageal echocardiogram (TEE) would be warranted. High LA pressure without an increase in flow would result in shortened IVRT and pressure half-time without an increase in the gradient. A good example of this is superadded restrictive cardiomyopathy. |
|
|
Risk of aortic dissection is increased in the following conditions except:
A. Marfan’s syndrome B. Bicuspid aortic valve C. Pregnancy D. Mitral stenosis |
Answer: D.
All conditions except mitral stenosis have weakened media predisposing to dissection. Hypertension can also increase the risk for dissection. |
|
|
In a person with suspected paravalvular (mechanical) mitral regurgitation, the following
transducer position has the best chance of revealing the mitral regurgitation jet: A. Parasternal long axis view B. Apical four-chamber C. Apical two-chamber D. Apical long axis |
Answer: A.
Shadowing in the left atrium is least with a parasternal long axis view, however TEE is the best technique to evaluate for paravalvular mitral leaks |
|
|
A patient with a bileaflet mechanical aortic valve has shortness of breath on exertion. An
echocardiogram revealed normal left ventricular systolic function and mitral valve function. The left ventricular outflow tract (LVOT) dimension was 2.2 cm, LVOT (V1) velocity was 1.5 m/s and aortic transvalvular velocity (V2) was 4.5 m/s, with no aortic regurgitation. Measurements obtained 2 years earlier when the patient was asymptomatic were: LVOT diameter 2.2 cm, V1 0.9 m/s and V2 2.7 m/s. Likely cause of this patient’s shortness of breath is: A. Prosthetic valve stenosis B. Patient–prosthesis mismatch C. High cardiac output state, patient may be anemic D. None of the above |
Answer: C.
An unchanged V1/V2 ratio compared to prior echo confirms the absence of prosthetic valve stenosis. An elevated V1 indicates elevated cardiac output and the transvalvular gradient is flow dependent. Anemia is a common problem secondary to blood loss due to anticoagulation, and less commonly due to mechanical hemolysis. Patient prosthesis mismatch occurs when the valve is too small for the cardiac output needs of the patient. The effective aortic orifice area in this patient is about 1.3 cm2. There is no change in the intrinsic valvular function in this patient. |
|
|
A patient with a mechanical prosthetic mitral valve has gastrointestinal bleeding and the
following measurements were obtained: diastolic mean gradient 11 mmHg, peak gradient 16 mmHg, pressure half-time 65 ms, heart rate 114/min. This increased gradient is likely to be: A. Normal B. Abnormal C. Cannot comment |
Answer: A.
The measurements are normal. Pressure half-time of 65 ms indicates normal valve function. Mean gradient is appropriately increased due to tachycardia (which shortens the diastolic filling period), anemia and possibly high cardiac output. Prosthetic valves are intrinsically mildly stenotic |
|
|
The following measurements were obtained in a patient with mitral regurgitation:
proximal isovelocity surface area (PISA) radius 1 cm at a Nyquist limit of 50 cm/s, peak mitral regurgitation velocity 5 m/sec and mitral regurgitation signal time velocity integral 100 cm. The regurgitant volume is: A. 63 cc/beat B. 31 cc/beat C. 63 cc/s D. 63% |
Answer: A.
Effective regurgitant orifice area is given by the formula 2 x 3.14 x r x r x Nyquist limit / MR velocity, i.e.( 2 x 3.14 x 1 x1 x 50) / 500 cm/s = 0.628 cm2. Regurgitant volume is effective regurgitant orifice area (in cm2)TVI (in cm). In this patient it is 0.628 x 100 =62.8 cc. This is per beat and not per second. |
|
|
Distribution of leaflet thickening and calcification in rheumatic mitral stenosis is:
A. More at the tip B. More at the base C. Uniform throughout the leaflets |
A. More at the tip
|
|
|
Leaflet calcification in degenerative mitral stenosis is:
A. More at the tip B. More at the base C. Uniform throughout the leaflets |
B. More at the base
|
|
|
The predominant mechanism of chronic ischemic mitral regurgitation is:
A. Restriction of mitral leaflet closure B. Papillary muscle dysfunction C. Ruptured chordae tendinae D. Ruptured papillary muscle |
Answer: A.
Restriction, tethering and tenting refer to the phenomenon of incomplete systolic closure due to apical traction on the mitral leaflets due to outward displacement of the papillary muscles. This causes tenting of the leaflets and the coaptation point is displaced apically. This is not due to contractile failure of the papillary muscles (papillary muscle dysfunction). Papillary muscle rupture causes acute MR, leading to pulmonary edema and hemodynamic compromise. |
|
|
In a person with chronic ischemic mitral regurgitation (MR) due to old inferior
myocardial infarction (MI) and an ejection fraction of 50%, the location of the MR jet would be: A. Medial commissure B. Lateral Commissure C. Central |
Answer: A.
Due to displacement of the posteromedial papillary muscle, there is tethering of medial portions of both leaflet (P3 and A3) segments causing a medial commissural jet. When the left ventricle (LV) is uniformly dilated, the jet could be central in origin. |
|
|
In the above patient the jet direction would be:
A. Posterior B. Anterior C. Central |
B posterior
|
|
|
In a patient with old anteroseptal MI with an ejection fraction of 28%, an ischemicMR
jet is likely to be: A. Central B. Lateral wall hugging C. Medial wall hugging |
Answer: A.
In an anterior MI, there is generally remodeling of the noninfarcted segments as well, causing dilatation of the whole LV cavity. This is reflected by a low ejection Chapter 7 | 43 fraction. This causes displacement of both papillary muscles and tenting of all segments of both leaflets, giving rise to central MR, although exceptions may occur. MR in dilated cardiomyopathy occurs because of a similar mechanism. |
|
|
Mitral regurgitation in aortic stenosis is related to which of these factors:
A. Degree of mitral annular calcification B. Severity of aortic stenosis C. An increase in LV end systolic dimension D. Degree of aortic leaflet calcification |
Answer: C.
The mechanism of MR is functional and is related to LV dilation and leaflet tethering. Aortic leaflet calcification, mitral annular calcification and severity of aortic stenosis contribute very little in the genesis of MR. A higher driving pressure in more severe degrees of aortic stenosis may increase the regurgitant volume and the jet area,but will not cause MR in the absence of a defect in the mitral coaptation mechanism. |
|
|
Left atrial myxoma may be differentiated from a left atrial thrombus by all of the
following characteristics except: A. Enhancement with transpulmonary contrast agent B. Presence of blood vessels on color flow imaging C. Attachment to the atrial septum D. Similar mass in the left ventricle (LV) with normal LV function |
Answer: C.
Myxomas are vascular: blood vessels may be seen on color flow imaging and enhanced mildly with transpulmonary contrast agent. Though left atrial thrombus is most commonly seen in the appendage, it may be attached to the atrial septum or may traverse through a patent foramen ovale from the right side (paradoxical embolism). The presence of a mass in the LV in the face of normal LV function makes a thrombus unlikely and points to a familial myxoma syndrome (Carney’s syndrome). |
|
|
The most common location of left atrial thrombus is:
A. Left atrial appendage B. Body C. Atrial septum D. Atrial roof |
LAA
|
|
|
141. The most common benign tumor in the heart is:
A. Left atrial myxoma B. Papillary fibroelastoma C. Lamble’s excrescences D. Fibroma |
Answer: B.
Papillary fibroelastoma is the most common benign tumor seen in the heart, followed by myxoma. |
|
|
142. The most common metastatic malignant tumor of the heart is:
A. Melanoma B. Fibrosarcoma C. Rhabdomyoma D. Liposarcoma |
Answer: A.
The most common metastatic malignancy of the heart is melanoma, followed by lung and breast cancer. Primary malignant tumors of the heart are rare, but rhabdomyoma is the commonest. |
|
|
143. In a person with flail P2 segment of the posterior mitral leaflet (PML), the mitral
regurgitation (MR) jet is likely to be: A. Posterior wall hugging B. Anterior wall hugging C. Central D. Cannot comment |
Answer: B.
The jet is away from the flail segment in contrast to a tethered segment. |
|
|
In a person with flail A2 segment of the anterior mitral leaflet (AML), the MR jet is
likely to be: A. Posterior wall hugging B. Anterior wall hugging C. Central D. Cannot comment |
Answer: A.
|
|
|
Total surface area of mitral leaflets is generally ———% of mitral annular area.
A. 100% B. 120% C. 150% D. 200% |
Answer: C.
Normally there is 50% more leaflet tissue than annular area to cause a 2–3mm leaflet overlap at the coaptation margin. The absolute leaflet area is increased in myxomatous mitral valve disease and hypertrophic cardiomyopathy. The normal annular area is roughly 7–8 cm2 and the leaflet area is 10–12 cm2. |
|
|
The PML compared to the AML is:
A. Shorter B. Longer C. Same length as the anterior leaflet D. Of variable length |
Answer: A.
The posterior leaflet length is 10–14mm and the anterior leaflet length is 20–24 mm. |
|
|
The length of the posterior leaflet attachment to the mitral annulus compared to that of
the AML is: A. Shorter B. Longer C. Same D. Variable |
Answer: B.
This results in equal surface area of anterior and posterior leaflets. |
|
|
In an apical long-axis view the following mitral leaflet segments are seen:
A. A2P2 B. A3P3 C. A1P1 D. A3P1 |
Answer: A.
This view cuts through the middle of both leaflets, i.e. A2 and P2. |
|
|
Apical two-chamber view is likely to show the following mitral leaflet segments:
A. P1A2P3 B. A2P2 C. A3P1 D. A1P1 |
Answer: A.
Two-chamber view goes through the intercommissural plane and cuts through P1 and P3, with A2 seen between them in systole. A medial tilt of the transducer will cut the AML entirely showing A1A2A3, and a lateral tilt will cut the PML entirely revealing P1P2P3. |
|
|
The major diameter of the mitral annulus is best imaged from:
A. Apical two-chamber view B. Apical long axis view C. Apical five-chamber view D. Parasternal long axis view |
Answer: A.
Equivalent to this on a TEE examination is the mid-esophageal view at 70–808. Apical long axis view gives the minor dimension of the mitral annulus. |
|
|
151. The MR jet is best visualized in parasternal long axis view when the transducer tip is
directed more inferomedially. The location of the MR jet in this patient is: A. Medial commissure B. Lateral commissure C. Central |
Answer: A.
Tilting the transducer from this location towards the left shoulder will reveal the lateral commissure. |
|
|
A continuous flow is visualized in the main pulmonary artery. This could be related to:
A. Patent ductus arteriosus (PDA) B. Coronary A–V fistula C. Idiopathic dilatation of main pulmonary artery D. None of the above |
Answer: A.
The PDA drains at the origin of the left pulmonary artery. Anomalous origin of coronary artery from pulmonary artery can cause continuous flow because of retrograde flow into the pulmonary artery. Both are examples of left to right shunts. Dilatation of the pulmonary artery can cause swirling of blood in the pulmonary artery in systole, giving a false impression of shunt flow because of reversed flow direction. |
|
|
Echocardiographic features of anatomic right ventricle in a congenitally corrected
transposition of great vessels are all of the following except: A. Trileaflet A–V valve B. Apical position of associated A–V valve C. Presence of moderator band D. Wall thickness <7 mm |
Answer: D.
Ventricles go with corresponding A–V valves, i.e. right ventricle with tricuspid valve and left ventricle with mitral valve. Wall thickness is not a reliable feature. In right ventricular hypertrophy the wall thickness may be >7mm and the pulmonary left ventricle may have a wall thickness of <7 mm. |
|
|
Problems encountered with congenitally corrected great arteries are all of the following
except: A. Failure of systemic ventricle B. Tricuspid regurgitation C. Atrial and ventricular arrhythmias D. Aortic regurgitation |
Answer: D.
The systemic RV has a high likelihood of failure. It may also have myocardial perfusion defects. Tricuspid valve failure is common secondary to annular dilatation. Atrial and ventricular arrhythmias are common due to dilatation of the left atrium and systemic RV. Presence of atrial arrhythmias may contribute to RV dysfunction. |
|
|
Features of Tetralogy of Fallot are all of the following except:
A. Overriding aorta B. Nonrestrictive ventricular septal defect (VSD) C. Pulmonary stenosis D. Right ventricular (RV) hypertrophy E. Atrial septal defect (ASD) |
Answer: E.
ASD is not a feature in classical Tetralogy of Fallot. Presence of ASD in Tetralogy has been referred to as the Pentalogy of Fallot. |
|
|
Associations of atrial septal aneurysm include all of the following except:
A. Patent foramen ovale B. Atrial arrythmias C. Transient ischemic attacks D. Pulmonary hypertension |
Answer: D.
The left to right shunt through the patent foramen ovale (PFO) is generally very small and hence pulmonary hypertension is not seen with an aneurysmal atrial septum. Risk of transient ischemic attack is highest when PFO and atrial septal aneurysm coexist and the shunt flow is large. Speculated mechanisms for this include paradoxical embolism, in situ thrombus formation and atrial arrhythmias. |
|
|
Echocardiographic findings in Ebstein’s anomaly may include all of the following exept:
A. Apical displacement of the septal leaflet of the tricuspid valve> 8mm compared to position of AML attachment B. Large, septal tricuspid leaflet with tethering to RV wall C. Tricuspid regurgitation D. Atrial septal defect E. Hypoplastic pulmonary arteries |
Answer: E.
Hypoplastic pulmonary arteries is not a feature. ASD may co-exist and in the presence of tricuspid regurgitation may result in right to left shunt, causing cyanosis. |
|
|
The most common location of the accessory pathway in Ebstein’s anomaly is:
A. Posteroseptal B. Anteroseptal C. Right lateral D. Left lateral |
Answer: C.
Right lateral. |
|
|
The following type of ventricular septal defect is likely to be associated with aortic
regurgitation: A. Perimembranous B. Muscular C. Supracristal D. Inlet |
159. Answer: C.
In this type of VSD there is loss of support to the right coronary cusp of the aortic valve, which will result in aortic regurgitation. |
|
|
In a patient with secundum ASD, the following features are consistent with amenability
of percutaneous closure except: A. Defect size of 22 mm B. Mitral rim of 8 mm C. Aortic rim of 2 mm D. Inferior vena cava rim of 1mm |
Answer: D.
The maximum stretched diameter of the defect that can be closed is 40mm with an Amplatzer device provided that the septum is large enough to allow the 8mm flange all around, making the total disc diameter on the left atrial side 56 mm. Aortic rim is the least important and not essential. Inadequacy of other rims may result not only in the impingement of the discs on related structures, but device instability and dislodgement. Proximity of superior vena cava rim to right upper pulmonary vein is important as well. |
|
|
Diagnostic sensitivity of stress echocardiography is higher with:
A. One-vessel disease B. Two-vessel disease C. Three-vessel disease D. All of the above |
Answer: C.
It is about 50% for one-vessel disease and 80% for three-vessel disease |
|
|
False-positive rate for stress echocardiography is high for which group of patients:
A. Low probability of coronary artery disease (CAD) B. Intermediate probability of CAD C. High probability of CAD D. Independent of CAD |
Answer: A.
Based on Baye’s theorem, the diagnostic accuracy is highest for intermediate probability, and in patients with extremely low probability most of the tests will be false positive, yielding a low positive predictive value. Positive predictive value (PPV) is the proportion of patients with positive tests who truly have disease. In other words, PPV=TP/(TP + FP). |
|
|
Negative predictive value of stress echo is lowest in this group of patients:
A. Low probability of CAD B. Intermediate probability of CAD C. High probability of CAD D. Independent of CAD |
Answer: C.
Testing this group of patients is likely to yield a high proportion of patients with a falsenegative test, hence lowering the negative predictive value (NPV). In other words, NPV=TN/TN+FN. |
|
|
False-positive wall motion abnormalities are most commonly seen in which of the
following myocardial segments? A. Posterior basal wall B. Anterior septum C. Lateral wall D. Apex |
Answer: A.
Wall motion abnormality in the posterior basal wall is most difficult to analyze due to a range of normalcy, proximity to valvular plane and the apical displacement of the wall during systole, which results in imaging of different parts of the inferior wall during systole and diastole in the short axis view. False positivity during stress echocardiography is in the range of 40–50% for this wall. |
|
|
The usual response of left ventricular (LV) end systolic size during exercise is:
A. Reduction B. Increase C. Variable response D. No change |
Answer: A.
Reduction due to a combination of reduced systemic vascular resistance (SVR) and increased LV contractility. |
|
|
An increase in LV end systolic volume during stress may occur in all of the situations
except: A. Multivessel CAD B. Left main CAD C. Hypertensive blood pressure response D. Left ventricular hypertrophy |
Answer: D.
This is the equivalent of transient ischemic dilatation of the LV on stress nuclear perfusion imaging. |
|
|
A 53-year-old patient is undergoing dobutamine stress echocardiography (DSE). At
20 mg dose the blood pressure drops from 140/80mmHg to 80/50mmHg associated with severe nausea, and the heart rate dropped from 110/min to 60/min. The most likely cause of this response is: A. Left ventricular cavity obliteration causing a vagal response B. Severe ischemic response due to multivessel CAD C. 2:1 A–V block produced by ischemia in right coronary artery territory D. None of the above |
Answer: A.
This is typical of a vasovagal response that is preceded by a hyperdynamic response, which triggers this. This may be exaggerated by volume depletion and may potentially be prevented by volume loading. When a hyperdynamic response with cavity obliteration is seen, instead of increasing the dobutamine dose, atropine should be administered to increase the heart rate. This will help to avert a vagal response. Drop in SVR is universal during DSE and may not be fully compensated by cardiac output increase. Systolic anterior movement can occur during DSE, especially in patients with LV hypotension who develop a hyperdynamic response, but LV outflow tract obstruction is rarely responsible for hypotension. Hypotension in such patients, when it occurs, is generally due to a vagal response produced by the hyperdynamic LV stimulating the vagal C type of fibers in the LV wall. This response could be partially prevented by volume loading before DSE. |
|
|
What proportion of normal patients undergoing DSE may have a drop in their blood
pressure: A. Zero B. 20% C. 50% D. 89% |
Answer: B.
Drop in blood pressure during DSE does not have the same clinical significance as in a regular exercise stress test. This is because normal cardiac output increase duringDSE is only 50–80%, which is far less than exercise. Dobutamine causes peripheral vasodilatation. |
|
|
All of the following factors affect pulmonary vein A-wave amplitude except:
A. LV end diastolic stiffness B. Left atrial function C. Pulmonary vein diameter D. Heart rate E. Pulmonary artery pressure |
Answer: E.
The amplitude is increased in the presence of a stiff LV and reduced in left atrial mechanical failure. The pulmonary A wave may disappear with heart rates in excess of 100/min, where flow may be entirely antegrade, and atrial contraction may produce a transient deceleration pulmonary flow without reversal. As velocity depends upon flow volume and cross-sectional area, a dilated pulmonary vein is likely to reduce the A-wave velocity and a collapsed vein in a dry patient can result in a giant A wave. |
|
|
The pulmonary vein S wave may be less prominent than the D wave in the following
situations except: A. Young children B. Moderate to severe mitral regurgitation C. Atrial fibrillation D. Elevated left atrial (LA) pressure E. Abnormal LV relaxation with normal left atrial (LA) pressure |
Answer: E.
Young children have very efficient LV relaxation properties, resulting in rapid early filling (mitral E wave) paralleled by an increase in D wave that might have rapid deceleration as well. As S1 is due to atrial relaxation, atrial fibrillation results in reduced S-wave amplitude. Systolic left atrial filling from mitral regurgitation will impede pulmonary vein flow in systole. High LA pressure renders the LA less compliant due to rightward shift of its pressure–volume curve and hence will impede atrial systolic filling, as LA is a closed chamber receiving only pulmonary venous flow during systole. Abnormal LV relaxation reduces E- and D-wave amplitudes, resulting in an increase in S-wave amplitude in the absence of elevated LA pressure. |
|
|
Normal pulmonary vein A-wave duration compared to mitral A-wave duration is:
A. Less B. More C. Same D. Variable |
Answer: A.
Increased pulmonary A-wave duration compared to mitral A-wave duration indicates high LV end diastolic pressure. Delta duration of more than 30 ms is very suggestive of high LV end diastolic pressure. |
|
|
Normal pulmonary vein D-wave deceleration in an adult is:
A. 50–100 ms B. 100–170 ms C. 170–260 ms D. Highly variable |
Answer: C.
Reduced D-wave deceleration time indicates high LA pressure very similar to mitral E-wave deceleration time |
|
|
Increased pulmonary vein D-wave decelaration time may be encountered in:
A. Mitral stenosis B. Mitral regurgitation C. High LA pressure D. Pulmonary stenosis |
Answer: A.
The D-wave deceleration time parallels mitral E-wave deceleration time and the slope is flatter in mitral stenosis. It may also be prolonged in patients with prosthetic mitral valves and abnormal LV relaxation. The deceleration time is reduced with high left atrial pressure. Pulmonary stenosis has no known effect on this slope. |
|
|
Normal mitral E-wave propagation velocity by color M mode inside the LV is:
A. 10–30 cm/s B. 30–50 cm/sec C. Greater than 50 cm/s D. Greater than 500 cm/s |
Answer: C.
Greater than 50 cm/s. |
|
|
A reduced mitral E-wave propagation velocity indicates:
A. High LA pressure B. Increased tau C. Reduced tau D. Increased modulus LV chamber stiffness |
Answer: B.
Slower propagation indicates abnormal LV relaxation, and this is reflected by increased tau by invasive measurement. |
|
|
A reduced A-wave transit time to the LV outflow tract is indicative of:
A. Low negative dp/dt B. Increased tau C. Reduced tau D. Increased modulus of LV chamber stiffness |
Answer: D.
The A-wave propagation is an end diastolic phenomenon and its propagation velocity is increased with increased LV end diastolic stiffness. Increased propagation velocity results in a shorter transit time. The modulus of chamber stiffness is a measure of operative LV stiffness |
|
|
Rate of acceleration of the early portion of the aortic regurgitation (AR) signal is
determined by: A. LV negative dp/dt B. LV positive dp/dt C. LV end diastolic pressure D. Aortic end diastolic pressure |
Answer: A.
Assuming a constant aortic pressure during this short time, rate of acceleration is principally determined by LV pressure decay after aortic valve closure during the LV isovolumic relaxation period. For example, at the 1 m/s point the aortic–LV pressure gradient is 4mmHg and at 2.5 m/s it is 25 mmHg. Assuming a constant aortic pressure, the drop in LV pressure between these two points is 25-4=21 mmHg. The rate of LV pressure drop would be 21/time taken for AR signal to increase from 1 m/s to 2.5 m/s. For example, if this time interval is 20 ms (0.02 s) the average negative LV dp/dt would be 21/0.02=1050 mmHg/s. |
|
|
Rapidly decelerating terminal portion of the AR signal is mainly influenced by:
A. LV negative dp/dt B. LV positive dp/dt C. LV end diastolic pressure D. Aortic end diastolic pressure |
Answer: B.
This rapidly decelerating terminal portion of AR occurs during the isovolumic contraction period and the LV positive dp/dt may be calculated with similar assumptions as for the LV negative dp/dt between 2.5 m/s and 1 m/s points of the AR signal. |
|
|
A patient has mild mitral regurgitation and the time taken for mitral regurgitation
velocity to drop from 3 m/s velocity to 1 m/s on continuous wave Doppler examination was 40 ms. The average rate of LV pressure decay in this patient is: A. 3600 mmHg/s B. 1280 mm/s C. 800 mm/s D. 400 mm/s |
Answer: C.
The LV negative dp/dt = 36-4/0.04 = 800 mmHg/s. This again assumes a constant LA pressure during this portion of LV isovolumic contraction time. This noninvasive measure has been validated against invasively derived negative dp/dt by high-fidelity LV pressure recordings. |
|
|
By tissue velocity imaging, the mitral annular Sm wave is produced by:
A. Annular descent during systole B. Annular ascent during systole C. Atrial contraction D. LV relaxation |
Answer: A.
Annular descent produced by LV long axis shortening causes a positive systolic deflection recorded from the apical view. |
