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EXAM II
CELLULAR AND MOLECULAR BIOLOGY
December 6, 2007
Name: ________________________________
1. (20 pts) You are interested in how a specific protein-encoding mouse gene named Hrt1 is regulated. Full
activity of the Hrt1 gene product is present in heart cells, but no activity of this gene product is present in
liver cells. You hypothesize that the Hrt1 gene product is regulated in one of the following ways:
-- 1) whether the mRNA is translated or not
-- 2) whether the protein product is stable or immediately completely degraded
-- 3) whether the gene is transcribed or not
-- 4) whether the protein product is phosphorylated or not
-- 5) whether the pre-mRNA is spliced to include all 5 exons or only 4 of those exons
You prepare four samples from cells: all of the mature mRNAs in heart cells, all of the mature mRNAs in
liver cells, all of the proteins in heart cells, all of the proteins in liver cells. You load each sample into a
separate lane in a gel, and separate the components by size via gel electrophoresis. You then detect
presence/absence of any band corresponding to Hrt1 RNA or protein.
(a, 4 pts) Towards which pole of the gel will the Hrt1 RNA run? Your choices are:
positive, negative, neither, both, or inconclusive. Explain your answer.
The positive pole. RNA is negatively charged so it will be attracted to the positive pole.
RNA is negative because it is made of nucleotides, and part of every nucleotide is a
phosphate group, and phosphate groups are very negative.
(b, 4 pts) Towards which pole of the gel will the Hrt1 protein run? Your choices are:
positive, negative, neither, both, or inconclusive. Explain your answer.
Inconclusive. Proteins can be of any total charge. Each protein’s charge depends upon the
amino acid composition of that protein. If the protein has more negatively charged amino
acids (glutamic acid and aspartic acid) than positively charged amino acids (histidine,
lysine, arginine), then the protein will be overall negative and will run towards the positive
pole. If the protein has more positively charged amino acids, it will run towards the
negative pole. If it has the same number of positive and negative amino acids, it will not
run either way because it will be neutral. Given that you have no information about the
makeup of Hrt1 protein, you do not know which way it will run.
You perform the type of experiment described above, and detect any Hrt1 mRNA that is present in heart
cells or in liver cells.
(c, 4 pts) In which of the five ways listed above do you think the Hrt1 gene product is regulated? Discuss the
likelihood of each possible way (i.e. LABEL #1 through #5) given this result:
RNA from heart cells:
a band is present
RNA from liver cells:
a band with the same intensity is present that
ran the same distance in the gel
1, 2, 4
#1 is possible because it may be that, although the same RNA is present in both heart cells
and liver cells, the way that Hrt1 is active only in heart cells is that the mRNA is translated
only in heart cells (and not in liver cells).
#2 is possible because all that you know is that mRNA is made in both cell populations.
Therefore the regulation of the gene must occur post-transcriptionally. Degradation of a
protein occurs post-translationally and is thus still an option for regulation.
#3 is NOT possible because there is mRNA produced from Hrt1 in both heart and liver
cells, so this Hrt1 gene is not regulated at the level of transcription.
#4 is possible because all that you know is that mRNA is made in both cell populations.
Therefore the regulation of the gene must occur post-transcriptionally. Phosphorylation
occurs post-translationally and is thus still an option for regulation.
#5 is NOT possible because, if the pre-mRNA from Hrt1 were regulated by splicing, the
mature mRNA would be a different length in heart cells than in liver cells, and this result
tells you that the mRNA is the same length in both cell populations.
Now say that, INSTEAD, you perform the type of experiment described above, but detect any Hrt1 protein
that is present in heart cells or in liver cells.
(d, 4 pts) In which of the five ways listed above do you think the Hrt1 gene product is regulated? Discuss the
likelihood of each possible way (i.e. LABEL #1 through #5) given this result:
Protein from heart cells:
a band is present
Protein from liver cells:
a band with the same intensity is present that
ran a different distance in the gel
4, 5
#1 is NOT possible because Hrt1 protein has clearly been translated in both cell
populations, because you can see protein in both lanes. Thus this gene is not regulated at
the level of translation.
#2 is NOT possible because, if the gene product were regulated by immediate and complete
protein degradation, then one of these two lanes would have no protein in it.
#3 is NOT possible because there is Hrt1 protein in both lanes, so Hrt1 mRNA must have
been produced from the Hrt1 gene in both heart and liver cells, so this Hrt1 gene is not
regulated at the level of transcription.
#4 is possible because, if the gene product was regulated by phosphorylation, then the
protein would be present in both cell types, but it would be a slightly different size and
charge in one cell type.
#5 is possible because, if the pre-mRNA from Hrt1 were regulated by splicing, then protein
would be present in both lanes, but the protein would be of a different size in one of the two
lanes.
(e, 4 pts) In which of the five ways listed above do you think the Hrt1 gene product is regulated? Discuss the
likelihood of each possible way (i.e. LABEL #1 through #5) given this result:
Protein from heart cells:
a band is present
Protein from liver cells:
no band is present
1, 2, 3
#1 is possible because it may be that, although the same RNA is present in both heart cells
and liver cells, the way that Hrt1 is active only in heart cells is that the mRNA is translated
only in heart cells (and not in liver cells).
#2 is possible because seeing no protein in liver cells could mean that the protein is never
made in liver cells, or it could mean that the protein is made but then immediately
completely degraded.
#3 is possible because it could be that there is no protein in liver cells because the Hrt1
gene
is simply never transcribed into mRNA in liver cells.
#4 is NOT possible because, if the gene product was regulated by phosphorylation, then the
protein would be present in both cell types, but it would be a slightly different size and
charge in one cell type.
#5 is NOT possible because, if the pre-mRNA from Hrt1 were regulated by splicing, then
protein would be present in both lanes, but the protein would be of a different size.
2. (6 pts.) The following is a map of several contiguous genes in a bacterium published recently. The genes
are simply named ORF with a number; ORF means "open reading frame", a name for a region that appears
to be a gene, but has not yet been specifically identified. Some of the ORFs were later shown to be part of
the pathway for synthesizing the required amino acid nutrient, arginine; these genes are also labeled with the
name arg. The bent arrows at the top identify transcriptional start sites.
The numbers below the top map (173, etc) are the number of nucleotides between genes.
a. (2 pts.) The original study showed that a regulatory protein binds a site overlapping the argF promoter.
Part of their evidence was a footprint experiment, which shows protection of a 29 base region, which
extends from the middle of the -10 region to somewhat upstream of the -35 region. Explain what the
footprint experiment is and sketch the results of the footprint experiment that would lead to this
conclusion. Be sure to include a control. Label key features of your sketch well enough to make the main
point. Relate the sketch to the map shown above. (That is, indicate what region of the top map is
involved in this part.)
The control lane should show a series of bands at all possible lengths. The experimental lane
should be similar, but with 29 bands missing, over the indicated region. The region involved here
is just in front of argF; given the spacing shown on the map, it would just about fit in the 43 base
space between ORF8 and argF.
b. (1 pt.; no credit without explanation) Judging from the information given in part a, is it most likely that
this protein is a positive control protein or a negative control protein? Explain; that is, how can you tell
from the given information? Be sure your explanation is clear enough to distinguish what the two terms
mean.
The region covered by the regulatory protein closely corresponds to where RNA Pol binds. It is
hard to imagine how a protein covering that whole region could help Pol bind (activator protein,
positive control). It is easier to imagine that it interferes with Pol binding (repressor protein;
negative control).
c. (2 pts.) The Figure below shows part of an RNA from a region near the beginning of ORF5. What is the
significance of this structure? Which transcript is it from? Point to two specific features that help you
identify its function.
It looks like a rho-independent transcriptional terminator, to mark the end of a transcript. The key
features are the GC-rich hairpin, followed immediately by a run of U. It is from the transcript for
ORF4; it is the end of a transcript "near the beginning of ORF5".
d. (1 pt.) Adding arginine to the bacteria increases transcription of ORF4. What does this suggest to you
about the function of the ORF4 protein? In particular, is ORF4 likely to be involved in making arginine?
Explain.
It is not likely to be involved in making arginine. Logically, one expects Arg to repress synthesis of
enzymes to make Arg, but the data says that Arg induces ORF4. This would be expected if the ORF4
gene product helps to utilize exogenous Arg, or helps to degrade excess Arg.
The function of this product is not known. From its sequence, the authors suggest that it is a
membrane protein, and speculate that it is involved in Arg transport.
Based on: R Sanchez et al, Organization and expression of a Thermus thermophilus arginine cluster:
presence of unidentified open reading frames and absence of a Shine-Dalgarno sequence. J Bacteriol
182:5911-5915, 10/00.
3. (6 pts.; 2 pts. per part) The following is an abstract of a recent paper.
The Escherichia coli proP P2 promoter, which directs the expression of an integral membrane transporter of
proline, glycine betaine, and other osmoprotecting compounds, is induced upon entry into stationary phase to
protect cells from osmotic shock. Transcription from the P2 promoter is completely dependent on RpoS (σ38)
and Fis. Fis activates transcription by binding to a site centered at -41, which overlaps the promoter, where it
makes a specific contact with the C-terminal domain of the α subunit of RNA polymerase (α-CTD). We
show here that Fis and cyclic AMP (cAMP) receptor protein (CRP)-cAMP collaborate to activate
transcription synergistically in vitro. Coactivation both in vivo and in vitro is dependent on CRP binding to a
site centered at -121.5, but CRP without Fis provides little activation. The contribution by CRP requires the
correct helical phasing of the CRP site and a functional activation region 1 on CRP. We provide evidence
that coactivation is achieved by Fis and CRP independently contacting each of the two α-CTDs. Efficient
transcription in vitro requires that both activators must be preincubated with the DNA prior to addition of
RNA polymerase. (end of abstract)
Answer the following parts based on the information in this abstract. The parts are substantially independent.
(The particular σ (sigma) factor discussed is of no concern for the questions below.)
a. Sketch the genome over the region discussed here. Your sketch should show the transcriptional start site,
the gene whose transcription is of interest, and all the features discussed in or clearly implied by the abstract.
You should show the gene (proP), the start site (+1), the promoter (presumed to be at -10 and -35), the
Fis binding site (at -41), and the CRP binding site (at -121).
b. Using the features of this system, give an example of a genetic test that would show cis dominance. That
is, describe a diploid strain for this region and describe the result that would show cis dominance. Be sure to
explain what is meant by cis dominance. (You can add a reporter gene, as needed.)
The key idea is that cis dominance refers to a mutation having an effect only on genes on its own
chromosome. This typically occurs for mutations in DNA "sites" that bind proteins (rather than for
genes that code for proteins). In this case, you can use the binding site for Fis or for CRP (or even for
σ). It is shown by using two marker genes, so you can tell which chromosome is functioning.
The system as provided only has one gene, but you can add your own reporter genes. In fact, the given
gene is not easy to measure (a transport protein), so I will choose to use two marker genes, the
common z and a genes from the lac operon. In the following diploid, all regulatory elements are from
the proP gene, but the proteins are from the lac operon.
There are many possible examples; here is one. + means that the gene or site is wild type (functional); means that it is not.
In this case, we have a defective Fis-binding site (F) on the lower chromosome. As a result, the a
enzyme cannot be made even when the Fis protein is present; the only a+ allele is behind a defective
Fis-binding site. However, at the same time, the z enzyme can be made, since the chromosome
containing the z+ allele has a Fis-binding site.
If you tried to present this without a drawing of the chromosomes, that makes it difficult.
c. Sketch the "pre-initiation complex" -- the DNA with appropriate proteins bound, ready to initiate
transcription. Try to incorporate as many features described in the abstract as possible. (The core RNA Pol
subunit composition is α2 β β'.)
The sketch below is for your general guidance; it shows the general (simplified) organization of the lac
operon:
The CAP (= CRP) and i proteins are the regulatory proteins, which act at the sites C or o, respectively. p is
the promoter; the remaining letters denote genes for the lac enzymes.
The complex should include RNA Pol with σ bound to the promoter, Fis bound to its site and to a Pol α
subunit, and CRP bound to its site and to the other Pol α subunit. The DNA should be bent, to allow
CRP to bind to its site and to Pol.
I didn't require that you show the DNA bent, but it is hard to imagine the complex without bending
the DNA. The distance from p to the CRP site is a long distance for a protein to cover.
4. (15 pts) You are working with a piece of DNA of the sequence:
5’-TATTGAGCTCCCGGAT-3’
3’-ATAACTCGAGGGCCTA-5’
(a, 3 pts) You cut the above piece of DNA with a restriction enzyme that recognizes the sequence 5’GAGCTC-3’ and cuts on the 3’ side of the A within this sequence. Draw all products that result from this
digestion. Make sure to draw the nucleotide sequence of both strands, and label all 5’ and 3’ ends.
5’-TATTGA-3’ 5’-GCTCCCGGAT-3’
3’-ATAACTCG-5’ 3’-AGGGCCTA-5’
(b, 3 pts) Now you ligate the DNA you produced in part (a) to the sequence below, which you have also cut
with the same restriction enzyme. Draw the shortest DNA product that could form from ligating a piece of
DNA from part (a) to a piece of DNA from part (b). Make sure to draw the nucleotide sequence of both
strands, and label all 5’ and 3’ ends.
5’-TAGAGCTCCGCAATG-3’
3’-ATCTCGAGGCGTTAC-5’
This sequence becomes the following when cut:
5’-TAGA-3’ 5’-GCTCCGCAATG-3’
3’-ATCTCG-5’ 3’-AGGCGTTAC-5’
The shortest product of ligation would be if you ligated the piece on the left from part (b) to
the piece on the left from part (a). To do this, you have to flip the piece from part (a) on the
left around so that its sticky ends line up with those of the piece from part (b).
5’-TAGA-3’ 5’-GCTCAATA-3’
3’-ATCTCG-5’ 3’-AGTTAT-5’
giving the final answer
5’-TAGAGCTCAATA-3’
3’-ATCTCGAGTTAT-5’
(c, 3pts) Restriction enzymes have DNA binding domains. Name three other different types of proteins (that
are not restriction enzymes) that have DNA binding domains.
Acceptable answers are any proteins whose job it is to act upon DNA. These include:
activators and/or repressors (which are both transcription factors), DNA polymerase, RNA
polymerase, helicase, primase, ligase, topoisomerase, telomerase, histones (part of the
nucleosome), nuclease, DNA repair enzymes, and methylase.
Which reaction is catalyzed by each of the enzymes listed below? Answer by stating which specific type of
bond is affected (e.g. “covalent” is not specific enough), and whether each enzyme catalyzes the formation or
the breaking of that type of bond.
(d, 3 pts) the T. aquaticus “Taq” polymerase used in PCR
Formation of phosphodiester bonds. (PCR is just many rounds of DNA replication.)
(e, 3 pts) a restriction enzyme
Breaking of phosphodiester bonds.
Multiple Choice. Choose the one best answer. 2 pts each
5. To study the function of the essential cytosolic Hsc70 genes in yeast, researchers constructed a shuttle
vector in which a copy of the Hsc70 gene was ligated to the GAL1 promoter. The vector was then introduced
into haploid yeast cells in which all four copies of the Hsc70 genes had been disrupted. Following
introduction of the vector, you would expect that
a. the yeast cells would grow on both glucose and galactose media.
b. the yeast cells would grow on glucose but not galactose medium.
c. the yeast cells would grow on galactose but not glucose medium.
d. on transfer to either glucose or galactose medium, the vector-carrying cells would eventually stop
growing because of insufficient Hsc70 activity.
6. In RNA interference studies, the double-stranded RNA
a. disrupts the target DNA sequence.
b. results in the destruction of the target mRNA.
c. destroys the target protein.
d. all of the above
7. A step in the generation of knockout mutations in mice includes selection of embryonic stem (ES) cells
that are
a. resistant to G418 and resistant to ganciclovir.
b. sensitive to G418 and resistant to ganciclovir.
c. resistant to G418 and sensitive to ganciclovir.
d. sensitive to G418 and sensitive to ganciclovir.
8. A mutation that changes the recognition sequence for the restriction enzyme EcoRI from GAATTC to
GATTTC is an example of a
a. restriction fragment length polymorphism (RFLP).
b. single nucleotide polymorphism (SNP).
c. simple sequence repeat (SSR).
d. a and b
e. all of the above
9. Southern blotting is used to detect a specific
a. DNA.
b. RNA.
c. protein.
d. carbohydrate.
e. all of the above
10. The chicken lysozyme gene is considered to be a solitary gene because
a. it contains no introns.
b. it is not present on a chromosome.
c. it is represented only once in the haploid genome.
d. none of the above
11. All the following statements about complex transcription units are true except
a. They can have multiple poly(A) sites.
b. They can generate multiple mRNAs.
c. They can generate multiple polypeptides.
d. They are common in bacteria.
12. In eukaryotes, tandemly repeated genes encode
a. rRNAs.
b. cytoskeletal proteins.
c. β-globin.
d. all of the above
13. Which of the following classes of repetitious DNA is most abundant in the human genome?
a. simple-sequence DNA
b. non-LTR transposons
c. LTR transposons
d. DNA transposons
14. SINES (short interspersed elements)
a. are approximately 300 base pairs long.
b. are LTR containing retrotransposons.
c. are present in over 1 million copies in the human genome.
d. a and c
e. all are correct
15. Operator constitutive mutants of the lac operon would
a. express the lac repressor constitutively.
b. block the binding of RNA polymerase to the promoter.
c. express β-galactosidase constitutively.
d. prevent the inducer from binding to the repressor.
16. Most eukaryotic genes are controlled at the level of
a. transcription initiation.
b. transcription elongation.
c. transcription termination.
d. translation initiation.
17. All of the following statements about the essential carboxy terminal domain (CTD) of RNA polymerase
are true except
a. The CTD is present in RNA polymerase I, II, and III.
b. The CTD can become phosphorylated.
c. The CTD is critical for viability.
d. The CTD of mammals contains more than 50 repeats of a heptapeptide.
18. An enhancer
a. is a DNA element that stimulates transcription of eukaryotic promoters.
b. binds to RNA polymerase and stimulates transcription.
c. acts as a binding site for RNA polymerase.
d. interacts with repressor proteins to enhance transcriptional repression.
19. The TATA box
a. serves as a promoter sequence for genes transcribed by RNA polymerase III.
b. is located approximately 100 base pairs upstream of the start site for mRNAs.
c. is present in all eukaryotic genes.
d. acts to position RNA polymerase II for transcription initiation.
20. Open reading frame (ORF) analysis is not effective in identifying genes in higher eukaryotes because of
the presence of
a. promoters.
b. enhancers.
c. introns.
d. repetitious DNA.
21. All the following statements are true about a nucleosome except
a. It contains an octamer core of histones
b It is about 10 nm in diameter
c. It is the “string” of the “beads-on-a-string” appearance
d. It contains approximately 150 base pairs of DNA
22. Which of the following statement(s) is (are) true of a eukaryotic chromosome?
a. It is a linear structure
b. It consists of a single DNA molecule
c. It can contain greater than a billion base pairs of DNA
d. all of the above
23. The karyotype for any particular species is characterized by
a. the number of metaphase chromosomes.
b. the size and shape of the metaphase chromosomes.
c. the banding pattern of the metaphase chromosomes.
d. all of the above
24. Chromosome painting involves
a. staining chromosomes with Giemsa reagent.
b. hybridizing fluorescent probes to chromosomes.
c. hybridizing radioactive probes to chromosomes.
d. a and b
26. All the following statements about heterochromatin except
a. It is a dark-staining area of a chromosome.
b. It is usually transcriptionally active.
c. It is often simple sequence DNA.
d. It is a region of condensed chromatin.
27. Telomeres
a. consist of repetitive sequences with high G content.
b. are a few hundred base-pairs long in vertebrates.
c. have specific proteins bound at the DNA ends.
d. a and c
28. A leucine zipper motif contains
a. a stretch of five leucine residues in a row.
b. a leucine residue at every seventh position.
c. a leucine residue complexed with a zinc ion.
d. an alternating leucine-alanine-proline structure.
29. Which of the following is not a structural motif found in a DNA-binding domain?
a. homeodomain
b. zinc-finger
c. helix-loop-helix
d. random-coil acidic domain
30. (5 pts.) Write down the two sequences that appear in the gels below, from 5’ to 3’.
After you have written down the two sequences on separate lines,write the amino acid sequence of the
protein represented by the normal and mutant sequences.
Normal DNA
GAG CAG GTC GAA GGG GAT GGC GGC CAC CAT GTC GAT GAG GAA CCA GCC
Deletion DNA
T GAG CAG GTC GAG GAA CCA GCC
Normal protein
Leu---Val---Gln---Leu---Pro—Leu---Pro—Pro—Val---Val---Gln---Leu---Leu—Leu---Gly--Arg
Deletion protein
Thr---Arg---Pro---Ala---Pro---Thr---Ser or Leu---Val---Gln---Leu—Leu---Gly--Arg