Constructing an ABG

All of the cases on this site are works of fiction. Having to construct cases which are consistent across test results has taught me quite a lot, particularly with regard to the arterial blood gas (ABG). So, I thought I’d take you through the construction of one such case. In this internet age, you do not need to be able to reproduce any of the equations illustrated here but being aware of this material will help you understand the ABG in practice.

Diabetic Emergencies Case 2 (a case of diabetic ketoacidosis) was supposed to teach you about the recognition of metabolic acidosis on the ABG. Diabetic ketoacidosis (DKA) is characterised by the accumulation of large amounts of organic acids (ketones) in the extracellular fluid. The cardinal feature of this condition is, therefore, the presence of a metabolic acidosis. For historical reasons, the analysis of changes in pH in the extracellular fluid is based on analysis of the concentration of components of the bicarbonate buffer system as measured on the arterial blood gas (ABG). pH disturbances produce characteristic patterns of changes (‘footprints’) in the concentration of components of this buffer system.

Case 9 image 1

The level of these components in any solution, including arterial blood, is dictated by the Henderson Hasselbalch equation.

Case 9 image 2

(In this equation, the paCO2 is measured in mmHg. 7.5 mmHg (US units) = 1 kPa (UK/SI units))

In fact, modern ABG analysers directly measure the pH and pCO2 in an arterial sample and then use this equation to calculate the bicarbonate concentration. The bicarbonate concentration (cHCO3-, c = calculated) is calculated by the machine, it is not directly measured by it. When constructing a case we have to do things the other way around.

Construction of a metabolic acidosis on the ABG

Ketones are organic acids. Addition of organic acid to the ECF lowers the level of metabolic base (HCO3-) in the ECF (measured in the arterial blood). This low bicarbonate is the hallmark of a metabolic acidosis on the ABG. This is our starting point. From clinical experience, we give our patient a bicarbonate concentration of 9 mmol/l (normal 20 – 24 mmol/l). In the construction of our case, this value dictates almost everything else.

So, in our fictitious case, the metabolic acidosis is sufficient to push the bicarbonate down to 9 mmol/l. We know that this metabolic acidosis will trigger respiratory compensation with a lowering of the level of respiratory acid (CO2) in the ECF. We want to construct a metabolic acidosis with respiratory compensation. For teaching purposes, we want this to be a simple pH disturbance, that is, we want a level of compensation that the body can deliver, no more, no less. Winter’s formula tells us that the expected level of paCO2 in a metabolic acidosis is determined by the bicarbonate concentration in the following way,

case 9 image 3

In our case

Predicted paCO2 = 1.5(9) +8 +/-2 = 20 – 24 mmHg (normal range 35 – 45 mmHg)

So, we opt for a paCO2 in this expected range predicted by Winter’s formula (say 23 mmHg). We can now use these two values (bicarbonate and paCO2) to calculate the pH of our imaginary sample using the Henderson Hasselbalch equation.

pH = 6.1 + log (HCO3-/0.03 x CO2)  = 6.1 + log (9/0.03 x 23) = 6.1 + 1.115  = 7.215

Winter’s formula is based on observations in multiple patients. This completes the acid-base component of our patient’s ABG.

case 9 1 ABGCSi

(Obviously, going to three decimal places on the pH here won’t get me an A in maths, but in reality, the pH will be close to this value and we want the result to look the same as on the printout from the blood gas analyser)

Our patient has an acidosis (low pH). This is a metabolic acidosis (low bicarbonate) and has an appropriate level of respiratory compensation (low paCO2 in the range predicted by Winter’s formula). The student will conclude that this is a simple metabolic acidosis with adequate respiratory compensation.

What about the paO2?

case 9 intermed ABG CSi

Can we throw in any old normal value? No. The paO2 is closely related to the values we have calculated above and if you can understand this relationship, as we saw in Course 3 Case 1 (missed PE) this can help you greatly in clinical practice.

The partial pressure of oxygen in the alveoli (pAO2) is determined by the percentage of inhaled oxygen (the FiO2) and the degree of alveolar ventilation. The degree of alveolar ventilation (the volume of air entering the alveoli per unit time) is controlled by centres in the brainstem. These brainstem centres set the alveolar ventilation at a level sufficient to eliminate all CO2 produced during metabolism.

In fact, it can be shown that the pAO2 is related to these variables by the following equation.

case 9 image 4

The pAO2 is the partial pressure of oxygen in the alveoli measured in mmHg. The FiO2 is the percentage of oxygen inhaled (expressed as a decimal fraction-see below). Patm is the pervading atmospheric pressure (at sea level = 760 mmHg), PH2O is the water vapour pressure in the airways (at sea level = 47 mmHg). The paCO2 is the partial pressure of carbon dioxide in the arterial blood as measured in mmHg on the ABG.

RQ is the respiratory quotient: the ratio of the volume of carbon dioxide produced during tissue respiration relative to the volume of oxygen consumed during tissue respiration. The ratio varies with dietary composition but is usually taken as 0.8.

So, in our patient on room air at sea level the alveolar pA02 is given by

pAO2 = 0.21(760 – 47) – 23/0.8 = 0.21(713) – 29 = 121 mmHg

This, of course, assumes that our patient is breathing room air (%O2: 21%, FiO2: 0.21). However, we want our case to be realistic. If a critically ill young woman is brought into ED/ER we can assume that, irrespective of her oxygen saturation on arrival, she will immediately be placed on oxygen. So, in the construction of our fictitious case we assume that she has been started on 4l oxygen per minute (36% oxygen, FiO2: 0.36). We indicate this on her observation chart and hope that the student spots it. Her (alveolar) pAO2 is therefore,

pAO2 = 0.36 (713) – paCO2/0.8 = 228 mmHg

How will this (alveolar) pAO2 translate into an (arterial) paO2? Oxygen diffuses from the alveoli into the alveolar capillaries, hence, oxygenating the blood returning to the left side of the heart. Therefore, there is a tight relationship between the pAO2 and paO2 on the ABG.

Transfer of oxygen between the alveolar lumen and blood in the alveolar capillaries is never 100% efficient. The pAO2 is always higher than the paO2. There is always a gap (a gradient) between the alveolar and arterial pO2. The magnitude of the difference between the two is known as the A-a gradient and varies with age. The normal A-a gradient at a given age on room air can be estimated by the expression Age/4 + 4. At a given age, the presence of disease elevates the A-a gradient. We decide that our fictitious case has normal lungs.

She is 22 years old. So, her predicted A-a gradient is 22/4 + 4 = about 9 – 10 mmHg

However, we must also remember that the normal A-a gradient rises by about 7.5 mmHg (1 kPa) for every 10% increase in FiO2 above room air. Our fictitious patient is on 36% O2. So, we will add about 11 mmHg to her gradient. A woman of her age on this level of oxygen will have a normal A-a gradient of about 20 mmHg (9 -10 + 11). So, her predicted paO2 is near pAO2 – 20 mmHg = 228 – 20 = 208 mmHg. So, a paO2 somewhere around the 200 mmHg mark is fine. We decide on 202 mmHg for no particular reason.

With this value our fictitious patient’s fictitious gas is complete and scientifically tight.

case 9 final ABGCSi

Note that we work in the US units of mmHg for partial pressure of a gas. Much of the research work, including useful expressions such as Winter’s formula,  has been done in these units. Converting to the units in use in Europe (kPa) is easy just divide the mmHg values by 7.5 (7.5 mmHg = 1 kPa). aP = arterial plasma, c = calculated.

Serum electrolytes

Absence of insulin is the cause of DKA. In the absence of insulin, glucose cannot enter cells and the concentration of glucose rises dramatically in the ECF. There is a switch to metabolism of ketones as an energy source, the level of ketones rise in the ECF. Ketones are organic acids so the patient develops a metabolic acidosis due to the addition of fixed acid to the ECF. You have learned that metabolic acidosis due to the addition of fixed acid to the ECF is usually accompanied by a rise in the anion gap.

Na+ + K+ > HCO3- + Cl-

By convention we leave out K+ and so,

Anion gap = Na+ – (Cl- + HCO3-) normal <12 mEq/l (mmol/l)

I wanted to illustrate a further teaching point in this case. Marked hyperglycemia, may be accompanied by an ‘appropriate’ hyponatremia. We want to take the opportunity to teach this point so we make the serum sodium concentration low. We need to be careful. This is a metabolic acidosis due to the addition of fixed acid to the ECF, it should be associated with a high anion gap (often > 30 mEq/l). We have just lowered a determinant of the anion gap (Na+ concentration). If we just set the Cl- concentration in the middle of the normal range, we find that the anion gap is unimpressive!

130 – 100 + 10 =20 mEq/l (normal < 12 mEq/l)

We need to drop the Cl- to 90 to generate the anion gap we want (30 mEq/l). This illustrates the fact that the anion gap is essentially a measure of the relationship between Na+ and Cl- levels in the ECF.

Case analysis

So, now when the student comes to analyse the case, based on the clinical history he or she strongly suspects that the patient has DKA. The student expects to see a metabolic acidosis and indeed the hallmark of metabolic acidosis is present; a low level of metabolic base (bicarbonate) on the ABG. He/she would expect a degree of respiratory compensation (a deliberate reduction by the body in the level of respiratory acid-paCO2) and indeed the paCO2 is low. If the student is clever and interested, she will next assess the adequacy of compensation. She assesses this by applying Winter’s formula and finds that the observed paCO2 is in the predicted range. Compensation is adequate, this is likely to be a simple metabolic acidosis. No factor, other than the metabolic acidosis, is interfering with ECF pH. When the student calculates the anion gap she finds it to be elevated consistent with a metabolic acidosis secondary to addition of fixed acid (in this case ketones) to the ECF.

As underlying infection can trigger DKA, she wonders if there is any evidence of pneumonia. She analyses the A-a gradient and finds that it is normal for the level of oxygen therapy. This combined with a normal CXR and absence of physical findings in the chest makes respiratory pathology very unlikely. She can come and work for us.

Most of the time in clinical practice this sort of careful analysis of the ABG is not carried out and it makes no difference to the outcome of the case. However, as you have seen in Course 3 Case 1 (missed PE) once in a blue moon it can make all the difference.