Biochem exam 1 - question Flashcards

1
Q

E. coli is known as a gram-negative bacterial species. (a) How is this determined? (b) How do
gram-negative bacteria differ structurally from gram-positive bacteria?

A

: (a) Gram-negative bacteria have little affinity for the dye gentian violet used in Gram’s stain, but grampositive bacteria retain Gram’s stain. (b) Gram-negative bacteria have an outer membrane and a peptidoglycan
layer; gram-positive bacteria lack an outer membrane and the peptidoglycan layer is much thicker

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2
Q

Define a chiral center

A

A chiral center is a carbon atom that has four different substituents attached, and cannot be
superimposed on its mirror image – as a right hand cannot fit into a left glove

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3
Q

Differentiate between configuration and conformation

A

Configuration denotes the spatial arrangement of the atoms of a molecule that is conferred by the
presence of either double bonds, around which there is no freedom of rotation, or chiral centers, which give
rise to stereoisomers. Configurational isomers can only be interconverted by temporarily breaking covalent
bonds. Conformation refers to the spatial arrangement of substituent groups that, without breaking any bonds,
are free to assume different positions in space because of the freedom of bond rotation.

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4
Q

Give examples of 5 types of isomers.

A

See HM1

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5
Q

Explain why living organisms are able to produce particular chiral forms of different
biomolecules while laboratory chemical synthesis usually produces a racemic mixture. Is using a
racemic good? If one wants to use only one enantiomer, how would he/she do it?

A

Laboratory syntheses usually use achiral reagents and thus produce racemic mixtures of products. In
contrast, because all enzymes are made of chiral precursors, all enzymes are inherently chiral catalysts. Thus,
they will show strong stereoselectivity in reactants and mechanisms, leading to the production of chiral
products
Using a racemic is not good as biomolecules such as as receptors for drugs are stereospecific, so each of the
two enantiomers of the drug may have very different effects on an organism. One may be beneficial, the other
toxic; or one enantiomer may be ineffective and its presence could reduce the efficacy of the other enantiomer

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6
Q

Describe the principle of hydrophobic interactions chromatography; now describe the principle of
hydrophilic interactions chromatography

A

Hydrophobic Interaction Chromatography: Sample molecules containing hydrophobic and hydrophilic
regions are applied to an HIC column in a high-salt buffer. The salt in the buffer reduces the solvation of
sample solutes. As solvation decreases, hydrophobic regions that become exposed are adsorbed by the
media. The more hydrophobic the molecule, the less salt is needed to promote binding. Usually a
decreasing salt gradient is used to elute samples from the column in order of increasing hydrophobicity.
Sample elution may also be assisted by the addition of mild organic modifiers or detergents to the elution
buffer
In hydrophilic interaction chromatography, is a liquid chromatography technique that uses a polar
stationary phase in conjunction with a mobile phase containing an appreciable quantity of water combined
with a higher proportion of a less polar solvent. Here, the hydrophilic, polar, and charged compounds are
retained preferentially compared with hydrophobic neutral compounds

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7
Q

Describe the principle of reversed phase chromatography; now describe the principle of normal
phase chromatography

A

In reversed phase chromatography, the most polar compounds elute first with the most nonpolar
compounds eluting last. The mobile phase is generally a binary mixture of water and a miscible polar
organic solvent like methanol, acetonitrile. Retention increases as the amount of the polar solvent (water)
in the mobile phase increases.
In normal-phase chromatography, the least polar compounds elute first and the most polar compounds
elute last. The mobile phase consists of a nonpolar solvent such as hexane or heptane mixed with a
slightly more polar solvent such as isopropanol, ethyl acetate or chloroform. Retention decreases as the
amount of polar solvent in the mobile phase increases.

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8
Q

Give the general Henderson-Hasselbalch equation and sketch the plot it describes (pH against
amount of NaOH added to a weak acid). On your curve label the pKa for the weak acid, and
indicate the region in which the buffering capacity of the system is greatest.

A

: The inflection point, which occurs when the weak acid has been exactly one-half titrated with NaOH,
occurs at a pH equal to the pKa of the weak acid. The region of greatest buffering capacity (where the titration
curve is flattest) occurs at pH values of pKa ±1. (See Fig. 2-17, p. 59.)

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9
Q

Briefly define “isotonic,” “hypotonic,” and “hypertonic” solutions. (b) Describe what happens
when a cell is placed in each of these types of solutions.

A

: (a) An isotonic solution has the same osmolarity as the solution to which it is being compared. A
hypotonic solution has a lower osmolarity than the solution to which it is being compared. A hypertonic
solution has a higher osmolarity than the solution to which it is being compared. (b) Higher osmolarity results
in osmotic pressure, which generally leads to movement of water across a membrane. In an isotonic solution,
in which the osmolarity of the solution is the same as the cell cytoplasm, there will be no net water movement.
In a hypotonic solution, water will move into the cell, causing the cell to swell and possibly burst. In a
hypertonic solution, water will move out of the cell and it will shrink.

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10
Q

Draw the structures of the amino acids tyrosine and aspartate in the ionization state you would
expect at pH 4.0.

A

low pH = COOH
high pH =NH2
neutral = zwitterion
See HM1

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11
Q

How does the shape of a titration curve confirm the fact that the pH region of greatest buffering
power for an amino acid solution is around its pK’s?

A

: In a certain range around the pKa’s of an amino acid, the titration curve levels off. This indicates that
for a solution with pH  pK, any given addition of base or acid equivalents will result in the smallest change in
pH—which is the definition of a buffer.

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12
Q
Leucine has two dissociable protons, one with a pKa of 2.3, the other with a pKa of 9.7. Sketch a
properly labeled titration curve for leucine titrated with NaOH; indicate where the pH = pK and
the region(s) in which buffering occurs.
A

See the titration curve for glycine in Fig. 3-10, p. 79

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13
Q

Draw the structure of Lys–Ser–Glu in the ionic form that predominates at pH 4.

A

See HW1
low pH = COOH
high pH =NH2
neutral = zwitterionHW1

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14
Q

Draw the structure of a) Arginine, b) Aspartate, 3) Tyrosine and 4) +H3N-Arg-Asp-Tyr-COO- in
zwiterionic form and their ionized forms at low pH, neutral pH and high pH

A
See HW1
High pH = Everything is deprotonated.
Low pH = Everything is protonated
low pH = COOH
high pH =NH2
neutral = zwitterion
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15
Q

Why do smaller molecules elute after large molecules when a mixture of proteins is passed
through a size-exclusion (gel filtration) column?

A

The column matrix is composed of cross-linked polymers with pores of selected sizes. Smaller
molecules can enter pores in the polymer beads from which larger molecules would be excluded. Smaller
molecules therefore have a larger three-dimensional space in which to diffuse, making their path through the
column longer. Larger molecules migrate faster because they pass directly through the column, unhindered by
the bead pores.

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16
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

A

: (a) Ion-exchange chromatography separates proteins on the basis of their charges. (b) Size-exclusion 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.)

17
Q

What factors would make it difficult to interpret the results of a gel electrophoresis of proteins in
the absence of sodium dodecyl sulfate (SDS)?

A

Without SDS, protein migration through a gel would be influenced by the protein’s intrinsic net
charge—which could be positive or negative—and its unique three-dimensional shape, in addition to its
molecular weight. Thus, it would be difficult to ascertain the difference between proteins based upon a
comparison of their mobilities in gel electrophoresis.

18
Q

A biochemist is attempting to separate a DNA-binding protein (protein X) from other proteins in a
solution. Only three other proteins (A, B, and C) are present. The proteins have the following
properties:
16
pI
(isoelectric Size
Point ) M Bind to DNA?
–––––––––––––––––––––––––––––––––––––––––––––––––––––
protein A 7.4 82,000 yes
protein B 3.8 21,500 yes
protein C 7.9 23,000 no
protein X 7.8 22,000 yes
–––––––––––––––––––––––––––––––––––––––––––––––––––––
(a) What type of protein separation techniques might she use to separate (b)
protein X from protein A?
(c) protein X from protein B?
(d) protein X from protein C?

A

: (a) size-exclusion (gel filtration) chromatography to separate on the basis of size; (b) ion-exchange
chromatography or isoelectric focusing to separate on the basis of charge; (c) specific affinity
chromatography, using immobilized DNA

19
Q

How can isoelectric focusing be used in conjunction with SDS gel electrophoresis?

A

: Isoelectric focusing can separate proteins of the same molecular weight on the basis of differing
isoelectric points. SDS gel electrophoresis can then separate proteins with the same isoelectric points on the
basis of differing molecular weights. When combined in two-dimensional electrophoresis, a great resolution of
large numbers of proteins can be achieved.

20
Q

In one or two sentences, describe the usefulness of each of the following reagents or reactions in
the analysis of protein structure:
(a) Edman reagent (phenylisothiocyanate)
(b) Sanger reagent (1-fluoro-2,4-dinitrobenzene, FDNB) (c) trypsin

A

(a) used in determination of the amino acid sequence of a peptide, starting at its amino terminus; (b)
used in determination of amino-terminal amino acid of a polypeptide; (c) used to produce specific peptide
fragments from a polypeptide.

21
Q

The following reagents are often used in protein chemistry. Match the reagent with the purpose
for which it is best suited. Some answers may be used more than once or not at all; more than one
reagent may be suitable for a given purpose.
(a) CNBr (cyanogen bromide) (e) performic acid
(b) Edman reagent (phenylisothiocyanate) (f) chymotrypsin
(c) FDNB (g) trypsin (d) dithiothreitol
___ hydrolysis of peptide bonds on the carboxyl side of Lys and Arg
___ cleavage of peptide bonds on the carboxyl side of Met
___ breakage of disulfide (—S—S—) bonds
___ determination of the amino acid sequence of a peptide
___ determining the amino-terminal amino acid in a polypeptide

A

g; a; d and e; b; c

22
Q

A biochemist wishes to determine the sequence of a protein that contains 123 amino acid residues.
After breaking all of the disulfide bonds, the protein is treated with cyanogen bromide (CNBr),
and it is determined that that this treatment breaks up the protein into seven conveniently sized
peptides, which are separated from each other. It is your turn to take over. Outline the steps you
would take to determine, unambiguously, the sequence of amino acid residues in the original
protein.

A

: (1) Use Edman degradation to determine the sequence of each peptide (2) Create a second set of
peptides by treatment of the protein with a specific protease (for example, trypsin), and determine the
sequence of each of these. (3) Place the peptides in order by their overlaps. (4) Finally, by a similar
analysis of the original protein without first breaking disulfide bonds, determine the number and location
of —S—S— bridges
An alternate answer is determining the sequence by Mass Spectrometry.

23
Q

You are trying to determine the sequence of a protein that you know is pure. Give the most likely
explanation for each of the following experimental observations. You may use a simple diagram
for your answer.
(a) The Sanger reagent (FDNB, fluorodinitrobenzene) identifies Ala and Leu as amino-terminal
residues, in roughly equal amounts.
(b) Your protein has an apparent Mr of 80,000, as determined by SDS-polyacrylamide gel
electrophoresis. After treatment of the protein with performic acid, the same technique reveals two
proteins of Mr 35,000 and 45,000.
(c) Size-exclusion chromatography (gel filtration) experiments indicate the native protein has an
apparent Mr of 160,000.

A

(a) The protein has some multiple of two subunits, with Ala and Leu as the amino-terminal residues. (b)
The protein has two subunits (Mr 35,000 and 45,000), joined by one or more disulfide bonds. (c) The native
protein (Mr 160,000) has two Mr 35,000 subunits and two Mr 40,000 subunits.

24
Q

Describe two major differences between chemical synthesis of polypeptides and synthesis of
polypeptides in the living cell

A

There are many such differences; here are a few: (1) Chemical synthesis proceeds from carboxyl
terminus to amino terminus; in the living cell, the process starts at the amino terminus and ends at the carboxyl
terminus. In the living cell, synthesis occurs under physiological conditions; chemical synthesis does not.
Chemical synthesis is only capable of synthesizing short polypeptides; cells can produce proteins of several
thousand amino acids.

25
Q

A polypeptide is hydrolyzed, and it is determined that there are 3 Lys residues and 2 Arg residues
(as well as other residues). How many peptide fragments can be expected when the native
polypeptide is incubated with the proteolytic enzyme trypsin?

A

6

26
Q

Any given protein is characterized by a unique amino acid sequence (primary structure) and threedimensional (tertiary) structure. How are these related?

A

Anfinsen showed that a completely denatured enzyme (ribonuclease) could fold spontaneously into its
native, enzymatically active form with only the primary sequence to guide it.

27
Q

How does one determine the three-dimensional structure of a protein? Describe the principle of
the method

A

The protein is crystallized, and the crystal structure is determined by x-ray diffraction. The pattern of
diffracted x-rays yields, by Fourier transformation, the three-dimensional distribution of electron density. By
matching electron density with the known sequence of amino acids in the protein, each region of electron
density is identified as a single atom. Sometimes, the three-dimensional structure of a small protein or peptide
can be determined in solution by sophisticated analysis of the NMR spectrum of the polypeptide. This
technique can also reveal dynamic aspects of protein structure such as conformational changes. Computer
analysis of two-dimensional NMR spectra can be used to generate a picture of the three-dimensional structure
of a protein.

28
Q

Each of the following reagents or conditions will denature a protein. For each, describe in one or
two sentences what the reagent/condition does to destroy native protein structure.
(a) urea
19
(b) high temperature
(c) detergent
(d) low pH

A

: (a) Urea acts primarily by disrupting hydrophobic interactions. (b) High temperature provides thermal
energy greater than the strength of the weak interactions (hydrogen bonds, electrostatic interactions,
hydrophobic interactions, and van der Waals forces, breaking these interactions). (c) Detergents bind to
hydrophobic regions of the protein, preventing hydrophobic interactions among several hydrophobic patches
on the native protein. (d) Low pH causes protonation of the side chains of Asp, Glu, and His, preventing
electrostatic interactions

29
Q

How can changes in pH alter the conformation of a protein?

A

Changes in pH can influence the extent to which certain amino acid side chains (or the amino and
carboxyl termini) are protonated. The result is a change in net charge on the protein, which can lead to
electrostatic attractions or repulsions between different regions of the protein. The final effect is a change in
the protein’s three-dimensional shape or even complete denaturation.

30
Q

Once a protein has been denatured, how can it be renatured? If renaturation does not occur, what
might be the explanation?

A

Because a protein may be denatured through the disruption of hydrogen bonds and hydrophobic
interactions by salts or organic solvents, removal of those conditions will reestablish the original aqueous
environment, often permitting the protein to fold once again into its native conformation. If the protein does
not renature, it may be because the denaturing treatment removed a required prosthetic group, or because the
normal folding pathway requires the presence of a polypeptide chain binding protein or molecular chaperone.
The normal folding pathway could also be mediated by a larger polypeptide, which is then cleaved (for
example, insulin). Denatured insulin would not refold easily.

31
Q

Describe the principles, similarities and differences between Blue Native PAGE and Colorless
Native PAGE

A

Blue Native-PAGE - Separation is according to the molecular mass of the native protein complexes, due
to the external charge induced by Coomassie dye.
Colorless Native-PAGE - Separation is according to the internal charge of the native protein complexes, there
is no external charge induced by Coomassie dye and no separation according to the molecular mass

32
Q

List at least four physico-chemical properties of proteins used in protein purification. Now list
five guidelines that are used for protein purification

A

Physico-chemical properties of proteins: charge, polarity, hydrophobicity, solubility, isoelectric point,
molecular mass, affinity, enantioselectivity, post-translational modifications
20
Guidelines that are used for protein purification: define objectives, define properties of the target protein and
critical impurities, develop analytical assays, minimize sample handling at every stage, minimize use of
additives, remove damaging contaminants early, use different technique at each step, minimize number of
steps.

33
Q

PROBLEM: A biochemist is attempting to isolate a cytosolic heterodimeric enzyme from MCF
breast cancer cells grown in cell culture. The heterodimer is held together through hydrophobic
interactions and is composed of a 129 kDa protein A (pI 4.5) and a 12 kDa protein B (pI 6.9).
Design a purification scheme for this heterodimeric protein and justify each step that you will use

A

Collection of the cells, cell lysis and extraction of proteins. Chromatography (hydrophobic interactions
chromatography) to separate subunits of proteins. Separation of subunits using 2D electrophoresis (first
according to isolectric point, then according to molecular weight).

34
Q

PROBLEM: Describe a purification procedure for a mitochondrial (membrane or soluble) protein
from 1) spinach leaves 2) dog brain and 3) cat fur.

A
  1. Take the leaves, put in a blender, and start with sufficient volume to have enough final product.
  2. The leaves should be subjected to gentle blending or grinding with neutral buffer, all the steps will be
    done on ice.
  3. Subject it to low speed centrifugation, discard pellet and collect supernatant.
  4. 2
    nd centrifugation at a higher speed, followed by 3rd very strong centrifugation, where all the
    chloroplast goes to the pellet.
  5. Add percoll or sucrose gradient which will separate the cells and chloroplasts in a gradient.
  6. Based on the gradient, cells will be at the top, intact mitochondria in the middle and intact chloroplasts
    at the bottom.
  7. Use a suction device and get each layer out into their respective vials.
  8. Dialyze them in a dialysis bag.
  9. Take the mitochondria fraction, break it by osmotic shock.
  10. Centrifuge and the membrane goes to the bottom and soluble part will be in the supernatant.
  11. To purify the soluble protein, do an ion exchange chromatography and gel filtration. Then to verify –
    do electrophoresis with the controls from every step.
  12. Finally do MS analysis for sequence information.
    (B)
     Dissect brains and, place on ice & cut into small pieces and grind it.
     Gently homogenize the tissue.
     Centrifuge for few minutes to separate the cytoplasm (supernatant).
     Rest of the steps are same as above.
    (C) Cat fur: Not possible