BCHM218 – Summer 2019 – PBA #1
This exercise will give you an opportunity to apply your understanding of
some of the topics covered in Module 1, Sections 1-3.
There are 9 questions in total consisting of multiple choi
...
BCHM218 – Summer 2019 – PBA #1
This exercise will give you an opportunity to apply your understanding of
some of the topics covered in Module 1, Sections 1-3.
There are 9 questions in total consisting of multiple choice and short answer.
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Please download, edit using Powerpoint to indicate your answers and upload.
Please fill in your name and student number:
Name: Marianne Bassil
Student #: 20194687
Mark: /46Multiple choice: (4 marks): For the following questions, highlight to indicate the
best answer. Mark =
1. Which of the following is a difference between weak chemical bonds and covalent or ionic
bonds?
a) Weak chemicals bonds are harder to break.
b) Weak chemical bonds are shorter than covalent or ionic bonds.
c) Weak chemical bonds are transient compared to covalent and ionic bonds.
d) Weak chemical bonds are not particularly useful in biological systems.
e) Weak chemical bonds require more energy to break than covalent and ionic bonds.
2. Hydrophobic interactions:
a) arise from the tendency of polar molecules to exclude water in solution.
b) destabilize proteins.
c) stabilize the base-stacking interactions of DNA helices.
d) occur between molecules having partial charges.
e) are a function of the repulsive forces generated when atoms are closer than their van der
Waals radii.3. Which of the following atoms CANNOT participate in biological hydrogen bonds?
a) nitrogen
b) oxygen
c) carbon
d) both nitrogen and oxygen
e) both oxygen and carbon
4. Which of the following amino acids is/are negatively charged at physiological pH?
a) serine
b) aspartate
c) glutamate
d) serine, aspartate, and glutamate
e) aspartate and glutamateQuestion 5: (3 marks) Mark
On the diagram: identify and label with appropriate terminology the following
structural features:
a) Regions that provide access to genetic information for a wide range of DNAbinding proteins.
b) Linkages that allow DNA to be stable as very long polymers.
c) Sites where positively charged amino acids would likely interact in a nonsequence-specific manner
You can duplicate, rearrange and fill in text boxes/arrows as required to complete
your answer.
label
a) Most of the DNA binding proteins bind to the major groove because it exposes most of the functional groups. The
bases are exposed and the backbone do not interact with the interaction of the DNA binding protein.
b) The phosphodiester bonds are the main reason why long polymers of DNA are formed. In fact, they form the
phosphate backbone in a DNA molecule. To add, the hydrogen bonds are the main linkages that help maintain stability
of the polymer. Two hydrogen bonds are formed between adenine and thymine and some are formed in between
guanine and cytosine.
c)The phosphate groups in the sugar-phosphate backbone serves the negative charge sites where positively charged
side chains of amino acids can bind. Only interactive charges between the two groups are considered.net charge of an amino acid depends on its pH environment. The schematic below outlines the different
ionization states of lysine as the pH ranges from 1.5 to 13.5. The states for different ionization groups of
lysine are shown at pH 7.0 (dotted box).
a) fill in the grey boxes (>right click>edit text) to indicate how these states changes as the pH becomes
more acidic or basic.
b) Indicate the pKa values (as discussed in Module 1) associated with the different changes in ionization
states.
c) Indicate net charges of each ionization state of lysine.
pKa:2.17
pH: 1.5 -------------------- pH: 7.0 ----------------------------------------------------- pH: 13.5
H
3N+ -
NH
3
+ NH
3
+ NH
2
H
H 2N - H2N
pKa:
8.95
pKa:
10.53
Net charge: +2 Net charge: +1 Net charge: 0 Net charge: -1d) What would be the isoelectric point (pI) for lysine? (1 mark). It would be
9.74
e) Explain your answer. (1 marks)
The isoelectric point is the pH at which the net charge of amino acid is 0. It is
calculated by averaging the pKa values for the ionizations states that straddle
the electrically neutral species.
Isoelectric Point
pI=(pKa1 +pKa2)/2
= (8.95 + 10.53)/2
= 9.74The schematic below outlines the different ionization states of histidine.
The states for different ionization groups of histidine are shown at pH 7.0 (dotted box).
a) fill in the grey boxes to indicate how these states changes as the pH becomes more acidic or basic.
b) Indicate the pKa values (as discussed in Module 1) associated with the different changes in ionization
states.
c) Indicate net charges of each ionization state of histidine.
pKa: 1.82
pH: 1.5 ---------------------------------------------------------- pH: 7.0 -------------------- pH: 13.5
H
3N+
-
H+ H+
H
3N+ H2N
H -
pKa: 9.17 pKa:6
Net charge: +2 Net charge:+1 Net charge: 0 Net charge: -1d) What would be the isoelectric point (pI) for histidine ? (1 mark). 7.6
e) Explain your answer. (1 marks)
The isoelectric point is the pH at which the net charge of amino acid is 0. It is
calculated by averaging the pKa values for the ionizations states that straddle
the electrically neutral species.
Isoelectric Point
pI=(pKa1 +pKa2)/2
= (9.17 + 6)/2
= 7.6The following line-angle diagram shows a potent neuronal signaling molecule which consists of a linear 11-mer
polypeptide. This polypeptide contains some of the 20 amino acids that you should be familiar with, as well
as an unusual amino acid.
a) On the diagram, please circle (>insert shapes) the amino (N) terminal amino acid, which will be
designated amino acid #1.
b) List the amino acids using their 3 letter code.
Indicate the unusual amino acid.
Amino acid 1: Arg
Amino acid 2: Pro
Amino acid 3: Lys
Amino acid 4: Pro
Amino acid 5: Gln
Amino acid 6: Gln
Amino acid 7: Phe
Amino acid 8: Phe
Amino acid 9: Gly
Amino acid 10: Leu
Amino acid 11: Met (unusual amino acid)Question 9: (8 marks) Mark =
The diagram shows a well characterized inhibitor that is used in the clinic to manage high blood pressure.
As the basis of action, this chemical structure binds to the active site pocket of the protein, angiotensin
converting enzyme, using a combination of multiple weak chemical interactions.
a) Label all H-bond acceptor atoms.
b) Label all H-bond donor atoms.
c) Label any other chemical groups that could also form interactions AND indicate the type of interaction
formed.
You can duplicate, rearrange and fill in text boxes/arrows as required to complete your answer.
In green is written =
hydrophobic interraction non
polar alkyl chain
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