Summary of equations used for blood gas interpretation

This is a summary of simple bedside rules to help interpret blood gases, as well as a synopsis of the equations CICM fellowship candidates are expected to remember.

A detailed explanation of the alveolar gas equation occurs elsewhere.

Briefly:

**PAO _{2} = (FiO_{2} × (760 - 47)) - (PaCO_{2} × 1.25)**

Or, at room air:

**PAO _{2} = (149 - (PaCO_{2} x 1.25)**

Thus, the patient with a relatively normal PaCO_{2} (say, 40) :

**PAO _{2} = (149 - 50)**

So, a normal person should have a PaO_{2} of around 99 mmHg.

We take a range of 75-100.

Or, for a patient with normal PaCO_{2} and an increased FiO_{2}:

**PAO _{2} = (FiO2 x 713) - 50**

The A-a gradient and all the other tension-based indices of oxygenation are discussed elsewhere.

Briefly:

**A-a gradient = [PAO _{2} - PaO_{2}]**

i.e. alveolar concentration minus arterial concentration.

The normal A-a gradient in a healthy young person should be aroung 5-10mmHg.

It changes with age.

**Age-adjusted A-a gradient = (age / 4) + 4**

Thus, an 80 yr old should have an A-a gradient of around 24.

**For every 10 mmHg increase in PaCO _{2}, the pH will decrease by 0.08**

*In other words, pH = 7.40 - ((PaCO_{2 }- 40) × 0.008))*

This relationship tends to overestimate the pH change. the reason is its failure to account for the increase in bicarbonate as a result of increasing CO_{2}.

**pH = 7.40 - ((PaCO _{2}-40) x 0.003))**

Thus, if the CO_{2} is chronically elevated at 50, one should expect the pH to drop from 7.40 to 7.37. Again, the change in bicarbonate due to dissolved CO_{2} is unaccounted for. The "0.003" multiplier accounts only for the renal compensation mechanisms.

This equation is a more accurate way of calculating what the pH should be for any given PaCO_{2} and HCO_{3}^{-} combination:

Note that this [H+] result is in nanomoles; so before calculating pH from it, one must convert it into moles first. Thus: pH = -log^{10}((24 × PaCO_{2} / HCO_{3}^{-}) × 10^{-8})

**For every 10 mmHg increase in PaCO _{2}, the pH will decrease by 0.08**

*pH = -log ^{10}((24 × PaCO_{2} / HCO_{3}^{-}) × 10^{-8})*

The HCO_{3}^{-} used for this equation should ideally be the "total bicarbonate" as measured by the enzymatic method, in an autoanalyser. If such a value is not available, one may estimate the "expected" bicarbonate value from the Boston Rules. The various merits and demerits of the Boston and Copenhagen acid-base compensation rules are discussed elsewhere.

**(Na ^{+} + K^{+}) - (Cl^{-} + HCO_{3}^{-})**

The normal anion gap is about 12.

It changes with changes in albumin concentration; for every 4g/L of albumin below 40 the normal anion gap value should decrease by 1mmol. The anion gap corrected for a given albumin level (reported as Ag_{c}) can be calculated in the following manner:

**Agc = 0.25 × (albumin in g/L) **

Thus for an albumin of 40 the normal AG is 12, but at an albumin level of 20 the AG is 7.

The precise correction factor for albumin, and whether one needs to correct for phosphate, is a matter of some debate. If one were to correct for phosphate, the Agc equation would look like this:

**Agc = 0.25 × (albumin in g/L) + 1.5 × (phosphate in mmol/L)**

**Delta ratio = (change in anion gap) / (change in bicarbonate)**

The normal anion gap is assumed to be 12, and the normal HCO_{3} is assumed to be 24.

**A delta ratio of less than 0.4 **suggests that none of the change in bicarbonate can be explained by the change in anion gap, and thus a normal anion gap acidosis prevails.

**A delta ratio of 0.4-0.8 **suggests that a mixed high and normal anion gap acidosis exists.

**A delta ratio of 0.8-1.0 **suggests that the disorder is purely due to a high anion gap (i.e. the change in bicarbonate is matched by a change in anion gap)

**A delta ratio of 1.0-2.0 **suggests that a high anion gap metabolic acidosis exists (i.e. the change in anion gap is somewhat greater than the change in bicarbonate)

**A delta ratio of over 2.0 **suggests that together with a high anion gap acidosis, a metabolic alkalosis is also present (i.e. despite the change in anion gap, the bicarbonate has barely shifted from the normal value)

The osmolal gap is the difference between measured osmolality and calculated osmolality. It alerts one to the presence of *unexpected* osmoles, which might represent something hideously toxic.

**Osmolar gap = measured osmolality - calculated osmolality**

**Calculated osmolality (Osm) = (Na ^{+} × 2 + urea + glucose)**

*A normal value being ~ 10 mOsm/Kg*

This is the "stripped down" equation which most people remember with relative ease.

However, turns out it is fairly inaccurate.

To complicate an already complicated subject, Oh's Manual presents us with another equation (by Bhagat et al, from 1984) which is much more acurate:

**Osm = (1.89 × Na ^{+ }) + (1.38 × K^{+} ) + ( 1.03 × urea ) + (1.08 × glucose)** + 7.45

The various merits and demerits of the Boston and Copenhagen acid-base compensation rules are discussed elsewhere.

**For every 10 mmHg increase in PaCO _{2}, the HCO_{3}^{-} will rise by 1 mmol/L**

*In other words, expected HCO_{3} = 24 + ((PaCO_{2}-40) / 10)*

**For every 10 mmHg increase in PaCO _{2}, the HCO_{3}^{-} will rise by 4 mmol/L**

*In other words, expected HCO_{3} = 24 + (4 × (PaCO_{2}-40) / 10)*

**For every 10 mmHg decrease in PaCO _{2}, the HCO_{3}^{-} will fall by 2 mmol/L**

*In other words, expected HCO_{3} = 24 + (2 ×(PaCO_{2}-40) / 10)*

**For every 10 mmHg decrease in PaCO _{2}, the HCO_{3}^{-} will fall by 5 mmol/L**

*In other words, expected HCO_{3} = 24 + (5 ×(PaCO_{2}-40) / 10)*

**For complete compensation, expected PaCO _{2} = (1.5 × HCO_{3}^{-}) + 8**

*This rule is also known as Winter's Rule.** An error magin of +/- 2mmHg is tolerated.*

**For complete compensation, expected PaCO _{2} = (0.7 × HCO_{3}^{-}) + 20**

*An error magin of +/- 5mmHg is tolerated.*

**An acute change in PaCO _{2} will not change the Standard Base Excess.**

*So, expected SBE = 0... in other words, *

*Working backwards, expected CO_{2} = 40 + (0 × SBE)*

**Expected change in SBE = 0.4 times the change in PaCO _{2}**

*In other words,* **expected SBE = 0.4 × (40 - PaCO _{2})**

*Working backwards, ***expected CO _{2} = 40 + (0.4 × SBE)**

**Compensatory change in PaCO _{2} will be proportional to the SBE**.

*In other words, expected CO_{2} = 40 + (1.0 × SBE)*

**Compensatory change in PaCO _{2} will be proportional to 0.6 times the SBE**.

*In other words, expected CO_{2} = 40 + (0.6 × SBE)*