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APPROACH TO ABG AND SO
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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.
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
KNa+
Na+
TUBULAR CELL BLOODFILTRATE
H2O + CO2
H2CO3
H+ + HCO3-
CA II
Na K ATPase
Na
KNa+
Na+
H+
Na+ / H+
Antiporter
HCO3-H+ATPase
TUBULAR CELL BLOODFILTRATE
H2O + CO2
H2CO3
CA II
Na K ATPase
Na
KNa+
Na+
H+
Na+ / H+ Antiporter
HCO3-
Na / KH+ATPase
TUBULAR CELL BLOODFILTRATE
H2O + CO2
H2CO3
CA II
Na K ATPase
Na
KNa+
Na+
H+
Na+ / H+ Antiporter
HCO3-
Na / KHCO3
-
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-
ExchangerCl-
H+
COLLECTING TUBULE CELL FILTRATEBLOOD
H2O + CO2
H2CO3
HCO3-
CA II
H+ ATPase
Cl- / HCO3-
ExchangerCl-
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 compensationMean "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 ANote:
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
24HCO
PaCOH
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 7.48PaCO2 34 mm Hg
PaO2 85 mm Hg
SaO2 95%
Hemoglobin 7 gm%
• Patient B: pH 7.32PaCO2 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 7.31PaCO2 26 mmHgHCO3 12 mEq/LPaO2 92 mm Hg
• Evaluate the acid-base disturbance(s)?
Serum Electrolytes:Na 140 mEq/LK 5.0 mEq/LCl 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 7.31PaCO2 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 7.31PaCO2 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.
• ABGpH 7.50PaCO2 48HCO3 32PaO2 90
• Evaluate the acid-base disturbance(s)?
Serum Electrolytes:Na 139K 3.9Cl 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 7.50PaCO2 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 7.50PaCO2 48 HCO3 32 PaO2 90
Na 139 K 3.9 Cl 85
Case 3: Varieties of Metabolic Acidosis
Patient A B CECF volume Low Low NormalGlucose 600 120 120pH 7.20 7.20 7.20Na 140 140 140Cl 103 118 118
HCO3- 10 10 10
AG 27 12 12Ketones 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 CU. Na 10 50U. K 14 47U. Cl 74 28Urine 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 7.30PaCO2 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 7.30PaCO2 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 7.50PaCO2 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 7.50PaCO2 25 HCO3 21 PaO2 100
Na 137 K 3.9 Cl 99Calcium 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 7.52PaCO2 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 7.52PaCO2 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 7.51PaCO2 50 HCO3 40 PaO2 62
Na 155 K 5.5 Cl 90BUN 121
Case 8: Solution
• Δ AG = 25 – 12= 13
• High AG metabolic acidosis
• Final Diagnosis:
Metabolic Alkalosis + High AG Metabolic Acidosis
pH 7.51PaCO2 50 HCO3 40 PaO2 62
Na 155 K 5.5 Cl 90BUN 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 ChemistrypH 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 7.45PaCO2 25 HCO3 18
Na 142K 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 ChemistrypH 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 7.20PaCO2 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 7.20PaCO2 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 InitialSubsequent
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 7.51PaCO2 47 HCO3 35
Na 137 K 3.1 Cl 90
Case 12: Solution
• Subsequently, she has developedpH HCO3 PaCO2
↓ ↓ ↓
pH 7.51 7.42PaCO2 47 39 HCO3 35 25
Na 137 138 K 3.1 2.8Cl 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 = 7.45PCO2 = 20 mmHgHCO3 = 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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