10.3.5: Questions

Deeper questions

The following questions require some thought and reaching the answer may require you to think beyond the contents of this TLP.

1. Crystallization is usually fatal to cells because:

Yes	No	a	Crystals can puncture cell walls
Yes	No	b	The formation of crystals changes the ionic ratios in the cytosol
Yes	No	c	Crystallization speeds up the cell reactions
Yes	No	d	Ice formation causes water to be drawn into the cell by osmosis causing the cell to swell and burst
Yes	No	e	Nucleation of the crystals requires large amounts of energy

Answer

a. Yes

b. Yes

C. No

D.Yes

e. No

2. Ice formation in cells can be limited by:

Yes	No	a	The presence of antifreeze proteins
Yes	No	b	The dispersion of water in the cellular structure, which limits heterogeneous nucleation
Yes	No	c	The formation of starch from sugar
Yes	No	d	The introduction of Ice Nucleating Agents (INAs) into the cells
Yes	No	e	The freezing of extracellular water

Answer

a. Yes

b. Yes

C. No

D.No

e. Yes

3. Crystallization of minerals in the cells can be limited by:

Yes	No	a	The presence of anti-freeze proteins
Yes	No	b	The hydrolysis of starch to sugar
Yes	No	c	The preferential formation of a glass
Yes	No	d	The increased activity of the Golgi Apparatus

Answer

a. No

b. Yes

C. Yes

D.No

In some alpine plants, extracellular ice formation occurs at around –2°C. What is the critical radius for ice nucleation at this temperature?

In these plants, the extracellular ice forms on ice nucleating agents. If an ice nucleating agent is a circular disc and is a perfect template for ice, what size must it be for nucleation to occur at –2°C?

(For the ice-water interface, the interfacial energy γ = 0.028 J m^-2; the latent heat of freezing of ice is ΔH_v = -3.34 x 10⁸ J m^-3.)

hint: The critical radius for nucleation is r*, where r* = -2γ /ΔG_v, and γ is the solid-liquid interfacial energy and ΔG_v is the free energy of freezing per unit volume. For further details see the section on nucleation and crystallization.

Answer

r*, the critical nucleus for nucleation is given by, r* = -2γ /ΔG_v where γ is the solid-liquid interfacial energy and ΔG_v is the free energy of freezing per unit volume. γ is given in the question and ΔG_v can be determined from ΔH_v. For freezing at a temperature T, the supercooling, ΔT is (T_m - T), where (T_m is the melting temperature. The entropy of fusion per unit volume is ΔS_v = ΔH_v/T_m. As shown in the section on Nucleation and Crystallization, ΔG_v = ΔS_vΔT. Thus:

\[r^{*}=-\frac{2 \gamma}{\Delta G_{v}}=-\frac{2 \gamma}{\Delta S_{v} \Delta T}=-\frac{2 \gamma T_{m}}{\Delta H_{v} \Delta T}\]

With γ and ΔH_v as given, T_m taken to be 273 K, and T = 271 K, we find r* = 22.9 nm

For ice to grow on the INA, the radius of the INA must be greater than or equal to r*, so, for ice to grow, the INA should have radius 22.9 nm.