it is fairly clear from Figure 10.14 that casein molecules expand much more
at the A–W interface than lysozyme does. It is, however, also obvious that
different proteins do not react in the same manner on a change in interface
type; the author is not aware of a reasonable explanation based on protein
structure. It is fairly certain that globular proteins tend to become more
strongly denatured on adsorption at an O–W interface than at an A–W; this
follows, e.g., from studies on the loss of activity of adsorbed enzymes. Heat
denaturation of a protein before adsorption also leads to a change in the
P–Grelation; see Figure 10.14, lysozyme. This may (partly) be due to the
aggregation that often occurs upon heat denaturation. Furthermore, the
surface equations of state of proteins also depend on conditions like pH,
ionic strength, temperature, etc., but few unequivocal results are available.
Typical results for the plateau value ofGat the O–W interface of
globular proteins range from 2 to 4 mg?m^2 , the higher values correspond-
ing to larger molecules. For nonglobular proteins (gelatin, caseins), values
between 3 and 5 are generally observed. As a rule of thumb,Gtends to be
slightly smaller at an A–W interface, and smaller still at a S–W interface, but
there are exceptions. For aggregated proteins, e.g., as caused by heat
denaturation, far higherGvalues can be obtained, often 10–15 mg?m^2 or
even higher.
Question
In a study on O–W emulsions made withb-casein, a plateau value ofG¼4mg?m^2
was observed. What would be the number of amino acid residues per casein molecule
at the interface? You may assume that fully unfolded peptide chains at the interface
would give a plateau value ofG¼1mg?m^2.
Answer
For a surfactant of molar massMg?mol^1 , a surface load ofGg?m^2 corresponds
to G?NAV=M molecules per m^2. The area occupied per molecule is thus
M=G?NAVm^2. Applying this to unfolded peptide chains and taking forMthe
average value of an amino acid residue, i.e., 115 (see Section 7.1), we calculate that
the surface area taken up by one residue will equal 0.115/6? 1023 m^2 &0.2 nm^2. For
b-casein (see Figure 7.1),M¼24,000 g?mol^1 andG¼ 4? 10 ^3 g?m^2 , which yields
a surface area per molecule of 10 nm^2. Dividing by 0.2 nm^2 results in a figure of about
50 for the number of residues at the interface. Comparing this with the total number
of residues forb-casein of 209, it means that about a quarter of the residues would be
directly involved in the adsorption. It should be realized that this is a rough figure,
because it is not known what fraction of the interface is covered by adsorbate in the
two cases.