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1D 2A 3B 4D 5E 6C 7B 8C 9B 10A

11A 12D 13D 14E 15C 16E 17A 18C 19C 20D

21B 22B 23D 24B 25A 26B 27E 28D 29B 30A

31D 32E 33A 34C 35A 36C 37E 38C 39D 40B

41A 42E 43D 44C 45B 46B 47E 48B 49E 50E

51C 52A 53D 54B 55C 56C 57B 58E 59B 60A

61E 62A 63B 64C 65C

1. Which of the following statements about the human genome is TRUE?

a. The human genome is one continuous stretch of DNA, 6m in length in each cell. b. The human genome was sequenced in 1980, and found to have 9-10 billion

nucleotides. c. The human genome has 22 pairs of chromosomes, where the 22nd pair is the sex

chromosomes. d. J. Craig Venter is one of the few people to have had the approximately 6 billion base

pairs of DNA of his personal genome sequenced. e. One half of each pair of chromosomes is paternally inherited, and in females, only

the paternal copy of all chromosomes is functional.

2. Which of the following statements about genome sequencing is FALSE?

a. The first plant genome to be sequenced was that of Mendel’s garden pea, Pisum sativa.

b. The first eukaryotic genome to be sequenced was that of yeast, Saccharomyces

cerevisiae. c. The first mammalian genome to be sequenced was that of the mouse, Mus

musculus. d. The first great ape genome to be sequenced was that of the human, Homo sapiens

sapiens. e. The first animal genome to be sequenced was that of the nematode worm,

Caenorhabditis elegans.

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3. A scientist has been working to characterise a novel organism from outer space. Recently, the organism’s chromatin was examined. Which of the following features of its chromatin would be novel relative to that of most eukaryotes on earth? That is, which one is UNEXPECTED?

a. 147 nucleotide pairs of DNA wrap around a nucleosome core. b. The nucleosome core is an octamer comprising pairs of histones H1, H2, H3A, and

H3B. c. The carboxy terminal tails of some of the histones can be post-translationally

modified by the addition of ubiquitin. d. There is a linker histone that functions between nucleosome cores. e. Over the 1.7 turns of DNA that wrap around the histone core, there are 142 hydrogen

bonds between the DNA and histone core proteins. 4. Which of the following statements ACCURATELY describes the cellular packaging of the

bacterial genome?

a. The bacterial genome is condensed approximately 1000 fold in order to fit in the cell.

b. The bacterial genomic DNA is combined with proteins to form the nucleoid.

c. The bacterial genome is supercoiled by enzymes called topoisomerases.

d. a, b, and c are correct.

e. b and c only

(i.e., not a) are correct.

5. A microarray experiment was conducted to examine the differences in the transcriptomes in the muscle cells of 4 breeds of dog. The transcriptome of the muscle cells of wolves was used as the reference for the experiment. A clusterogram was generated based on the transcriptome microarray data. What might one EXPECT to see in the clusterogram?

a. Genes with increased transcript abundance in the dog breeds relative to wolves would be denoted by a red colour.

b. Genes with decreased transcript abundance in the dog breeds relative to wolves

would be denoted by a green colour. c. Genes with the most similar patterns of transcript abundance across the dog breed

would be in rows that are closest to each other in the clusterogram. d. One should be able to identify genes that distinguish the transcriptome of one dog

breed from another. e. All of the above.

6. Which of the following statements about transcriptome microarrays is FALSE?

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a. Transcriptome microarrays can be thought of as gigantic “reverse northern blots”. b. For transcriptome microarrays, DNA “probe” in excess is affixed to a solid support,

like a glass slide. c. Transcriptome microarrays reveal that the transcriptome is relatively static and

unchanging in comparisons between different cells or tissues. d. Transcriptome microarrays can report on transcript abundance for a large number of

genes. e. The amount of fluorescent label that is reported by each DNA spot on the

transcriptome microarray provides an indication of transcript abundance.

7. While examining the dynamic nature of chromatin in a novel organism from outer space, a scientist makes a startling discovery. Which of the following is most likely to be the startling discovery? That is, which is UNEXPECTED?

a. Unwrapped nucleosomes exist for 10-50 milliseconds, enabling sequence-specific gene regulatory proteins to bind to the DNA.

b. In GTP-dependent reactions involving the hydrolysis of GTP to GDP, chromatin

remodelling complexes catalyse nucleosome sliding. c. Exchange of histone dimers occurs in the nucleosome core to remodel the chromatin

structure. d. Energy-dependent binding of histones by histone chaperones results in removal of

nucleosomes from specific regions of the chromatin. e. Decondensed regions of chromatin have higher levels of gene expression.

8. You have been asked to conduct analyses of histones from active and inactive regions of chromatin, where genes are expressed and repressed respectively. While conducting your analyses, which of the following observations would be UNEXPECTED?

a. The amino terminal tails of some of the histones have covalent, post-translational modifications, involving addition of particular chemical groups.

b. Serine residues in the tails of some histones have been phosphorylated, and this is

associated with changes in chromatin activity. c. Lysine residues in the histone tails have been serinated, resulting in inactive

chromatin, where genes are not expressed. d. Acetylation of lysine residues in some histones is associated with active regions of

chromatin, where genes are expressed. e. Methylation of lysine residues in some histones is associated with establishment of

heterochromatin, and results in gene silencing.

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9. Which of the following statements is TRUE?

a. DNA regulatory proteins always “read” the minor groove because it provides the maximum amount of information pertaining to functional groups in the DNA.

b. The minor groove provides only two possible configurations of hydrogen bond donor,

hydrogen bond acceptor, hydrogen atom and methyl group for a gene regulatory protein to “read”.

c. Hydrophobic amino acid residues, like glutamate, have R-groups that allow gene

regulatory proteins to bind to DNA. d. Gene regulatory proteins bind to DNA through their transcriptional activation domain,

allowing them to bind to stretches of DNA >100bp long. e. DNA sequence motifs bound by gene regulatory proteins tend to contain only one

type of nucleotide, generally polyA. 10. A group of molecular biologists discover that a gene that should normally be expressed is no

longer expressed, and that the chromosomal region where it resides is now heterochromatin, but should be euchromatin. The scientists have put forward some hypotheses to test, to provide a molecular explanation for their observation. Which of the following is NOT a reasonable hypothesis to test?

a. There is a mutation in a gene activator protein, making it more active and recruiting histone acetylase, resulting in heterochromatin formation.

b. There is a mutation in a barrier sequence making it dysfunctional, and allowing

heterochromatin to spread into the normally active gene. c. There is a mutation in a barrier protein so that it no longer binds to the barrier

sequence, resulting in spread of heterochromatin into the normally active gene. d. There is a mutation in a gene repressor protein so that it binds to gene regulatory

sequences in the normally active gene, resulting in heterochromatin formation at that gene.

e. There is a mutation in a barrier protein so that it no longer interacts with the nuclear

pore, resulting in spread of heterochromatin into the normally active gene.

11. Using microarray analysis, scientists have discovered a gene that is only expressed in muscle cells. When the gene regulatory regions of this gene are fused to a reporter gene, reporter gene activity is detected in muscle. Promoter deletion analysis identified a segment of the gene regulatory region that, when deleted, resulted in a loss of expression of the reporter gene in muscle. Fusion of this gene regulatory region to a minimal promoter and a reporter gene resulted in no expression of the reporter gene in any tissue. What do these findings indicate about the segment of the gene regulatory region?

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a. The segment of the gene regulatory region is necessary but insufficient for expression of the gene in muscle cells.

b. The segment of the gene regulatory region is unnecessary and insufficient for

expression of the gene in muscle cells. c. The segment of the gene regulatory region is unnecessary and sufficient for

expression of the gene in muscle cells. d. The segment of the gene regulatory region is necessary and sufficient for expression

of the gene in muscle cells. e. The findings indicate nothing about the segment of the gene regulatory region.

12. Working with a newly discovered, novel extraterrestrial life form, scientists believe they have found an excellent new reporter gene. Which of the following statements would be INCONSISTENT with the gene being an excellent reporter gene?

a. The potential reporter gene encodes a purple pigmented protein that can be viewed with the naked eye once it is active.

b. The potential reporter gene has codon usage that is similar to that of model

eukaryotic organisms, including Drosophila. c. Fusion proteins between the purple protein gene product and other proteins of

interest are functional, and localise to the subcellular location of the protein of interest.

d. In order to detect the purple protein encoded by the potential reporter gene, one

conducts an enzymatic reaction using the rare extraterrestrial substrate purpleum. e. The potential reporter gene has no deleterious effect on organisms in which it is

expressed.

13. When conducting an electrophoretic mobility shift assay using nuclear extracts from muscle cells, you find that you have three shifted bands. Why might this be the case?

a. Three different DNA-binding proteins have bound to the DNA probe. b. One DNA-binding protein binds to the DNA probe, but it can bind as a monomer, a

dimer, or a trimeric complex. c. One DNA-binding protein binds to the DNA probe, but it exists as three different

post-translationally modified forms in the nuclear extract. d. a, b, and c could explain the observation. e. a and b only

(i.e., not c) could explain the observation.

14. You have used affinity chromatography to identify proteins that bind to a DNA gene regulatory motif that you have characterised. You conduct a two-step purification of proteins

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that bind to the DNA motif, as discussed in lecture. At the end of the protocol you discover that the protein(s) you have purified bind to any DNA sequence and not to your DNA motif specifically. Why might this be the case?

a. You accidentally used a low salt wash only for both stages of the purification process.

b. You accidentally used the same column from the first step of the protocol in the

second stage of the protocol. c. A random pool of DNA sequences was used in both stages of the protocol. d. You conducted the protocol properly, but there are no proteins that bind to your

motif. e. a, b, and c only

(i.e., all but d) could explain the outcome of the experiment.

15. At the conclusion of a binding site selection assay, when all of the DNA sequences were compared, no consensus sequence was identified. Why might this be the case?

a. The protein used in the assay does not bind to DNA. b. The specific sequence to which the protein binds was not present in the original

mixture of oligonucleotides. c. The protein used in the assay has no DNA binding specificity, and binds to any

sequence. d. a and b only

(i.e., not c) could explain the observation.

e. b and c only

(i.e., not a) could explain the observation.

16. Working on a novel, extraterrestrial organism, a group of scientists make what they believe to be some remarkable observations pertaining to gene regulatory proteins in this organism. Which one of the following findings is the SAME as what we know about gene regulatory proteins in terrestrial organisms?

a. Helix-turn-helix motifs in DNA-binding proteins are uncommon. b. Recognition helices from two different helix-turn-helix motifs tend to bind DNA in the

minor groove, some 5.4nm apart. c. Homeodomain proteins contain a homeodomain that is 16 amino acids long. d. Zinc-finger motifs function in isolation, using serine residues to coordinate zinc

atoms. e. Leucine zippers can dimerise to give rise to different DNA-binding specificities.

17. At the conclusion of a chromatin immunoprecipitation (ChIP) experiment making use of antibodies raised against a transcriptional repressor protein, the DNA that is obtained is

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subjected to high throughput sequencing and found to contain the sequence CCACCTACC. Which of the following hypotheses is JUSTIFIED on the basis of this finding?

a. The sequence CCACCTACC is likely to be bound by the transcriptional repressor in vivo.

b. The sequence CCACCTACC is necessary and sufficient to support transcription of

genes with which it is associated. c. The sequence CCACCTACC is necessary and sufficient to repress transcription of

genes with which it is associated. d. The sequence CCACCTACC is sufficient but unnecessary to support transcription of

genes with which it is associated. e. The sequence CCACCTACC is sufficient but unnecessary to repress transcription of

genes with which it is associated. 18. Which of the following statements about the lac operon is TRUE?

a. lacZ encodes a lactose permease, which enables lactose to enter the cell. b. lacZ, but not lacY, transcription is high when glucose is absent and lactose is present

in the growth medium. c. CAP binding to the CAP-binding site is an example of positive regulation by an

inducer, cAMP. d. The production of cAMP is highest when glucose is present in the growth medium. e. lac repressor binding to the operator site, is an example of positive regulation by a

co-repressor, allolactose.

19. Which of the following statements about the nuclear import as it relates to eukaryotic gene regulation is FALSE?

a. Gene regulatory proteins are frequently larger than the size exclusion for entry into the nucleus, and must have nuclear localisation signals to enter the nucleus.

b. Gene regulatory proteins can be retained in the cytoplasm by inhibitor proteins,

which then release the gene regulatory proteins in response to specific signalling molecules.

c. The nuclear localisation signals on eukaryotic gene regulatory proteins are rich in

leucine and arsiline residues. d. The hydrolysis of GTP to GDP is part of the cycle related to the transit of proteins to

and from the nucleus. e. A small protein, Ran, is a component of the cycle related to the transit of proteins to

and from the nucleus.

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20. Which of the following statements about eukaryotic gene repressor proteins is FALSE?

a. Eukaryotic gene repressor proteins can mediate their effect through competitive DNA binding with gene activator proteins.

b. Eukaryotic gene repressor proteins can mediate their effect by masking the

activation surface of a gene activator protein. c. Eukaryotic gene repressor proteins can mediate their effect through direct interaction

with general transcription factors. d. Eukaryotic gene repressor proteins can mediate their effect through recruitment of

histone acetylase. e. Eukaryotic gene repressor proteins can mediate their effect through recruitment of

histone methyltransferase. 21. A mutant rat has a mutation in the glucocorticoid receptor that prevents the inhibitor proteins

from binding to the receptor. Two different ChIP assays were run using antibodies raised against the glucocorticoid receptor: one where cells from the mutant rat were first exposed to glucocorticoid and one where cells were not exposed to glucocorticoid. The DNA samples obtained at the end of the two ChIP assays were sequenced and compared. What findings are EXPECTED?

a. No DNA was obtained from either assay as the mutant protein cannot enter the nucleus.

b. The DNA sequences from both assays were the same, and enriched in

glucocorticoid response elements. c. The DNA sequences from both assays were the same, and were lacking

glucocorticoid response elements. d. The DNA sequences obtained from the ChIP assays from cells exposed to

glucocorticoid lacked glucocorticoid response elements entirely and DNA sequences from the cells that were not exposed to glucocorticoid were random.

e. The DNA sequences obtained from the ChIP assays from cells exposed to

glucocorticoid were depleted in glucocorticoid response elements relative to the DNA sequences from the cells that were not exposed to glucocorticoid.

22. Critical examination of a novel, extraterrestrial form of life suggests that it has many features of eukaryotic transcriptional regulation. Which of the follow features would make it DIFFERENT from most eukaryotes?

a. The RNA polymerase II of the extraterrestrial organism cannot initiate transcription on its own, and instead requires ancillary factors.

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b. The extraterrestrial gene regulatory proteins act nearby the minimal promoter, with most gene regulatory sequences residing within 100bp of the minimal promoter.

c. Individual genes in the genome of the extraterrestrial organism are regulated by

many different, sometimes hundreds, of different gene regulatory proteins. d. The extraterrestrial gene regulatory proteins tend to be modular in structure, with a

DNA-binding domain and a transcriptional-activation domain. e. The extraterrestrial gene activator proteins tend to exhibit transcriptional synergy,

where binding of two proteins at the gene regulatory region has greater than an additive effect.

23. The RUDOLF locus encodes a transcription factor that confers reindeer with a glowing red nose. Loss of RUDOLF function in rudolf mutant reindeer results in reindeer with no discernable nose, red glowing or not. A dubious elf genetically engineered some reindeer so that the RUDOLF gene was expressed at the tip of their tails. This resulted in not only a glowing red tip of the tail, but a tip of the tail that had all of the features of a reindeer nose. These findings illustrate which of the following?

a. The RUDOLF locus has the features of a master control gene.

b. RUDOLF function is necessary for nose development.

c. RUDOLF function is sufficient for nose development.

d. a, b, and c are illustrated by the findings.

e. None of the above is illustrated by the findings.

24. Gene A and Gene B reside next to each other on a mouse chromosome. Normally, Gene A is transcribed completely independently of Gene B. That is, Gene A is transcribed under one set of conditions, and Gene B under a different set of conditions. Scientists discover a mutant mouse where Gene A is now transcribed under the same set of conditions that Gene B is normally transcribed. Which of the following hypotheses is MOST LIKELY to explain the observation in the mutant mouse?

a. Gene A and Gene B are now part of the Gene B operon, transcribed as a polycistronic message with Gene C.

b. An insulator element found between Gene A and Gene B has been mutated so that

a gene regulatory enhancer element that normally acts to regulate Gene B now also regulates Gene A.

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c. A rare chromosomal inversion has occurred so that DNA where Gene A and Gene B are located is now heterochromatin all of the time.

d. A barrier sequence adjacent to Gene A has been mutated so that heterochromatin

spreads to both Gene A and Gene B. e. An insulator element found between Gene A and Gene B has been mutated so that

a gene regulatory enhancer element that normally acts to regulate Gene A now also regulates Gene B.

25. Which of the following is NOT an example of epigenetic inheritance?

a. Gene A is mutated so that a CG dimer is changed to CC, and this mutation is transmitted through meiosis.

b. Gene A is in a heterochromatin state, and is in a heterochromatin state in both

mitotic daughter cells following mitosis. c. Protein A is able enhance the transcription of the gene that encodes it, Gene A.

Mitotic daughter cells that contain Protein A continue to express Gene A. d. Gene A is methylated, and this methylated state is maintained in both mitotic

daughter cells following mitosis. e. Gene A is located on the human X chromosome, and is silenced through X-

chromosome inactivation.

26. A scientist thinks that she has made a startling, unexpected discovery with regards to X-chromosome inactivation in a human cell line she is examining. Which of the following observations related to human X-chromosome inactivation would be UNEXPECTED relative to what is currently known?

a. The X-inactivation centre seeds and facilitates the spread of heterochromatin.

b. The XIST RNA is translated into a protein involved in heterochromatin formation.

c. X-chromosome heterochromatin formation involves histone hypoacetylation.

d. X-chromosome heterochromatin formation involves DNA methylation.

e. DNA methylation of the X chromosome occurs at CG dimers, and involves DNA methyltransferase.

27. Apple trees are frequently propagated by making “cuttings”, which involves taking segments of a branch to generate a large number of genetically identical individuals. Trees produced from cuttings of the same tree are effectively clones of each other, like monozygotic twins. If

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an apple grower were to examine such clones in an orchard at the molecular level over time, what observation would be UNEXPECTED?

a. The differences in DNA 5-methyl cytosine content between any two clones of equivalent age will increase.

b. The differences in histone acetylation between any two clones of equivalent age will

increase. c. The differences in DNA hypermethylation between any two clones of equivalent age

will increase. d. The differences in DNA hypomethylation between any two clones of equivalent age

will increase. e. The differences in gene expression between any two clones of equivalent age will

decrease.

28. Which of the following statements about the interplay between intron splicing and the regulation of gene expression is FALSE?

a. A single transcript can produce two different protein products by splicing out an optional exon.

b. A single transcript can produce two different protein products by failing to splice out

an optional intron. c. The Drosophila DSCAM transcript can be spliced to potentially produce more than

35000 different proteins. d. Positive regulation of intron splicing involves binding of a repressor to enhance intron

skipping. e. Blocking of a splice site in the transformer (tra) transcript is one step in the

repression of male differentiation in Drosophila.

29. Characterisation of a novel extraterrestrial organism has revealed that it has many features of eukaryotic transcriptional and post-transcriptional regulation. Which of the following would be an UNEXPECTED observation with regards to such aspects of gene expression?

a. Degradation of mRNA can occur in a deadenylation-dependent manner, involving shortening of the poly-A tail in the 3’ to 5’ direction.

b. The context of the AUG is unimportant with respect to the initiation of translation. c. Some transcripts are localised to regions of the cell by active transport via the

cytoskeleton. d. RNA fragments (introns, 3’ cleaved sequences, incomplete/damaged RNA) are

degraded in the nucleus in the RNA exonuclease-containing exosome.

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e. A sequence in the 3’UTR of some transcripts results in retention of the transcript in

the nucleus.

30. A farmer has prized tomato plants that look very unhealthy. She is going to have her plants tested by the Canadian Food Inspection Agency (CFIA) to determine if the plants are infected by tomato ringspot virus (TRSV), a virus with a single-stranded RNA genome. What test might CFIA conduct to determine if TRSV is present?

a. A western blot to detect TRSV RNA-dependent RNA-polymerase protein.

b. A northern blot to detect TRSV DNA-dependent DNA-polymerase transcripts.

c. A Southern blot to detect the TRSV genome.

d. All three of the preceding tests could be used to determine if TRSV is present.

e. None of the preceding tests could be used to determine if TRSV is present.

31. As part of a biodiversity assessment, scientists believe they have discovered a new species of plant with a novel form of virus-induced gene silencing (VIGS). Which of the following features of VIGS in the new species is NOT novel?

a. A DNAse cleaves double-stranded viral genomic DNA into 21-23nt segments. b. Double-stranded DNA segments are recognised by a DNA-induced silencing

complex. c. The DNA-induced silencing complex directs further cleavage of the viral genome. d. The VIGS system in the new species can result in silencing of related DNA

sequences in the plant’s genome, through DNA methylation. e. Transcription of the viral genome is enhanced by the VIGS system in the new

species.

32. RNA interference (RNAi) of a gene regulatory protein was conducted. Following RNAi, the transcriptome of the treated cells was compared to the transcriptome of untreated cells, using microarray analysis. Surprisingly, no difference in transcriptome activity was observed between treated and untreated cells. Why might this be the case?

a. The protein is not a gene regulatory protein. b. The protein requires co-regulatory proteins that are not present in the cell type

tested. c. The protein was unable to enter the nucleus on account of the absence of a signal

that allows transit to the nucleus. d. The double-stranded RNA used in the assay was ineffective at silencing the

expression of the gene regulatory protein. e. a, b, c and d could account for the observation.

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33. A typical microRNA (miRNA) is expressed in neurons so that it is abundant during the night but absent during the day. Which of the following statements would be UNEXPECTED based on this information?

a. Target sequences identical to the miRNA would be expected to have high transcript abundance during the night and lower abundance during the day.

b. Target transcript sequences with less extensive similarity to the miRNA would be

expected to have low translation during the night and higher translation during the day.

c. Target transcript sequences with less extensive similarity to the miRNA would be

expected to accumulate in P-bodies during the night. d. Target transcript sequences with less extensive similarity to the miRNA would be

expected to be associated with the RISC during the night. e. Target transcript sequences identical to the miRNA would be expected to be

associated with the RISC during the night. 34. Which of the following statements about two-dimensional polyacrylamide gel electrophoresis

(2D-PAGE) is FALSE?

a. In the first dimension of 2D-PAGE, proteins migrate to a location in the gel that is contingent on their isoelectric point.

b. A protein with a high isoelectric point would be found nearest the cathode in the

isoelectric focusing dimension. c. In the SDS-PAGE dimension of 2D-PAGE, proteins migrate to a region of the gel that

is contingent on the charge of their R-groups. d. In the SDS-PAGE dimension of 2D-PAGE, proteins have the same charge to mass

ratio, and separate on the basis of their molecular weight. e. Mercaptoethanol is generally only used in the SDS-PAGE dimension of 2D-PAGE.

35. You have been asked to PCR amplify a specific sequence from cDNA that was synthesised from mRNA isolated from brain tissue. After you run your potential PCR product on an agarose gel containing ethidium bromide, you observe no bands when you visualise the gel using ultraviolet light. Why might this be the case?

a. The gene you are interested in is not expressed in brain tissue. b. You used poly dT instead of poly dA to prime the first strand cDNA synthesis. c. You used RNAse H instead of DNAse H to leave behind small segments of RNA to

prime the second strand cDNA synthesis reaction. d. You used reverse transcriptase instead of DNA polymerase to synthesise the first

strand of cDNA. e. All of the four preceding answers are potential explanations for the outcome of the

PCR reaction.

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36. Examination of an extraterrestrial organism suggests that it monitors protein folding in a novel manner. Which of the following observations would set the extraterrestrial apart from terrestrial organisms, in terms of UNEXPECTED differences in protein folding?

a. Proteins begin their existence as molten globules. b. Protein folding occurs co-translationally. c. Misfolded proteins are captured by hydrophilic interaction with the barrel rim of

hsp60. d. Molecular chaperones like hsp70 refold proteins in an energy-dependent process

involving hydrolysis of ATP. e. Refolding of proteins by hsp60 is an energy-dependent process involving hydrolysis

of ATP.

37. In order to determine the location of expression of a protein of interest you take two different approaches: immunolocalisation using an antibody against the protein of interest, and generation of a fusion protein between the protein of interest and the green fluorescent protein (GFP) reporter. Immunolocalisation shows that the protein is expressed in both the cytoplasm and nucleus; whereas the fusion protein between the protein of interest and GFP is observed in the cytoplasm only. What might explain this discrepancy?

a. The molecular weight of the protein of interest is below the exclusion limit for the nuclear pore; whereas, the fusion protein is larger than the exclusion limit and has no nuclear localisation signal.

b. The promoter used to synthesise the transcripts for fusion between GFP and the

protein of interest is only expressed in the cytoplasm. c. The GFP fusion masks amino acids on the protein of interest that normally allow the

protein of interest to move into and out of the nucleus. d. a, b and c could explain the discrepancy between the observations. e. Only a or c could explain the discrepancy between the observations.

38. In the yeast two-hybrid assay, what is the role of the TRP gene?

a. The TRP gene provides the upstream activating sequences that allow the expression of the LEU gene when the “bait” and “prey” fusion proteins have interacted.

b. The TRP gene provides the coding sequences for the TRP protein DNA-binding

domain that comprise part of the “bait” fusion protein, and the coding sequences for the TRP protein transcriptional activation domain that comprise part of the “prey” fusion protein.

c. The TRP gene is included on the “bait” plasmid as a selectable marker, as the yeast

strain that is used is trp-. d. The TRP gene encodes an enzyme that is a reporter that indicates when the “bait”

and “prey” fusion proteins have interacted and activated transcription.

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e. The TRP gene is used to indicate the extent of transformation efficiency when the yeast strain has been transformed with the “prey” plasmid.

39. In cancer cells, a protein that is normally degraded through the ubiquitin pathway accumulates to high levels. What might explain this accumulation of the protein in cancer cells?

a. The amino acid residue that is normally poly-ubiquitinated on the target protein has been mutated so that it is no longer recognised by E2.

b. The degradation signal on the target protein has been mutated so that it is no longer recognised by E3.

c. A mutation has occurred so that the E2 and E3 pair that normally ubiquitinate the target protein no longer interact.

d. a, b, and c could explain the observation. e. Only a or b only

40. Which of the following statements related to the yeast two-hybrid assay is FALSE? (i.e., not c) could explain the observation.

a. If two proteins never occur together in the same cell in vivo, and the yeast two-hybrid assay shows that they interact, this result is likely to be a false positive.

b. A false positive result may arise if one of two proteins that normally do not interact in

vivo has a transcriptional activation domain and it is the prey protein. c. If two proteins require post-translational modification to interact in vivo, and the

yeast-two hybrid assays shows that they do not interact, this result is likely to be a false negative.

d. If two proteins that normally interact in vivo are unable to enter the nucleus in yeast

cells, this is likely to give rise to a false negative result. e. A false positive result may arise if one of two proteins that normally do not interact in

vivo has a transcriptional activation domain and it is the bait protein.

41. Which of the following statements concerning proteins is CORRECT?

a. Alpha helices are stabilized by hydrogen bonding between the carbonyl oxygen and the amide hydrogen of amino acids.

b. Proteins are the only molecules that can catalyze reactions in the cell.

c. The peptide bond is formed through a hydrolysis reaction.

d. Beta sheets are always anti-parallel, NOT parallel.

e. Disulfide bonds are formed from ionic interactions between charged amino acids.

42. What is FALSE concerning translation in prokaryotes and eukaryotes?

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a. In bacteria, tmRNA’s enter the A-site of stalled ribosomes, which enable the ribosome to add a special 11 amino acid tag, targeting the truncated protein for degradation.

b. In eukaryotes, broken mRNA’s are not translated as intact 5’ and 3’ ends of mRNA

need to be bound by the translation-initiation machinery before translation can begin. c. In bacteria, translation often begins before synthesis of the RNA molecule is

completed. d. Prokaryotic ribosomes initiate translation at Shine-Dalgarno sequences which do not

have to occur near the 5’ end of the RNA, and can be located anywhere along an mRNA molecule.

e. In eukaryotes, the initiating methionine is located by a process in which the large and

small subunits of the ribosome scan the RNA starting from the 5’ end, moving in the 5’ to 3’ direction until the first AUG sequence is found.

43. You are sequencing an amplified fragment of genomic DNA that you suspect has been subject to damage. The sequence you obtained for the template strand of your fragment in the 5’ to 3’ direction is: GTCAATCGGTA. Your fragment could be the result of damage due to deamination, if the CORRECT template sequence was:

a. 5’- GTCAATCGGTA -3’

b. 5’- TACCGATTGAC -3’

c. 5’- GTGAATCGGTA-3’

d. 5’- GTCAACCGGTA -3’

e. None of the above.

44. Which of the following statements concerning hemoglobin are TRUE?

(1) The disease sickle cell anemia results from a mutated hemoglobin gene.

(2) Hemoglobin is composed of 4 identical subunits.

(3) Each subunit of hemoglobin contains a heme group in the site that binds O2.

(4) The same mutation causing sickle cell anemia is thought to be advantageous (in heterozygotes) against malaria infection.

a. 1, 2

b. 1, 2, 3

c. 1, 3, 4

d. 1, 2, 3, 4

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e. None of the statements are true.

45. Which statement concerning phylogenetic analyses is TRUE?

a. Parsimony algorithms minimize the number of gaps introduced into the sequences.

b. Phylogenetic analyses produce a bifurcating tree with branch lengths as output.

c. Software packages that perform phylogenetic analyses such as PAUP* do NOT require a specific input format for the sequence alignment.

d. Altering the alignment of a set of sequences will NOT change the tree topology.

e. Phylogenetic analyses require sequence alignments without any gaps.

46. What is the CORRECT order of events during DNA replication on the lagging strand?

(1) Formation of a new phosphodiester bond by DNA polymerase.

(2) Binding of single-strand DNA binding proteins.

(3) Separation of double-stranded DNA by helicase.

(4) Exonucleolytic proofreading by DNA polymerase.

(5) RNA primer synthesized by DNA primase.

a. 1, 3, 2, 5, 4

b. 3, 5, 2, 1, 4

c. 3, 1, 5, 2, 4

d. 3, 2, 1, 4, 5

e. 5, 3, 2, 1, 4

47. Which statement about mitochondrial genomes is FALSE?

a. They use a genetic code different from the standard eukaryotic genetic code.

b. They are circular genomes.

c. Their genes do NOT have introns.

d. They probably arose as a result of a symbiotic event with formerly free living organisms.

e. They contain genes necessary for ALL mitochondrial functions.

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48. Which statement is FALSE concerning mechanisms that ensure accuracy during translation?

a. When an incorrectly coupled amino acid enters the editing site of an aminoacyl-tRNA synthetase, it is hydrolyzed and released.

b. The A- and E- sites of the ribosome are close enough for their two associated tRNA

molecules to be forced to form base pairs with adjacent codons, helping to maintain the correct reading frame on the mRNA transcript.

c. At the A-site, if the codon-anticodon match between the tRNA and mRNA template is

NOT correct, EF-Tu does NOT dissociate from the tRNA. d. At the A-site, rRNA in the small subunit of the ribosome forms a series of hydrogen

bonds with the codon-anticodon pair between the incoming tRNA, and the mRNA template, to determine if the basepairing is correct.

e. Incorrectly matched tRNA’s at the A-site whose anti-codon/codon interactions are

weaker tend to dissociate more quickly from the mRNA template than those which are correctly basepaired.

49. The formation of a phosphodiester bond between two nucleotides:

a. is a hydrolysis reaction and requires a hydroxyl group attached to the 5’ carbon of the sugar, a phosphate attached to the 3’ carbon of the sugar on another nucleotide and an ester bond.

b. is a condensation reaction and requires a hydroxyl group attached to the 5’ carbon of the sugar, a phosphate attached to the 3’ carbon of the sugar on another nucleotide, and an ester bond.

c. is a hydrolysis reaction and requires a hydroxyl group attached to the 3’ carbon of the sugar, a phosphate attached to the 5’ carbon of the sugar on another nucleotide, and an ester bond.

d. is a condensation reaction and requires a hydroxyl group attached to the 3’ carbon of the sugar, a phosphate attached to the 5’ carbon of the sugar on another nucleotide, and an ether bond.

e. is a condensation reaction and requires a hydroxyl group attached to the 3’ carbon of the sugar, a phosphate attached to the 5’ carbon of the sugar on another nucleotide, and an ester bond.

50. Imagine that tube A contains 1000uL of water. If you use each of the three micropipettors with the settings shown below once to transfer water to tube B, then how much water would REMAIN in tube A?

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P1000 P200 P20

0 1 0

3 0 2

0 0 0

a. 850uL

b. 402uL

c. 496uL

d. 480uL

e. 598uL

51. In a particular double-stranded DNA molecule, G and C together account for 46% of the total number of bases. In one chain of the DNA molecule, A and C account for 28% and 22%, respectively, of the chain’s bases. Therefore, the other chain of the same molecule must contain what percentage of A and C?

a. A=28%, C=22%

b. A=22%, C=28%

c. A=26%, C=24%

d. A=23%, C=27%

e. A=20%, C=30%

52. In an alpha helix, the carbonyl oxygen hydrogen bonds to the amide nitrogen to stabilize the alpha helix structure. Therefore, the shortest, stable alpha helix that can be detected by measuring circular dichroism is:

a. 6 amino acids long.

b. 5 amino acids long.

c. 4 amino acids long.

d. 3 amino acids long.

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e. 2 amino acids long.

53. In lab three, you combined your wheat germ DNA sample with water and loading dye. What is the function of loading dye in electrophoresis?

a. It is used as a way of estimating how far your DNA sample has migrated in the gel.

b. It is used to visualize your DNA sample.

c. It increases the density of your DNA sample causing it to sink to the bottom of the well in the gel.

d. a and c

e. b and c

54. In lab three, you prepared a Master Mix solution for PCR. Which of the ingredients listed below was NOT included in the Master Mix

a. Taq polymerase

?

b. EDTA

c. dNTPs

d. MgCl2

e. sterile distilled water

55. You are in the final steps of extracting DNA from wheat germ. In your final spin, you discover that you have a large, white pellet. You dissolve the pellet in TE (pH 8) and then use the spectrophotometer and get an A260:A280 ratio of 1.4. Which of the following BEST explains your observations?

a. The pH of the TE altered the properties of the DNA sample.

b. The DNA is contaminated with RNA.

c. The DNA is contaminated with protein.

d. The spectrophotometer must be broken.

e. The EDTA in the TE absorbs light maximally at 260 nm.

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56. Which of the following is TRUE with regard to BLAST?

a. A BLAST search involves an input sequence that is called the subject sequence, to which query sequences will be matched.

b. The bit score can be used to calculate both sequence similarity and homology.

c. The bit score is normalized so that it is independent of both sequence length and size of the database.

d. BLAST is used to make alignments between more than 2 sequences.

e. Because BLAST is an alignment search tool, only sequences of identical size can be compared.

57. Which of the following is TRUE about ClustalW?

a. It is a commonly used Single Sequence Alignment (SSA) program.

b. ClustalW can successfully align two sequences, even those that are not related.

c. Like BLAST, ClustalW compares one input sequence against a database of sequences.

d. ClustalW can only accurately align protein sequences.

e. ClustalW can not be used to create hypotheses about functional similarities between protein sequences.

58. You are given a DNA sequence from a sample of retinal tissue. You BLAST the sequence and part of the BLAST result is shown below.

>gb|M92036.1|GECP521A Gecko gecko green sensitive visual pigment (P521) mRNA, complete cds Length=2118 Score = 398 bits (215), Expect = 2e-107 Identities = 570/737 (77%), Gaps = 42/737 (5%) Strand=Plus/Plus Query 7 CCCGTTT-GAAGGTCCAAATTATCATATTGCACCACGATGGGTCTACAACATAACT-TCA 64 ||| ||| ||||||||||| ||||| ||||||||||| ||||| |||||| | | || Sbjct 177 CCC-TTTCGAAGGTCCAAACTATCACATTGCACCACGGTGGGTGTACAAC-TTGGTCTCT 234 . . . Query 646 CCAGCAGAAAGAGTCTGAATCAACACAGAAGGCTGAGAAGGAAGTGTCAAGAATGGTAGT 705 |||||||||||| || || || || |||||||| |||| ||| ||||| |||||||| ||

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Sbjct 816 CCAGCAGAAAGAATCCGAGTCCACGCAGAAGGCCGAGAGGGAGGTGTCCAGAATGGTGGT 875

Based on the results shown above, you are SURE that (choose the BEST answer):

a. the sequence you have been given is identical to the one from Gecko gecko (GECP521A).

b. the sequence you have been given is 2118 nucleotides long.

c. the Gecko gecko sequence is 875 nucleotides long.

d. given the size of the database, it is highly likely that matches as good as this one could be found by chance.

e. the Gecko gecko sequence is 2118 nucleotides long.

59. If you were going to make a 2% agarose gel and knew that you needed a final volume of 50 mL, how much agarose would you need?

a. 2.0 grams

b. 1.0 grams

c. 0.1 grams

d. 0.2 grams

e. 0.5 grams

60. Which of the following is TRUE about the restriction enzyme BamH1 (5-GGATCC-3)? The arrow indicates where the enzyme cuts the DNA.

a. When it cuts DNA, it creates a 5’ overhang.

b. When it cuts DNA, it creates a 3’ overhang.

c. It is classed as an exonuclease.

d. It can cut both single- and double- stranded DNA.

e. Both a and d.

61. Which of the following sequences does not contain a 6-base restriction enzyme cut site?

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a. ATTCTGCAGGGC

TAAGACGTCCCG

b. CTGCAGGGCAA GACGTCCCGTT

c. AAGCTTAAA TTCGAATTT

d. CGGCCGGG GCCGGCCC

e. ATCGGCA TAGCCGT

62. You have isolated a bacterium that exhibits very poor growth in the presence of low

tryptophan. You then hypothesized that a mutation occurred in the trp repressor causing it to bind too strongly to the operator. Which of the following experiments would BEST

a. Sequence the trp repressor, perform a ClustalW alignment to compare the new sequence with wild-type sequences, and then perform an EMSA (Gel Mobility shift assay).

confirm your hypothesis?

b. Sequence the trp operator, perform a ClustalW alignment to compare the new

sequence with wild-type sequences, and then perform a Western blot.

c. Generate a trp repressor-luciferase recombinant bacterium to check for expression levels in cell cultures.

d. Sequence CAP, perform a ClustalW alignment to compare the new sequence with wild-type sequences, and then perform an EMSA (Gel Mobility shift assay).

e. Perform a Northern blot to assess the expression levels of the trp operon.

63. You have just performed a Northern Blot to check for β-galactosidase transcript levels of a

novel bacterium that you have identified to have a mutation in CAP. You then discovered the β-galactosidase transcript was detected in your bacterium, the positive control (RNA from a wild-type bacterium), and in your negative control (RNA from an organism that you are sure does not express β-galactosidase). You included an additional control containing water only, but the transcript was not detected in this control. Which of the following BEST explains your results?

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a. There were no possible experimental errors conducted and therefore, the results are valid.

b. The probe was designed incorrectly and this caused it to cross-hybridize with non-specific transcripts.

c. Instead of RNA, protein was isolated from the three different bacteria and then used to hybridize with the probe.

d. The RNA of the negative control (RNA from an organism that does not express β-galactosidase) was not denatured enough prior to the electrophoresis and blotting.

e. The reagents were contaminated.

64. Which of the following would you use to study when a gene of interest is transcribed?

a. Southern Blot

b. EMSA (Gel Mobility Shift Assay)

c. Luciferase Assay

d. Chromatin Immunoprecipitation

e. DNA footprinting

65. What property of ethidium bromide makes it a potential mutagen?

Its

a. ability to fluoresce.

b. invisibility when dissolved in the running gel buffer.

c. ability to intercalate between the bases of nucleic acids.

d. ability to emit ultraviolet light.

e. ability to remove histones thereby exposing nucleic acids.

THE END!