Case 1

• • TR has a velocity of 4m/s. There is also a pulmonary valvular stenosis with a gradient of 2 m/s

    • What is the RV systolic pressure = 64+5 = 69 mmHg

    • What is the PA systolic pressure = 69 - 16 = 53 mmHg

  • If the Aortic pressure is 98 mmHg, what would be the directionality and  the expected peak velocity through the ductus (assumes peak velocity occurs at peak aortic pressure):

    • 98 - 53 = 45 mmHg difference. Bernouilli: Delta-P ~ 4V2.

    • Peak systolic velocity through the PDA would be: 3.35 m/s, left to right. This assumes that the peak velocity through the ductus represents the point of peak aortic pressure.

Case 2

• Blood pressure is 100/55 mmHg in Aorta

• VSD gradient restrictive is left to right and 45 mmHg of gradient.

• There is an Aortic Valvular Gradient because of aortic valve stenosis of 65 mmHg. 

• What is the RV systolic pressure. What is the PA systolic pressure.

  • LV systolic pressure is 100 + 65 mmHg = 165 mmHg

  • RV systolic pressure is 165 mmHg - 45 mmHg = 120 mmHg

  • The RV systolic pressure is assumed to be the same as the PA systolic pressure 120 mmHg unless there is an obstruction of the RVOT

  • Using VSD gradient assumes that peak velocity occurs at the time of peak pressure in RV/LV, although this may not be true as a Doppler gradient provides you with instantaneous velocity differences which may not be exactly at the time of the peak LV and/or RV pressure as these timings may not be synchronous to the moment in the cardiac cycle where the difference between RV and LV pressure are the highest (which is the moment where the VSD gradient velocity is the highest).

Case 3

• The Aortic blood pressure is 110/60

• The LVOT gradient through aortic stenosis is 60 mmHg

• What is the LV systolic pressure?

• If you assume a LA pressure of around 10 mmHg. What would be the MR velocity gradient?

• What is the LV systolic pressure. It is 110+60 = 170 mmHg

• Because the LV systolic pressure is 170 and the LA pressure is 10, the LV-LA gradient will be 160 mmHg. MR velocity gradient is 6.32 m/s as per Bernouilli. 

  • Again, these calculations assume that the Doppler gradient occurs at the time of peak aortic pressure (i.e. 110), when it may actually not occur exactly at that time point, and may happen during the ascending limb of the aortic pressure build-up. See aortic stenosis section and Understanding PDA Spectral Doppler.

Case 4

• A PDA is left to right with a gradient of 80 mmHg

• The Aortic pressure (systolic) is 100 mmHg

• The TRJ gives a RV systolic pressure of 130 mmHg of RV-RA gradient.

• What is the PA systolic pressure? (100 - 80 = 20 mmHg)

  • Assumes that the peak PDA gradient occurs approximately at the time of peak Aortic and Peak pulmonary systolic pressure. The sPAP and systolic systemic BP may not occur at the same time. More importantly, these peak pressures may not be at the moment where there is the biggest gradient between the Aortic and Pulmonary pressure curves. Learn more in the Understanding PDA Spectral Doppler section.

• What is the RV systolic pressure? (130 mmHg + 5 mmHg of RA pressure) = 135 mmHg

• What is the gradient via the RVOT? (135 - 20 mmHg = 115 mmHg)

Case 5

• • If Ao saturation=100%, Mixed venous saturation=70%, Pulmonary Venous saturation=100%, and Pulmonary arterial saturation=85% (indicating oxygen enrichment from the shunt), then Qp/Qs is ?

  • Answer: Qp/Qs is (100-70)/(100-85) = 30/15 = 2/1. This means the pulmonary circulation receives twice the systemic flow. 

Case 6

Large left to right VSD:

• Aortic pressure is 90/60 (mean 70)

• The LA pressure is 15 mmHg. RA pressure is 5 mmHg.

• mPAP is 30 mmHg

• There is no hepatomegaly (low IVC pressure).

• The LV is dilated on echocardiography (LV end-diastolic diameter with a Z-score 3.5) due to the high pulmonary blood flow by the left to right shunt. 

• Qp >> Qs because the LA and LV are dilated. 

• If Qp/Qs = 3, what is the pulmonary vascular resistance to systemic vascular resistance ratio?

• Answer:

  • The Aortic pressure is assumed to be the same as LV pressure (unless aortic stenosis). The PA pressure is isosystemic in systole because the VSD is large so pressure between RV and LV will equalize in systole (unless there is outflow tract obstruction). The PA pressure in systole is thus 90 mmHg as well. A large VSD often presents with a systolic pulmonary artery pressure equal to systemic pressure. A dilated left ventricle indicates significant pulmonary hyper-flow (Qp > Qs) and very low PVR.

  • Pulmonary Vascular Resistance: Rp = ΔP = (Mean PAP – Mean PCWP) / Qp

  • Systemic Vascular Resistance: Rs = ΔP = (Mean Ao Pressure – Mean RAP) / Qs 

  • Pulmonary/Systemic Resistance Ratio = Rp / Rs = (ΔPp × Qs) / (ΔPs × Qp) 

  • Rp / Rs = (ΔPp / ΔPs) × (Qs / Qp)

    • Here we know LA pressure is 15 and we know that Mean Aortic pressure is 70. We also know that the RA pressure is around 5 mmHg. We also mentioned that the Qs/Qp is 1/3

    • Rp/Rs = (mPAP-15)/(70-5) x 1/3. If mPAP is around 30 mmHg, we get Rp/Rs = (30-15)/65 x 1/3 = 0.076 ~ 0.08 ~ 1/12 of Rp/Rs

Case 7

Large VSD on echocardiography:

• • The patient has an aortic blood pressure of 110/60 (85 mean). By cath, the PA saturation is 70%. The pulmonary venous saturation is 100%. The Aortic saturation is 100%. The RA saturation is 70%. Mean PAP is 65 mmhg, Mean LA pressure is 10 mmHg, Mean Aortic pressure is 70 and Mean RA pressure is 5. Of note: Right atrial saturation should not be considered a true “mixed” venous value (such as for example in the presence of a PFO). However, for the purpose of this question, assume that the RA saturation represents mixed venous saturation.

  • What is the Rp/Rs?

  • What is the Qp/Qs?

    • 1. The systolic PA pressure will also be 110 mmHg because of the large VSD equalizing pressure in systole.

         1. Qp/Qs = (Aortic Sat - RA Sat)/(PA Sat - Pulm Vein sat) = (100 - 70) / (100 - 70) = 1 of Qp/Qs

         2. Rp / Rs = (ΔPp / ΔPs) × (Qs / Qp) = (Mean PAP – Mean LA pressure)/ (Mean Ao Pressure – Mean RA Pressure) x Qs/Qp = (65 - 10) / (85 - 5) x 1 = 55/80 x 1 = 0.69 (almost 2/3). 

         3. Of note, a prior version of the question listed a mean PAP of 45 mmHg, which is likely too low in the context of a VSD, as the diastolic PAP would be expected to be higher and unlikely to yield a mean of 45 mmHg. This resulted in an Rp/Rs of 0.44.

Case 8

Large right to left VSD:

• Is the Rp/Rs ratio less or more than 1?

• Ao saturation is 85%, SVC saturation is 55 (AV difference of 30%). PA saturation is 55%. There is no VQ mismatch. What is the Qp/Qs?

• The patient has an aortic blood pressure of 110/60 (85 mean). PA pressure is: 110/38 (mean 70). RA pressure is 5, LA pressure is 10.

• The Rp/Rs ratio will be greater than 1 by definition because the pulmonary vascular resistance is suprasystemic.

  • • 1. PA saturation is 55% because it is the same as the SVC saturation (no oxygen enrichement since the shunt is right to left). The pulmonary venous saturation is 100% if there is no V/Q mismatch at the pulmonary level. Qp/Qs = (Aortic Saturation - RA saturation)/(Pulm vein saturation - Pulm Artery saturation) = (85-55)/(100-55) , leading to a Qp/Qs ratio less than 1. Here 2/3.

      2. The patient has an aortic blood pressure of 110/60 (85 mean). The systolic PA pressure will also be 110 mmHg because of the large VSD equalizing pressure in systole.

      3. If the PA pressure is: 110/38 (mean 70). 

      4. We have already outlined shunt by the shunt direction that the Rp/Rs is > 1. 

         1. Rp / Rs = (ΔPp / ΔPs) × (Qs / Qp) = (Mean PAP – Mean LA pressure)/ (Mean Ao Pressure – Mean RA Pressure) x Qs/Qp = (70 - 10) / (85 - 5) x 3/2 = 60/80 x 3/2 = 1.13

Case 9

• Central venous line reads a pressure of 10 mmHg

• The PFO gradient is restrictive and estimating a peak gradient of 15 mmHg

• There is a prolonged retrograde flow in the pulmonary vein.

• What would be the peak LA pressure?

• What could explain this rise in LA pressure?

• Estimated peak LA pressure is 15 + 10 mmHg. The rise could be explained by restrictive cardiomyopathy, mitral stenosis, mitral insufficiency (severe), LV hypertrophy for various causes (aortic stenosis, coarctation, systemic hypertension, LV fibrosis post-ischemia). If there was a large ASD, it would decompress to the RA and the LA pressure would not necessarily rise. A patient with such increase in LA pressure is at risk of pulmonary edema due to post-capillary phenomenon. Peak LA pressure is usually during ventricular systole when the mitral valve is closed and the LA continues to fill. As such, the QRS and timing of the peak gradient also can pin-point to an etiology. Usually severe mitral insufficiency will increase LA pressure during ventricular systole. Severe mitral stenosis will lead to an increase in LA pressure throughout the cardiac cycle, and the left to right velocity gradient may become more pronounced when the RA pressure drops at the time of tricuspid valve opening and RA emptying into the RV. This occurs in ventricular diastole and usually before the p-wave.

Case 10

• At baseline, the patient is evaluated in room air with iNO at 10 ppm. 

  • RA pressure is 5 mmHg with a RA saturation of 69% (SVC saturation 72%, IVC satuation 67%). The saturation at the LPA is also 69%. 

  • Is there a left to right shunt at a VSD or PDA level? 

    • Since the RA sat and LPA sat are the same, this outlines there is no left to right shunt in the post-tricuspid area (VSD or PDA that would bring oxygenated blood in the PA). Mixed venous saturation is the same as PA saturation.

  • The RV pressures are 44/4 compared to LV pressures of 82/8. Is the RV pressure infra, supra, or isosystemic?

    • The RV pressure is half systemic. 

  • The MPA pressure is 45/17 (mean of 30). Is this pulmonary hypertension?

    • The mean PA ressure corresponds to pulmonary hypertension (≥20 mmHg). The RV systolic pressure (44) and MPA systolic pressure (45) are the same. There is no RVOT obstruction

  • The RPA and LPA pressure are similar (46/19; mean 30 for the RPA; 48/19; mean 30 for LPA). Is there obstruction of the RPA or LPA, could it be a sadle emboli? 

    • There is no obstruction at the level of the pulmonary arterial branches or MPA since all the pressure are the same. 

  • The wedge pressure on both side is similar (9 and 10 mmHg). Is there pulmonary venous hypertension? 

    • The wedge pressure less than 15 mmHg indicates that there is acceptable drainage pressure - no sign of post-capillary restriction. The LV end-diastolic pressure is similar at 8 mmHg. As such, no signs of PV stenosis, pulmonary venous occlusive disease (PVOD), mitral valvular disease or LV diastolic impairment (HFpEF). 

Case 11

See graphic for answers

• Blood pressure is 29/10 and saturation post-ductal is 65%

• Pre-ductal saturation is 85%

• There is minimal V/Q mismatch on the chest radiography with nice aeration of the lungs = this tells you that pulmonary venous saturation will be good. 

• On ECHO the TRJ gives RV-RA of 25 mmHg

• The PDA is large and strictly right to left

• There is retrograde flow in the Ascending Arch.

• The PFO is shunting strictly left to right. 

• What would be your management of this patient? What is the phenotype of the patient?

  • Phenotype of severe LV dysfunction with poor output. The Retrograde flow in the arch and L-R shunt at atrial level explains it. 

• What is the expected saturation in the left atrium?

• What would be the expected saturation in the LV?

• What would be the expected saturation in the RA?

• What would be the expected saturation in the RV? What is the expected blood presure in the RV?

• What is the expected sPAP and diastolic PAP?

• On echocardiography, there is 

Case 12

• Blood pressure is 29/10 and saturation post-ductal is 75%

• Pre-ductal saturation is 75%

• There is minimal V/Q mismatch on the chest radiography with nice aeration of the lungs. This tells you the Pulm Vein saturation will be good. 

• The PDA is large and strictly left to right. Means that the Aorta is filling the PA despite the low BP in the Aorta (29/10) - first clue.

• The PFO is shunting strictly right to left. The patient has significant hepatomegaly. This tells you the RA pressure is significantly high. Either because the RA is not able to empty in RV (i.e. Tricuspid stenosis, atresia, severe TR), and/o RV obstruction. Here there is RV hypocontractility (RV is almost dead). There is no other CHD as outlined, so it must be RV function pulmonary valvular atresia or stenosis with significant TR. 

• There is no congenital heart defect (outside of the inter-atrial shunt and PDA presence). 

• What would be your management of this patient? What is the phenotype of the patient?

  • PV stenosis - atresia.

  • Inotropy, keep the duct open, support the pulmonary vasculature with iNO and/or Oxygen to vasodilate to reduce afterload on RV that is trying to recover. 

• What is the expected saturation in the left atrium? see graphic. (mix from PFO and pulmonary venous flow = 75%, same as LV and Aortc saturation)

• What would be the expected saturation in the LV?

• What would be the expected saturation in the RA? Likely about 20-30% point lower than the Aortic saturation if there is good extraction in the periphery. 

• What would be the expected saturation in the RV? Same as the RA saturation. 

• What is the expected sPAP and diastolic PAP? Same as the aortic pressure as the PDA is large. 

Case 13

• If Qp/Qs is 3 with a arterio-venous difference (O2 consumption) of 30%, what will be the pulmonary artery saturation in:

  • Large ASD

  • Truncus Arteriosus

  • Sinus Venosus ASD

  • Ventricular Septal Defect

  • The answer is that the PA saturation is the same for all these scenarios: Qp/Qs = (Aortic Sat - Mixed Venous Sat) / (Pulm Vein Sat - Pulmonary Artery Sat) = 3

    • Aortic saturation will be 100 in large ASD, Sinus Venosus ASD, VSD and there is an assumed 30% oxygen extraction. The pulm vein sat will be the same as Aortic Sat since no VQ mismatch. AV diff = 30 = (Ao sat 100- RA sat 70). The Pulm Vein sat = 100. So Qp/Qs = 3 = (100-70)/(100- X). X has to be 90. 

    • For Truncus arteriosus: the Aorta and PA saturations are the same. The AV difference is assumed to be 30 again. The Pulm vein sat assume at 100. 

      • 3 = (X - (X-30))/(100 - X) = 30/(100-X) = X must be 90 (which is the same as the saturation in the Aorta. 

Case 14

• A patient has pulmonary atresia with a large VSD and a large PDA. The aortic saturation is 85%. The patient has good cardiac function and is well perfused. You assume the AV-difference of 30%. There is no signs of V-Q mismatch or parenchymal lung disease. What is the Qp/Qs in this patient?

• Aortic saturation is 85%, Pulmonary Artery saturation is the same because fed by the left to right PDA (85%)

• Assuming a normal systemic A-V difference (oxygen consumption in the systemic compartment) of 30%, then vena cava saturations will be approximately 55% (85-30). 

• Pulmonary vein saturation is 100% if no V/Q mismatch 

• Then Qp/Qs = (Aortic Sat - Mixed Venous Sat) / (Pulm Vein Sat - Pulmonary Artery Sat) = (85-55)/(100-85) = 30/15 = 2/1. 

• Clinical Paradox: Patients can be cyanotic (blue) yet present with symptoms of overcirculation (e.g., dyspnea) due to elevated pulmonary blood flow by the PDA and high filling pressures. This emphasizes that cyanosis does not directly correlate with low pulmonary blood flow in all congenital heart diseases.

• Morphological Consequences: High pulmonary flow can lead to dilated left atrium and left ventricle. The management of oxygen administration in these children is complex since it can further vasodilate the pulmonary vasculature, and requires careful consideration once the diagnosis is established

Case 15

• You are called at bedside for a TnECHO. There is a large non-restrictive left to right PDA bigger than the size of the LPA. The Aortic Pressure is 85/65. 

• If you would do a cath, what would likely be the blood pressure in the pulmonary artery. Same as aortic pressure 85/65 (no restriction)

• If you obtain a saturation in the LPA, what would it be if the Qp:Qs is 4/1, knowing that the saturation in the Aorta is 100%. 

  • 4 = (100 - 70)/(100 - X) = 30/(100-X). 100-X = 30/4 = 7.5. X =100-7.5 = 92.5%

• What will be the septal curve on TnECHO? Flat because by definition the PDA creates iso-systemic PA pressure. 

Case 16

• A patient has a large VSD that shunts left to right at low-velocity.

• The Aortic pressure is 85/65. What would be the systolic RV pressure? Same as the Aortic systolic pressure unless there is Aortic valvular stenosis. It would be 85 mmHg.

• If there is a pulmonary stenosis with a gradient of 20 mmHg at peak systole, what would be the systolic PA pressure. RVSP - 20 = 85-20 = 65 mmHg

• What would be the septal curve position at peak of systole? Flat because the large VSD will equalize pressure accross septum. 

Case 17

• A patient has a large unrestrictive VSD that shunts right to left at low-velocity.

• The Aortic pressure is 85/65. What would be the systolic RV pressure? Same as before = equalization of pressure in 2 chambers that are connected by unrestrictive channel, gives 85 mmHg. 

• If there is a pulmonary stenosis with a gradient of 75 mmHg at peak systole, what would be the systolic PA pressure. Here the sPAP is barely 10 mmHg, probably because it is a severe PS with almost no output. 

• What would be the septal curve position at peak of systole? It would still be flat because of equal pressure on both side of the VSD. 

• If there is no V/Q mismatch and the Qp:Qs is 1/2, what would be the saturation in the Pulmonary Artery, Right Atrium and Left Atrium. Aortic saturation is 75%.

  • 0.5 = (75 - 45)/(100-X) = 30/(100-X). 60 = 100-X. X = 40%.

Case 18

• You are seeing a patient with TnECHO, the septal curvature is flat. 

• There is a large PDA that is left to right. What is higher, PVR or SVR? What is higher sPAP or sBP? Why is the septal curve flat and not bowing to RV? If PDA was very small, left to right, and restrictive, what would be the septal position?

  • PVR is lower than SVR

  • sPAP = sBP because PDA equalize pressure - trick question. 

  • Flat because the RVsp = LVsp; that is if there is no pulm stenosis or aortic stenosis. 

  • If PDA small and restrictive, septal position would be bowing towards RV. 

• There is a large PDA that is right to left. What is higher, PVR or SVR? What is higher sPAP or sBP? Why is the septal curve flat and not bowing to LV? If PDA was very small, right to left, and restrictive, what would be the septal position?

  • SVR higher than PVR.

  • sPAP = sBP because equalization. 

  • Flat because the pressure peak in LV = RV systolic pressure. 

  • It would be bowing towards the LV in systole (pancaking LV). 

Case 19

• You are assessing a patient by TNE

• The mitral insufficiency jet gives a LV-LA gradient of 60 mmHg

• The LA-RA peak gradient by PW-Doppler is 7 mmHg. Here we assume the peak gradient occurs at peak ventricular systole (atrio-ventricular valve are closed and atriums are still getting filled by venous return).

• The gradient via the LVOT is 14 mmHg

• What is the Aortic systolic blood pressure, if the RA is 5 mmHg

  • If the LA-RA gradient is 7 and the RA pressure is 5, the LA pressure is 5+7 = 12 mmHg

  • If the LV-LA gradient is 60 mmHg and the LA pressure is 12, the LV systolic pressure is 60+12 = 72 mmHg

  • If the LV-Ao gradient is 14 mmHg, and the LV systolic pressure is 72, the sBP in the Aorta is 72-14 = 58 mmHg. 

Case 20

• You are assessing a patient with Tricuspid Atresia, normally related great vessels and a VSD. There is no PDA.

• The aortic saturation is 75%. The pulmonary venous saturation is 100%.

• Explained how this patient could have a Qp/Qs of 0.5:1, a Qp/Qs of 1:1 or a Qp/Qs of 2:1?

  • Qp:Qs: 0.5 = (Aortic Sat - RA sat)/(Pulm venous sat - PA sat). 

    • By definition the PA sat is the same as the Aortic saturation.

    • 0.5 = (75 - X)/(100 - 75) 

    • X = 62.5 % in the Right Atrium (Mixed venous saturation). 

    • The AV difference is of 75 - 62.5 = 12.5% (much less than the usual 30%)

    • This patient has less pulmonary blood flow and would have merited possibly a source of pulmonary blood flow such as a PDA or BTT-Shunt. Otherwise the pulmonary vasculature will not grow appropriately due to insufficiency in the pulmonary blood flow. 

    • The is too much VSD restriction and/or RVOT obstruction leading to imbalance between pulmonary and systemic blood flow, with diminished PBF.

  • Qp:Qs: 1:1

    • 1 = (75 - X)/(100 - 75) 

    • X = 50. The AV-Difference is of 25% which is appropriate. 

    • The is some degree of VSD restriction and/or RVOT obstruction allowing for a balance between pulmonary and systemic blood flow.

  • Qp:Qs: 2:1

    • 2 =  (75 - X)/(100 - 75) 

    • X = 25%. This tells you there is a lot less systemic blood flow at the expense of pulmonary blood flow via the flow going through the VSD into the PA. As such, the systemic vasculature gets less flow and there is more oxygen extraction to meet metabolic demand. This is a situation that is high risk because this patient can have systemic hypoperfusion and too much flow in the pulmonary vasculature which may injure the endothelium and induce remodelling by shear stress. These patients will need pristine pulmonary vasculature to achieve an eventual univentricular palliation based on passive pulmonary blood flow from the caval veins to the pulmonary arteries. 

Case 21

• You are doing a TnECHO on a preterm infant at 36 weeks. The PDA is left to right and unristrictive, pulsatile in pattern. There is a low velocity gradient accross the duct that is completely left to right. There is retrograde holodiastolic flow in the descending aorta, celiac, SMA, ACA and MCA. The LA and the LV are dilated. There is mitral insufficiency. The Systolic BP at the time of the ECHO is 76 mmHg, diastolic is 50 mmHg. There is no RVOT or LVOT obstruction. There is no VSD. The PFO is stretched, restrictive and left to right with a peak gradient of about 10 mmHg. 

• The septum is flat. The TRJ gives a RV-RA of 70 mmHg

• Is the PA pressure infra-systemic, iso-systemic or supra-systemic? 

  • The PA pressure is iso-systemic. There is pressure equalization accross the ductus with pressure transmission in systole and diastole. 

• What is the expected diastolic BP in the pulmonary artery.

  • Should be about the systemic diastolic BP (50 mmHg). 

• If you put CW-Doppler accross the MR, what would be your gradient by Bernouilli?

  • It would be about 60 mmHg. The sBP is 76. The LA-RA gradient is 10. We assume a 5 mmHg of RA pressure, giving about 15 mmHg in the LA. If Aortic pressure is 76, the LV systolic pressure is 76. The Gradient is 76-15 = 61 mmHg. 

Case 22

A newborn presents with refractory cyanosis and oxygen saturations around 90% (regardless of oxygen supplementation) in both pre- and post-ductal sites. Chest radiography shows cardiomegaly with oligemic (“black”) lung fields. What is your differential diagnosis?

• Key features to interpret: 

  • Refractory cyanosis: Poor response to oxygen → suggests structural heart disease with reduced pulmonary blood flow or admixture lesion. 

  • Equal pre- and post-ductal saturations: Excludes differential shunting at the ductus → not typical of persistent pulmonary hypertension of the newborn (PPHN). 

  • Cardiomegaly + oligemic lungs: Indicates decreased pulmonary blood flow with enlarged cardiac silhouette, pointing toward a right-sided obstructive lesion or severe tricuspid valve disease.

• Differential Diagnosis

  • Ebstein Anomaly

    • Pathophysiology: Apical displacement of the tricuspid valve → severe tricuspid regurgitation, massive right atrial enlargement, and functional pulmonary atresia.

    • Imaging clues: Marked cardiomegaly (“wall-to-wall” heart) with decreased pulmonary vascularity (“black lungs”).

    • Oxygen response: Poor; pre- and post-ductal saturations similar.

  • Pulmonary Atresia with Intact Ventricular Septum (PA/IVS)

    • Pathophysiology: No flow to pulmonary artery; lungs perfused via ductus arteriosus.

    • CXR: Normal to enlarged heart, oligemic lungs. The left heart may be dilated due to increased Qp:Qs by the PDA.

    • Clinical: Duct-dependent pulmonary circulation; severe cyanosis early after birth.

  • Critical Pulmonary Stenosis (Severe)

    • Pathophysiology: Severe obstruction to RV outflow → markedly reduced pulmonary flow, ductal dependence.

    • CXR: Cardiomegaly (RV hypertrophy) with dark lungs.

    • O₂: Poor improvement despite 100% FiO₂.

  • Tricuspid Atresia (especially with normally related great arteries)

    • Pathophysiology: No direct RA-RV connection; flow limited to pulmonary artery via VSD or ductus.

    • CXR: Variable cardiomegaly; decreased pulmonary vascular markings if restrictive VSD.

    • O₂: Refractory cyanosis, equal sats.

  • Tetralogy of Fallot with Severe RVOT Obstruction

    • Pathophysiology: Severe RVOT or pulmonary valve obstruction → decreased pulmonary flow.

    • CXR: Boot-shaped heart occasionally, often normal size or mildly enlarged; oligemic lungs.

    • O₂: Minimal improvement.

  • Less Likely (but Consider)

    • Transposition of the Great Arteries (TGA) with restrictive ASD/VSD —usually does not have a large pericardial silhouette. 

    • Total Anomalous Pulmonary Venous Return (TAPVR) with obstruction — typically pulmonary congestion (“white lungs”), not black lungs. 

    • Persistent Pulmonary Hypertension (PPHN) — usually shows normal or small heart size and differential saturations (pre-ductal > post-ductal).

If it is a PAIVS - What is the Qp/Qs?

• Qp/Qs = (Sat Aorta - Sat RA)/(Sat LA - Sat Pulm Artery) = (90 - 60)/(100 - 90) = 30 / 10 = 3:1 - PA saturation will be the same than Aortic saturation because you are providing flow via the ductus arteriosus. 

• In PA/IVS, the right ventricle (RV) is not connected to the pulmonary artery (no ejection). The pulmonary blood flow (Qp) comes retrogradely through the ductus arteriosus from the aorta. When the ductus is large and systemic pressure is high, pulmonary flow can greatly exceed systemic flow — hence Qp:Qs = 3:1. This produces pulmonary overcirculation → increased venous return to the left atrium and volume overload of the left heart. The LV is hyperdynamic and dilated, handling ~4× normal stroke volume.

• Pressure Volume Loop: 

  • X-axis: Volume (preload) Y-axis: Pressure

  • For the LV: large stroke volume (SV) due to massive pulmonary venous return. The curve shows a shift to the right (higher EDV → volume overload) and a higher upstroke (enhanced contractility due to Starling effect).

  • For the RV: No true ejection loop. The RV has no ejection — it is essentially isolated from the pulmonary artery due to atresia. Its curve ends prematurely: there is filling (diastolic portion) but no pressure rise leading to ejection.

  • Conclusion: In PA/IVS with high ductal shunt (Qp:Qs = 3:1), the right ventricle does not eject, so its PV curve cannot be completed. The left ventricle, overloaded by excessive pulmonary venous return, exhibits a Starling-type curve shifted to the right, generating a stroke volume four times normal — explaining the cardiomegaly and pulmonary overcirculation.

PGE are weaned off and the saturation drops to 70%. How come? What is the Qp:Qs? Which treatment can improve the saturation?

• As prostaglandin is gradually withdrawn and the PDA closes, systemic oxygen saturation may fall from 85% to 70%. This desaturation reflects the true severity of the raised RV compliance and the limited antegrade pulmonary flow.

• Pulmonary blood flow becomes low (Qp ↓) because the RV cannot sustain forward ejection. 

• Systemic venous desaturation (SaO₂ ~70%) mirrors the residual right-to-left atrial shunt. The calculated Qp:Qs ratio approaches 0.5:1, consistent with a hypoperfused pulmonary circuit. Qp:Qs = (70%-40%)/(100%-40%) = 30 / 60 = 1/2

• Management focuses on improving RV relaxation with beta-blockers and/or providing more adequate pulmonary blood flow — PGE, modified Blalock–Taussig-Thomas shunt (1.5 ventricular repair), PDA stenting or RV outflow re-intervention. This depends on context. 

At 6 months, the saturation is 85%. What is the Qp:Qs? What is the pressure-volume loop?

• Qp/Qs = (85 - 55)/(100 - 55) = 30/45 = 2/3

• PV loop: The compliance of the RV got better but is still less than the LV, hence why there is a fraction of the venous return going to the RA and then to the LA via the inter-atrial shunt. 

If you do a partial cavo-pulmonary shunt (BTT shunt) from SVC to the PA, what would be the expected estimated saturation after the surgery? Assuming that the SVC and IVC contribution to the overall venous return is equivalent (50% - 50%). 

• Qp/Qs was 2/3 (or 4/6). This means that 1/3 is going from the RA to the LA (2/6). From this 2/6, there is 1/6 from the IVC return and 1/6 from the SVC return. This 1/6 shunting right to left at the inter-atrial level from this IVC return will remain stable after the surgery.

• 1/2 of Qs is coming from SVC and going into the pulmonary artery (3/6). 

• After the surgery, IVC drains 1/2 (or 3/6) to the RA. But there is still a portion (1/6) shunting from the RA to the LA by the inter-atrial shunt due to poor RV compliance. This means that the RA contributes to 3/6-1/6=2/6 to the RV and eventually PA. The PA is fed by 3/6 from SVC and 2/6 from IVC = 5/6

• New Qp/Qs = 5/6 = (Aortic Sat - RA sat)/(100 - PA sat). In this setup, the oxygen systemic consumption will still be 30% and the PA sat is the same as the RA sat; as such 5/6 = 30 / (100 - PA sat). 6/5 = (100 - PA sat)/30. 36 = 100 - PA sat. PA Sat = 64. As such, Aortic sat 64+30 = 94%

Case 23

3 months old baby with a VSD. BP 90/50 mmHg and you are told the Qp/Qs is 3. The VSD is described as somewhat restrictive. The cardiologist is quite reassured because the cardiology team estimates the PA pressure is less than 1/3 systemic. They will follow only in 1 month. What would be the saturation in the Pulmonary Artery? What is the Max Velocity via the VSD? How is the Pressure-Volume loop?

• If Qp​:Qs=3, this implies the Left Ventricle must generate 3 stroke volume to sustain 1 systemic stroke volume.

• The systemic pressure in systole is 90 mmHg. If sPAP is <30 mmHg (safe threshold). The required pressure gradient across the VSD is ΔP>90−30=60 mmHg. Using the modified Bernoulli equation (ΔP=4V2 ), it means the velocity is V≈3.87 m/s.

• The saturation in the PA will be 3 = 30/(100 - X). 10 = 100 - X. X = 90%.

For a left-to-right shunt producing Qp:Qs = 3:1, describe and compare the total left-ventricular output in a ventricular septal defect (VSD) and a patent ductus arteriosus (PDA). Include in your answer which chambers handle the excess volume and why the LV ejection differs (generated LV stroke volume differs from the VSD vs the PDA)?

A ratio of Qp:Qs = 3:1 means that the lungs receive three times as much flow as the systemic circulation. That is: Pulmonary blood flow (Qp) = 3 parts Systemic blood flow (Qs) = 1 part So, there’s a large left-to-right shunt, meaning blood that should go to the body recirculates through the lungs. But the location of that shunt (pre- or post-ductal, intra- vs extra-cardiac) determines which chambers handle the excess volume.

Qp:Qs = 3:1 via a VSD (intra-cardiac shunt)

Pathway:

• LV → VSD → RV → Pulmonary artery → Lungs → LA → LV

• In a VSD, blood flows from the LV to the RV during systole, entering the pulmonary circuit and returning again to the left atrium and ventricle. Thus, the LV handles both the systemic output (Qs) and the recirculated shunt flow (Qshunt).

Implications:

• In a ventricular septal defect (VSD), the left ventricle ejects both the systemic flow (Qs) and the shunt flow (Qshunt) through the interventricular communication, thereby handling the entire combined output. The total left ventricular output equals Qs + Qshunt. For a shunt ratio of Qp:Qs = 3, this implies that Qs = 1 and Qp = 3, so the shunt flow is Qshunt = Qp − Qs = 2. Consequently, the left ventricle ejects 1 (systemic) + 2 (shunt) = 3 units of flow, or three times the normal systemic output. This excessive recirculation of blood through the pulmonary circuit returns to the left atrium, leading to volume overload and dilatation of the left atrium and left ventricle. 

  1. Qp:Qs=3:1

  2. Qp=3,Qs=1

  3. Therefore, Qshunt=Qp−Qs=2

  4. LV total output = Qs + Qshunt = 1 + 2 = 3

  5. The LV ejects 3 times the systemic flow, because part of its output re-enters the pulmonary circuit. → Causes volume overload of LA and LV

Qp:Qs = 3:1 via a PDA (extra-cardiac shunt)

Pathway:

• Aorta → PDA → Pulmonary artery → Lungs → LA → LV → Aorta

Implications:

• The LV still receives all the pulmonary venous return, but unlike the VSD case, the shunted volume does not recirculate through the RV; it comes directly from the aorta.

• Some of the LV’s work is lost as ductal runoff 

• In a patent ductus arteriosus (PDA), the left ventricle ejects its entire output into the aorta, and the shunt occurs distal to the aortic valve, from the aorta to the pulmonary artery. Thus, the left ventricle must sustain the total combined flow of systemic output (Qs) plus pulmonary shunt flow (Qp). For a Qp:Qs = 3, this means the left ventricle ejects Qs + Qp = 1 + 3 = 4 units of flow, or approximately four times the normal workload. The regurgitant return from the pulmonary circulation similarly causes marked volume overload and dilatation of the left atrium and ventricle.

  1. In PDA, LV ejects all its blood into aorta. The shunt to the pulmonary artery occurs after the aortic valve (i.e., extracardiac). Therefore, the entire LV stroke volume must include: The systemic flow (Qs), and The additional pulmonary recirculated flow (Qp) that leaks back from the aorta to the PA and returns via pulmonary veins to the LV.

  2. Qp:Qs=3:1 

  3. Because the LV must generate the entire flow that passes into the aorta (systemic + pulmonary): LV output = Qp + Qs = 3 + 1 = 4

  4. The LV ejects four times the normal systemic volume, since it must supply both circuits before the shunt point. → Produces severe volume overload of LA and LV, plus diastolic runoff and systemic steal

• “In VSD, the LV must maintain both systemic and shunt flow across the defect; in PDA, the LV output equals the sum of systemic and ductal flows since the shunt is distal to the valve.” (Congenital Diseases of the Heart, Ch. 6–7)

• Lai et al., Echocardiography in Pediatric and Congenital Heart Disease (2016): “The degree of left-sided volume overload depends on the ratio of pulmonary to systemic flow. For the same Qp:Qs, the left ventricular volume load is greater in PDA than in VSD because the shunt is distal to the aortic valve.” 

• Noori & Acherman, Practical Neonatal Echocardiography (2020): “In large ductal shunts, LV output can be up to four times normal systemic flow, leading to progressive left heart dilation and systemic hypoperfusion.”

• Summary: In a VSD (intra-cardiac shunt), the flow returning from the lungs is Q P ​ (3 units), which must be ejected along with the systemic demand (Q S ​ =1). • In a case where the PDA carries the hyper-pulmonary flow (as seen in the ductus-dependent PAIVS example), the total volume ejected by the Left Ventricle must accommodate both the systemic demand and the large shunt flow diverted to the lungs. The source specifies that with Q P ​ :Q S ​ =3:1 via a ductus shunt, the Left Ventricle volume ejection is four times that of a normal ejection volume (1 volume for Aorta, 3 volumes for the canal, total 4 volumes returning to the Left Ventricle).

Case 24

Baby with a Truncus Arteriosus and is saturating at 90%. The Blood pressure is 120/40 (mean 80). What is the Qp/Qs? What is the PVR/SVR ratio? After surgery, if PVR and SVR are unchanged, what is the mPAP if the systemic BP is 125/60 - mean 80?

• Qp/Qs = (90 - 60)/(100-90) = 3:1. The aortic saturation is the same as the pulmonary arterial saturation. 

• ΔP=R×Q. mPAP = mBP because the pulmonary arteries are coming out of the aorta (unless there is stenosis at the branches).

  • (mPAP-LAp)=PVR×Qp

    • 80 - 5 = 75 = PVR x Qp

  • (mBP-RAp)=SVR×Qs

    • 80 - 5 = 75 = SVR x Qs

• Qp/Qs = 3:1 = (75/PVR)/(75/SVR)

• Qp/Qs = 3:1 = SVR/PVR. Hence, PVR/SVR = 1/3.

After surgery:

• PVR/SVR is still 1/3 but now the Qp:Qs is 1:1. 

• (mBP-RAp)=SVR×Qs; 

  • SVR = (mBP-RAp)/Qs = (80 - 5)/Qs = 75/Qs

• (mPAP-LAp)=PVR×Qp

  • (mPAP - 5)/Qp = PVR. Because Qp = Qs: (mPAP - 5)/Qs = PVR

• PVR/SVR = 1/3 = ((mPAP - 5)/Qs)/(75/Qs) = (mPAP - 5)/75. mPAP = 30

Answer:

• Qp/Qs=(95−65)​/(98-78)=30/20​=1.5. There must be a left to right shunt somewhere. Mild left-to-right shunt, but not large enough to explain pulmonary hypertension (PH).

• Mean PA pressure (mPAP): 40 mmHg. Definition of PH: mPAP ≥ 20 mmHg at rest (2022 ESC/ERS guideline). → This patient clearly has pulmonary hypertension. We can also check RV pressure (80/5) ≈ equal to systemic systolic (90 mmHg), confirming systemic-level pulmonary hypertension. This is consistent with the TRJ which gives a RV systolic pressure of 81+5, which is about the value we got by cath (80 of sPAP). Now, we must determine whether it is pre-capillary, post-capillary, or combined.

• PCWP = 21 mmHg (elevated). LVEDP = 5 mmHg (normal). 

  • Values 15 or more are considered consistent with pulmonary venous hypertension as per guidelines. In neonatal life, values 10 or more are sometimes quoted although literature is not as consistent. This is post-capillary PH.

  • Increased PCWP can occur in: 

    • Pulmonary venous hypertension due to pulmonary venous obstruction (e.g., pulmonary veno-occlusive disease, pulmonary vein stenosis, Pulmonary capillary hemangiomatosis), 

    • Technical overestimation of PCWP

    • Mitral disease (example: mitral valve stenosis or mitral insufficiency.

  • PCWP is higher than LVEDP, which is discordant. 

    • This post-capillary PH is not due to LV diastolic dysfunction.

    • There’s a clear step-up from IVC (65%) to PA (78%) — that is, an oxygen saturation increase of 13% between IVC and PA. The left to right shunt must be either at the atrial or ventricular level. 

    • In the presence of a left-to-right atrial shunt, the wedge pressure can be elevated because the pulmonary veins drain into a high-pressure left atrium that is decompressing into the right atrium via a restrictive inter-atrial shunt (hence the RA-LA pressure difference).

    • Post-capillary PH with elevated wedge not due to LV disease - likely a component of Group 2 PH and might be some mitral valve disease.

• PVR/SVR ratio = 

  • (mPAP−PCWP)​ = PVR x Qp

  • (mAo−RA)​ = SVR x Qs

  • Qp/Qs = 1.5 = ((mPAP−PCWP)/PVR)/((mAo−RA)​/SVR) = ((40-21)/PVR)/((60-4)/SVR) = (19/PVR)/(56/SVR) 

    • 19/56 x SVR/PVR = 1.5

    • SVR/PVR = 4.42

    • PVR/SVR = 0.23

    • Normal < 0.25 in children; >0.5 indicates severe pulmonary vascular disease. → This is borderline-mild elevation in PVR relative to SVR.

      • Strictly normal: Less than 0.3.  Borderline: A range of 0.3-0.5 can be considered borderline.  General good outcome: A ratio of less than 0.75 may be acceptable in some cases, such as for patients with pulmonary hypertension undergoing certain procedures.

• PVR= (mPAP−PCWP) / Qp​ = (40−21) ​/ 1 =19 Wood units 

  • Qp (calculated using Fick method) ≈ 0.98 L/min

    • VO₂=182×0.22=40.04 mL/min

    • O₂ content=1.36×Hemoglobin×O₂ saturation

    • Pulmonary vein (arterial): O₂pv=1.36×150×0.98=199.92 mL O₂/L 

    • Pulmonary artery (venous): O₂pa​=1.36×150×0.78=159.12 mL O₂/L 

    • Fick principle: Qp = VO2 / (O2pv - O2pa) = 40.04 / (199.92 - 159.12) = 40.04/40.8 = 0.98 L/min

  • PVR = 19 WU; BSA = 0.22 m²; PVR index = 19 ×0.22 ≈ 4.18 Wood units x m²

• SVR= (mAo−RA) / Qs ​ = (60−4) / 0.66 ​ = 84.8 Wood units

• Severely elevated PVR. Normal PVR in infants is <2–3 WU. This confirms pulmonary vascular disease. Patient has a mixed pre- and post-capillary pulmonary hypertension.

• For Qp:Qs - If we use the detailed Fick Formula: Q=VO2​​/ΔO2, where ΔO2=(Ca−Cv)×10

  • Cardiac output (CO) is the blood volume pumped by the heart per minute (measured in L/min). 

  • Tissue oxygen consumption (VO2) is the rate at which oxygen is consumed by the body tissues per minute (measured in mL/min). 

  • Arterial oxygen content (CaO2) is the oxygen content in arterial blood (measured in mL/L). 

  • Venous oxygen content (CvO2) is the oxygen content in venous blood (measured in mL/L).

  • VO2 (mL/min) = ____ mL/min/m2 × body surface area (BSA; measured in m2). ____ can be found in specific tables, here it is 182. 

    • Here it is 182 x 0.22 = 40.04

  • In the systemic circulation - CaO2: Arterial oxygen content can be determined by measuring the peripheral arterial blood gas oxygen saturation (or Aorta) and multiplying it by the hemoglobin oxygen-carrying capacity. CvO2: Venous oxygen content (RA or IVC saturation) can be determined by measuring the oxygen saturation from a central venous catheter and multiplying it by the hemoglobin oxygen-carrying capacity.

    • Assume Hb = 15 g/dL → 1.34 × 15 = 20.1 mL O₂/dL at 100%.

    • ΔO2 (sys)=(SaO2 ​ −SvO2 ​ )×1.34 x Hgb = (95−65)%×20.1/100=6.03 mL/dL=60.3 mL/L. 

    • Qs=VO2/ΔO2​(sys) = 40.04/60.3 ​= 0.66 L/min

  • In the pulmonary circulation CaO2 = pulmonary venous saturation; CvO2 = PA saturation. 

    • (Ca - Cv)pulmonary = (SpvO₂ - SpaO₂) × 1.34 × Hb. Assume Hb = 15 g/dL → 1.34 × 15 = 20.1 mL O₂/dL at 100%.

    • ΔO2​(pulm)=(98−78)%×20.1/100=4.02 mL O₂/dL

    • Convert to mL O₂/L → 4.02 × 10 = 40.2 mL/L. 

    • Qp=VO2/ΔO2​(pulm)​​=182×0.22/40.2​=40.04/40.2​=0.996 L/min - Qp ≈ 1.0 L/min

  • Confirms Qp/Qs = 1.5 = 1 / 0.66 

Case 26

In aortic stenosis or pulmonary stenosis, what is the instantaneous pressure gradient? What is the peak-to-peak pressure gradient? What do you measure by Doppler echocardiography? What do you measure by cath? Which one tends to be higher?

This schematic compares right ventricular (RV) and pulmonary artery (PA) pressure tracings in pulmonary valve stenosis. The X-axis represents time, and the Y-axis shows pressure in mmHg. The RV pressure curve peaks earlier and higher than the PA pressure curve, illustrating the obstruction across the pulmonary valve. A denotes the instantaneous pressure gradient, the maximal moment-to-moment difference between RV and PA pressures during systole. This is what Doppler echocardiography measures (via the Bernoulli equation from jet velocity). B denotes the peak-to-peak gradient, the difference between peak RV pressure and peak PA pressure, even though these peaks occur at different times. This is what is typically reported in cardiac catheterization. Since A measures simultaneous points and B compares two peaks at different times, the instantaneous gradient (A) is usually about 20% greater than B. In summary: Doppler (A) → instantaneous, physiologic gradient, higher value. Catheter (B) → peak-to-peak gradient, lower, slightly underestimates the true instantaneous obstruction.

After normal extra-uterine transition is complete:

• In neonates, indexed values (PVRI & SVRI) are preferred. Absolute Wood Units appear numerically higher due to very small cardiac output.

• Pulmonary Vascular Resistance (PVRI – preferred in neonates)

• PVRI = (mPAP − LAP) / CI × 80

  • Normal PVRI after 48–72 hours: < 3 Wood Units·m²

  • At birth / early transition: higher — near-systemic or supra-systemic

  • Mature neonatal state: ~1–2 Wood Units·m²

  • Adult reference (non-indexed PVR): 100–250 dyn·s/cm⁵ (≈ 1–3 Wood Units)

• Systemic Vascular Resistance (SVRI – preferred in neonates)

  • SVRI = (MAP − RAP) / CI × 80

  • Normal SVRI (term neonate post-transition): ~7–12 Wood Units·m² (≈ 600–1000 dyn·s/cm⁵·m²)

  • Adult reference (non-indexed SVR): 900–1400 dyn·s/cm⁵ (≈ 11–18 Wood Units)

• Calculate the PVRi and the SVRi for this infant

  • PVRI = (mPAP − LAP) / CI = (15 - 10) / 3.5 = 1.43 Wood Units·m²

  • SVRI = (MAP − RAP) / CI = (30 - 5) / 3.5 = 7.14 Wood Units·m²

• What is the Coronary Perfusion Pressure (CPP): 

  • The formula for CPP is: CPP=Diastolic Blood Pressure−Left Ventricular End Diastolic Pressure (LV EDP) 

  • For a normal infant with pressures of 60/25 (aorta) and an LV diastolic pressure of 10, the CPP is calculated as 25 minus 10, yielding 15 mmHg. A myocardial perfusion pressure of at least 15 mmHg is generally desired, with 20–25 mmHg being preferred to prevent ischemia and rhythm issues.

• In systole - what is the CPP for the left ventricle:

  • 60 (Aortic systolic pressure) - 60 (LV systolic pressure) = 0. Indicating no forward flow in systole of the LV. 

• In systole, what is the CPP of the right ventricle

  • 60 (Aortic systolic pressure) - 25 (RV systolic pressure) = 35 mmHg. Indicating there is still some forward flow insystole for the RV. 

  • Systolic Perfusion: Net perfusion into the left ventricular myocardium during systole is minimal because the aortic pressure (60 mmHg) is counteracted by the systolic pressure inside the LV (60 mmHg). The right ventricle, however, still receives some net perfusion during systole because its systolic pressure (25 mmHg) is much lower than the aortic pressure (60 mmHg).

Case 28

• 25-week infant, weight 0.85 kg, HR 165 

• LVOT diameter (inner-edge to inner-edge): 0.55 cm 

• LVOT VTI: 10.5 cm 

• Echo shows a large PDA with unrestrictive left-to-right shunting 

• Pre-ductal SpO₂ 94%

• Hemoglobin 150 g/L 

• Questions: 

  • Calculate LVO (mL/kg/min) using LVOT diameter, VTI, and HR. 

  • If Qp/Qs is estimated clinically at 2.5:1 (large L→R PDA), estimate the systemic blood flow (Qs) in mL/kg/min. 

  • Using Hb and SpO₂, estimate arterial oxygen content (CaO₂) (assume PaO₂ contribution negligible for the calculation) and calculate systemic oxygen delivery (DO₂) in mL O₂/kg/min using your estimated Qs.

Answers for Case 28:

1) LVOT cross-sectional area (CSA)

• Radius r = D/2 = 0.55/2 = 0.275 cm

• CSA = πr² = π × (0.275)²
  (0.275)² = 0.075625
  CSA ≈ 3.1416 × 0.075625 = 0.238 cm²
  
  

2) Stroke volume (SV)

• SV = CSA × VTI = 0.238 × 10.5 = 2.50 mL/beat (since cm³ = mL)
  
  

3) Cardiac output (CO, i.e., LVO in mL/min)

• CO = SV × HR = 2.50 × 165 = 412 mL/min
  
  

4) LVO indexed (mL/kg/min)

• LVO = 412 / 0.85 = 485 mL/kg/min (≈ 484–485 depending on rounding)
  
  

5) Estimate systemic flow (Qs) using Qp/Qs
In a large unrestrictive L→R PDA, LV output approximates Qp (systemic flow + ductal runoff to PA).

• Qs = Qp / (Qp/Qs) = LVO / 2.5 = 485 / 2.5 = 194 mL/kg/min

• The LVO represents the total volume leaving the left ventricle. However, a significant portion of this blood "shunts" back into the lungs via the PDA. In neonatal echo, LVO is often used as a surrogate for pulmonary flow) when a large PDA is present.

• The reason we say LVO = Qp (and RVO = Qs) is based on where the blood returns from. In a steady state, the amount of blood a ventricle pumps out must equal the amount of blood it receives. Pulmonary Flow (Qp): This is the total volume of blood that passes through the lungs. Every drop of blood that passes through the lungs eventually drains into the pulmonary veins, enters the left atrium, and fills the left ventricle. Therefore, the Left Ventricular Output (LVO) is exactly equal to flow going through the mitral valve (unless there is a significant left to right inter-atrial shunt) and must be equal to the total pulmonary blood flow (Qp). Systemic Flow (Qs): This is the total volume of blood that reaches the body's tissues. Every drop of blood that services the body (brain, gut, kidneys) eventually returns via the Vena Cava (and the coronary sinus) into the right atrium and fills the right ventricle. Therefore, the Right Ventricular Output (RVO) is exactly equal to the flow crossing the tricuspid valve, which is equal to the systemic blood flow (Qs). Qs is the blood flow feeding the coronary arteries and the systemic vessels.
  
  

6) Arterial oxygen content (CaO₂)
Use simplified: CaO₂ ≈ 1.34 × Hb × SaO₂ (ignore dissolved O₂ for simplicity)

• CaO₂ ≈ 1.34 × 15 × 0.94
  = 1.34 × 14.1 = 18.9 mL O₂/dL

• Convert to mL O₂/mL blood: 18.9/100 = 0.189 mL O₂/mL
  
  

7) Systemic oxygen delivery (DO₂)

• DO₂ = Qs × CaO₂ = 194 × 0.189 = 36.7 mL O₂/kg/min

Case 29: 

An ex 22-week gestational age preterm infant is now 28 week post-menstural age and weighing 1.2 kg with a body surface area of 0.10 m² presents with documented inability to be weaned from mechanical ventilation. The physical examination reveals bounding peripheral pulses. Blood pressure is 45/15 mmHg with a mean of 25 mmHg. Right atrial pressure is estimated at 5 mmHg. Echocardiography confirms a large left-to-right patent ductus arteriosus with an echo-derived left ventricular output of 480 mL/kg/min. The inter-atrial left to right PW-Doppler derived mean gradient is 8 mmHg. 

1. What would be the left ventricular coronary perfusion pressure? 

2. What would be the right ventricular coronary perfusion pressure?

3. What is the pulse pressure?

Answer:

• Left sided CPP is Aortic diastolic pressure - LV end diastolic pressure = 15 - (5+8) = 2 mmHg

  • Step 1 — Estimate LA (≈ LVEDP) from the inter-atrial gradient 

    • Mean LA→RA PW Doppler gradient = 8 mmHg 

    • Estimated RA pressure = 5 mmHg 

    • So: LA pressure ≈ RA + 8 = 5 + 8 = 13 mmHg 

    • In this context (large L→R shunt, LA hypertension), LA pressure is a reasonable surrogate for LVEDP. 

  • Step 2 — Left ventricular coronary perfusion pressure (LV CPP) 

    • LV CPP ≈ Aortic diastolic − LVEDP (≈ LA pressure) Aortic diastolic = 15 mmHg 

    • LVEDP ≈ LA = 13 mmHg 

    • LV CPP ≈ 15 − 13 = 2 mmHg; Left ventricular coronary perfusion pressure ≈ 2 mmHg

• Right sided CPP is Aortic diastolic pressure - RV end diastolic pressure = 15 - 5 mmHg = 10 mmHg. 

  • RV CPP ≈ Aortic diastolic − RVEDP (≈ RA pressure) 

  • Aortic diastolic = 15 mmHg 

  • RVEDP ≈ RA = 5 mmHg 

  • RV CPP ≈ 15 − 5 = 10 mmHg 

  • Right ventricular coronary perfusion pressure ≈ 10 mmHg

• Pulse pressure is 45-15 mmHg = 30 mmHg

Oximetry provides the following data: Aorta 96%, Mixed Venous 66%, Pulmonary Vein 96%, and Pulmonary Artery 87%. The infant’s current haemoglobin is 13.5 g/dL. 

1. Calculate the total left ventricular output (LVO) in L/min. 

2. Determine the pulmonary to systemic flow ratio (Qp/Qs). 

3. Calculate the systemic vascular resistance (SVR) in Wood units and the indexed systemic vascular resistance (SVRi). 

4. Is the systemic flow sufficient based on normative values and calculated systemic cardiac index (Qsi)

Answer 

1. The total left ventricular output, which represents total pulmonary flow (Qp) when assuming the right ventricular output equals systemic return, is 0.576 L/min, calculated by multiplying 480 mL/kg/min by 1.2 kg and dividing by 1000. 

2. The Qp/Qs ratio is determined by the saturation difference formula: (96 - 66) divided by (96 - 87), which results in a ratio of 3.33. 

3. Total systemic blood flow (Qs) is found by dividing the total pulmonary flow by this ratio: 0.576 L/min divided by 3.33 equals 0.173 L/min. 

4. Systemic vascular resistance is the difference between mean arterial and central venous pressures divided by systemic flow: 

   1. (25 - 5) divided by 0.173, which equals 115.6 Wood units. 

   2. The indexed systemic vascular resistance is 115.6 Wood units multiplied by the BSA of 0.10 m², resulting in 11.56 Wood units · m². 

5. Finally, the systemic cardiac index is 0.173 L/min divided by 0.10 m², which equals 1.73 L/min/m². This index is significantly lower than the normal neonatal range of 4 to 5 L/min/m², indicating that systemic flow is decreased by the massive left-to-right shunt and steal, despite a high LVO.

"Measurement of cardiac output can be performed in the catheterization laboratory and offers insights to the patient’s hemodynamic status. Cardiac output refers to the volume of blood pumped by the heart in 1 min, expressed in liters per minute, and often normalized for patient size by dividing by the body surface area (BSA) to obtain the so-called cardiac index (CI) measured in liters per minute per square meter (L/min/m²). The normal value is 5–8 L/min in adults at rest or a cardiac index of >2.4 L/min/m². Different techniques are available for calculating cardiac output, and it is important to bear in mind the strengths and weaknesses intrinsic to each." (Reference: Jones, Juan Pablo Sandoval, and Lee Benson. "Hemodynamics: pressures and flows." Cardiac Catheterization for Congenital Heart Disease: From Fetal Life to Adulthood. Milano: Springer Milan, 2014. 125-148.)

Case 30

A child with a large ventricular septal defect undergoes cardiac catheterization to assess pulmonary blood flow, pulmonary vascular resistance, and vasoreactivity.

• Baseline data in room air 

  • Hemoglobin 100 g/L;

  • Indexed oxygen consumption VO2 150 mL/min/m2

  • Aortic oxygen saturation 95%

  • Pulmonary artery oxygen saturation 80%

  • Mixed venous oxygen saturation 72.5 %

  • Mean pulmonary artery pressure 60 mmHg

  • Mean left atrial pressure 6 mmHg

• After administration of 100% oxygen

  • Mean pulmonary artery pressure 60 mmHg

  • Mean left atrial pressure 8 mmHg

  • Pulmonary artery saturation 95 percent

  • Pulmonary artery PaO2 95 mmHg

  • Aortic saturation 100 percent

  • Aortic PaO2 600 mmHg

Questions

1. Calculate the Qp to Qs ratio in room air.

2. Calculate the oxygen carrying capacity of blood in mL O₂ per liter.

3. Calculate the oxygen content in the pulmonary artery and pulmonary vein.

4. Calculate the pulmonary arteriovenous oxygen difference in room air.

5. Calculate the indexed pulmonary blood flow Qp in room air.

6. Calculate the pulmonary vascular resistance index PVRi in room air.

7. During 100 percent oxygen administration, calculate the pulmonary arteriovenous oxygen difference ignoring dissolved oxygen.

8. During 100 percent oxygen administration, calculate the pulmonary arteriovenous oxygen difference including dissolved oxygen.

9. Calculate indexed pulmonary blood flow during 100 percent oxygen ignoring dissolved oxygen.

10. Calculate indexed pulmonary blood flow during 100 percent oxygen including dissolved oxygen.

11. Calculate PVRi during 100 percent oxygen ignoring dissolved oxygen.

12. Calculate PVRi during 100 percent oxygen including dissolved oxygen.

13. Based on the correct calculations, is there evidence of pulmonary vasoreactivity
    
    

Answer Key

1. Qp to Qs ratio in room air

   • Qp/Qs equals (Ao sat - mixed venous sat)/(Ao sat - PA sat)
     Qp/Qs equals (95 - 72.5) / (95 - 80)
     Qp/Qs equals 22.5/15
     Qp/Qs equals 1.5 to 1
     
     

2. Oxygen carrying capacity

   • You may see 1.36 or 1.39 used for Hüfner's constant depending on the textbook or clinical setting (1.34 is the most common clinical value).

   • If we use highest constant - Oxygen capacity equals hemoglobin (10 g/dL) x 1.39 x 10.; or hgb (g/L) x 1.39.
     Oxygen capacity equals 100 times 1.39
     Oxygen capacity equals 139 mL O2 per liter
     
     

3. Oxygen content  in the pulmonary artery and pulmonary vein.

   • With 80% saturation in the PA

     • Pulmonary artery oxygen content equals 139 times 0.80 equals 111.2 mL O₂ per liter

   • With 95% aortic saturation, assumint it is the same as pulmonary venous oxygen saturation

     • Pulmonary vein oxygen content equals 139 times 0.95 equals 132.05 mL O₂ per liter
       
       

4. Pulmonary arteriovenous oxygen difference in room air 

   • Pulmonary AV O₂ difference equals pulmonary vein (assumed here to be the same as aortic saturation) minus pulmonary artery

   • Pulmonary AV O₂ difference equals 132.05 minus 111.20

   • Pulmonary AV O₂ difference equals 20.85 mL O₂ per liter
     
     

5. Indexed pulmonary blood flow in room air
   Qp equals VO₂ divided by pulmonary AV O₂ difference
   Qp equals 150 divided by 20.85
   Qp equals 7.19 L per minute per m2
   
   

6. Pulmonary vascular resistance index in room air
   PVRi equals (mean pulmonary artery pressure minus mean left atrial pressure) divided by Qp indexed to BSA
   PVRi equals (60 minus 6) divided by 7.19 L per minute per m2
   PVRi equals 54 divided by 7.19 L per minute per m2
   PVRi equals 7.23 Wood units x m2

   • Here PVRi and no PVR because the Qp is  already indexed to BSA
     
     

7. Pulmonary AV oxygen difference during 100 percent oxygen ignoring dissolved oxygen

   • Pulmonary AV O₂ difference equals oxygen capacity x (arterial saturation (assumed to be same as pulmonary vein saturation) - PA saturation)

   • Pulmonary AV O₂ difference equals 139 x (1.00 - 0.95)

   • Pulmonary AV O₂ difference equals 6.95 mL O₂ per liter
     
     

8. Pulmonary AV oxygen difference during 100 percent oxygen including dissolved oxygen

   • For pulmonary venous oxygen content you need to add the dissolved oxygen, which is 0.003 times PaO₂

     • Pulmonary vein oxygen content equals (139 x 1.00) + (0.003 x 600)

     • Pulmonary vein oxygen content equals 139 + 1.8 = 140.8 mL O₂ per liter

   • Pulmonary artery oxygen content = (139 x 0.95) + (0.003 x 95)

     • Pulmonary artery oxygen content = 132.05 + 0.285 = 132.335 mL O₂ per liter

   • Pulmonary AV O₂ difference = 140.8 - 132.335

     • Pulmonary AV O2 difference = 22.1 mL O₂ per liter

9. Indexed pulmonary blood flow during 100% oxygen ignoring dissolved oxygen

   • Qp = 150 /6.95

   • Qp = 21.53 L per minute per m2
     
     

10. Indexed pulmonary blood flow during 100% oxygen including dissolved oxygen

    • Qp = 150 /22.1

    • Qp = 6.78 L per minute per m2
      
      

11. Pulmonary vascular resistance index during 100 percent oxygen ignoring dissolved oxygen
    PVRi equals (60 - 8) / 21.53
    PVRi = 52 /21.53
    PVRi equals 2.4 Wood units x m2
    
    

12. Pulmonary vascular resistance index during 100 percent oxygen including dissolved oxygen
    PVRi = (60 - 8) / 6.78
    PVRi = 52 / 6.78
    PVRi = 7.66 Wood units x m2
    
    

13. Vasoreactivity interpretation
    There is no true pulmonary vasoreactivity.
    Mean pulmonary artery pressure does not decrease and pulmonary vascular resistance index remains elevated when dissolved oxygen is correctly included.
    Ignoring dissolved oxygen falsely suggests vasoreactivity and leads to incorrect interpretation of operability.

Case 31

A preterm infant born at 25+2 weeks’ gestation is now day of life 5. The infant is mechanically ventilated and receiving caffeine but no inotropes. Blood pressure is monitored intermittently by oscillometry. Echocardiography is performed to assess pulmonary hemodynamics and ductal shunting.

Key findings:

• Large, unrestrictive left-to-right PDA

• Laminar systolic and diastolic left-to-right ductal flow

• No evidence of ductal constriction

• Mild tricuspid regurgitation with a well-defined continuous-wave Doppler envelope

Measured values:

• Oscillometric systolic blood pressure: 52 mmHg

• Diastolic blood pressure: 24 mmHg

•  Mean blood pressure: 34 mmHg

PDA Doppler:

• Peak systolic PDA velocity: 2.8 m/s

• Timing of peak PDA velocity: occurs early in systole, before the end of the T wave

TR jet Doppler:

• Peak TR velocity: 3.2 m/s

• Peak TR velocity occurs late in systole, near the end of the T wave

Assume right atrial pressure = 5 mmHg.

1. What is the difference between an instantaneous pressure gradient versus a peak (or end-systolic) pressure difference. Do they occur necessarily at the same time? 

2. Using the simplified Bernoulli equation, calculate the instantaneous systolic pressure gradient across the PDA at the time of peak PDA velocity. Why is this pressure gradient a falacy?

3. If one were to estimate systolic pulmonary arterial pressure (sPAP) using the common shortcut: sPAP ≈ systolic aortic pressure − PDA systolic gradient; what value would be obtained in this infant?

4. Calculate the estimated right ventricular systolic pressure (RVSP) using the TR jet velocity.

5. Compare the sPAP estimated from the PDA velocity method (Question 2) with the RVSP estimated from the TR jet. Are they the same? If not, which is higher?

6. Explain why these two estimates differ, focusing specifically on timing within systole and the relationship between pressure and velocity.

7. True or False (justify your answer): “The peak systolic PDA velocity must occur at the same time as peak aortic pressure and peak pulmonary arterial pressure.”

8. In this infant, does using the peak systolic PDA velocity to estimate sPAP most likely underestimate or overestimate the true peak pulmonary arterial systolic pressure? Explain why.

Answer Key:

Question 1:

• An instantaneous pressure gradient is the pressure difference between two locations measured at a specific moment in time during the cardiac cycle. In Doppler echocardiography, this is what is derived from velocity using the Bernoulli equation: it reflects the real-time difference in pressure across a structure (for example, across a PDA, valve, or regurgitant orifice) at the exact instant when that velocity is recorded. When we speak of a peak instantaneous pressure gradient, we mean the largest velocity of flow (and pressure) difference observed at any single moment, which corresponds to the moment of maximal velocity difference—not necessarily to maximal pressure in either chamber or vessel. A peak (or end-systolic) pressure difference, in contrast, refers to the difference between absolute pressures at their own peaks, typically near end-systole. This concept comes from invasive pressure measurements or pressure tracings, where one can identify the maximal aortic pressure and maximal pulmonary or ventricular pressure and then compare them at a defined phase of systole. It is tied to pressure peaks, not to flow velocity. These two quantities do not necessarily occur at the same time. 

• The instantaneous pressure gradient peaks when the difference between the two pressure curves is greatest, which may occur early, mid, or late in systole—or even in late diastole—depending on the shape and timing of the pressure waveforms. Peak or end-systolic pressures, on the other hand, occur when each individual pressure reaches its maximum, which may happen later in systole and not simultaneously between the two vascular beds. 

• In a normal heart, the peak aortic pressure and the peak pulmonary artery pressure occur at roughly the same phase of systole, but the right ventricle reaches its peak and relaxes slightly earlier, so pulmonary artery systolic pressure typically peaks a bit earlier and falls sooner than aortic pressure. This difference in timing may becoming more or less marked in disease states or in the context of presence of shunts or congenital heart defects. In sinus rhythm with normal valves, both ventricles start ejecting shortly after the QRS, and aortic and pulmonary artery pressures rise together during early systole. The right ventricular outflow and pulmonary artery pressure waves tend to have a shorter acceleration and slightly earlier peak than the left ventricular outflow and aortic pressure wave, so pulmonary artery systolic pressure may reach its maximum a bit before the aortic systolic peak and then decline earlier in late systole.

• The ventricles initiate mechanical contraction shortly after the QRS (depolarization of the ventricles). The ventricle is still in mechanical systole during the beginning of the T wave. Mechanical relaxation (the start of diastole) typically begins around the peak of the T wave and is completed by the time the T wave ends. While we often think of the T wave simply as "relaxation," there is a slight lag between the electrical signal (repolarization) and the physical movement of the muscle.

• In short, Doppler measures a time-specific pressure difference (instantaneous gradient), whereas peak or end-systolic pressure differences compare absolute pressure maxima. Because pressure and velocity are time-dependent and not temporally locked, these two measures often occur at different moments in the cardiac cycle.

Question 2:

• Bernoulli equation:  ΔP = 4 × V²

• ΔP = 4 × (2.8)² = 4 × 7.84 = 31 mmHg

• This is the instantaneous pressure gradient between the aorta and pulmonary artery at the moment of peak PDA velocity gradient, not necessarily at peak systolic pressure. This gradient may occur before, at, or after peak systolic gradient in the great vessels. Importantly, the peak aortic pressure may or not occur at the same time as the peak pulmonary arterial pressure.

Question 3:

• Estimated sPAP using PDA shortcut: 

• sPAP ≈ systolic aortic pressure − PDA gradient

• sPAP ≈ 52 − 31 = 21 mmHg

Question 4:

• RVSP from TR jet:

• RVSP = 4 × (3.2)² + RA pressure

• Here we use 5 mmHg as estimated RA pressure 

• RVSP = 4 × 10.24 + 5 = 41 + 5 = 46 mmHg

Question 5:

• The two estimates are not the same.

• PDA-derived sPAP ≈ 21 mmHg

• TR-derived RVSP ≈ 46 mmHg

• The TR-derived RVSP is substantially higher.

Question 6:

• These estimates differ because velocity reflects an instantaneous pressure gradient, not absolute pressure, and the timing of peak velocity does not necessarily coincide with peak ventricular or arterial pressure. Moreover, the peak aortic pressure may also not occur at same time as the peak pulmonary arterial pressure, and these may also not occur at the same time as the maximal pressure gradient between the 2 pressures, which will drive the peak velocity gradient obtained at the PDA level. 

• Key concepts:

  • PDA velocity reflects the difference between aortic and pulmonary artery pressures at a specific moment in time throughout the cardiac cycle.

  • Peak PDA velocity occurs when the pressure difference is greatest, not when either pressure is maximal

  • In this case, peak PDA velocity gradient occurs early in systole, before peak aortic pressure

  • Later in systole, aortic pressure continues to rise, but the PA pressure rises in parallel, reducing the gradient and therefore the velocity. The velocity gradient at peak aortic and PA pressure may be smaller than the peak systolic velocity gradient at another time of the cardiac cycle. 

  • In contrast: TR velocity reflects peak RV systolic pressure, which occurs near end-systole and TR-derived RVSP corresponds to the true peak RV pressure, not an early systolic value. Thus, the two Doppler signals are measuring different physiologic moments.

Question 7: False.

• Peak PDA velocity does not have to occur at peak aortic pressure or peak pulmonary arterial pressure.

• Velocity peaks when the instantaneous pressure gradient is maximal, which can occur: Early systole; Mid-systole; Late systole; Even in late diastole (presystolic acceleration).

• Pressure and velocity are related but not temporally locked.

Question 8: In this infant, using the peak systolic PDA velocity most likely underestimates the true peak pulmonary arterial systolic pressure.

• Peak PDA velocity occurs early in systole, before PA pressure reaches its maximum. The gradient at that moment is larger than later in systole, leading to a large Bernoulli subtraction. Subtracting this early systolic gradient from peak aortic pressure mixes non-simultaneous events. This results in a falsely low estimated sPAP.

• The TR jet, by contrast, captures peak RV pressure near end-systole and therefore provides a more physiologically accurate estimate of peak RVSP.

Core Teaching Take-Home Message:

• PDA Doppler measures instantaneous pressure gradients, not absolute pressures

• Peak velocity ≠ peak pressure

• Timing matters

• Mixing peak values from non-simultaneous physiologic events introduces systematic error

• TR-derived RVSP and PDA-derived sPAP are not interchangeable and answer different questions

Case 32

You are in the catheterization laboratory and obtain the following pressure tracing in a patient with a patent ductus arteriosus (PDA). Simultaneously, you are performing live echocardiography.

1. Explain the difference between an instantaneous pressure gradient and a peak-to-peak pressure gradient. Which of these reflects the true driving force across the PDA at any given moment in the cardiac cycle?

   • Answer: An instantaneous pressure gradient is the pressure difference between two chambers or vessels measured at the same moment in time during the cardiac cycle. It reflects the real-time difference between aortic and pulmonary arterial pressures at that exact point — for example, early systole, late diastole. A peak-to-peak pressure gradient, in contrast, is the difference between the peak pressure in one chamber and the peak pressure in another chamber, even though those peaks do not occur at the same time. It is simply the numerical difference between the highest measured pressures in each chamber, without accounting for timing. 

   • The true driving force across a PDA at any given moment is the instantaneous pressure gradient, because blood flow is determined by the pressure difference that exists at that specific instant. This is the gradient that determines the direction of shunt (right-to-left vs left-to-right), and the velocity of flow (via the Bernoulli equation: ΔP = 4v²). Peak-to-peak gradients do not represent the actual force generating flow and can be misleading when interpreting shunt physiology.

2. In the cath lab, based on the pressure curves, what would typically indicates the end of systole and beginning of diastole?

   • In the cath lab, when looking purely at pressure tracings, the end of systole and the beginning of diastole are identified by specific features of the arterial pressure waveform. The most reliable marker is the dicrotic notch on the aortic pressure tracing. During systole, ventricular ejection causes a rapid rise in aortic pressure (ascending limb), followed by a peak and then the descending limb. When the left ventricle finishes ejecting and the aortic valve closes, there is a brief retrograde flow toward the valve. This produces a small notch or incisura on the aortic waveform — the dicrotic notch. The dicrotic notch marks aortic valve closure, end of mechanical systole, beginning of mechanical vascular diastole. Immediately after this notch, the pressure decline represents diastolic runoff. In pulmonary artery tracings, a similar notch may be seen corresponding to pulmonic valve closure, though it is often less distinct than in the aorta. So in summary: The dicrotic notch on the arterial pressure tracing (aortic or pulmonary) typically indicates the transition from systole to diastole in cath lab pressure recordings.

Based on the pressure tracing provided: