Respirometer data, mass correction and respiratory quotient

What you’ll learn

• Convert capillary movement into a gas-volume rate.
• Use a control correction and express respiration per unit mass.
• Calculate RQ and state the assumptions behind gas measurements.

Understand the idea

Core practical 16 uses a simple respirometer to investigate respiration and respiratory quotient. In the usual oxygen-uptake arrangement, carbon dioxide is absorbed, so a gas-volume decrease can reflect oxygen consumption. Changes in temperature or external pressure can also move a marker. A comparable control and stable, equilibrated conditions help distinguish these physical changes from biological gas exchange.

The capillary's cross-sectional area converts a marker displacement into a volume. Apply any stated control correction with the appropriate direction, then divide by elapsed time. A rate per gram makes samples of different mass more comparable, although age, species and physiological state may still differ. Check units carefully: a radius in millimetres and displacement in millimetres give volume in cubic millimetres.

Respiratory quotient is carbon dioxide production divided by oxygen consumption, using matched amounts or gas volumes under comparable conditions. A value near one is consistent with carbohydrate as a respiratory substrate, but a mixed substrate or changing metabolism can complicate interpretation. A single oxygen-uptake trace with carbon dioxide absorbed cannot on its own supply both quantities. Suitable plant material can avoid animal-welfare concerns; any live-organism or absorbent use remains part of an approved supervised method.

Key formulae

$$\Delta V=\pi r^2\Delta x$$

$$r_{\mathrm{mass}}=\frac{\Delta V_{\mathrm{corrected}}}{m\Delta t}$$

$$\mathrm{RQ}=\frac{V_{\mathrm{CO_2\ produced}}}{V_{\mathrm{O_2\ consumed}}}$$

Worked example

A capillary of radius $0.50\ \mathrm{mm}$ shows a $14\ \mathrm{mm}$ displacement in $5.0\ \mathrm{min}$. A matched control shows $2\ \mathrm{mm}$ in the same direction, so subtract it. The respiring material has mass $0.50\ \mathrm{g}$. Calculate the corrected oxygen-uptake rate per gram, assuming carbon dioxide is absorbed.

1. The corrected displacement is $14-2=12\ \mathrm{mm}$.
2. The corrected volume is $\pi(0.50)^2(12)=3\pi\approx9.425\ \mathrm{mm^3}$.
3. Divide by time and mass: $9.425/(5.0\times0.50)=3.77\ \mathrm{mm^3\,g^{-1}\,min^{-1}}$.

Answer: Approximately $3.77\ \mathrm{mm^3\,g^{-1}\,min^{-1}}$.