AS. 410.610 – Biochemistry –Fall 2019 Problem Set #1

AS. 410.610 – Biochemistry –Fall 2019
Problem Set #1

- Answers: red.
- Correct answers: Green (i.e., 100% credit)

Question 1
1. Which of the following statements is true about water? Select all answers that apply.
A. it is non-polar
B. it fully ionizes at neutral pH
C. liquid-phase water is more dense than solid-phase water
D. it is a weak acid
E. it is a weak base
0.5 points
Question 2
1. The Ka for a weak acid (HA) is known to be 1.9 x 10-5. What is the ratio of [A-] to [HA] at a pH of 6?
Explain your answer & show all mathematical work.
Answer: [A-] / [HA] is 19:1 at pH = 6.
Reasoning:
Using Henderson-Hasselbach Equation:
pH = pkA + log10 [conjugate base] / [weak acid]
pH = pKa + log ([A¯] / [HA])
pH = (pKa = -log Ka)+ log ([A¯] / [HA])

pH = 6
Ka = 1.9 x 10-5 (weak acid)

6 = (-log (1.9 x 10-5)) + log ([A¯] / [HA])
6 = (4.721246399) + log ([A¯] / [HA])
6 - (4.721246399) = + log ([A¯] / [HA])
1.278753601 = + log ([A¯] / [HA]), cancel the log (right side) and base-log left
101.278753601 = ([A¯] / [HA])
19.0000000021 or
Concentration ratio: At pH = 6, the ratio of [A-] / [HA] is 19:1.
1 points
Question 3
1. For questions 3-6, consider the titration curve below. The chemical species being titrated by a
strong base must be a , since it is capable of donating proton(s).
Reasoning: Three steps on the graph, lent 3 H+’s, and the pH is high, so triprotic acid.
A. weak base; no
B. strong base; three 0.5 points
Question 4
C. monoprotic acid; one
1. The chemical gr oup being titrated around the point labeled (C) in the
D. diprotic acid; two graph is most lik ely a(n)__________________(basic or aci dic) group.
basic
E. triprotic acid; three Blank 1
Reasoning: (c) pKa is ~ 12.5 at point (C), then this must be asic group
(like ab
0.5 points an amino grp).
Question 5


/

, 1. The titration curve shown in Question #3 is that of a free amino acid in aqueous solution. Based on the
graph, which particular amino acid is most likely represented? Simply name the amino acid. Blank 1
Arginine

Reasoning:
-Titration curve clearly =3 ionizable groups,
-must be Tyr, Cys, Lys, His, Arg, Asp, or Glu (only free amino acids w/ionizable side
chains).
-(A) pKa = 2+ something (2.25 maybe),
-(B) pKa = 9
-(C) pKa = 12.5
-Only amino acid within this pKa is arginine.
0.5 points
Question 6
1. Assume that the equilibrium represented around point (A) in the titration can generically be
described as:
H3A + OH- ---> H2A- + HOH. What is the pH at which the ratio of [HA2-] to [H2A-] is 25:1? Show all
work & clearly explain your answer.

Answer: pH at [HA2-] / [H2A-] is 25:1 = 10
Reasoning:
1. Equilibrium (A) = [HA2-] to [H2A-] is 25:1
2. Equilibrium (B) = pKa = 9 (or 9.04 if it is really arginine)
3. Triprotic acid
H2A- + OH----> HA2- + HOH
Per the H/H equation: pH = pKa + log([HA2-]/[H2A-])
-pKa at point (B) is approximately 9 (or precisely 9.04 if arginine), so:
-pH = 9 + log(25/1) = 9 + 1.4 = 10.

1 points
Question 7
1. For Questions #7-13, consider the following peptide:
His-Met-Asp-Tyr-Phe-Ser
The code for this peptide, using one-letter symbols, is .
HMDYFS

0.5 points
Question 8
1. How many distinct "ionizable" groups exist within this peptide?
His-Met-Asp-Tyr-Phe-Ser = 5
+3 Ionizable side chain
+1 Ionizable carboxul grp
+1 Ionizable amino terminal grp
A. 2

B. 3

C. 4

D. 5

E. 6

0.5 points