chap. 2 problem 3

6
Chap. 2 Problem 3 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. F X Phe (F)

Upload: emi-washington

Post on 30-Dec-2015

18 views

Category:

Documents


0 download

DESCRIPTION

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. - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Chap. 2 Problem 3

Chap. 2 Problem 3

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.

F

X Phe (F)

Page 2: Chap. 2 Problem 3

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.

Page 3: Chap. 2 Problem 3

Chap. 2 Meaning of the Kd

Example: Protein (P) binding to DNA (D)

P + D P.D

Kd = [P][D]/[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.

Page 4: Chap. 2 Problem 3

Chap. 2 Derivation of the HH Equation

HA H+ + A-

Ka = [H+][A-]/[HA]

Take the log of both sides of the equation.

log Ka = log [H+][A-]/[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]

Page 5: Chap. 2 Problem 3

Chap. 2 Problem 11What 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+ + HPO4

2-

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.

Page 6: Chap. 2 Problem 3

pKa2

Fig. 2.28