Our daughter of 4 yrs has 2 ASDs of 10mm and 4mm each. With a distance of 6mm between the 2 ASDs.
How can the relative plane of the 2 ASD be determined. i.e. are they side by side or how much lower in relative sence.
As we think this will be vitial to determine the success rate to perform a device based procedure to close both ASDs. Let say using a 14mm or 16mm device.
It is very difficult to determine the exact location of multiple ASDs, as they look one way on a transthoracic (2D) echocardiogram, but in reality can be in slightly different locations than they appear. 3 D echo can better look at the entire atrial septum, but that is an emerging technology; not only is it not widely available, but the doctors are still building experience in reading these studies. Transesophageal echocardiography (TEE) is the standard used to assess whether ASDs have a good chance of being device closed. Unfortunately, even in situations where we think it will work, there are instances when a device will not "seat" itself properly and must be withdrawn. When multiple defects are present, then the cath doctor needs to figure out if there is enough room along the atrial septum to accomodate 2 separate devices (without blocking nearby veins and valves) or whether one device could cover most of both holes and leave only a small residual defect. Additionally the size of the device(s) used is not a function of the size measured by the echo, but rather by the size of the ASD in the cath lab when it is "sized" by a balloon catheter. So it may actually be a larger defect than expected. In summary, doing a TEE may help the team decide if device closure is likely to work, but in reality, you have to try and see. Good luck.
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