Temperature, development and ethical use of organisms

What you’ll learn

• Calculate growth rates, proportions and Q10 from supplied data.
• Distinguish developmental speed from final success.
• Plan comparable observations with suitable welfare limits.

Understand the idea

Core practical 12 investigates temperature and development, using suitable contexts such as seedlings or hatching data. A growth rate is a change in a defined measurement per unit time. A proportion hatched at one time is a cumulative outcome, not automatically a rate. Use repeated observations over time if the question asks how quickly development proceeds, and distinguish faster development from a greater final proportion surviving or hatching.

Q10 describes the factor by which a rate changes for a 10-degree temperature rise over a specified interval. It is not assumed to be constant over every temperature range. Development depends on many enzyme-controlled processes, and high temperatures can disrupt proteins or stress organisms. For a fair comparison, match starting stage, genetic or source group, density, water or oxygen availability and observation criteria as appropriate to the organism.

An ethical plan uses a justified range of non-harmful conditions, the minimum number of organisms compatible with meaningful independent replication, and agreed stopping or welfare criteria. Existing datasets or plant material may answer some questions without exposing animals to additional stress. The actual exposure conditions belong to an approved supervised method; data interpretation should not become a justification for testing damaging extremes.

Key formulae

$$\mathrm{growth\ rate}=\frac{L_2-L_1}{t_2-t_1}$$

$$Q_{10}=\left(\frac{r_2}{r_1}\right)^{10/(T_2-T_1)}$$

$$\mathrm{hatched}\ (\%)=\frac{N_{\mathrm{hatched}}}{N_{\mathrm{initial}}}\times100$$

Worked example

A supplied seedling dataset gives growth rates of $0.60\ \mathrm{mm\,day^{-1}}$ at $15\ ^\circ\mathrm{C}$ and $2.40\ \mathrm{mm\,day^{-1}}$ at $35\ ^\circ\mathrm{C}$. Calculate Q10 for this interval.

1. The rate ratio is $2.40/0.60=4$.
2. The temperature difference is $20\ ^\circ\mathrm{C}$, so the exponent is $10/20=1/2$.
3. Thus $Q_{10}=4^{1/2}=2$. This describes the supplied interval, not every possible temperature.

Answer: $Q_{10}=2.0$.