DNB OSCE ON ABG
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DNB OSCE ON ABG

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APPROACH TO ABG AND SO

APPROACH TO ABG AND SO

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

DNB OSCE ON ABG DNB OSCE ON ABG Presentation Transcript

  • 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.
  • TUBULAR CELL BLOODFILTRATE
  • TUBULAR CELL BLOODFILTRATE H2O + CO2 H2CO3 H+ + HCO3 - CA II
  • TUBULAR CELL BLOODFILTRATE H2O + CO2 H2CO3 H+ + HCO3 - CA II
  • TUBULAR CELL BLOODFILTRATE H2O + CO2 H2CO3 H+ + HCO3 - CA II Na K ATPase Na K Na+ Na+
  • TUBULAR CELL BLOODFILTRATE H2O + CO2 H2CO3 H+ + HCO3 - CA II Na K ATPase Na K Na+ Na+ H+ Na+ / H+ Antiporter HCO3 -H+ATPase
  • TUBULAR CELL BLOODFILTRATE H2O + CO2 H2CO3 CA II Na K ATPase Na K Na+ Na+ H+ Na+ / H+ Antiporter HCO3 - Na / K H+ATPase
  • TUBULAR CELL BLOODFILTRATE H2O + CO2 H2CO3 CA II Na K ATPase Na K Na+ Na+ H+ Na+ / H+ Antiporter HCO3 - Na / K HCO3 - H2CO3 H2O + CO2 CA IV H+ATPase
  • TUBULAR CELL BLOODFILTRATE H2O + CO2 CA II H2O CA IV HCO3 -H+
  • COLLECTING TUBULE CELL FILTRATEBLOOD H2O + CO2 H2CO3 HCO3 - CA II H+ ATPase Cl- / HCO3 - Exchanger Cl- H+
  • COLLECTING TUBULE CELL FILTRATEBLOOD H2O + CO2 H2CO3 HCO3 - CA II H+ ATPase Cl- / HCO3 - Exchanger Cl- H+
  • COLLECTING TUBULE CELL FILTRATE BLOOD H+ ATPase H+ HPO4 =
  • COLLECTING TUBULE CELL FILTRATE BLOOD H+ ATPase H+ HPO4 =H2PO4 -
  • COLLECTING TUBULE CELL FILTRATE BLOOD H+ ATPase H+ SO4 = HSO4-
  • COLLECTING TUBULE CELL FILTRATE BLOOD H+ ATPase H+ NH3 NH3 NH4 +
  • 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 pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2
  • Step 3: Identification of the dominant disorder Primary disorder pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2 Metabolic alkalosis ↑ ↑ HCO3 ↑ PCO2
  • Step 3: Identification of the dominant disorder Primary disorder pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2 Metabolic alkalosis ↑ ↑ HCO3 ↑ PCO2 Respiratory acidosis ↓ ↑ PCO2 ↑ HCO3 Respiratory alkalosis ↑ ↓ PCO2 ↓ HCO3
  • • WHERE THE PROBLEM START
  • Calculation of compensation Mean "whole body" response equations for simple acid-base disturbances. Note: The formula calculates the change in the compensatory parameter. 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
  • 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 that would have existed IF the high AG metabolic acidosis were to be absent, i.e., “Pre-existing Bicarbonate.”  Bicarbexistinge BicarbCurrent AGDelta __Pr _ _            
  • SOME FORMULA •THAT YOU SHOULD KNOW
  • CALCULATION OF H+ 20 – 7.70 30 – 7.50 40(H+) – 7.40 (PH) 50 – 7.30 65 – 7.20      3 2 24 HCO PaCO H
  • 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 corresponds to SaO2 60mm Hg 90% 50mm Hg 80% 40mm Hg 70% 30mm Hg 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 x 100 POST DUCTAL PAO2
  • 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 7.31 PaCO2 26 mmHg HCO3 12 mEq/L PaO2 92 mm Hg • Evaluate the acid-base disturbance(s)? Serum Electrolytes: Na 140 mEq/L K 5.0 mEq/L Cl 100 mEq/L
  • 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 7.31 PaCO2 26 HCO3 12 PaO2 92 Na 140 K 5.0 Cl 100
  • Case 1: Solution • Δ AG = 28 – 12 = 16 • Δ HCO3 = 24 – 12 = 12 • Δ AG > Δ HCO3 - • Final Diagnosis: High AG Met. Acidosis + Met. Alkalosis pH 7.31 PaCO2 26 HCO3 12 PaO2 92 Na 140 K 5.0 Cl 100
  • 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 PaCO2 48 HCO3 32 PaO2 90 • Evaluate the acid-base disturbance(s)? Serum Electrolytes: Na 139 K 3.9 Cl 85
  • 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 7.50 PaCO2 48 HCO3 32 PaO2 90 Na 139 K 3.9 Cl 85
  • Case 2: Solution • Δ AG = 22 – 12 = 10 • High AG metabolic acidosis • Final Diagnosis: Metabolic Alkalosis + High AG Met. Acidosis pH 7.50 PaCO2 48 HCO3 32 PaO2 90 Na 139 K 3.9 Cl 85
  • Case 3: Varieties of Metabolic Acidosis Patient A B C ECF volume Low Low Normal Glucose 600 120 120 pH 7.20 7.20 7.20 Na 140 140 140 Cl 103 118 118 HCO3 - 10 10 10 AG 27 12 12 Ketones 4+ 0 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 A B C U. Na 10 50 U. K 14 47 U. Cl 74 28 Urine AG –50 +69 Dx: Diarrhea RTA In Normal anion gap Metabolic Acidosis, Positive Urine AG suggests distal Renal Tubular Acidosis Negative Urine AG suggests non-renal cause for Metabolic Acidosis. Urine Anion Gap = (U. Na + U. K – U. Cl)
  • 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, pulse- 100/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 7.30 PaCO2 60 mmHg HCO3 28 mEq/L PaO2 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 7.30 PaCO2 60 HCO3 28 PaO2 68 Na 136 K 4.5 Cl 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 7.50 PaCO2 25 mmHg HCO3 21 mEq/L PaO2 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: Δ HCO3 = 0.2 × Δ PaCO2 = 0.2 × 15 = 3 HCO3 = 24 – 3 = 21 Compensation is appropriate. • Anion Gap = 137 – (99 + 21) = 17 AG is slightly high which can be seen in respiratory alkalosis. pH 7.50 PaCO2 25 HCO3 21 PaO2 100 Na 137 K 3.9 Cl 99 Calcium 9.0
  • 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 Serum Electrolytes pH 7.52 Na+ 145 mEq/L PaCO2 30 mm Hg K+ 2.9 mEq/L PaO2 62 mm Hg Cl - 98 mEq/L HCO3 - 21 mEq/L
  • Case 7: Solution • Dominant disorder is Respiratory Alkalosis • Compensation formula: Δ HCO3 = 0.2 × Δ PaCO2 = 0.2 × 10 = 2 HCO3 = 24 – 2 = 22 Compensation is appropriate. • Anion Gap = 145 – (98 + 21) = 26 AG is very high suggestive of metabolic acidosis. pH 7.52 PaCO2 30 HCO3 21 PaO2 62 Na 145 K 2.9 Cl 98
  • Case 7: Solution • Δ AG = 26 – 12 = 14 • Δ HCO3 = 24 – 21 = 3 • Δ AG > Δ HCO3 - High AG Met Acidosis + Met. Alkalosis • Final Diagnosis: Respiratory Alkalosis + High AG Metabolic Acidosis + Metabolic Alkalosis pH 7.52 PaCO2 30 HCO3 21 PaO2 62 Na 145 K 2.9 Cl 98
  • Case 8 • The following values are found in a 65-year-old patient. Evaluate this patient's acid-base status? ABG Serum Chemistry pH 7.51 Na + 155 mEq/L PaCO2 50 mm Hg K+ 5.5 mEq/L HCO3 - 40 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 = 155 – (90 + 40) = 25 AG is high. pH 7.51 PaCO2 50 HCO3 40 PaO2 62 Na 155 K 5.5 Cl 90 BUN 121
  • Case 8: Solution • Δ AG = 25 – 12 = 13 • High AG metabolic acidosis • Final Diagnosis: Metabolic Alkalosis + High AG Metabolic Acidosis pH 7.51 PaCO2 50 HCO3 40 PaO2 62 Na 155 K 5.5 Cl 90 BUN 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 Serum Chemistry pH 7.45 Na + 142 mEq/L PaCO2 25 mm Hg K+ 4.0 mEq/L Cl- 100 mEq/L HCO3- 18 mEq/L
  • Case 9: Solution • Dominant disorder is Chronic Respiratory Alkalosis • Compensation formula: Δ HCO3 = 0.5 × Δ PaCO2 = 0.5 × 15 = 7.5 HCO3 = 24 – 8 = 16 Compensation is appropriate. • Anion Gap = 142 – (100 + 18) = 24 AG is very high suggestive of metabolic acidosis. pH 7.45 PaCO2 25 HCO3 18 Na 142 K 4.0 Cl 100
  • Case 9: Solution • Δ AG = 24 – 12 = 12 • Δ HCO3 = 24 –18 = 6 • Δ AG > Δ HCO3 - High 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 Serum Chemistry pH 7.20 Na + 140 mEq/L PaCO2 24 mm Hg 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 7.20 PaCO2 24 HCO3 10 Na 140 K 5.6 Cl 110
  • Case 11: Solution • Δ AG = 20 – 12 = 8 • Δ HCO3 = 24 –10 = 14 • Δ AG < Δ HCO3 - High AG Met Acidosis + Normal-AG Met. Acidosis • Final Diagnosis: Mixed Metabolic Acidosis pH 7.20 PaCO2 24 HCO3 10 Na 140 K 5.6 Cl 110
  • 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 = 137 – (90 + 35) = 12 AG is normal. pH 7.51 PaCO2 47 HCO3 35 Na 137 K 3.1 Cl 90
  • Case 12: Solution • Subsequently, she has developed pH HCO3 PaCO2 ↓ ↓ ↓ pH 7.51  7.42 PaCO2 47  39 HCO3 35  25 Na 137  138 K 3.1  2.8 Cl 90  102
  • Case 12: Solution • Subsequently, she has developed The decrease in bicarbonate is almost same as the rise in chloride. • Final Diagnosis: pH HCO3 PaCO2 ↓ ↓ ↓ Metabolic acidosis 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.
  • ALL THE BEST