Exam 2 - questions Flashcards

1
Q
  1. Describe how you would determine the Ka (association constant) for a ligand and a protein.
A

An experiment would be carried out in which a fixed amount of the protein is incubated with
varying amounts of ligand (long enough to reach equilibrium). The fraction of protein molecules that
have a molecule of ligand bound is then determined. A plot of this fraction () vs. ligand concentration
[L] should yield a hyperbola. The value of [L] when = 0.5 is equal to 1/Ka

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2
Q

For the binding of a ligand to a protein, what is the relationship between the Ka (association
constant), the Kd (dissociation constant), and the affinity of the protein for the ligand?

A

Ka = 1/Kd. The larger the Ka (and hence the smaller the Kd), the higher the affinity of the protein
for the ligand.

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3
Q

What fraction of ligand binding sites are occupied () when [ligand] = Kd? Show your work.

A
From equation (5 - 8) on page 156 :
 = [Ligand]/ [Ligand] + Kd
thus, when [Ligand] = Kd , the equation becomes :
 = Kd 1
 Kd + Kd 2
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4
Q

Describe briefly the two principal models for the cooperative binding of ligands to proteins with
multiple binding sites

A

In the concerted model, binding of a ligand to one site on one subunit results in an allosteric
effect that converts all of the remaining subunits to the high-affinity conformation. As a result, all of
the subunits are either in the low- or high-affinity conformation. In the sequential model, each
subunit is changed individually to the high affinity conformation. As a result, there are many possible
combinations of low- and high-affinity subunits.

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5
Q

Describe briefly the basic structure of an IgG protein molecule.

A
An IgG protein contains two copies of a large polypeptide (heavy chain) and two copies of a
small polypeptide (light chain). structure contributes significantly to the tertiary structure of
domains of both chains. Disulfide bonds link the heavy chains to one another and to the light chains.
The chains are arranged in a Y-shaped structure where the two arms are linked to the base by a
protease sensitive (“hinge”) region
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6
Q

What is the concept of “induced fit” as it applies to antigen-antibody binding?

A

The conformations of the antigen and antigen-binding site of the antibody are influenced by
each other and change as binding occurs. These conformational changes increase the chemical
complementarity of the sites and result in tighter binding.

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7
Q

Describe how immunoaffinity chromatography is performed.

A

The specific antibody is covalently attached to an inert supporting material, which is then
packed into a chromatography column. The protein solution is passed through the column slowly;
most proteins pass directly through, but those for which the antibody has specific affinity are
adsorbed. They can subsequently be eluted by a buffer of low pH, a salt solution, or some other agent
that breaks the antibody-antigen association.

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8
Q

What properties of antibodies make them useful biochemical reagents? Describe one biochemical
application of antibodies (with more than just the name of the technique)

A

The important properties are the high specificity of protein recognition, and the high affinity of
the antibody-antigen association. These make possible immunoaffinity chromatography,
immunocytochemistry, enzyme-linked immunosorbent assay (ELISA), and immunoblotting, all of
which are described on pp. 173-175.

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9
Q

Draw and label a reaction coordinate diagram for an uncatalyzed reaction, S  P, and the same
reaction catalyzed by an enzyme, E

A

See Fig. 6-3, p. 187.

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10
Q

Write out the equation that describes the mechanism for enzyme action used as a model by
Michaelis and Menten. List the important assumptions used by Michaelis and Menten to derive a
rate equation for this reaction.

A

The two equations are
k1 k2
E + ES E + P
k-1 k-2
One assumption is that [P] = 0, so that the rate of the reaction depends exclusively on the breakdown
of ES and is not influenced by the reverse reaction; that is, k-2 can be ignored and V0 = k2 [ES]. This
condition is possible only if early reaction times are measured; the velocity, therefore, is an initial
velocity. A second assumption is that the rate of ES formation equals the rate of ES breakdown; in
other words, the reaction is at a steady state. A third assumption is [S]&raquo_space; [Et], so that total [S],
which equals free substrate and enzyme-bound substrate, is essentially equal to [S].

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11
Q

For the reaction E + S  ES  P, the Michaelis-Menten constant, Km, is actually a summary of
three terms. What are they? How is Km determined graphically?

A

Km = (k2 + k-1)/ k1, where k-1 and k1 are the rate constants for the breakdown and association,
respectively, of the ES complex and k2 is the rate constant for the breakdown of ES to form E + P. Km
can be determined graphically on a plot of V0 vs. [S] by finding the [S] at which V0 = 1/2 Vmax. More
conveniently, on a double-reciprocal plot, the x-axis intercept = –1/ Km.

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12
Q

Give the Michaelis-Menten equation and define each term in it. Does this equation apply to all
enzymes? If not, to which kind of enzymes it doesn’t apply?

A

The Michaelis-Menten equation is: V0 = Vmax [S]/( Km + [S]), in which V0 is the initial velocity at
any given concentration of S, Vmax is the velocity when all enzyme molecules are saturated with S, [S] is
the concentration of S, and Km is a constant characteristic for the enzyme. This equation does not apply
to enzymes that display sigmoidal V0 vs. [S] curves, but only to those giving hyperbolic kinetic plots.

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13
Q

Methanol (wood alcohol) is highly toxic because it is converted to formaldehyde in a reaction
catalyzed by the enzyme alcohol dehydrogenase:
NAD+ + methanol  NADH + H+ + formaldehyde
Part of the medical treatment for methanol poisoning is to administer ethanol (ethyl alcohol) in
amounts large enough to cause intoxication under normal circumstances. Explain this in terms of
what you know about examples of enzymatic reactions.

A

Ethanol is a structural analog of methanol, and competes with methanol for the binding site of
alcohol dehydrogenase, slowing the conversion of methanol to formaldehyde, and allowing its
clearance by the kidneys. The effect of ethanol is that of a competitive inhibitor.

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14
Q

What is a zymogen (proenzyme)? Explain briefly with an example

A

A zymogen is an inactive form of an enzyme that is activated by one or more proteolytic cleavages
in its sequence. Chymotrypsinogen, trypsinogen, and proelastase are all zymogens, becoming
chymotrypsin, trypsin, and elastase, respectively, after proper cleavage.

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15
Q
This compound is L-glyceraldehyde. Draw a stereochemically correct representation of C-1 and C2 of D-glucose.
CHO
|
HO—C—H
|
CH2OH
A

In D-glucose, the positions of the —H and —OH on C-2 are the reverse of those for C-2 of Lglyceraldehyde. (Compare Fig. 7-1, p. 236, with Fig. 7-2, p. 236.)

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16
Q

Define each in 20 words or fewer:

(a) anomeric carbon
(b) enantiomers
(c) furanose and pyranose
(d) glycoside
(e) epimers
(f) aldose and ketose

A

: (a) The anomeric carbon is the carbonyl carbon atom of a sugar, which is involved in ring
formation. (b) Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.
(c) Furanose is a sugar with a five-membered ring; pyranose is a sugar with a six-membered ring. (d) A
glycoside is an acetal formed between a sugar anomeric carbon hemi-acetal and an alcohol, which may be
part of a second sugar. (e) Epimers are stereoisomers differing in configuration at only one asymmetric
carbon. (f) An aldose is a sugar with an aldehyde carbonyl group; a ketose is a sugar with a ketone
carbonyl group

17
Q

. (a) Draw the structure of any aldohexose in the pyranose ring form. (b) Draw the structure of the
anomer of the aldohexose you drew above. (c) How many asymmetric carbons (chiral centers) does
each of these structures have? (d) How many stereoisomers of the aldohexoses you drew are
theoretically possible?

A

(a) Any of the hexoses drawn with a six-membered ring, as shown in Fig. 7-7 on p. 239, is
correct. The hydroxyls at C-2, C-3, and C-4 can point either up or down. (b) For the anomer, the
structure should be identical to the first, except that the hydroxyl group at C-1 should point up if it
pointed down in your first structure, and vice versa. (c) The number of chiral centers is 5; all are
carbons except C-6. (d) The number of possible stereoisomers for a compound with n chiral centers
is 2n; in this case, 25, or 32 possible isomers.

18
Q

In the following structure:
PICTURE

(a) How many of the monosaccharide units are furanoses and how many are pyranoses? (b) What is the
linkage between the two monosaccharide units? (c) Is this a reducing sugar?
Explain.

A

(a) Any of the hexoses drawn with a six-membered ring, as shown in Fig. 7-7 on p. 239, is
correct. The hydroxyls at C-2, C-3, and C-4 can point either up or down. (b) For the anomer, the
structure should be identical to the first, except that the hydroxyl group at C-1 should point up if it
pointed down in your first structure, and vice versa. (c) The number of chiral centers is 5; all are
carbons except C-6. (d) The number of possible stereoisomers for a compound with n chiral centers
is 2n; in this case, 25, or 32 possible isomers

19
Q

(a) Define “reducing sugar.” (b) Sucrose is a disaccharide composed of glucose and fructose
(Glc(1  2)Fru). Explain why sucrose is not a reducing sugar, even though both glucose and
fructose are.

A

(a) A reducing sugar is one with a free carbonyl carbon that can be oxidized by Cu2+ or
Fe3+. (b) The carbonyl carbon is C-1 of glucose and C-2 of fructose. When the carbonyl carbon is
involved in a glycosidic linkage, it is no longer accessible to oxidizing agents. In sucrose (Glc(1
2)Fru), both oxidizable carbons are involved in the glycosidic linkage.

20
Q

The number of structurally different polysaccharides that can be made with 20 different
monosaccharides is far greater than the number of different polypeptides that can be made with 20
different amino acids, if both polymers contain an equal number (say, 100) of total residues.
Explain why.

A

Because virtually all peptides are linear (in other words, are formed with peptide bonds between
the -carboxyl and -amino groups), the variability of peptides is limited by the number of different
subunits. Polysaccharides can be linear or branched, can be - or -linked, and can be joined 1 4,
1 3, 1 6, etc. The number of different ways to arrange 20 different sugars in a branched
oligosaccharide is therefore much larger than the number of different ways a peptide could be made
with an equal number of residues.

21
Q

Draw the structure of the repeating basic unit of (a) amylose and (b) cellulose

A

(a) For the structure of amylose, see Fig. 7-14a, p. 245. The repeating unit is -D-glucose linked
to -D-glucose; the glycosidic bond is therefore (1 4). (b) Cellulose has the same structure as amylose,
except that the repeating units are -D-glucose and the glycosidic bond is (1 4). (See Fig. 7-15a, p. 246.

22
Q

Describe the differences between a proteoglycan and a glycoprotein.

A

Both are made up of proteins and polysaccharides. In proteoglycans, the carbohydrate moiety
dominates, constituting 95% or more of the mass of the complex. In glycoproteins, the protein
constitutes a larger fraction, generally 50% or more of the total mass.

23
Q

What are lectins? What are some biological processes that involve lectins?

A

Lectins are proteins that bind to specific oligosaccharides. They interact with specific cellsurface glycoproteins thus mediating cell-cell recognition and adhesion. Several microbial toxins and
viral capsid proteins, which interact with cell surface receptors, are lectins.

24
Q
The composition (mole fraction) of one of the strands of a double-helical DNA is [A] = 0.3, and [G]
= 0.24. Calculate the following, if possible. If impossible, write "I."
For the same strand:
(a) [T] = \_\_\_\_
(b) [C] = \_\_\_\_
(c) [T] + [C] = \_\_\_\_
For the other strand:
(d) [A] = \_\_\_\_
(e) [T] = \_\_\_\_
(f) [A] + [T] = \_\_\_\_
(g) [G] = \_\_\_\_
(h) [C] = \_\_\_\_
(i) [G] + [C] = \_\_\_\_
A

(a) I; (b) I; (c) 0.46; (d) I; (e) 0.3; (f) I; (g) I; (h) 0.24; (i) I

25
Q

Describe briefly what is meant by saying that two DNA strands are complementary.

A

The nucleotide sequences of complementary strands are such that wherever an A occurs in one
strand, there is a T in the other strand with which it can form a hydrogen-bonded base pair. Wherever
a C occurs in one strand, a G occurs in the other. A is the base complementary to T, and C is the base
complementary to G.

26
Q

Write a double-stranded DNA sequence containing a six-nucleotide palindrome.

A

Any double-stranded sequence that has the form:
1-2-3-4-5-6-6’-5’-4’-3’-2’-1’
1’-2’-3’-4’-5’-6’-6-5-4-3-2-1
where each number and its prime represent correctly paired bases (A with T, C with G) all
along the double-stranded molecule.

27
Q

Describe qualitatively how the tm for a double-stranded DNA depends upon its nucleotide composition

A

In general, the higher the proportion of G and C, the higher the melting temperature, tm. More
thermal energy is required to break the three hydrogen bonds holding GC pairs than to break the two
hydrogen bonds holding A=T pairs.

28
Q

Draw Serine and Asparagine; now connect a sugar molecule to each of those amino acids in alpha or
beta o-glycosidic bond.

A

Look at Fig. 7-29 on page 255. Asparagine does not form o-glycosidic bond, it forms Nglycosidic bond.

29
Q

Draw lactose, fructose, cellobiose, trehalose and table sugar. Now connect them in alpha or beta
glycosidic form on one end and in 2OH, 3 OH, 4OH or 6OH on the another end. If you can’t do it,
argue why it is not possible.

A

PICTURE

30
Q

You have to investigate the distribution of two small proteins in the liver and blood tissues of 20 rats.
You have available antibodies against these proteins. You also have equipment ranging from SDSPAGE, Western blotting, ELISA, mass spectrometer, etc. You also have an unlimited amount of money.
Design an experiment for investigation of these proteins from an academic, biotech/pharma and
government perspective. Justify your rationale.

A

Reasonable answers will be given points.

31
Q

Draw Beta-L-2N-acetyl glucosamine-(1-4) alpha-D-2galactosamine 3Sulphate (1-3) alpha-D-2N-acetyl6-sulphate mannosamine

A

PICUTURE

32
Q

Draw Alpha-D-2N-acetyl-4S-glucosamine-(1-6)-beta-L-4S mannosamine (1-2) beta-L-fructofuranose

A

PICTURE

33
Q

Draw Adenine N9 – x non-glycosidic ribofuranose beta (1-1) alpha 3N-acetylglucosamine-4-serine

A

PICTURE

34
Q

Draw a chemical structure that contains a combination of amino acids, saccharides, furanose, pyranose,
phosphates, O-glycosidic bonds, N-glycosidic bonds, nucleotides, nucleosides, alpha anomer, beta
anomer, furanose, pyranose, sulphate, fructofuranose, glucopyranose.

A

Any structure that contains all mentioned biomolecules and correct bonds.

35
Q

Explain the difference between ESI-MS, ESI-MS/MS and LC-MS/MS

A

ESI-MS: Electrospray ionization mass spectrometry. “ES” shows the ionization technique.
ESI-MS/MS: Electrospray ionization tandem mass spectrometry. Two steps of mass spectrometry are
done with fragmentation in between the steps. In the first MS, precursor ion is obtained, after
fragmentation, the product ion is obtained in second MS.
LC-MS/MS: Liquid chromatography coupled with tandem mass spectrometry

36
Q

For each of these methods of separating proteins, describe the principle of the method, and tell what
property of proteins allows their separation by this technique.
(a) ion-exchange chromatography
(b) size-exclusion (gel filtration) chromatography
(c) affinity chromatography
(d) reversed phase chromatography
(e) atomic change chromatography

A

(a) Ion-exchange chromatography separates proteins on the basis of their charges. (b) Sizeexclusion or gel filtration chromatography separates on the basis of size. (c) Affinity chromatography
separates proteins with specific, high affinity for some ligand (attached to an inert support) from other
proteins with no such affinity. (See Fig. 3-17, p. 87.)

37
Q

You overexpressed a protein in bacteria and then purified it by biochemical methods. However, the
purified protein is not soluble, so you are thinking about increasing its solubility. The fastest way to
increase the protein’s solubility was chemical linking of polyethylene-glycol. However, while through
this procedure, the solubility of the polyethylene-glycol-modified protein increased, its hepatotoxicity
also increased. So, alternative strategies are needed. Use your biochemical knowledge to design an
experiment that will increase the solubility of this overexpressed protein, which will ultimately allow it
to be purified with high solubility. List the controls that you will perform during the purification
procedure.

A

For example:
Biologically, glycosylation site can be introduced into the protein to improve the solubility and it can be
purified by lectin based affinity purification. Tags can also be introduced into the plasmid before protein
expression. The proteins can be secreted. Chemically buffers can be altered to make it more soluble in
small scale. Chemical engineering corresponds to the large scale production and purification of the
protein and small laboratory techniques is involved small scale.
Controls: A standard protein is chosen to compare the purity and variability during experimental
procedures. Also, the solubility, toxicity and stability of the protein must be monitored.

38
Q

You want to purify and characterize three cytoplasmic protein complexes from cells grown in cell
culture: one is a 480 kDa homotetramer (complex A), the second is a 520 kDa 12 subunit heteromer
(Complex B) and the third one is a 540 kDa complex that has an equal number of two subunits with a
mass of 52 kDa and 14 kDa (complex C). Complex C has oxidoreductase activity. Design a purification
scheme for these protein complexes, keeping in mind that they have similar isoelectric points.

A

For example:
Use size exclusion to separate complex A from complex C. Complex C also can be isolated from the
other two complexes by using affinity purification using its oxidoreductase property. Blue native PAGE
can be used to separate the other two complexes. And they can be purified by electro elution.
Alternatively, ion exchange chromatography and SDS PAGE can also be performed

39
Q

Order the following peptides from the most hydrophobic to the most hydrophilic. Now order them from
the most negative to the most positive charges. Now indicate the order of their elution in anion
exchange chromatography. Now indicate the order of their elution in reversed phase chromatography.
Now order them from the ones that absorb the most UV to the ones that absorb the least UV. For all
questions, assume neutral pH. Now titrate one peptide of your choice.
A) KYGDHDK
B) DYASNTY
C) GYGKLDS
D) WYAFLIG
E) WYAFGTI

A

1) Hydrophobic to Hydrophillic – DECBA
2) Negative to Positive – B/C D/E A
3) Anion Exchange – A D/E B/C
4) Reverse Phase – A B/C E/D
5) Size Exclusion – ADEBC
6) UV – D/E B C/A

PICTURE of titrate graohs