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Scribe Notes for October 24, 2012
Page 5: the point at which the graph reaches a steady % repression represents a stable threshold
Page 6: The graph on left shows a plot for a relationship that is not 1:1….what is the
molecularity of the process??
Page 8: n represents 1,2,3,…etc, any number that could explain a relationship that isn’t 1:1.
Therefore Kd=Mn (the units of Kd reflect the molecularity of the process). For the equation ln(y /
1 – y) = n ln[R] – ln[Kd], you’re interested in n R and Kd. (By taking the log of the equation,
you’re linearizing the data)
Page 9: in the graph shown, the slope = n = which in this case is 2 or the number of interacting
molecules. However, if R dimerizes at a smaller concentration then R binds to O then you would
see a 1:1 relationship enough though two R molecules are present.
Page 10: If the Hill coefficient = 2 mean, what does that say about the number of R molecules
bound to one O molecule? There are at least two bound, minimum value. Can you think of how
a hill plot will not give n=2 even though the R needs to dimerize to bind O? If the dimer is
formed at really low concentration, then you’re looking at the dimer binding to O in a 1:1
relationship. You may also get a number from the hill plot that doesn’t make sense, ex. 2.5,
getting these types of numbers signifies that there are probably intermediates species. However,
this number gives the lower limit, at least 2 molecules bind, could be more but the hill plot won’t
give you that answer.
Page 11: [O]total >>> [R]total this is opposite of measuring binding constants. Fraction(Rbound) = 1
because [O] is much greater then Kd
Page 12: If you don’t know the Kd vary the concentration of [O] to make sure you have the same
data set, aka same stoichiometric relationship. If [O] is below the Kd then the plot will look like a
binding curve. Someone else using their own prep of R found that 400nM of R was required to
titrate O. What could be happening here? Half of your protein is inactive…how much of your R
is actually competent to bind to your operator?
Page 13: in the bottom figure, the cellular concentration of c1 (100nM) is no longer enough for
what the system requires (1000nM). In a normal cellular environment the concentration of c1 is
10 fold below the kd.
Page 17: specificity comes from the ability of these proteins to discrimination between O binding
sites. What happens under low [c1] and [cro]? C1 binds OR1 first and cro binds OR3. Therefore
at low concentrations, they inhibit each other.
Page 19: The plot is looking a binding in context of the whole operator sequence.
Page 22: Why don’t we see pairwise cooperativity between OR2 and OR3? OR2 is tied up in
other interactions with OR1….therefore binding between OR2 is a competition between OR2 &
OR1 and OR3.