Renal physiology and acid–base
導入済み| English | 意味 |
|---|---|
| clearance/ˈklɪərəns/ | clearance |
| filtration/fɪlˈtreɪʃn/ | filtration |
| aldosterone | aldosterone |
| antidiuretic hormone/ˌæntɪˌdɪjuːˈretɪk ˈhɔːməʊn/ | A hormone that increases water reabsorption in the kidneys, helping regulate body water balance. |
| Henderson–Hasselbalch equation | Henderson–Hasselbalch equation |
| Winter's formula | Winter's formula |
The body's chemist
- The kidney filters, takes back, secretes and regulates. Its clearance arithmetic is the language of drug dosing and kidney function.
- The same kidney, with the lungs, sets the blood's acid–base balance.
Filtration
- Filtration 滤过 at the glomeruli runs at about 125 mL per minute (the GFR): about 180 L a day.
- Only about 1.5 L a day leaves as urine, so more than 99% of the filtrate is taken back.
180 L are filtered in a day and 1.8 L of urine are made. What percentage of the filtrate was taken back?
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(180 − 1.8) / 180 = 0.99, so 99% is reabsorbed.
Segment by segment
- The proximal tubule takes back about 65% of the filtered Na⁺ and the loop of Henle about 25%.
- The distal nephron fine-tunes: aldosterone 醛固酮 swaps Na⁺ for K⁺ and H⁺, and antidiuretic hormone 抗利尿激素 (ADH) inserts water channels.
Match each part of the nephron to its role.
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Bulk work early, fine-tuning late, with hormones acting at the end.
Clearance
- Clearance 清除率 is the volume of plasma fully cleared of a substance each minute: $C_x = \dfrac{U_x V}{P_x}$.
- Creatinine is filtered and barely reabsorbed, so its clearance estimates GFR. Example: $U = 1.2$ mg/mL, $V = 1.4$ mL/min, $P = 0.012$ mg/mL gives $C = 140$ mL/min (slightly high, since some creatinine is secreted).
Urine creatinine is 0.9 mg/mL, urine flow 1.0 mL/min and plasma creatinine 0.010 mg/mL. What is the creatinine clearance, in mL/min?
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0.9 × 1.0 / 0.010 = 90 mL/min: an estimate of GFR.
Worked example: when glucose spills
- Plasma glucose 5 mmol/L × GFR 0.125 L/min = 0.625 mmol/min filtered, all taken back by the proximal tubule. Its maximum rate (Tm) is about 2 mmol/min.
- Tm/GFR predicts spill at 2/0.125 = 16 mmol/L, but glucose appears in urine from about 11 mmol/L, because nephrons differ and the weakest saturate first.
Glucose in the urine always means the kidney is damaged.
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Usually it means plasma glucose is above the threshold (about 11 mmol/L), so the filtered load exceeds what the tubules can take back.
Henderson–Hasselbalch
- Blood pH comes from the bicarbonate buffer, the Henderson–Hasselbalch equation 亨德森-哈塞尔巴尔赫方程:
- Normal: $[HCO_3^-] = 24$ mmol/L and $P_{CO_2} = 40$ mmHg give $6.1 + \log_{10}(24/1.2) = 7.40$. The kidney sets the top; the lungs set the bottom.
$[HCO_3^-]$ is 24 mmol/L and $P_{CO_2}$ is 60 mmHg. What is the pH? (Give two decimals.)
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0.03 × 60 = 1.8; 24/1.8 = 13.3; log₁₀ 13.3 = 1.12; pH = 6.1 + 1.12 = 7.22: an acute respiratory acidosis.
Worked example: is the compensation right?
- (Fictional patient) Metabolic acidosis with $[HCO_3^-] = 12$, and the lungs have lowered $P_{CO_2}$ to 24: $pH = 6.1 + \log_{10}(12/0.72) = 7.32$.
- Winter's formula 温特公式 predicts the expected $P_{CO_2} = 1.5 \times 12 + 8 = 26 \pm 2$ mmHg. 24 fits; a value outside the range means a second disorder hiding in the first.
In a metabolic acidosis with $[HCO_3^-]$ = 10 mmol/L, what $P_{CO_2}$ does Winter's formula expect?
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1.5 × 10 + 8 = 23 ± 2 mmHg. Higher means the breathing is not keeping up: a respiratory acidosis is hiding as well.
Three common slips
- "Normal creatinine, normal GFR." Creatinine ∝ 1/GFR: losing the first half of GFR only doubles a low creatinine, which can still look near normal.
- Clearance as excretion rate. Clearance is a volume of plasma per minute; excretion is an amount per minute ($U \times V$).
- Wrong CO₂ units. Dissolved CO₂ is 0.03 × $P_{CO_2}$ in mmHg.
Put the steps of reading a blood gas in order.
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The pH says which way, the primary change says why, and the compensation check catches mixed disorders.
Recap
- GFR about 125 mL/min; more than 99% taken back; proximal 65%, loop 25% of Na⁺.
- $C = UV/P$: creatinine estimates GFR.
- $pH = 6.1 + \log_{10}([HCO_3^-]/0.03 P_{CO_2})$; Winter's formula checks the breathing response to metabolic acidosis.