Fractional Excretion of Sodium

The fractional excretion of sodium (FeNa) calculation predicts the likelihood that a patient with acute renal failure has acute tubular necrosis (ATN). In ATN, the kidneys have difficulty reabsorbing sodium and concentrating urine. The following formula takes advantage of this by using plasma sodium and creatinine concentrations (P[Na] and P[Cr]) and spot urine sodium and creatinine levels (U[Na] and U[Cr]):FeNa = (U[Na] × P[Cr])/(P[Na] × U[Cr]) × 100

FeNa > 2% suggests post-glomerular failure
FeNa < 1% suggests prerenal azotemia (volume depletion).

ARF大致可以分為prerenal, post renal 跟intrarenal , prerenal的特徵是Na(u)<20 FeNa <1 為什麼會這樣?因為prerenal發生的原因是perfusion不夠,所以可能volume太差,cardiac output不夠或是sepsis引起的血管擴張,總之就是腎臟內的血流不夠,那腎臟怎麼自救? 1. 節流:留住Na跟水(藉由腎小管的再吸收), 2. 開源:讓入球小動脈擴張增加腎臟內血流 ,所以尿中的Na不會太多(相對於intrinsic的 Na(u)>40),FeNa則是比單純的看尿中排出多少Na外,計算被排出的Na是濾過的Na的多少比率也就是  FENa = (排出Na / 濾出Na)*100 = (Na(u) * 尿量/ Na (p)* GFR)*100 而導出這個公式 ,所以FENa越小,表示腎臟功能越完整,因為他可以把Na回收。
1. hepatorenal syndrome的腎臟是完整的,並沒有壞(這也是hepatorenal syndrome的要件),做biopsy都是好的,所以不算是intrinsic renal failure
2. drug(NSAID, ACEI,amphotericin,移植抗排斥藥)以及顯影劑造成的ARF,就是前面說過開源不力,以NSAID來說,因為要讓入球小動脈擴張就是藉由prostagladin,但是被NSAID給抑制了,所以腎臟內的血管沒有擴張,繼續在灌流不足的狀態下,就會出現如同prerenal的情況,而這種情形久了就造成缺血而變成ATN,那FENa就不會再繼續小於一了。
3.obstruction算是postrenal,因為小便出不來,所以當然尿中的Na也不會多,基本上腎臟本身也沒壞,除非真的拖太久,導致無法回復的腎元破壞
4.那為什麼AGN(例如SLE,Wegener's granulomatosis)的FENa也是<1? 那是因為renin刺激aldosterone所以會留鈉排鉀,尿中的Na也會少,這類的urine routine大多還有proteinuria, hematuria等等 ,病史也可以看到水腫,高血壓,所以也很好辨別,不會混淆。(此內容取自http://elc.skh.org.tw/elc/er119forum/viewtopic.php?t=695&sid=667682c4e9f40c2f1eb58128376c1a11)

以下這個例子可供練習喔~~~~Example: A middle-aged male was found lying on the ground and has ulcers on all pressure points on his back, buttocks, and legs. A urinary catheter is placed and yields a small amount of dark urine, which tests positive for blood on dipstick. He is found to have acute renal failure. Plasma and urine electrolytes and creatinine analyses are performed. The results are P[Na] = 134 mmol/L, P[Cr] = 3.5 mg/dL, U[Na] =108 mmol/L, U[Cr] = 68 mg/dL. What is the most likely cause of his renal failure, ATN/myoglobinuria, or dehydration?FeNa = [(108 × 3.5)/(134 × 68)] × 100 = 4.1 The patient's FeNa is 4.1 (which is >2%) and suggests that ATN is the most likely the cause of his renal failure and myoglobinuria.

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  • 花花
  • S.O.S

    為了專師的考試,我已經快喪失了鬥志.無意中進入你的blog,讓我有在沙漠中發現綠洲的驚喜與希望,希望可以分享你的考題讓我學習;因為無法成功下載,所以可以寄給我嗎?銘謝在心,祝平安!
  • triyei
  • 抱歉想跟你要專師的講義密碼-->我也是考生之ㄧ..3QQ..
  • 寄給你囉

    msicu1 於 2010/06/10 23:17 回覆

  • 小么
  • 跟你要專師的講義密碼

    跟你要專師的講義密碼,我也是考生.本院位於苗栗.錄取率一向不高.拜託
  • 孟
  • 拜讀您的網站有很多有用的資料
    能否請您寄密碼給我
    以便開啟內容閱讀
    謝謝您
  • 訪客
  • FeNa is 4.1 (which is >2%) 是如何將4.1換算成%?謝謝~^^
  • 公式得出來的值 x 100= %

    UNa * PCr
    ---------------- x 100= FENa %
    PNa * UCr

    msicu1 於 2010/11/30 21:51 回覆

  • 訪客
  • 說的好清楚!!