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The document discusses acid-base regulation by the kidney. It does this through three main processes: 1) reabsorption of filtered bicarbonate in the proximal tubule, 2) secretion of hydrogen ions into the filtrate in the distal tubule, and 3) use of urinary buffers. It then provides detailed diagrams and explanations of the transport mechanisms involved in bicarbonate and hydrogen ion regulation throughout the renal tubular system. These include carbonic anhydrase, sodium-hydrogen antiporters, and the role of the collecting duct in final acid-base adjustments.

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APPROACH TO BLOOD GAS ANALYSIS Dr. MANDAR HAVAL D.C.H D.N.B
How does the kidney do it? • The kidney does it in three ways: – Total reabsorption of filtered bicarbonate (proximal). – Controlled secretion of H+ into filtrate (distal). – Judicious use of urinary buffers.
FILTRATE TUBULAR CELL BLOOD
FILTRATE TUBULAR CELL H2O + CO2 CA II H2CO3 H+ + HCO3- BLOOD
FILTRATE TUBULAR CELL H2O + CO2 CA II H2CO3 H+ + HCO3- BLOOD
FILTRATE TUBULAR CELL BLOOD H2O + CO2 CA II H2CO3 H+ + HCO3Na+ Na + Na Na K ATPase K
TUBULAR CELL FILTRATE BLOOD H2O + CO2 CA II H2CO3 H+ATPase Na+ Na+ / H+ Antiporter + HH+ + HCO33-HCO Na Na+ Na K ATPase K
TUBULAR CELL FILTRATE BLOOD H2O + CO2 CA II H2CO3 Na / K H+ HCO3- H+ATPase Na+ / H+ Antiporter Na+ Na Na+ Na K ATPase K
TUBULAR CELL FILTRATE H2O + CO2 BLOOD H2O + CO2 CA II H2CO3 CA IV H2CO3 HCO3- Na / K H+ HCO3- H+ATPase Na+ / H+ Antiporter Na+ Na Na+ Na K ATPase K
FILTRATE H2O TUBULAR CELL BLOOD H2O + CO2 CA II CA IV H+ HCO3-
COLLECTING TUBULE CELL BLOOD FILTRATE H2O + CO2 CA II H2CO3 Cl- / HCO3- Cl- Exchanger HCO3- H+ H+ ATPase
COLLECTING TUBULE CELL BLOOD FILTRATE H2O + CO2 CA II H2CO3 HCO3- H+ ATPase Cl- / HCO3- Exchanger Cl- H+
COLLECTING TUBULE CELL FILTRATE BLOOD HPO4= H+ ATPase H+
COLLECTING TUBULE CELL FILTRATE BLOOD H+ ATPase H+ HPO4= H2PO4
COLLECTING TUBULE CELL FILTRATE BLOOD SO4= H+ ATPase H+ HSO4-
COLLECTING TUBULE CELL FILTRATE BLOOD NH3 NH3 H+ ATPase NH4+ H+
Evaluation of Systemic Acid Base Disorders 1. Comprehensive history and physical examination. 2. Evaluate simultaneously performed ABG & serum electrolytes. 3. Identification of the dominant disorder. 4. Calculation of compensation. 5. Calculate the anion gap and the Δ. 1. Anion Gap 2. Δ AG 3. Δ Bicarbonate
Step 3: Identification of the dominant disorder Primary disorder Metabolic acidosis pH Initial change Compensatory change ↓ ↓ HCO3 ↓ PCO2
Step 3: Identification of the dominant disorder pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2 Metabolic alkalosis ↑ ↑ HCO3 ↑ PCO2 Primary disorder
Step 3: Identification of the dominant disorder pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2 Metabolic alkalosis ↑ ↑ HCO3 ↑ PCO2 Respiratory acidosis ↓ ↑ PCO2 ↑ HCO3 Respiratory alkalosis ↑ ↓ PCO2 ↓ HCO3 Primary disorder
• WHERE THE PROBLEM START
Calculation of compensation Mean "whole body" response equations for simple acid-base disturbances. Disorder pH Primary change Compensatory Response Equation Metabolic Acidosis   [HCO3-]  PCO2 ΔPCO2  1.2  ΔHCO3 Metabolic Alkalosis   [HCO3-]  PCO2 ΔPCO2  0.7  ΔHCO3 Respiratory Acidosis   PCO2  [HCO3-] Acute: ΔHCO3-  0.1  ΔPCO2 Chronic: ΔHCO3-  0.3  ΔPCO2 Respiratory Alkalosis   PCO2  [HCO3-] Acute: ΔHCO3-  0.2  ΔPCO2 Chronic: ΔHCO3-  0.5  ΔPCO2 Note: The formula calculates the change in the compensatory parameter.
Simple compensation Disorder pH Primary problem Compensation Metabolic acidosis ↓ ↓ in HCO3- PaCO2 =1.5xHCO3+8(+/-2) Metabolic alkalosis ↑ 10↑ in HCO3- 7↑ in PaCO2 Respiratory acidosis ↓ ACUTE -10↑ in PaCO2 CHRONIC -10↑ in PaCO2 1↑ in [HCO3-] 3.5↑ in [HCO3-] Respiratory alkalosis ↑ ACUTE-10↓ in PaCO2 CHRONIC-10↓ in PaCO2 2↓ in [HCO3-] 4↓ in [HCO3-]
Calculate the “gaps” Anion gap = Na+ − [Cl− + HCO3−] Δ AG = Anion gap − 12 Δ HCO3 = 24 − HCO3 Δ AG = Δ HCO3 −, then Pure high AG Met. Acidosis Δ AG > Δ HCO3 −, then High AG Met Acidosis + Met. Alkalosis Δ AG < Δ HCO3 −, then High AG Met Acidosis + Normal AG Met A Note:  Add Δ AG to measured HCO3− to obtain bicarbonate level  Delta _ AG    Pr e _ existing _ Bicarb  that would have existed IF the high AG metabolic acidosis Current _ Bicarb   were to be absent, i.e., “Pre-existing Bicarbonate.”
SOME FORMULA •THAT YOU SHOULD KNOW
CALCULATION OF H+ H   24  HCO  PaCO2  3 20 – 7.70 30 – 7.50 40(H+) – 7.40 (PH) 50 – 7.30 65 – 7.20
pH H+ pH H+ 6.70 200 7.40 40 6.75 178 7.45 35 6.80 158 7.50 32 6.85 141 7.55 28 6.90 126 7.60 25 6.95 112 7.65 22 7.00 100 7.70 20 7.05 89 7.75 18 7.10 79 7.80 16 7.15 71 7.85 14 7.20 63 7.90 13 7.25 56 7.95 11 7.30 50 8.00 10 7.35 45
CAO2= directly reflects the total number of oxygen molecules in arterial blood, both bound and unbound to hemoglobin • CaO2 = (1.34 x HB x SPO2) +(0.003 x PaO2) Normal CaO2 ranges from 16 to 22 ml O2/dl
Which patient is more hypoxemic, and why? • Patient A: pH 7.48 PaCO2 34 mm Hg PaO2 85 mm Hg SaO2 95% Hemoglobin 7 gm% • Patient B: pH 7.32 PaCO2 74 mm Hg PaO255 mm Hg SaO2 85% Hemoglobin 15 gm% www.dnbpediatrics.com
ANS CONT….. • Patient A: Arterial oxygen content = .95 x 7 x 1.34 = 8.9 ml O2/dl • Patient B: Arterial oxygen content = .85 x 15 x 1.34 = 17.1 ml O2/dl • Patient A, with the higher PaO2 but the lower hemoglobin content, is more hypoxemic www.dnbpediatrics.com
PaO2 • Factors affecting the PaO2 include alveolar ventilation, FIO2, altitude, age, and the oxyhemoglobin dissociation curve • Relation between PaO2 and SaO2: PaO2 60mm Hg 50mm Hg 40mm Hg 30mm Hg corresponds to SaO2 90% 80% 70% 60%
True or False: The pO2 in a cup of water open to the atmosphere is always higher than the arterial pO2 in a healthy person (breathing room air) who is holding the cup www.dnbpediatrics.com
ANS • The PO2 in the cup of water is always higher. This is for several reasons. First, there is no barrier to oxygen diffusing into the water; thus the PO2 in the cup will be the same as the atmosphere, at sea level approximately 160 mm Hg. • Second, there is no CO2 coming from the cup to dilute the oxygen, as there is in people. • Third, there is no V-Q inequality or shunt; even healthy people have a difference between alveolar PO2 and arterial PO2 for this reason. Thus a healthy person and a cup of water exposed to the atmosphere at sea level would have PO2 values of about 100 mm Hg and 160 mm Hg, respectively. www.dnbpediatrics.com
A-a Gradient • Determines the degree of lung function impairment • The A-a gradient is the partial pressure of alveolar oxygen minus the partial pressure of arterial oxygen (PAO2-PaO2) • Normal is 2-10mm Hg or 10 plus one tenth the person’s age
A-a Gradient • [(713*FIO2)-(PaCO2/0.8)] – PaO2 INTERPRETATION NORMAL – 10-20 (>30 is SINGNIFICANT) Seen in – Shunt Low V/Q Hypoventilation
A-a Gradient • PAO2-PaO2 of 20-30mm Hg on room air indicates mild pulmonary dysfunction, and greater than 50mm Hg on room air indicates severe pulmonary dysfunction • The causes of increased gradient include intrapulmonary shunt, intracardiac shunt, and diffusion abnormalities
a/A Ratio • Pao2/PAo2 NAORMAL LEVEL IS >0.75 • <0.60 IS INCOMPATIBLE WITH SPONTANIOUS BREATHING
PaO2/FIO2 Ratio • To estimate the impairment of oxygenation, calculate the PaO2/FIO2 ratio • Normally, this ratio is 500-600 • Below 300 is acute lung injury* • Below 200 is ARDS* *Along with diffuse infiltrates, normal PCWP, and appropriate mechanism
OXYGEN INDEX • OI =MAP X FIO2 POST DUCTAL PAO2 x 100
INTERPRETATION • OI >40 that is unresponsive to iNO predict a high mortality rate (>80%) and are indications for ECMO.
VENTILATORY INDEX • VI =PIP X PCO2 X RR 1000 VI > 65% INDICATE PREDICTIVE DEATH IN ARDS
RELATION OF ALBUMIN IN ABG  AG corrected = AG + 2.5[4 – albumin] (AG= Anion gap)
DELTA GAP      Delta gap = (actual AG – 12) + HCO3 Adjusted HCO3 should be 24 (+_ 6) {18-30} If delta gap > 30 -> additional metabolic alkalosis If delta gap < 18 -> additional non-gap metabolic acidosis If delta gap 18 – 30 -> no additional metabolic disorders
SOME CASE DISCUSSION
Case 1 • A 15 yr old juvenile diabetic presents with abdominal pain, vomiting, fever & tiredness for 1 day. He had stopped taking insulin 3 days ago. Examination revealed tachycardia, BP- 100/60, signs of dehydration. Abdominal examination was normal. • ABG: pH PaCO2 HCO3 PaO2 7.31 26 mmHg 12 mEq/L 92 mm Hg Serum Electrolytes: Na 140 mEq/L K 5.0 mEq/L Cl 100 mEq/L • Evaluate the acid-base disturbance(s)?
Case 1: Solution • Dominant disorder is Metabolic Acidosis • Compensation formula: Δ PaCO2 = 1.2 × Δ HCO3 = 1.2 × 12 = 14.4 PaCO2 = 40 – 14 = 26 Compensation is appropriate. • Anion Gap = 140 – (100 + 12) = 28 AG is high. pH PaCO2 HCO3 PaO2 7.31 26 12 92 Na K Cl 140 5.0 100
Case 1: Solution • Δ AG = 28 – 12 = 16 • Δ HCO3 • Δ AG > Δ = 24 – 12 = 12 HCO3 pH PaCO2 HCO3 PaO2 7.31 26 12 92 Na K Cl 140 5.0 100 • Final Diagnosis: High AG Met. Acidosis + Met. Alkalosis
Case 2 • A 14 yr old boy presents with continuous vomiting of 3 days duration, mental confusion, giddiness, and tiredness for 1 day. • Examination revealed tachycardia, hypotension and dehydration. • ABG pH 7.50 Serum Electrolytes: PaCO2 48 Na 139 HCO3 32 K 3.9 PaO2 90 Cl 85 • Evaluate the acid-base disturbance(s)?
Case 2: Solution • Dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 8 = 5.6 PaCO2 = 40 + 6 = 46 Compensation is appropriate. • Anion Gap = 139 – (85 + 32) = 22 AG is high. pH PaCO2 HCO3 PaO2 7.50 48 32 90 Na K Cl 139 3.9 85
Case 2: Solution • Δ AG = 22 – 12 = 10 • High AG metabolic acidosis • Final Diagnosis: pH PaCO2 HCO3 PaO2 7.50 48 32 90 Na K Cl 139 3.9 85 Metabolic Alkalosis + High AG Met. Acidosis
Case 3: Varieties of Metabolic Acidosis Patient ECF volume Glucose pH Na Cl HCO3 AG Ketones A Low 600 7.20 140 103 10 27 4+ B Low 120 7.20 140 118 10 12 0 C Normal 120 7.20 140 118 10 12 0 High-AG Met. Acidosis Non-AG Met. Acidosis Non-AG Met. Acidosis
Renal handling of Hydrogen in Metabolic Acidosis • In the setting of metabolic acidosis, normal kidneys try to increase H+ excretion by increasing titratable acidity and ammonia. The latter is excreted as NH4+. • When NH4+ is excreted, it also causes increased chloride loss, to maintain electrical neutrality. • Chloride loss, therefore, will be in excess of Na and K. • Urine Anion-Gap = Na + K – Cl • In metabolic acidosis, if Urine anion gap is negative, it suggests that the kidneys are excreting H+ effectively.
Urine Electrolytes in Metabolic Acidosis Patient U. Na U. K U. Cl Urine AG A Dx: B 10 14 74 –50 Diarrhea C 50 47 28 +69 RTA Urine Anion Gap = (U. Na + U. K – U. Cl) In Normal anion gap Metabolic Acidosis, Positive Urine AG suggests distal Renal Tubular Acidosis Negative Urine AG suggests non-renal cause for Metabolic Acidosis.
Case 4 • A 17 yr old boy presented with history of progressive dyspnoea with wheezing for 4 days. • He also had fever, cough with yellowish expectoration. • He had increased sleepiness for 1 day. • On examination, he was tachypnoeic, pulse100/min bounding, BP-160/96, central cyanosis +, drowsy, asterixis +, RS – B/L extensive wheezing +. • CXR- hyperinflated lung fields with tubular heart.
Case 4: Laboratory data • ABG: pH PaCO HCO PaO 2 3 2 7.30 60 mmHg 28 mEq/L 68 mm Hg • Serum Electrolytes: Na 136 mEq/L K 4.5 mEq/L Cl 98 mEq/L • Evaluate the acid-base disturbance(s)?
Case 4: Solution • Dominant disorder is Respiratory Acidosis • Compensation formula: Δ HCO3 = 0.3 × Δ PaCO2 = 0.3 × 20 =6 HCO3 = 24 + 6 = 30 Compensation is appropriate. • Anion Gap = 136 – (98 + 28) = 10 AG is normal. pH PaCO2 HCO3 PaO2 7.30 60 28 68 Na K Cl 136 4.5 98
Case 5 • 12 year old girl presented with complaints of difficulty in breathing and upper abdominal discomfort for the past 1 hr. • On examination, vitals normal, patient hyperventilating, RS – normal, Abdomen – normal.
Case 5: Laboratory data • ABG: pH PaCO HCO PaO 2 3 2 7.50 25 mmHg 21 mEq/L 100 mm Hg • Serum Electrolytes: Na 137 mEq/L K 3.9 mEq/L Cl 99 mEq/L Calcium 9.0 mEq/L • Evaluate the acid-base disturbance(s)?
Case 5: Solution • Dominant disorder is Respiratory Alkalosis • Compensation formula: pH 7.50 Δ HCO3 = 0.2 × Δ PaCO2 PaCO2 25 HCO3 21 = 0.2 × 15 PaO2 100 =3 HCO3 = 24 – 3 = 21 Na 137 K 3.9 Compensation is appropriate. Cl 99 Calcium 9.0 • Anion Gap = 137 – (99 + 21) = 17 AG is slightly high which can be seen in respiratory alkalosis.
Case 7 • Explain the acid-base status of a 18-year-old boy with history of chronic renal failure treated with high dose diuretics admitted to hospital with pneumonia and the following lab values: ABG pH 7.52 PaCO2 30 mm Hg PaO2 62 mm Hg Serum Electrolytes Na+ 145 mEq/L K+ 2.9 mEq/L Cl 98 mEq/L - HCO3 21 mEq/L
Case 7: Solution • Dominant disorder is Respiratory Alkalosis • Compensation formula: pH Δ HCO3 = 0.2 × Δ PaCO2 PaCO2 HCO3 = 0.2 × 10 PaO2 =2 HCO3 = 24 – 2 = 22 Na K Compensation is appropriate. Cl • Anion Gap = 145 – (98 + 21) = 26 AG is very high suggestive of metabolic acidosis. 7.52 30 21 62 145 2.9 98
Case 7: Solution • Δ AG = 26 – 12 = 14 • Δ HCO3 = 24 – 21 =3 • Δ AG > Δ HCO3High AG Met Acidosis + Met. Alkalosis • Final Diagnosis: Respiratory Alkalosis + High AG Metabolic Acidosis + Metabolic Alkalosis pH PaCO2 HCO3 PaO2 7.52 30 21 62 Na K Cl 145 2.9 98
Case 8 • The following values are found in a 65-year-old patient. Evaluate this patient's acid-base status? ABG pH 7.51 Serum Chemistry Na + 155 mEq/L PaCO2 50 mm Hg HCO3- 40 mEq/L K+ 5.5 mEq/L Cl- 90 mEq/L CO2 40 mEq/L BUN 121 mg/dl Glucose 77 mg/dl
Case 8: Solution • Dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 16 = 11.2 PaCO2 = 40 + 11 = 51 Compensation is appropriate. • Anion Gap AG is high. = 155 – (90 + 40) = 25 pH PaCO2 HCO3 PaO2 7.51 50 40 62 Na K Cl BUN 155 5.5 90 121
Case 8: Solution • Δ AG = 25 – 12 = 13 • High AG metabolic acidosis • Final Diagnosis: Metabolic Alkalosis + High AG Metabolic Acidosis pH PaCO2 HCO3 PaO2 7.51 50 40 62 Na K Cl BUN 155 5.5 90 121
Case 9 • A 52-year-old woman has been mechanically ventilated for two days following a drug overdose. Her arterial blood gas values and electrolytes, stable for the past 12 hours, show: ABG pH 7.45 PaCO2 25 mm Hg Serum Chemistry Na + 142 mEq/L K+ 4.0 mEq/L Cl- 100 mEq/L HCO3- 18 mEq/L
Case 9: Solution • Dominant disorder is Chronic Respiratory Alkalosis • Compensation formula: pH 7.45 Δ HCO3 = 0.5 × Δ PaCO2 PaCO2 25 = 0.5 × 15 HCO3 18 = 7.5 HCO3 = 24 – 8 = 16 Na 142 K 4.0 Compensation is appropriate. Cl 100 • Anion Gap = 142 – (100 + 18) = 24 AG is very high suggestive of metabolic acidosis.
Case 9: Solution • Δ AG = 24 – 12 = 12 • Δ HCO3 = 24 –18 =6 • Δ AG > Δ HCO3High AG Met Acidosis + Met. Alkalosis • Final Diagnosis: Chronic Respiratory Alkalosis + High AG Metabolic Acidosis + ? Metabolic Alkalosis
Case 11 • A 21 year old male with progressive renal insufficiency is admitted with abdominal cramping. He had congenital obstructive uropathy with creation of ileal loop for diversion. On admission, ABG pH 7.20 PaCO2 24 mm Hg Serum Chemistry Na + 140 mEq/L K+ 5.6 mEq/L Cl- 110 mEq/L HCO3- 10 mEq/L
Case 11: Solution • Dominant disorder is Metabolic Acidosis • Compensation formula: Δ PaCO2 = 1.2 × Δ HCO3 = 1.2 × 14 = 16.8 PaCO2 = 40 – 17 = 23 Compensation is appropriate. • Anion Gap = 140 – (110 + 10) = 20 High anion-gap metabolic acidosis. pH PaCO2 HCO3 7.20 24 10 Na K Cl 140 5.6 110
Case 11: Solution • Δ AG = 20 – 12 =8 • Δ HCO3 = 24 –10 = 14 • Δ AG < Δ HCO3- pH PaCO2 HCO3 7.20 24 10 Na K Cl 140 5.6 110 High AG Met Acidosis + Normal-AG Met. Acidosis • Final Diagnosis: Mixed Metabolic Acidosis
Case 12 • A 15 year old female with hypertension was treated with low salt diet and diuretics. BP 135/85. Otherwise normal. See initial lab values. • She developed profound watery diarrhea, nausea and weakness. • On exam, HR = 96, T=100.6 F, BP 115/70. Abdominal tenderness with guarding on palpation. Parameter Initial Subseq uent Na 137 138 K+ 3.1 2.8 Cl- 90 102 HCO3 35 25 pH 7.51 7.42 PaCO2 47 39
Case 12: Solution • Initally, dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 11 = 7.7 PaCO2 = 40 + 8 = 48 Compensation is appropriate. • Anion Gap AG is normal. = 137 – (90 + 35) = 12 pH PaCO2 HCO3 7.51 47 35 Na K Cl 137 3.1 90
Case 12: Solution • Subsequently, she has developed pH HCO3 PaCO2 ↓ ↓ ↓ pH PaCO2 HCO3 7.51  47  35  7.42 39 25 Na K Cl 137  3.1  90  138 2.8 102
Case 12: Solution • Subsequently, she has developed pH HCO3 PaCO2 ↓ ↓ ↓ Metabolic acidosis The decrease in bicarbonate is almost same as the rise in chloride. • Final Diagnosis: Metabolic Alkalosis + Hyperchloremic (non-AG) Metabolic Acidosis
Case 13 • A patient with salicylate overdose. pH = 7.45 PCO2 = 20 mmHg HCO3 = 13 mEq/L • Dominant disorder: Respiratory alkalosis • Appropriate Compensation would have been HCO3 of 20 (24 – 4) • Lower than expected HCO3 suggests presence of metabolic acidosis as well.
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Presentation1

  • 1.
    APPROACH TO BLOODGAS ANALYSIS Dr. MANDAR HAVAL D.C.H D.N.B
  • 3.
    How does thekidney do it? • The kidney does it in three ways: – Total reabsorption of filtered bicarbonate (proximal). – Controlled secretion of H+ into filtrate (distal). – Judicious use of urinary buffers.
  • 4.
  • 5.
    FILTRATE TUBULAR CELL H2O +CO2 CA II H2CO3 H+ + HCO3- BLOOD
  • 6.
    FILTRATE TUBULAR CELL H2O +CO2 CA II H2CO3 H+ + HCO3- BLOOD
  • 7.
    FILTRATE TUBULAR CELL BLOOD H2O +CO2 CA II H2CO3 H+ + HCO3Na+ Na + Na Na K ATPase K
  • 8.
    TUBULAR CELL FILTRATE BLOOD H2O +CO2 CA II H2CO3 H+ATPase Na+ Na+ / H+ Antiporter + HH+ + HCO33-HCO Na Na+ Na K ATPase K
  • 9.
    TUBULAR CELL FILTRATE BLOOD H2O +CO2 CA II H2CO3 Na / K H+ HCO3- H+ATPase Na+ / H+ Antiporter Na+ Na Na+ Na K ATPase K
  • 10.
    TUBULAR CELL FILTRATE H2O +CO2 BLOOD H2O + CO2 CA II H2CO3 CA IV H2CO3 HCO3- Na / K H+ HCO3- H+ATPase Na+ / H+ Antiporter Na+ Na Na+ Na K ATPase K
  • 11.
    FILTRATE H2O TUBULAR CELL BLOOD H2O +CO2 CA II CA IV H+ HCO3-
  • 12.
    COLLECTING TUBULE CELL BLOOD FILTRATE H2O +CO2 CA II H2CO3 Cl- / HCO3- Cl- Exchanger HCO3- H+ H+ ATPase
  • 13.
    COLLECTING TUBULE CELL BLOOD FILTRATE H2O +CO2 CA II H2CO3 HCO3- H+ ATPase Cl- / HCO3- Exchanger Cl- H+
  • 14.
  • 15.
  • 16.
  • 17.
  • 18.
    Evaluation of Systemic AcidBase Disorders 1. Comprehensive history and physical examination. 2. Evaluate simultaneously performed ABG & serum electrolytes. 3. Identification of the dominant disorder. 4. Calculation of compensation. 5. Calculate the anion gap and the Δ. 1. Anion Gap 2. Δ AG 3. Δ Bicarbonate
  • 19.
    Step 3: Identification ofthe dominant disorder Primary disorder Metabolic acidosis pH Initial change Compensatory change ↓ ↓ HCO3 ↓ PCO2
  • 20.
    Step 3: Identification ofthe dominant disorder pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2 Metabolic alkalosis ↑ ↑ HCO3 ↑ PCO2 Primary disorder
  • 21.
    Step 3: Identification ofthe dominant disorder pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2 Metabolic alkalosis ↑ ↑ HCO3 ↑ PCO2 Respiratory acidosis ↓ ↑ PCO2 ↑ HCO3 Respiratory alkalosis ↑ ↓ PCO2 ↓ HCO3 Primary disorder
  • 22.
    • WHERE THEPROBLEM START
  • 23.
    Calculation of compensation Mean"whole body" response equations for simple acid-base disturbances. Disorder pH Primary change Compensatory Response Equation Metabolic Acidosis   [HCO3-]  PCO2 ΔPCO2  1.2  ΔHCO3 Metabolic Alkalosis   [HCO3-]  PCO2 ΔPCO2  0.7  ΔHCO3 Respiratory Acidosis   PCO2  [HCO3-] Acute: ΔHCO3-  0.1  ΔPCO2 Chronic: ΔHCO3-  0.3  ΔPCO2 Respiratory Alkalosis   PCO2  [HCO3-] Acute: ΔHCO3-  0.2  ΔPCO2 Chronic: ΔHCO3-  0.5  ΔPCO2 Note: The formula calculates the change in the compensatory parameter.
  • 24.
    Simple compensation Disorder pH Primary problem Compensation Metabolicacidosis ↓ ↓ in HCO3- PaCO2 =1.5xHCO3+8(+/-2) Metabolic alkalosis ↑ 10↑ in HCO3- 7↑ in PaCO2 Respiratory acidosis ↓ ACUTE -10↑ in PaCO2 CHRONIC -10↑ in PaCO2 1↑ in [HCO3-] 3.5↑ in [HCO3-] Respiratory alkalosis ↑ ACUTE-10↓ in PaCO2 CHRONIC-10↓ in PaCO2 2↓ in [HCO3-] 4↓ in [HCO3-]
  • 25.
    Calculate the “gaps” Aniongap = Na+ − [Cl− + HCO3−] Δ AG = Anion gap − 12 Δ HCO3 = 24 − HCO3 Δ AG = Δ HCO3 −, then Pure high AG Met. Acidosis Δ AG > Δ HCO3 −, then High AG Met Acidosis + Met. Alkalosis Δ AG < Δ HCO3 −, then High AG Met Acidosis + Normal AG Met A Note:  Add Δ AG to measured HCO3− to obtain bicarbonate level  Delta _ AG    Pr e _ existing _ Bicarb  that would have existed IF the high AG metabolic acidosis Current _ Bicarb   were to be absent, i.e., “Pre-existing Bicarbonate.”
  • 26.
  • 27.
    CALCULATION OF H+ H  24  HCO  PaCO2  3 20 – 7.70 30 – 7.50 40(H+) – 7.40 (PH) 50 – 7.30 65 – 7.20
  • 28.
  • 29.
    CAO2= directly reflectsthe total number of oxygen molecules in arterial blood, both bound and unbound to hemoglobin • CaO2 = (1.34 x HB x SPO2) +(0.003 x PaO2) Normal CaO2 ranges from 16 to 22 ml O2/dl
  • 30.
    Which patient ismore hypoxemic, and why? • Patient A: pH 7.48 PaCO2 34 mm Hg PaO2 85 mm Hg SaO2 95% Hemoglobin 7 gm% • Patient B: pH 7.32 PaCO2 74 mm Hg PaO255 mm Hg SaO2 85% Hemoglobin 15 gm% www.dnbpediatrics.com
  • 31.
    ANS CONT….. • PatientA: Arterial oxygen content = .95 x 7 x 1.34 = 8.9 ml O2/dl • Patient B: Arterial oxygen content = .85 x 15 x 1.34 = 17.1 ml O2/dl • Patient A, with the higher PaO2 but the lower hemoglobin content, is more hypoxemic www.dnbpediatrics.com
  • 32.
    PaO2 • Factors affectingthe PaO2 include alveolar ventilation, FIO2, altitude, age, and the oxyhemoglobin dissociation curve • Relation between PaO2 and SaO2: PaO2 60mm Hg 50mm Hg 40mm Hg 30mm Hg corresponds to SaO2 90% 80% 70% 60%
  • 33.
    True or False: ThepO2 in a cup of water open to the atmosphere is always higher than the arterial pO2 in a healthy person (breathing room air) who is holding the cup www.dnbpediatrics.com
  • 34.
    ANS • The PO2in the cup of water is always higher. This is for several reasons. First, there is no barrier to oxygen diffusing into the water; thus the PO2 in the cup will be the same as the atmosphere, at sea level approximately 160 mm Hg. • Second, there is no CO2 coming from the cup to dilute the oxygen, as there is in people. • Third, there is no V-Q inequality or shunt; even healthy people have a difference between alveolar PO2 and arterial PO2 for this reason. Thus a healthy person and a cup of water exposed to the atmosphere at sea level would have PO2 values of about 100 mm Hg and 160 mm Hg, respectively. www.dnbpediatrics.com
  • 35.
    A-a Gradient • Determinesthe degree of lung function impairment • The A-a gradient is the partial pressure of alveolar oxygen minus the partial pressure of arterial oxygen (PAO2-PaO2) • Normal is 2-10mm Hg or 10 plus one tenth the person’s age
  • 36.
    A-a Gradient • [(713*FIO2)-(PaCO2/0.8)]– PaO2 INTERPRETATION NORMAL – 10-20 (>30 is SINGNIFICANT) Seen in – Shunt Low V/Q Hypoventilation
  • 37.
    A-a Gradient • PAO2-PaO2of 20-30mm Hg on room air indicates mild pulmonary dysfunction, and greater than 50mm Hg on room air indicates severe pulmonary dysfunction • The causes of increased gradient include intrapulmonary shunt, intracardiac shunt, and diffusion abnormalities
  • 39.
    a/A Ratio • Pao2/PAo2NAORMAL LEVEL IS >0.75 • <0.60 IS INCOMPATIBLE WITH SPONTANIOUS BREATHING
  • 40.
    PaO2/FIO2 Ratio • Toestimate the impairment of oxygenation, calculate the PaO2/FIO2 ratio • Normally, this ratio is 500-600 • Below 300 is acute lung injury* • Below 200 is ARDS* *Along with diffuse infiltrates, normal PCWP, and appropriate mechanism
  • 41.
    OXYGEN INDEX • OI=MAP X FIO2 POST DUCTAL PAO2 x 100
  • 42.
    INTERPRETATION • OI >40that is unresponsive to iNO predict a high mortality rate (>80%) and are indications for ECMO.
  • 43.
    VENTILATORY INDEX • VI=PIP X PCO2 X RR 1000 VI > 65% INDICATE PREDICTIVE DEATH IN ARDS
  • 44.
    RELATION OF ALBUMININ ABG  AG corrected = AG + 2.5[4 – albumin] (AG= Anion gap)
  • 45.
    DELTA GAP      Delta gap= (actual AG – 12) + HCO3 Adjusted HCO3 should be 24 (+_ 6) {18-30} If delta gap > 30 -> additional metabolic alkalosis If delta gap < 18 -> additional non-gap metabolic acidosis If delta gap 18 – 30 -> no additional metabolic disorders
  • 46.
  • 47.
    Case 1 • A15 yr old juvenile diabetic presents with abdominal pain, vomiting, fever & tiredness for 1 day. He had stopped taking insulin 3 days ago. Examination revealed tachycardia, BP- 100/60, signs of dehydration. Abdominal examination was normal. • ABG: pH PaCO2 HCO3 PaO2 7.31 26 mmHg 12 mEq/L 92 mm Hg Serum Electrolytes: Na 140 mEq/L K 5.0 mEq/L Cl 100 mEq/L • Evaluate the acid-base disturbance(s)?
  • 48.
    Case 1: Solution •Dominant disorder is Metabolic Acidosis • Compensation formula: Δ PaCO2 = 1.2 × Δ HCO3 = 1.2 × 12 = 14.4 PaCO2 = 40 – 14 = 26 Compensation is appropriate. • Anion Gap = 140 – (100 + 12) = 28 AG is high. pH PaCO2 HCO3 PaO2 7.31 26 12 92 Na K Cl 140 5.0 100
  • 49.
    Case 1: Solution •Δ AG = 28 – 12 = 16 • Δ HCO3 • Δ AG > Δ = 24 – 12 = 12 HCO3 pH PaCO2 HCO3 PaO2 7.31 26 12 92 Na K Cl 140 5.0 100 • Final Diagnosis: High AG Met. Acidosis + Met. Alkalosis
  • 50.
    Case 2 • A14 yr old boy presents with continuous vomiting of 3 days duration, mental confusion, giddiness, and tiredness for 1 day. • Examination revealed tachycardia, hypotension and dehydration. • ABG pH 7.50 Serum Electrolytes: PaCO2 48 Na 139 HCO3 32 K 3.9 PaO2 90 Cl 85 • Evaluate the acid-base disturbance(s)?
  • 51.
    Case 2: Solution •Dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 8 = 5.6 PaCO2 = 40 + 6 = 46 Compensation is appropriate. • Anion Gap = 139 – (85 + 32) = 22 AG is high. pH PaCO2 HCO3 PaO2 7.50 48 32 90 Na K Cl 139 3.9 85
  • 52.
    Case 2: Solution •Δ AG = 22 – 12 = 10 • High AG metabolic acidosis • Final Diagnosis: pH PaCO2 HCO3 PaO2 7.50 48 32 90 Na K Cl 139 3.9 85 Metabolic Alkalosis + High AG Met. Acidosis
  • 53.
    Case 3: Varietiesof Metabolic Acidosis Patient ECF volume Glucose pH Na Cl HCO3 AG Ketones A Low 600 7.20 140 103 10 27 4+ B Low 120 7.20 140 118 10 12 0 C Normal 120 7.20 140 118 10 12 0 High-AG Met. Acidosis Non-AG Met. Acidosis Non-AG Met. Acidosis
  • 54.
    Renal handling ofHydrogen in Metabolic Acidosis • In the setting of metabolic acidosis, normal kidneys try to increase H+ excretion by increasing titratable acidity and ammonia. The latter is excreted as NH4+. • When NH4+ is excreted, it also causes increased chloride loss, to maintain electrical neutrality. • Chloride loss, therefore, will be in excess of Na and K. • Urine Anion-Gap = Na + K – Cl • In metabolic acidosis, if Urine anion gap is negative, it suggests that the kidneys are excreting H+ effectively.
  • 55.
    Urine Electrolytes inMetabolic Acidosis Patient U. Na U. K U. Cl Urine AG A Dx: B 10 14 74 –50 Diarrhea C 50 47 28 +69 RTA Urine Anion Gap = (U. Na + U. K – U. Cl) In Normal anion gap Metabolic Acidosis, Positive Urine AG suggests distal Renal Tubular Acidosis Negative Urine AG suggests non-renal cause for Metabolic Acidosis.
  • 56.
    Case 4 • A17 yr old boy presented with history of progressive dyspnoea with wheezing for 4 days. • He also had fever, cough with yellowish expectoration. • He had increased sleepiness for 1 day. • On examination, he was tachypnoeic, pulse100/min bounding, BP-160/96, central cyanosis +, drowsy, asterixis +, RS – B/L extensive wheezing +. • CXR- hyperinflated lung fields with tubular heart.
  • 57.
    Case 4: Laboratorydata • ABG: pH PaCO HCO PaO 2 3 2 7.30 60 mmHg 28 mEq/L 68 mm Hg • Serum Electrolytes: Na 136 mEq/L K 4.5 mEq/L Cl 98 mEq/L • Evaluate the acid-base disturbance(s)?
  • 58.
    Case 4: Solution •Dominant disorder is Respiratory Acidosis • Compensation formula: Δ HCO3 = 0.3 × Δ PaCO2 = 0.3 × 20 =6 HCO3 = 24 + 6 = 30 Compensation is appropriate. • Anion Gap = 136 – (98 + 28) = 10 AG is normal. pH PaCO2 HCO3 PaO2 7.30 60 28 68 Na K Cl 136 4.5 98
  • 59.
    Case 5 • 12year old girl presented with complaints of difficulty in breathing and upper abdominal discomfort for the past 1 hr. • On examination, vitals normal, patient hyperventilating, RS – normal, Abdomen – normal.
  • 60.
    Case 5: Laboratorydata • ABG: pH PaCO HCO PaO 2 3 2 7.50 25 mmHg 21 mEq/L 100 mm Hg • Serum Electrolytes: Na 137 mEq/L K 3.9 mEq/L Cl 99 mEq/L Calcium 9.0 mEq/L • Evaluate the acid-base disturbance(s)?
  • 61.
    Case 5: Solution •Dominant disorder is Respiratory Alkalosis • Compensation formula: pH 7.50 Δ HCO3 = 0.2 × Δ PaCO2 PaCO2 25 HCO3 21 = 0.2 × 15 PaO2 100 =3 HCO3 = 24 – 3 = 21 Na 137 K 3.9 Compensation is appropriate. Cl 99 Calcium 9.0 • Anion Gap = 137 – (99 + 21) = 17 AG is slightly high which can be seen in respiratory alkalosis.
  • 62.
    Case 7 • Explainthe acid-base status of a 18-year-old boy with history of chronic renal failure treated with high dose diuretics admitted to hospital with pneumonia and the following lab values: ABG pH 7.52 PaCO2 30 mm Hg PaO2 62 mm Hg Serum Electrolytes Na+ 145 mEq/L K+ 2.9 mEq/L Cl 98 mEq/L - HCO3 21 mEq/L
  • 63.
    Case 7: Solution •Dominant disorder is Respiratory Alkalosis • Compensation formula: pH Δ HCO3 = 0.2 × Δ PaCO2 PaCO2 HCO3 = 0.2 × 10 PaO2 =2 HCO3 = 24 – 2 = 22 Na K Compensation is appropriate. Cl • Anion Gap = 145 – (98 + 21) = 26 AG is very high suggestive of metabolic acidosis. 7.52 30 21 62 145 2.9 98
  • 64.
    Case 7: Solution •Δ AG = 26 – 12 = 14 • Δ HCO3 = 24 – 21 =3 • Δ AG > Δ HCO3High AG Met Acidosis + Met. Alkalosis • Final Diagnosis: Respiratory Alkalosis + High AG Metabolic Acidosis + Metabolic Alkalosis pH PaCO2 HCO3 PaO2 7.52 30 21 62 Na K Cl 145 2.9 98
  • 65.
    Case 8 • Thefollowing values are found in a 65-year-old patient. Evaluate this patient's acid-base status? ABG pH 7.51 Serum Chemistry Na + 155 mEq/L PaCO2 50 mm Hg HCO3- 40 mEq/L K+ 5.5 mEq/L Cl- 90 mEq/L CO2 40 mEq/L BUN 121 mg/dl Glucose 77 mg/dl
  • 66.
    Case 8: Solution •Dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 16 = 11.2 PaCO2 = 40 + 11 = 51 Compensation is appropriate. • Anion Gap AG is high. = 155 – (90 + 40) = 25 pH PaCO2 HCO3 PaO2 7.51 50 40 62 Na K Cl BUN 155 5.5 90 121
  • 67.
    Case 8: Solution •Δ AG = 25 – 12 = 13 • High AG metabolic acidosis • Final Diagnosis: Metabolic Alkalosis + High AG Metabolic Acidosis pH PaCO2 HCO3 PaO2 7.51 50 40 62 Na K Cl BUN 155 5.5 90 121
  • 68.
    Case 9 • A52-year-old woman has been mechanically ventilated for two days following a drug overdose. Her arterial blood gas values and electrolytes, stable for the past 12 hours, show: ABG pH 7.45 PaCO2 25 mm Hg Serum Chemistry Na + 142 mEq/L K+ 4.0 mEq/L Cl- 100 mEq/L HCO3- 18 mEq/L
  • 69.
    Case 9: Solution •Dominant disorder is Chronic Respiratory Alkalosis • Compensation formula: pH 7.45 Δ HCO3 = 0.5 × Δ PaCO2 PaCO2 25 = 0.5 × 15 HCO3 18 = 7.5 HCO3 = 24 – 8 = 16 Na 142 K 4.0 Compensation is appropriate. Cl 100 • Anion Gap = 142 – (100 + 18) = 24 AG is very high suggestive of metabolic acidosis.
  • 70.
    Case 9: Solution •Δ AG = 24 – 12 = 12 • Δ HCO3 = 24 –18 =6 • Δ AG > Δ HCO3High AG Met Acidosis + Met. Alkalosis • Final Diagnosis: Chronic Respiratory Alkalosis + High AG Metabolic Acidosis + ? Metabolic Alkalosis
  • 71.
    Case 11 • A21 year old male with progressive renal insufficiency is admitted with abdominal cramping. He had congenital obstructive uropathy with creation of ileal loop for diversion. On admission, ABG pH 7.20 PaCO2 24 mm Hg Serum Chemistry Na + 140 mEq/L K+ 5.6 mEq/L Cl- 110 mEq/L HCO3- 10 mEq/L
  • 72.
    Case 11: Solution •Dominant disorder is Metabolic Acidosis • Compensation formula: Δ PaCO2 = 1.2 × Δ HCO3 = 1.2 × 14 = 16.8 PaCO2 = 40 – 17 = 23 Compensation is appropriate. • Anion Gap = 140 – (110 + 10) = 20 High anion-gap metabolic acidosis. pH PaCO2 HCO3 7.20 24 10 Na K Cl 140 5.6 110
  • 73.
    Case 11: Solution •Δ AG = 20 – 12 =8 • Δ HCO3 = 24 –10 = 14 • Δ AG < Δ HCO3- pH PaCO2 HCO3 7.20 24 10 Na K Cl 140 5.6 110 High AG Met Acidosis + Normal-AG Met. Acidosis • Final Diagnosis: Mixed Metabolic Acidosis
  • 74.
    Case 12 • A15 year old female with hypertension was treated with low salt diet and diuretics. BP 135/85. Otherwise normal. See initial lab values. • She developed profound watery diarrhea, nausea and weakness. • On exam, HR = 96, T=100.6 F, BP 115/70. Abdominal tenderness with guarding on palpation. Parameter Initial Subseq uent Na 137 138 K+ 3.1 2.8 Cl- 90 102 HCO3 35 25 pH 7.51 7.42 PaCO2 47 39
  • 75.
    Case 12: Solution •Initally, dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 11 = 7.7 PaCO2 = 40 + 8 = 48 Compensation is appropriate. • Anion Gap AG is normal. = 137 – (90 + 35) = 12 pH PaCO2 HCO3 7.51 47 35 Na K Cl 137 3.1 90
  • 76.
    Case 12: Solution •Subsequently, she has developed pH HCO3 PaCO2 ↓ ↓ ↓ pH PaCO2 HCO3 7.51  47  35  7.42 39 25 Na K Cl 137  3.1  90  138 2.8 102
  • 77.
    Case 12: Solution •Subsequently, she has developed pH HCO3 PaCO2 ↓ ↓ ↓ Metabolic acidosis The decrease in bicarbonate is almost same as the rise in chloride. • Final Diagnosis: Metabolic Alkalosis + Hyperchloremic (non-AG) Metabolic Acidosis
  • 78.
    Case 13 • Apatient with salicylate overdose. pH = 7.45 PCO2 = 20 mmHg HCO3 = 13 mEq/L • Dominant disorder: Respiratory alkalosis • Appropriate Compensation would have been HCO3 of 20 (24 – 4) • Lower than expected HCO3 suggests presence of metabolic acidosis as well.

Editor's Notes

  • #46 Must memorize how to calculate the delta gapJust read off the slide