2
$\begingroup$

I understand that it's also found elsewhere, such as in renal capillaries, but I can't see logic behind it being located in the lungs. Isn't ACE's function, through making more angiotensin II, causing increased fluid and sodium retention and increasing peripheral resistance? What benefit does that purpose get from ACE being mainly in the lungs?

$\endgroup$
1
  • $\begingroup$ I guess you do mean ACE, i.e you do not mean ACE2. It looks like some coincidence that you started your question when "Corona" began. ACE2 (in spite of homology: not ACE) is target receptor of CoV and COV-2 in lung: Pneumocytes type II. - Your question makes me re-research on ACE not ACE2 in lung epithelia. In fact, as answered: ACE2 but not ACE in lung epithelia. Highly interesting: in evolution, ACE2 might have mutated from ACE as both receptors are almost identical. ACE might be the more ubiquitious protein in the sense of answer given. $\endgroup$ Feb 10 at 19:33

1 Answer 1

1
$\begingroup$

ACE is present on the luminal surface of vascular endothelia throughout the body and is abundantly present in the endothelium-rich lungs.

ACE in the kidney—particularly in the endothelial cells of the afferent and efferent arterioles— can produce enough ANG II to exert local vascular effects.

Thus, the kidney receives ANG II from two sources:

✓ Systemic ANG II comes from the general circulation and originates largely from the pulmonary region, and

✓ Local ANG II forms from the renal conversion of systemic ANG I.

• In addition, the proximal tubule secretes ANG II into its lumen and thus achieves intraluminal concentrations in excess of those in the general circulation. ANG II in the circulation has a short half-life (~2 min).

$\endgroup$
2
  • $\begingroup$ Oh, so the reason ACE is abundant in the lungs is simply a product of the fact that the lung has a lot of endothelium? $\endgroup$
    – Dahen
    Feb 2, 2020 at 3:45
  • 1
    $\begingroup$ Yes, you are right. $\endgroup$
    – JM97
    Feb 2, 2020 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.