_______________________________
1) NAME: In class we discussed four different levels at which the concentration of proteins is regulated
in eukaryotic cells. List four of them. (4pts)
2) Name three mechanisms by which eukaryotic gene activation is facilitated. (3pts)
3) Name three features a plasmid vector should contain to clone a gene (3pts).
4) The first enzymatic step in making a cDNA uses which of the following? (2 pt)
a. EcoRI nuclease
b. Taq polymerase
c. DNA polymerase
d. RNA-dependent DNA polymerase
5) You are given the single stranded DNA template below, which is suitable for DNA
sequencing.
5-TTGCGTACGCTGCATATGGTCGTAGCGTATGCTGACTGACTGATCGCTC-3
You are also given the following radioactively labeled primer: 3-CATACGACTGACTG-5
A) If you carried out sequences reaction in the presence of E. coli DNA polymerase and the
four dNTPs, what would be the length of the radiolabeled product that is formed? (2pts)
B) If you now carry out the same sequencing reaction (template, primer, dNTPs,
polymerase), but include a small amount of dideoxy CTP in the mix, what would be the
lengths of the radiolabeled products that would be formed? (2pts)
1
NAME: _______________________________
6) It is difficult to introduce DNA into plant cells because of the plant cell walls. Mention two
methods that get around this problem. (2 pts)
7) Approximately 40% of the human genome are simple sequence DNA. For much of this DNA
its function is not known. However, simple sequence DNA is used in biotechnology
applications. Name an example. (2pt)
8) The human genome is approximately 1000 times larger than the genome of E. coli. However,
the human genome contains only approximately 10 times more genes than the E. coli genome.
Name two reasons? (3pts)
9) The figure shows a two-locus DNA fingerprinting gel using DNAs from a crime victim (V),
blood taken from the scene of the crime (E), and two suspects (S1, S2, and S3). Can you rule out
any of the suspects, and if so, which one? Based on the evidence, could you convict any of these
suspects beyond a reasonable doubt? If you are unable to draw firm conclusions state what other
information you would require for a firm conclusion. (3pts)
V
E
S1 S2 S3
2
NAME: _______________________________
10) The consensus sequence of the E. coli -10 promoter element it TATAAT. You are
comparing two promoters, one of which as a -10 element sequences of TATGAT and the other a
-10 element sequence of CATGAT. Can you make any predictions about how well the two
promoters will support the initiation of transcription? If yes, which is the stronger promoter and
why? (3pts)
11)In class we talked about how bacterial cells can take up the amino acid tryptophan from their
surroundings, or if the external supply is insufficient, they can synthesize trytophan by using
enzymes in the cell. In some bacteria, the control of glutamine synthesis is similar to that of
tryptophan synthesis, such that the glutamine repressor is used to inhibit the transcription of the
glutamine operon, which contains the genes that code for the enzymes required for glutamine
synthesis. Upon binding to cellular glutamine, the glutamine repressor binds to a site in the
promoter of the operon.
A. Why is glutamine-dependent binding to the operon a useful property for the glutamine
repressor? (2pt)
B. What would you expect to happen to the regulation of the enzymes that synthesize
glutamine in cells that express a mutant form of the glutamine repressor that cannot bind
to DNA? (2pts)
C. What would you expect to happen to the regulation of the enzymes that synthesize
glutamine in cells that express a mutant form of the glutamine repressor that binds to
DNA even when no glutamine is bound to it? (2pts)
3
NAME: _______________________________
12)
In the absence of glucose, E. coli can proliferate using the pentose sugar arabinose. The
ability of E. coli to utilize the sugar arabinose is regulated via the arabinose operon, depicted in
the figure below. The araA, araB, and araD genes encode enzymes for the metabolism of
arabinose. The araC gene encodes a gene regulatory protein that binds adjacent to the promoter
of the arabinose operon. To understand the regulatory properties of the AraC protein, you
engineer a mutant bacterium in which the araC gene has been deleted and look at the effect of
the presence or absence of the AraC protein on the AraA enzyme.
A.
If the AraC protein works as a gene repressor, would you expect araA RNA levels
to be high or low in the presence of arabinose in the araC mutant cells? What
about in the araC mutant cells in the absence of arabinose? (2pt)
B.
Your findings from the experiment are summarized in the table below.
Genotype
araC+ (normal
cells)
araC (mutant
cells)
in the absence of
arabinose
low
low
araA RNA Levels
in the presence of
arabinose
high
low
Do the results in the table indicate that the AraC protein regulates arabinose metabolism
by acting as a gene repressor or a gene activator? (2pts)
13)
List three ways that the process of eucaryotic transcription differs from the process of
bacterial transcription. (3pts)
4
NAME: _______________________________
14)
Label the following structures on the figure below (4pts).
A.
B.
C.
D.
15)
Activator protein
RNA polymerase
General transcription factors
Mediator
Consider a gene with a particular function. Mutation X and mutation Y each cause
defects in the function of the encoded protein. Yet a gene containing both mutations X
and Y encodes a protein that works even better than the original protein. The odds that a
single mutational event will generate both mutations X and Y are exceedingly small.
Explain a simple way that an organism with a mutant gene containing both mutations X
and Y could arise during evolution. (4pts)
5
NAME: _______________________________
16)
The gene for a hormone necessary for insect development contains binding sites for three
gene regulatory proteins called A, B, and C. Because the binding sites for A and B
overlap, A and B cannot bind simultaneously. You make mutations in the binding sites
for each of the proteins and measure hormone production in cells that contain equal
amounts of the A, B, and C proteins. The results of your studies are summarized in the
figure below. In each of the following sentences, choose one of the phrases within square
brackets to make the statement consistent with the above results.
A.
B.
Protein A is a [stronger/weaker] activator of transcription than protein B. (2pts)
C.
17)
Protein A binds to its DNA binding site [more tightly/less tightly] than protein B
binds to its DNA binding site. (2pts)
Protein C is able to prevent activation by [protein A only/protein B only/both
protein A and protein B]. (2pts)
A.
When a mutation arises in the human genome, it can have three possible
consequences: beneficial to the individual, selectively neutral, or detrimental.
Order these from most likely to least likely. (2pts)
B.
The spread of a mutation in subsequent generations will, of course, depend on its
consequences to individuals that inherit it. Order the three possibilities above
from that which is most likely to spread and become over-represented in
subsequent generations to that which is most likely to become under-represented
6
NAME: _______________________________
or disappear from the population. (2pts)
18)
The number of proteins found in humans and other organisms can vastly exceed the
number of genes. This is largely due to (2 pt)
(a)
protein degradation.
(b)
alternative splicing.
(c)
homologous genes.
(d)
synteny.
(e)
mutation.
19)
You have accidentally torn the labels off two tubes, each containing a different plasmid,
and now do not know which plasmid is in which tube. Fortunately, you have restriction
maps for both plasmids, shown in the figure below. You have the opportunity to test just
one sample from one of your tubes. You have equipment for agarose gel electrophoresis,
a standard set of DNA size markers, and the necessary restriction enzymes. (4 pts)
Outline briefly the experiment you would do to determine which plasmid is in which
tube. Which restriction enzyme or combination of restriction enzymes would you use in
this experiment?
7
NAME: _______________________________
20)You have sequenced a fragment of DNA and produced the gel shown. Near the top of the
gel, there is a section where there are bands in all four lanes (indicated by the arrow). Which of
the following mishaps would account for this phenomenon? (3pts)
(a)
(b)
(c)
(d)
(e)
You mistakenly added all four dideoxynucleotides to one of the reactions.
You forgot to add deoxynucleotides to the reactions.
You forgot to add dideoxynucleotides to each of the reactions.
A fraction of the DNA you are sequencing was cut by a restriction nuclease.
Your primer hybridizes to more than one area of the fragment of DNA you are
sequencing.
21)
What is the sequence of the DNA template in the gel in the previous question, starting
from the 5 end of the template and up to the position of the arrow? (4pts)
22)
In situ hybridization can be used to determine the (2 pt)
(a)
sequence of a cloned gene.
(b)
distribution of proteins in tissues.
(c)
position of a cloned fragment of DNA on a plasmid.
(d)
size of a gene.
(e)
distribution of a given type of mRNA in different tissues.
8
23)
24)
NAME: _______________________________
What is the main reason for using a cDNA library rather than a genomic library to isolate
a human gene from which you wish to make large quantities of the human protein in
bacteria? (2pts)
You have an oligonucleotide probe that hybridizes to part of gene A from a eucaryotic
cell. Indicate whether a cDNA library or a genomic DNA library will be more
appropriate to use for the following applications.
A.
You want to study the promoter of a gene A. (1pt)
B.
C.
You discover that Gene A is alternatively spliced and you want to see which
predicted alternative splice products are actually produced in a cell. (1pt)
D.
You want to find both gene A and the genes located near gene A on the
chromosome. (1pt)
E.
25)
Gene A encodes a tRNA and you wish to isolate a piece of DNA containing the
full-length sequence of the tRNA. (1pt)
You want to express gene A in bacteria to produce lots of protein A. (1pt)
You want to amplify the DNA between the two stretches of sequence shown in the figure
below. Of the listed primers, choose the pair that will allow you to amplify the DNA by
PCR. (4pts)
9
NAME: _______________________________
26)
You have been asked to consult for a biotech company that is seeking to understand why
some fungi can live in very extreme environments, such as the high temperatures inside naturally
occurring hot springs. The company has isolated two different fungal species, F. cattoriae and
W. gravinius, both of which can grow at temperatures exceeding 95C. The company has
determined the following things about these fungal species:
Property
F. cattoriae
W. gravinius
Genome size
Repetitive DNA?
1 MB
20% of genome contains
large stretches of CG repeats
3 MB
< 0.1% of genome
By sequencing and examining their genomes, the biotech company hopes to understand
why these species can live in extreme environments. However, the company only has the
resources to sequence one genome, and would like your input as to which species should be
sequenced and whether you believe a shotgun strategy will work in this case. (Be sure to explain
your answer). (2pts)
27)
Transposable elements litter the genomes of primates and a few of them are still capable
of moving to new regions of the genome. If a transposable element jumped into an important
gene in one of your cells when you were a baby and caused a disease, is it likely that your child
would also have the disease? Explain. (2pts)
28)
For each statement below, indicate if it is TRUE or FALSE.
(a)
To address a challenge or develop a new function, evolution essentially builds
from first principles, like an engineer, to find the best possible solution. (1pt)
(b)
Nearly every instance of DNA duplication leads to a new functional gene. (1pt)
(c) A pseudogene is highly homologous to a functional gene but cannot be expressed
due to mutation(s). (1pt)
(d) Most genes in vertebrates are unique, and only a few genes are members of
multigene families. (1pt)
(e) Horizontal transfer is very rare and thus has had little influence on the genomes of
bacteria. (1pt)
10
NAME: _______________________________
29) The sketch below shows an agarose gel with a plasmid that has been digested with different
restriction enzymes as indicated. Draw the restriction map of the plasmid. B = BamH1; E =
EcoR1; H = HindIII. The numbers on the left are the approximately molecular weights in
kilobasepairs. So for example, cutting the plasmid with BamH1 gives you one piece of DNA
that is 6000 basepairs long. (5 pts)
B
6
5
4
3
2
1
11
E
H
E
B
H
B
H
E
H
E
B
NAME: _______________________________
Answers
1) In class we discussed four different levels at which the concentration of proteins is regulated
in eukaryotic cells. List all four of them. (4pts)
Initiation of transcription
mRNA stability and modification
Initiation of translation
Protein stability/degradation
2) Name three mechanisms by which eukaryotic gene activation is facilitated. (3pts)
stabilizing the (pre)initiation complex of RNA polymerase
acetylation (modification is okay) of histone tails
recruiting of chromatin remodeling factors
3) Name three features a plasmid vector should contain to clone a gene (3pts).
origin of replication.
restriction nuclease cut site.
Selective marker (antibiotic resistance gene or similar is okay).
4) The first enzymatic step in making a cDNA uses which of the following?
a. EcoRI nuclease
b. Taq polymerase
c. DNA polymerase
d. RNA-dependent DNA polymerase
(1 pt)
5) You are given the single stranded DNA template below, which is suitable for DNA
sequencing.
5-TTGCGTACGCTGCATATGGTCGTAGCGTATGCTGACTGACTGATCGCTC-3
You are also given the following radioactively labeled primer: 3-CATACGACTGACTG-5
A) If you carried out sequences reaction in the presence of E. coli DNA polymerase and the
four dNTPs, what would be the length of the radiolabeled product that is formed? (2pts)
The product will be 40 nucleotides in length (14 from the primer and 26 newly snthesized).
B) If you now carry the out same sequencing reaction (template, primer, dNTPs,
polymerase), but include a small amount of dideoxy CTP in the mix, what would be the
lengths of the radiolabeled products that would be formed? (2pts)
16, 19, 22,23,29,32,36,38 nuleotides.
12
NAME: _______________________________
6) It is difficult to introduce DNA into plant cells because of the plant cell walls. Mention two
methods that get around this problem. (2 pts)
Agrobacterium tumefaciens
Genegun
7) Approximately 40% of the human genome are simple sequence DNA. For much of this DNA
its function is not known. However, simple sequence DNA is used in biotechnology
applications. Name an example. (1pt)
Answer:
DNA finger printing (DNA identification is okay) using VNTRs (variable numbers of tandem
repeats)
8) The human genome is approximately 1000 times larger than the genome of E. coli. However,
the humane genome contains only approximately 10 times more genes than the E. coli genome.
Name two reasons? (2pts)
Answer: repetitive DNA and introns (allow pseudo genes as one answer)
9) The figure shows a two-locus DNA fingerprinting gel using DNAs from a crime victim (V),
blood taken from the scene of the crime (E), and two suspects (S1, S2, and S3). Can you rule out
any of the suspects, and if so, which one? Based on the evidence, could you convict any of these
suspects beyond a reasonable doubt? If you are unable to draw firm conclusions state what other
information you would require for a firm conclusion. (3pts)
V
E
S1 S2 S3
Answer:
Rule out S1 and S3.
Cannot convict S2.
Need to know frequency of the alleles in the population.
13
NAME: _______________________________
10) The consensus sequence of the E. coli -10 promoter element it TATAAT. You are
comparing two promoters, one of which as a -10 element sequences of TATGAT and the
other a -10 element sequence of CATGAT. Can you make any predictions about how
well the two promoters will support the initiation of transcription? If yes, which is the
stronger promoter and why? (2pts)
The second promoter has diverged more from the consensus sequence and therefore will be the
weaker promoter.
11)
A.
B.
C.
If sufficient glutamine is present in cells, the glutamine repressor will block the
synthesis of enzymes that would make more glutamine. Likewise, if cells are
starved for glutamine, the unoccupied repressor would not bind to the DNA and
the enzymes that synthesize glutamine would be induced. This allows for a direct
connection between the levels of glutamine and the expression of glutamine
synthesizing enzymes.
The glutamine synthesis enzymes would be permanently ON, regardless of the
level of glutamine in the cells.
The glutamine synthesis enzymes would be always OFF, regardless of the level of
glutamine in the cells, since the repressor is always bound to the DNA. These
cells will not be able to grow unless glutamine is added to the medium.
14
NAME: _______________________________
12)
A.
B.
13)
If the AraC protein acted as a gene repressor for the arabinose operon, araA RNA
levels should be high in the presence or absence of arabinose when there is no
AraC protein around. In fact, the araA RNA levels should be high all the time,
regardless of the presence or absence of arabinose, since the AraA gene should be
transcribed under all conditions in the absence of AraC.
The results are consistent with AraC acting as a gene activator for the arabinose
operon. A gene activator must bind to the promoter regions of the arabinose
genes in order to stimulate their transcription. Thus, if the gene for the regulatory
protein is deleted, the arabinose genes cannot be turned on.
Any three of these are acceptable.
1.
Bacterial cells contain a single RNA polymerase whereas eucaryotic cells have
three.
2.
Bacterial RNA polymerase can initiate transcription without the help of additional
proteins whereas eucaryotic RNA polymerases need general transcription factors.
3.
In eucaryotic cells, gene regulatory proteins can influence transcriptional
initiation thousands of nucleotides away from the promoter whereas bacterial
regulatory sequences are very close to the promoter.
4.
Eucaryotic transcription is affected by chromatin structure and nucleosomes
whereas bacterial transcription is not.
14)
15)
A.
Protein A binds to its DNA binding site more tightly than protein B binds to its
15
NAME: _______________________________
B.
C.
DNA binding site.
Protein A is a weaker activator of transcription than protein B.
Protein C is able to prevent activation by both protein A and protein B.
16)
The simplest way to evolve the new gene is by duplication and divergence. If the gene is
duplicated, then the cell or lineage can maintain one functional, intact old copy of the
original gene and can thus tolerate the disabling mutations in the other copy. The second
copy can first be modified by the X or Y mutation that impairs its function; second, at
some later time, the gene with the single mutation can acquire the additional mutation to
yield the doubly mutant X+Y gene with the new or improved function.
17)
A.
B.
Selectively neutral, detrimental, beneficial. Most nucleotide changes in the
genome, or mutations, will have little to no effect on the fitness of the individual
because many changes are not located in regions that encode a protein or regulate
expression of a gene. Even changes within a coding region may not change the
amino acid encoded or may cause a conservative amino acid change, for example
from one small nonpolar amino acid to another. Most changes that have a
functional consequence will interfere with the regulation of a gene or the behavior
of the encoded protein, usually rendering it useless and occasionally making it
harmful or yielding a new function. Only very rarely will a mutation improve the
performance of the gene or its encoded protein.
Beneficial, selectively neutral, detrimental. Individuals bearing beneficial
mutations will be more likely to have more offspring than others in the
population, and thus the beneficial mutations will become over-represented in the
population in subsequent generations. Individuals bearing detrimental mutations
will be likely to have fewer children and grandchildren, and thus these mutations
will be culled from the population, though perhaps not eliminated.
18)
Choice (b) is the correct answer. Alternative splicing can produce several different
mRNA transcripts from a single gene, and these transcripts can be translated into several
different but related proteins. Choices (c), (d), and (e) do not yield more protein species
than genes. Protein degradation (choice (a)) can produce several proteins from a single
gene, but this mechanism is used sparingly.
19)
You would first digest your sample with a combination of restriction enzymes selected so
that they give a set of fragment sizes that could have come from only one of the
plasmids. Then you would run the resulting mixture of DNA fragments on a gel
alongside a set of size markers and determine the sizes of each fragment. By
looking at the restriction maps, you should then be able to match your results to
16
NAME: _______________________________
one of the plasmids.
Digestion with any of the following combinations will enable you to distinguish which
plasmid you have: Hind III + Bgl II; Eco RI + Bgl II; Eco RI + Bgl II + Hind III. The
plasmids are the same size, so you cannot distinguish them simply by making a single cut
(with Hind III) and determining the size of the complete DNA by gel electrophoresis. Nor
can you distinguish them by cutting with all four restriction nucleases, since the set of
fragment sizes produced from both plasmids will be the same. Cutting with Bam HI or
Eco RI on their own is not sufficient because you will get bands of the same size from
both plasmid A and plasmid B. The only difference between the two plasmids is the
location of the Bgl II site relative to the two Bam HI sites, so if you cut with an enzyme
that cuts outside the Bam HI fragment and with Bgl II, you will get different-sized
fragments from the two plasmids.
20)
21)
Choice (d) is the correct answer. If some of the DNA templates you are sequencing are
cut at one specific site (as would be the case if the DNA were cut by a restriction
enzyme), the polymerase will stop when it comes to the end of the DNA, giving rise to at
least some product of one particular size in all the reaction mixtures. If this were the case,
all four lanes will have a band of this particular size. In addition, you would get normal
sequence from the full-length templates, and normal sequence from those templates in
which the polymerase incorporated a dideoxynucleotide before encountering the end. The
other options are incorrect: if you added all four dideoxynucleotides to one of the
reactions (choice (a)), that lane would have a band at every position because the
polymerase would stop at As, Cs, Gs, and Ts instead of at only one type of nucleotide.
If you forgot to add deoxynucleotides to the reactions (choice (b)), you would not get any
polymerization, and all of your lanes would be blank. If you forgot to add
dideoxynucleotides (choice (c)), the reactions would not stop until the end of the DNA
fragment, and all of the products would be full-length and would all be at the top of the
gel. If your primer hybridized to more than one part of the fragment of DNA you were
sequencing (choice (e)), your gel would look as though two different sequences had been
superimposed onto each other.
5-CGTGCAAAGTGCCTAAGGGCGT-3
{JOHN PLEASE CHECK WETHER CORRECT}
22)
A.
B.
All four oligonucleotide probes.
Oligonucleotide probe B and oligonucleotide probe D. Both the upper and lower
strands of DNA are present in genomic Southern blots, so all four oligos will
hybridize to either Southern blot. (Oligonucleotides A and B will still be able to
hybridize to genomic DNA cut with Bgl II, since they can still base-pair to the
individual fragments that result from the digest.) Northern blots contain only
RNA, which has the sequence of the upper strand of the DNA. Hence, only
oligonucleotide probe B and oligonucleotide probe D will hybridize to a Northern
blot.
17
NAME: _______________________________
23)
(e)
24)
The gene isolated from a genomic library would still contain introns, and bacteria do not
contain the biochemical machinery for removing introns by RNA splicing. The same
gene isolated from the cDNA library will have already had its introns removed.
25)
A.
B.
C.
D.
E.
26)
Genomic library. (cDNAs are produced from mRNAs; therefore, the promoters
will not be included in a cDNA library.)
Genomic library. (cDNAs are usually constructed using an oligo dT primer;
tRNAs do not have poly(A) tails. If the cDNA library were made using random
primers, it would be unlikely to contain the full-length version of the tRNA.)
cDNA library. (Since cDNAs are produced from mRNAs, isolating cDNAs
would tell you which splice variants are produced in a cell.)
Genomic library. (A genomic DNA fragment can contain the genes next to your
gene of interest; a cDNA will not.)
cDNA library. (Bacteria do not have the ability to remove introns, which may
exist in DNA isolated from a genomic library.)
The appropriate PCR primers are primer 1 (5 -GACCTGTGGAAGC) and primer 8
(5 -TCAATCCCGTATG). The first primer will hybridize to the bottom strand and
prime synthesis in the rightward direction. The second primer will hybridize to the top
strand and prime synthesis in the leftward direction. (Remember that strands pair
antiparallel.)
The middle two primers in each list (primers 2, 3, 6, and 7) would not hybridize to either
strand. The remaining pair of primers (4 and 5) would hybridize, but would prime
synthesis in the wrong directionthat is, outward, away from the central segment of
DNA. Each of these wrong choices has been made at one time or another in most
laboratories that use PCR. In most cases the confusion arises because the conventions for
writing nucleotide sequences have been ignored. By convention, nucleotide sequences
are written 5 to 3 , with the 5 end on the left. For double-stranded DNA, the 5 end of
the top strand is on the left.
27)
Even though the genome of F. cattoriae is smaller, the fact that the W. gravinius genome
contains less repetitive DNA makes it more attractive to sequence. Repetitive DNA
makes the assembly of sequenced fragments difficult. Shotgun sequencing would not be
an unrealistic approach for W. gravinius, because the genome of W. gravinius contains
little repetitive DNA and is relatively small. The genome of H. influenzae is 1.83 MB
and was successfully sequenced using the shotgun approach. (For comparison, the
18
NAME: _______________________________
genome of S. cerevisiae is 14 MB.)
28)
It is not likely that child would have the disease, because it is unlikely that the mutation is
carried in the germ line. Probably the mutation occurred in a cell that gave rise to
somatic cells and not germ cells. Only mutations in germ cells are passed onto progeny.
29)
(a)
(b)
(c)
(d)
(e)
False. Evolution can work only by tinkering with the tools and materials on hand,
not by starting from scratch to make completely new genes or pathways. New
functions arise from the ancestral functions by a process of gradual mutational
change, and thus may not represent the best possible solution to a problem.
False. Many duplications are subsequently lost or become pseudogenes, and only
a few evolve into new genes.
True. Pseudogenes look very similar to normal genes but cannot produce a fulllength protein due to one or more disabling mutations.
False. A large proportion of the genes in vertebrates (and many other species) are
members of multigene families.
False. By some estimates, 20% of the genomic DNA in some bacterial species
arose by horizontal gene transfer.
19
NAME: _______________________________
30) The sketch below shows an agarose gel with a plasmid that has been digested with different
restriction enzymes as indicated. Draw the restriction map of the plasmid. B = BamH1; E =
EcoR1; H = HindIII. The numbers on the left are the approximately molecular weights in
kilobasepairs. So for example, cutting the plasmid with BamH1 gives you one piece of
DNA that is 6000 basepairs long. (5 points)
HindIII
EcoR1
BamH1
6 kb Plasmid
HindIII
EcoR1
20