DNB OSCE ON ABG

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

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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

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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.

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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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