Download Exam I Cell and Molecular Biology September 26, 2007 This exam

Exam I
Cell and Molecular Biology
September 26, 2007
This exam is for «surname» «Name»
I. Multiple choice. «surname», please mark the correct answer or answers to each of the following
questions. Did you read carefully the above statement…these are some multiple multiple choice
questions. There are only 17, «surname», but they will take careful analysis.
1. The percentage of G-C pairs in DNA is related to the Tm (melting temperature) because:
→a. the stability of G-C and A-T base pairs is intrinsically different.
b. A-T base pairs require a higher temperature for denaturation.
c. the triple bonds of G-C base pairs are less stable than the double bonds of A-T base pairs.
d. the G-C content equals the A-T content.
2. Multiple mRNAs can arise from a primary transcript by use of alternative
→a. splicing
→c. promoters
→b. poly A sites
→d. ribosome binding sites
3. If the genome of a new virus displays the base composition at right, the virus most likely consists of
a. single-stranded DNA
A = 27%
G = 30%
→b. double-stranded DNA
T = 26%
C = 31%
c. single-stranded RNA
d. double-stranded RNA
e. a double-stranded DNA-RNA hybrid
4. A gene contains information for the following polypeptide: met-lys-ala-arg-val. The DNA sequence
encoding this polypeptide is
a. U A C U U C C G G G C C C A G.
→b. T A C T T C C G G G C C C A G.
c. T T C T A C G C C C G G C A G.
d. A T G A A G G C C C G G G T C.
5. Which of the following DNA sequences can be used to screen a cDNA library for the gene that
encodes a protein with the amino acid sequence Phe-Gly-Cys-Ser?
a. 5’ TTTGAGTGCAGA 3’
b. 5’ TTCCAATGGTCG 3’
→c. 5’ TTCGGGTGCAGC 3’
d. 5’ TTTGGATGTTCA 3’
6. A 1-mL solution of 0.05 M H2SO4 is diluted to 100 mL at 25°C. What is the pH of the resulting
solution?
→c. 3
a. 1
b. 2
d. 4
7. In a transmembrane protein, which of the following amino acids is most likely to be found in contact
with the phospholipid tails?
a. alanine
d. asparagine
c. aspartate
→b. phenylalanine e. methionine
8. In a biochemical reaction in which ∆H < 0 and ∆S > 0,
→a. the reaction is spontaneous.
c. the reaction is endergonic.
b. the reaction is endothermic.
d. ∆G is positive.
9. If the equilibrium constant for the reaction A → B is 0.5 and the initial concentration of A is 25 mM
and of B is 12.5 mM, then the reaction
a. will proceed in the direction it is written, producing a net increase in the concentration of B.
b. will produce energy, which can be used to drive ATP synthesis.
c. will proceed in the reverse direction, producing a net increase in the concentration of A.
→d. is at equilibrium.
10. An open reading frame (ORF) is defined as a DNA sequence that
a. begins with a start codon
→b. ends with a stop codon
c. contains 50 codons
d. contains approximately an equal frequency of A, T, G, and C
11. Which buffer maintains the physiologic pH of the cytosol?
a. acetate
c. lactate
b. carbonate →d. phosphate
12. The effect of an enzyme on the end equilibrium concentration of reactants and products is to
a. greatly increase the equilibrium concentration of reactants
b. greatly increase the equilibrium concentration of product
c. greatly increase the equilibrium concentration of both
d. →there will be no effect on the equilibrium concentration of either
e. none of the above is correct
13. Addition of phosphate residues to proteins is a common and important reaction with a large positive
delta G value. In a cell, this reaction is driven forward by coupling to a reaction with a delta G value
a. larger positive
c. smaller positive
→b. larger negative
d. smaller negative
14. In the above question, the coupled reaction that drives the energetically unfavorable phosphorylation
of proteins is most likely to be:
a. formation of a peptide bond
b. formation of a disulfide bond
c. epimerization of glucose
→d. hydrolysis of ATP
15. «surname», DNA replication
a. utilizes Okazaki fragments on the leading strand.
b. requires the addition of deoxynucleotides to the 5’ free hydroxl group.
c. can be initiated de novo.
d. occurs on only 1 strand of DNA.
→e. requires a DNA template, deoxynucleotides, primers and DNA polymerase.
16. Which of the following lead(s) to a point mutation?
a. deamination of a cytosine base into a uracil base
b. benzo(a)pyrene conversion of guanine to a thymine base
c. deamination of 5-methyl cytosine into thymine
→d. all of the above
e. only b and c.
17. Which of the following is a protein that is involved in translation?
a. topoisomerase
b. ribosomal RNA
c. RNA polymerase
→d. aminoacyl-tRNA synthetase
Section II. Short Answer. «surname», give the information required for each of the following questions
in the space provided (use reverse if needed).
1. The pKa for the dissociation of acetic acid is 4.74.
a. Calculate the amounts of acetic acid (HAc) and sodium acetate (Ac-) which you would need to
use to make a buffer that is 0.1 M in total acetate concentration and at pH 5.
A classic Henderson Hasselbach scenario: pH = pKa + log [A-]/[HA]. So 5 – 4.74 = log [A-]/[HA] antilog .26 = antilog
[A-] /(.1M - [A-]) and [A-] = 0.0645M so [HAc] = (.1 - 0.0645) = 0.0355.
b. Calculate the pH of this buffer after the addition of 0.01ml of 1 M HCl.
You are adding 0.01M H+ and [HAc] will thus be 0.0455 and [A-] = 0.0545M so we can write
pH = 4.74 + log [0.0545]/[0.0455] = 4.82.
2. Proteins may be affinity purified (i.e. using affinity chromatography…remember, «surname»?) from a
complex cellular extract by absorption to an antibody that binds that protein specifically. Antibodies bind
proteins based on molecular complementarity through a variety of noncovalent interactions. Name three
noncovalent bonds that might mediate antibody-protein binding. Suggest three mechanisms to separate
the protein from the antibody after the absorption step. Use reverse if necessary.
Hydrogen bonds, ionic bonds and hydrophobic bonds all play a part in the formation of affinity bonds. We can use salts, pH
changes, heat, or organic solvents to interfere with these bonds, allowing the proteins to separate.
3. Two enzymes, enzyme X and enzyme Z, can both use substrate D to produce product E. The assay
results are shown below. Using these data, plot the reaction velocity (umol E produced/min) versus
substrate concentration ([D] mM) for both enzyme X and enzyme Z using the graph below.
Reaction velocity
umol E produced/min
Enzyme X
[D] mM
1.0
0.4
2.0
0.7
3.0
1.6
4.0
3.3
5.0
5.5
7.5
5.9
10.0
6.0
Enzyme Z
1.2
2.4
3.3
4.3
4.9
5.7
6.0
a. What type of enzyme kinetics do enzymes X and Z
show?
First and second degree kinetics – the red dots are indicative of
cooperativity…multiple subunits; the green dots are the normal
path for a standard enzyme with a single subunit.
b. Estimate the Km for both enzyme X and enzyme Z for substrate D.
For X, KM = ~3.7, for Y, KM = ~2.3
c. How do the activities of enzyme X and enzyme Z differ in the range of 3 to 5 mM D?
About 35% increase for Z, but about 4 fold for X, so for X a small change in substrate yields a huge change in activity.
d. Based on this data, what can you conclude about the binding of substrate D to enzymes X and Z?
The straightforward behavior of enz Z suggests only one site is involved. But for X there must be multipbe active sites that are
controlled in a cooperative fashion, such that a small increase in substrate makes accessible all sites.
4. «surname», what is one conclusion that can be drawn from the observation that the genetic code is
nearly identical in all cells on earth?
Either that organisms are related by descent, or that all life had a common mind behind it. One possibility that it was designed
top-down as a computer programmer might do with man designed first and other organisms are a subset of the totality.
5. «surname», protein A and protein B were
analyzed using two-dimensional gel electrophoresis
(Isoelectric Focusing and SDS) as shown below.
a. What can you conclude about the size and net
charge of protein A and protein B at neutral pH?
Both proteins have weights of about 30KD
b. Can you separate protein A from protein B using
an ion exchange column that consists of positively
charged beads (an anion exchanger)? How would
you proceed?
A has a net negative charge, B has a net positive charge
So at neutral pH, A will bind, B will pass through. After the
column binds A, you can elute the protein by increasing the
salt concentration.
c. Using gel filtration chromatography, a technique that separates molecules on the basis of size,
protein B was found to elute from the column at an apparent molecular weight of 60 kDa. How can
you explain the discrepancy between the molecular weight of protein B when determined by twodimensional gel electrophoresis and that determined by gel filtration chromatography?
Whereas gel electrophoresis alters the proteins and disassembles the subunits, column electrophoresis will leave the
proteins in the polymeric state. The change in molecular weight is indicative of the fact that B functions as a structural
dimer.
6. Triacylglycerol and cholesterol esters are nonpolar; in contrast, phospholipids are amphipathic
molecules. Biomembranes are based on phospholipids rather than on triacylglycerols. Why?
Biomembranes are based on phospholipids rather than on triacylglycerols because phospholipids as amphipathic
molecules can form planar lipid bilayers, whereas the nonamphipathic nonpolar triacylglycerols cannot. The
amphipathic property, the presence of a polar and nonpolar domain at opposite ends of the same molecule, allows
phospholipids to form hydrophilic associations with water at the same time as forming hydrophobic associations with
each other through their hydrophobic tails. Triacylglycerols are strictly hydrophobic in nature and hence in an
aqueous environment tend to associate with one another to form lipid droplets. This minimizes the contact of
triacylglycerol with water i.e. oil and water do not mix.
7. The enzyme alcohol dehydrogenase is capable of catalyzing the oxidation of a number of different
substances, including ethanol, ethylene glycol, and methanol, to an aldehyde. The metabolic
products of both ethylene glycol and methanol are highly toxic to humans. A standard medical
treatment for prevention of ethylene glycol or methanol poisoning is the administration of a dose of
ethanol. Why is this treatment effective?
The administration of a dose of ethanol is effective because the ethanol-like ethylene glycol and methanol are capable of
binding to the enzyme, alcohol dehydrogenase, and in binding can compete with the other substrates. A sufficient dosage
of ethanol can out-compete the other substrates, and hence the ethylene glycol and methanol are not metabolized to toxic
products. Gradually the ethylene glycol or methanol will be excreted from the body.
8. In the experiments that led to the deciphering of the genetic code, synthetic mRNAs such as
polyuridylate were incubated with a cell-free E. coli translation system. Although these synthetic
polynucleotides were translated slowly (relative to the rates observed for biological mRNAs), the
corresponding peptides were produced in sufficient quantity to be analyzed. You are probably wondering
why these synthetic mRNAs were ever translated, since they do not contain start codons. The answer,
++
«surname» lies in the relatively high conc. Mg (0.02 M) used by Nirenberg and his coworkers in these
++
experiments. The effects of Mg were demonstrated by incubating bacterial ribosomes, a synthetic
polyribonucleotide, initiation and elongation factors, tRNAs, and nucleotide triphosphates with 0.005 M
++
++
Mg or 0.02 M Mg for 2 min. A portion of each mixture was then centrifuged for 2 h at 100,000g on a
15-40 percent sucrose density gradient. This procedure separates macromolecules on the basis of mass, or
S value. After centrifugation, the centrifuge tubes were punctured and the contents allowed to drip slowly
into a series of collection tubes. These fractions were assayed for RNA content by measuring the
absorbance at 260 nm, a wavelength at which RNA absorbs quite strongly (the bulk of the RNA in these
preparations is rRNA). Results of such an analysis, called a shift assay, are shown below. Translation
assays also were conducted by adding amino acids to another portion of each incubation mixture and
measuring the amount of protein formed. Protein synthesis was observed at the higher magnesium
concentration; no protein synthesis could be detected at the lower magnesium concentration.
a. What can you tell me about the meaning of the terms 30S, 50S, 70S? What does “S” stand for, how is
it determined and what information does it contain?
S stands for the Swedberg constant, a term derived from consideration of the forces affecting the movement of a particle in a
sedimentary force field. 30, 50, 70 refer to the rate of movement of the small, large and combined ribosomal masses during
sedimentation. Since S is a function of the viscosity of the medium, the density of the medium (both knowns), the diameter of
the particle and the density of the particle, we can determine the diameter or density fo any protein from its rate of
sedimentation.
b. Explain the two curves illustrated above.
When the Mg2+ concentration is high, the ribosomal subunits aggregate into functional ribosomes. When the concentration is
low, the ribosomes are found as individual small and large subunits. Thus at high enough concentrations of Mg2+ the
ribosomes will combine and function even in the absence of an intact mRNA etc.
9. The replication of DNA is quite complicated and requires the participation of many different
enzymatic activities. These include an RNA polymerase (DNA primase) that synthesizes short
segments of RNA, called primers, which are base paired to the DNA template. The 3´-OH ends of
these RNA primers serve as initiation sites for the actual DNA polymerase activity. Obviously, DNA
primase can synthesize polynucleotides without the benefit of a 3´-OH primer; indeed, it can
catalyze the hydrolysis and subsequent linkage of two nucleoside triphosphates without any
complementary strand whatsoever. The short segments of RNA must be eliminated and replaced
with DNA before replication can be completed. Why do you think that RNA, rather than DNA,
primers are employed in the DNA replication process?
Because DNA is the molecule of inheritance, replication errors must be carefully avoided. Base pairing without a 3´-OH
primer, as necessarily performed by any RNA polymerase, is very error prone. If a DNA polymerase performed this
function, errors in the DNA sequence (mutations) would be introduced during replication and then transmitted to future
generations. An error rate of 1 base in 105, which is not unusual for RNA polymerases, would result in an enormous
increase in the mutation rate. However, the RNA primers made during DNA replication are erased and replaced with
high-fidelity DNA copies; any mismatched bases will be replaced before being passed on to the next generation.
10. When the CAU anticodon of a tRNAMet was modified to UAC, the anticodon for tRNAVal, valine
aminoacyl-tRNA synthetase recognized the altered tRNAMet and added valine rather than
methionine to it. When the converse modification was made, the altered tRNAVal containing a CAU
anticodon (rather than UAC) was recognized and activated by methionine aminoacyl-tRNA
synthetase. What do these data suggest about the mechanism by which aminoacyl-tRNA synthetases
recognize their cognate tRNAs?
These data suggest that the anticodon region of a tRNA is recognized by the corresponding aminoacyl-tRNA
synthetase. However, some tRNAs apparently contain other identity elements that are of primary importance in
recognition of these species by the appropriate aminoacyl-tRNA synthetase.
III. Draw the chemical structures of the three nucleotides that comprise a trp codon on the reverse of
this page, «surname». Indicate where start codon(s) and stop codons occur in the molecule below by
circling them. Then indicate the amino acid sequence for the protein coded for by this piece of DNA
by giving the abbreviations for the amino acids below the appropriate codons AND DRAWING THE
STRUCTURES OF EACH AMINO ACID INCLUDED IN THE RESULTANT PEPTIDE IN THE
SPACE BELOW.
AAATCCTTACAAAGGATTACCGATCTATCAAATTTACCTAT
UUUAGGAAUGUUUCCUAAUGGCUAGAUAGUUUAAAUGGAUA