Biochem exam 1 - question Flashcards
(34 cards)
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) 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
Define a chiral center
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
Differentiate between configuration and conformation
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.
Give examples of 5 types of isomers.
See HM1
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?
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
Describe the principle of hydrophobic interactions chromatography; now describe the principle of
hydrophilic interactions chromatography
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
Describe the principle of reversed phase chromatography; now describe the principle of normal
phase chromatography
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.
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.
: 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.)
Briefly define “isotonic,” “hypotonic,” and “hypertonic” solutions. (b) Describe what happens
when a cell is placed in each of these types of solutions.
: (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.
Draw the structures of the amino acids tyrosine and aspartate in the ionization state you would
expect at pH 4.0.
low pH = COOH
high pH =NH2
neutral = zwitterion
See HM1
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?
: 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.
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.
See the titration curve for glycine in Fig. 3-10, p. 79
Draw the structure of Lys–Ser–Glu in the ionic form that predominates at pH 4.
See HW1
low pH = COOH
high pH =NH2
neutral = zwitterionHW1
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
See HW1 High pH = Everything is deprotonated. Low pH = Everything is protonated low pH = COOH high pH =NH2 neutral = zwitterion
Why do smaller molecules elute after large molecules when a mixture of proteins is passed
through a size-exclusion (gel filtration) column?
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.
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) 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.)
What factors would make it difficult to interpret the results of a gel electrophoresis of proteins in
the absence of sodium dodecyl sulfate (SDS)?
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.
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) 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
How can isoelectric focusing be used in conjunction with SDS gel electrophoresis?
: 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.
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) 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.
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
g; a; d and e; b; c
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.
: (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.
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) 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.
Describe two major differences between chemical synthesis of polypeptides and synthesis of
polypeptides in the living cell
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.
