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5.7: Questions

  • Page ID
    20948
  • 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 ΔHv = -3.34 x 108 J m-3.)

    hint: The critical radius for nucleation is r*, where r* = -2γ /ΔGv, and γ is the solid-liquid interfacial energy and ΔGv 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 /ΔGv where γ is the solid-liquid interfacial energy and ΔGv is the free energy of freezing per unit volume. γ is given in the question and ΔGv can be determined from ΔHv. For freezing at a temperature T, the supercooling, ΔT is (Tm - T), where (Tm is the melting temperature. The entropy of fusion per unit volume is ΔSv = ΔHv/Tm. As shown in the section on Nucleation and Crystallization, ΔGv = ΔSvΔT. 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 ΔHv as given, Tm 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.