Chap. 2 Problem 3

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Chap. 2 Problem 3
F
X
Phe (F)
The structure of the peptide and the charges of
all functional groups at pH 7 are shown in the
figure. The net charge is -1.
The addition of a phosphate to the tyrosine adds
2 negative charges, so the net charge would
become -3. Most phosphorylation reactions in
cells result from transfer the ?-phosphate of ATP
to the acceptor, with a large negative ?G.
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Chap. 2 Problem 9
Ammonia is a weak base that can pick up protons
at acidic pH. Thus ammonia is protonated inside
lysosomes forming ammonium ion. The H of the
lysosome compartment decreases, and thus the pH
of the lysosome increases.
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Chap. 2 Meaning of the Kd
Example Protein (P) binding to DNA (D) P D ?
P.D Kd PD/P.D What is the ratio of
D/P.D for different values of P? P 0.1
x Kd D/P.D 10/1 P Kd D/P.D
1/1 P 10 x Kd D/P.D 1/10 This shows
that the DNA binding site is about 10 occupied
when the concentration of P is 10-fold lower
than the Kd, 50 occupied when P is the same as
the Kd, and 90 occupied when P is 10-fold
greater than the Kd.
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Chap. 2 Derivation of the HH Equation
HA ? H A- Ka HA-/HA Take the log of
both sides of the equation. log Ka log
HA-/HA Rearrange log Ka log H log
A-/HA Rearrange again log H log Ka - log
A-/HA Multiply both sides by -1 -log H
-log Ka log A-/HA Substitute pH and pKa pH
pKa log A-/HA
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Chap. 2 Problem 11
What is the ionization state of phosphoric acid
in the cytoplasm why is phosphoric acid a
physiologically important compound?
The pH of the cytosol is 7.2. This happens to be
equivalent to pKa2 of phosphoric acid. (Refer to
Fig. 2.28, next slide) The relevant equilibrium
reaction therefore is H2PO4- ? H HPO42- The
Henderson Hasselbach Eq is pH pKa log
HPO42-/H2PO4- Since pH pKa2 7.2, the HH
Eq becomes 100 HPO42-/H2PO4- And 1/1
HPO42-/H2PO4- Thus, these 2 compounds are
present at a 50/50 ratio in solution.
Phosphoric acid is physiologically important
because it serves as the buffering agent in the
cytosol.
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Fig. 2.28
pKa2