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Part A: Multiple Choice

Please circle the letter of the single best answer to each question. Each question is worth 3 points.

Answer any nineteen of the following 22 questions, for a total of 57 points. (I will grade the first 19 questions answered.)

1. A single-stranded, radiolabeled DNA probe would not be useful for which of the following assays?

• a. Southern blot
• b. Northern blot
• c. Primer extension
• d. Dideoxy sequencing
• e. such a probe would be useful for all of the above assays

2. If you wanted to know the precise location of a transcriptional start site, which of the following assays would be most useful?

• a. Southern blot
• b. Northern blot
• c. Primer extension
• d. Dideoxy sequencing
• e. none of the above would be useful

3. Which assay would you perform if you wanted to know the level of expression of a protein?

• a. Southern blot
• b. Northern blot
• c. Primer extension
• d. Dideoxy sequencing
• e. none of the above

NOTE: I also gave credit for the answer "Northern blot," which is the closest of the available options. A Northern blot does measure expression, but it does so at the level of mRNA. mRNA levels do not necessarily correlate with protein levels.

4. For studying bacterial cells, genomic libraries are generally preferable to cDNA libraries because…

• a. bacterial mRNA is notoriously unstable
• b. bacteria do not polyadenylate their mRNA, so standard methods of mRNA purification are not useful
• c. the bacterial genome is very compact and efficient
• d. both a and b
• e. a, b, and c
• f. none of the above. cDNA libraries are preferable for most purposes.

5. You are studying a mutant strain of yeast that grows unusually slowly. When you retrieve a streak plate of your mutants from the incubator, you notice that one of the colonies in the streak is much larger than the others. You streak this individual colony out, compare it to your original mutant, and confirm that cells in this colony grow much faster than the original mutants, thus forming much larger colonies when incubated for the same amount of time. You conclude that you have isolated a suppressor. If you're right and it is a suppressor, which of the following types is most likely?

• a. genomic
• b. bypass
• c. multicopy
• d. interaction

6. You are studying a single-gene human genetic illness, attempting to find the gene responsible. After much work, you identify a group of related individuals that have a high rate of occurrence of this illness. They agree to provide blood samples, and you collect blood from affected and non-affected individuals (people with the illness and people who don't have it). You send these samples off to the Center for Inherited Disease Research (CIDR) for VNTR analysis. What information are you hoping to obtain?

• a. the precise location of the illness-causing mutation
• b. a statistical correlation between a particular fragment size at one VNTR region and inheritance of the illness
• c. CIDR's double-blind prediction of which individuals in your cohort have inherited the illness
• d. information about the dominance or recessiveness of the illness

7. The term "satellite" DNA refers to

• a. extrachromosomal DNA fragments that are found close to (orbiting) the full-length chromosomes
• b. mitochondrial DNA, which is circular in nature
• c. long tandem repeats of simple DNA sequences
• d. mobile DNA elements such as transposons and insertion sequences

8. Knowing what you do about the genetic makeup of mammals and bacteria, imagine that you were to perform the following experiment. You purify the same mass amount of genomic DNA from a dog and a culture of bacteria, break that DNA into fragments by mechanical shearing, denature it with heat, and then cool it and allow it to reanneal. What prediction would you make about the rate of reannealing?

• a. a fraction of the dog DNA will reanneal more quickly than the bacterial DNA, and another fraction will reanneal more slowly
• b. bacterial DNA will reanneal more slowly than dog DNA, because more bacteria are required to generate the same mass amount of DNA
• c. bacterial DNA will reanneal more quickly than dog DNA, because the bacterial genome is much smaller and simpler than the dog genome
• d. the two types of DNA will reanneal at very similar rates, because all DNA has the same molecular content

9. Which of the following accurately describes a difference between LINES and SINES?

• a. LINES are common in mammalian genomes, while SINES are more common in bacteria
• b. SINES move via a DNA mechanism, while LINES move via an RNA intermediate
• c. LINES contain a protein-coding region, while SINES do not
• d. LINES are typically shorter than SINES

10. A typical function of an ankyrin repeat region in a protein is to…

• a. mediate the regulated degradation of the protein
• b. mediate DNA or RNA binding, usually to facilitate transcriptional regulation
• c. mediate protein-protein interactions
• d. contribute to protein stability

11. In the regulation of the lac operon in E. coli, the function of CAP is to…

• a. activate transcription of the operon in response to high levels of glucose
• b. activate transcription of the operon in response to high levels of cAMP
• c. repress transcription of the operon unless lactose is present
• d. activate transcription of the operon in response to falling levels of cAMP

12. A mutation in the lac operator that prevents it from serving its normal function will result in…

• a. expression of the operon even in the absence of lactose
• b. repression of the operon even in the presence of lactose
• c. expression of the operon even in the presence of glucose
• d. repression of the operon even in the absence of glucose

13. In splicing of pre-mRNA in a eukaryotic cell nucleus, a "lariat" is formed. Which of the following statements about this "lariat" is most accurate?

• a. The RNA in the lariat is an exon in the process of being spliced--the loop of the lariat forms when the branch point adenosine attacks an intron-exon junction and attaches itself to the exon RNA
• b. Formation of the lariat is an autocatalytic event that does not require the participation of the spliceosome
• c. The lariat is formed by two successive transsolidification reactions
• d. The RNA making up the lariat consists solely of intron RNA

14. What's a snRNP?

• a. a nuclear complex of proteins and catalytic RNA
• b. a particle required for recognition of the N-terminal signal sequence at the beginning of proteins destined for the cell membrane
• c. an enzymatic complex required for transcription initiation in eukaryotes
• d. a blue-skinned cartoon character popular in the 80s

15. The GTPase activity of Ran is important for nuclear transport because…

• a. GTP hydrolysis by Ran provides the energy necessary for active transport by the nuclear pore complex
• b. Ran undergoes a conformational change upon GTP hydrolysis, and this change affects its ability to bind to a cargo protein
• c. The GTP hydrolysis and nucleotide exchange activities of Ran are regulated by proteins whose cellular distribution is restricted to either the nucleus or the cytosol
• d. all of the above are true
• e. b and c, but not a

16. Topo II is important in DNA replication because…

• a. it induces positive supercoils ahead of the replication fork
• b. it induces negative supercoils, facilitating unwinding of the DNA helix
• c. it separates the two template strands from each other, providing a single-stranded template for DNA polymerase
• d. topo II is not important for DNA replication--only topo I is required

17. Which of the following best describes thymine dimers?

• a. intrastrand crosslinks induced by UV radiation
• b. most commonly produced through oxidative damage to DNA, which is caused by the by-products of mitochondrial respiration
• c. recognized by MutS and excised by MutH, when MutH is activated by MutL
• d. byproducts of DNA replication, formed when one dTTP molecule is mistakenly linked to another rather than to the 3' end of the growing strand

18. Which of the following is true of non-homologous end-joining, as used in the repair of double strand breaks?

• a. a molecule called Ku binds to the DNA ends and makes an endonucleolytic cut
• b. regions of at least 50 nucleotides of homology are required for joining
• c. strand invasion of one double-stranded molecule by the 3' end of the other is involved
• d. DNA synthesis is not required for this repair pathway

19. Which of the following is not an activity of HIV-1's viral protease?

• a. separating the pol gene products from the gag gene products
• b. separating the env gene products from each other
• c. maturation of the budded viral particle from a noninfectious to an infectious form
• d. removing itself from its polyprotein precursor molecule
• e. all of the above are activities of HIV-1 viral protease

20. In an HIV-infected cell, the host cell's Golgi apparatus is required for which of the following?

• a. maturation of reverse transcriptase to its final dimeric form
• b. packaging of a specific host cell tRNA into budding viral particles
• c. the addition of sugar residues to the extracellular surface of gp120
• d. the transition from early phase to late phase viral gene expression

21. T cells derived from mice and grown in culture are not susceptible to HIV infection, even if they have been engineered to express human CD4 on their cell surface. If the gene for one additional human protein is added and that molecule is expressed on their surface, they become susceptible to HIV. What is that one additional molecule?

• a. gp120
• b. a T cell receptor, or TCR
• c. a chemokine receptor such as CCR5
• d. NFkappaB

22. Which of the following is not a mechanism used by HIV to obtain 14 proteins via expression from its very short genome?

• a. translation initiation at "start" codons other than the first AUG of a processed message
• b. synthesis of more than one protein from a single translation initiation site
• c. continuous translation by a single ribosome in more than one reading frame
• d. alternate transcription initiation sites

23. Below is the data table from Jacob et al: "The Operon." In the table, circle an entry that would have been different if the o^c mutations had been found to act in trans rather than in cis. (An "entry" is a single value. If there is more than one entry that would be different, choose one.)

Briefly explain how you would have expected this entry to be different (what would it have been instead?) and why.

(entry circled was row 5, column 1) If the operator's consitutive function was trans then it could have acted on the functional lac Z gene located on the other allele. This would have given non-induced production of beta-gal, comparable to that seen in rows 3 and 4.

[Other answers were possible. For example, column 2, rows 3 and 4. Protein "Cz" would have been constitutively expressed in these cells.]

24. Below is figure 1B from Hoshijima et al: "Transcriptional regulation of the sex-lethal gene." This "promoter-bashing" experiment was designed to identify regions of the Sex-lethal early promoter that are required for da/sis-b-dependent activation. How would you have expected the data to be different if the "E box" had been the only region important for this activation? Indicate all of the readings that would have been different, and explain (or sketch) what you would have expected to see. (Assume that SE3.4K would remain the same.)

If the E box were the only region necessary for activation then all of the promoter regions that contained an E box would be at or around 100% galactosidase activity. In the SE385(deltaE) we would have seen the % of galactosidase activity significatly lower, possibly even zero. In the SE83+E region, the reading would have been near 100% since the E box is the only thing necessary for activation. In the SE83+Q region no activity would have been seen because there is no E box in the reading.

25. This is figure 1 from Shibahara et al: "Replication-dependent marking of DNA by PCNA." Explain what would have been different about lanes 10-15 if CAF-1 could induce supercoiling of both replicated and nonreplicated DNA.

If CAF-1 could induce supercoiling of nonreplicated DNA as well as replicated DNA, the band pattern in lanes 10-15 would look the same for both the top and bottom images. The top image is an autoradiogram that shows only the replicated DNA in the gel, which is shown on the bottom in its entirety, including all replicated and nonreplicated DNA. "I" corresponds to the negatively supercoiled DNA. If CAF-1 could induce supercoils in non-replicated DNA, the majority of the DNA in the bottom image would correspond to "I" as it does in the top image. 

26. This is figure 4 from Shibahara et al. This figure is titled "Removal of PCNA from replicated DNA by RFC and ATP." What is it about this assay that allows the authors to conclude that PCNA, and not some other cis-acting mark, is removed from the DNA by RFC? (The bands in the gel represent radiolabeled, newly synthesized DNA.)

A sample is incubated with the indicated amount of RFC, with or without ATP, at the indicated temperature. A PCNA antibody is then used to precipitate the PCNA out of the solution. The immunoprecipiated material is then run out on a gel (along with any DNA attached to it). Since increasing RFC levels correlate with less PCNA-DNA precipitate, it shows RFC is removing PCNA. This removed PCNA is still immunoprecipitated, but it is not seen since it is no longer attached to radiolabeled DNA.

27. Shibahara et al used micrococcal nuclease digestion of DNA to draw conclusions about its structure. What is this assay designed to assess, and (briefly) how does it work?

Micrococcal nuclease digests the linker region between nucleosomes and leaves the DNA in the nucleosome intact. Digestion results in mono-, di-, and tri-nucleosomes, etc. The DNA in a nucleosome has a specific length (the length required for two turns around the nucleosome, approximately 150 bp). When the intact DNA is run on a gel, there will be bands at ~150 bp and multiples thereof if the DNA was in the form of nucleosomes. This assay is designed to detect the presence of nucleosomes in DNA structure.

28. This is figure 2A from Zennou et al: "HIV-1 genome nuclear import is mediated by a central DNA flap." In the assay represented here, what stages of the HIV-1 viral life cycle would be required in order for RT activity to be found? List the stages, and provide a one-sentence explanation.

Stages required:

Binding of HIV-1 viral particles to the host cell and infection of the host cell, reverse transcription, nuclear import, integration into the host cell genome, expression of early and late phase genes, assembly at the membrane, budding of the virus, and protease activity.

Brief explanation:

All of the stages of the life cycle of the HIV-1 virus would be required because RT activity is measured from newly produced viral particles.

29. Below is figure 3A from Zennou et al. In this assay, what stages of the viral life cycle are required for "viral production" to result? List the stages and provide a brief explanation.

Stages:

Expression of viral RNA and proteins, budding of viral particles

Brief explanation:

In this assay, plasmids containing HIV DNA with a mutated (or WT) cPPT sequence were transiently transfected into the nucleus of host cells. Transcription and translation were allowed to take place via the host cells' mechanisms and production of viral particles was measured. This assay showed that cPPT mutants could produce viable viral proteins once the DNA was in a stable form in the nucleus. This indicated that their deficiency came at a previous stage in the life cycle.

Part C: Mechanisms and Experimental Design. 15 points per question. Answer five, for a total of 75 points. I'll grade the first five answered.

30. Below is the beginning of the mechanism of integration of the HIV-1 proviral genome into the host cell's DNA. The first enzymatic step has already occurred--integrase has used its 3' to 5' exonuclease activity to trim back the 3' ends of the viral DNA by 2 nucleotides at each end.

a. What's the other enzymatic activity of integrase?

Strand transferase (transesterification reaction)

b. Draw the rest of the mechanism of integration, ending with a fully continuous double-stranded hybrid DNA molecule. Briefly explain what is happening at each step. Use arrows to indicate the 3' ends of each strand.

The 3' ends of the viral DNA attack the host cell genome--transesterification reactions occur where -OH on 3' ends attack.

Viral DNA is incorporated into host cell DNA

Host cell machinery (DNA polymerase and then ligase) repairs gaps in DNA [and the 2 nt 5' overhang on each end of the viral genome is excised].

31. Below is a diagram showing the elongation of the minus strand DNA during reverse transcription of the HIV-1 viral genome. The "first jump" has already occurred. Minus strand DNA is shown on the bottom, oriented with its 3' end on the left. The genomic RNA is shown on the top, with its 3' end on the right.

The next step (which is probably concurrent with this one) is RNaseH digestion. Beginning with this step, diagram the completion of the synthesis of the HIV-1 proviral genome. Use arrows to indicate the 3' end of each strand, and be sure to indicate which are "+" strands and which are "-" strands. Don't forget to include the formation of the central DNA flap.

32. Diagram the mechanism of meiotic homologous recombination, up to the formation of the second Holliday junction. Label the processes occurring at each step. Use arrows to indicate the 3' ends of each strand.

 

33. Identify the six different protein molecules in this drawing of a bacterial replication fork. Provide a brief explanation of the role of each.

Letter	Name	Description of function
A	beta clamp	binds to polymerase to give rise to a processive beta-clamp/pol complex (that synthesizes DNA efficiently and processively)
B	DNA polymerase	synthesizes the new DNA, adding nucleotides to the growing 3' end
C	tau protein	functions to coordinate synthesis of the leading/lagging strands. Connects the 2 polymerases at each replication fork.
D	helicase	Separates the two template strands from each other. "Melts" the DNA double helix.
E	primase	synthesizes short RNA oligonucleotide primers for DNA polymerase to initiate DNA synthesis from
F	single-strand binding protein	binds to regions of single-stranded DNA and protects them from damage, and from reannealing to the other template strand

34. Explain the role of ribosomal slippage in the synthesis of HIV-1 viral proteins. Explain which viral genes are involved, which form of HIV-1 mRNA is necessary, how often slippage occurs, and what protein products result.

Ribosomal slippage is responsible for the formation of the gag-pol polyprotein. If while translating the unspliced mRNA encoding gag and pol the ribosome "slips back" at the end of the gag gene so that it then reads the pol gene as well, the two are translated together as a fusion protein. Ribosomal slippage only occurs 1 of 20 times [I accepted a range from 5 - 10%] and therefore the structural proteins of gag (capsid, matrix, and nucleocapsid) get produced much more often than the enzymatic proteins of pol (reverse transcriptase, protease, and integrase).

35. Explain how you could use Sanger dideoxy sequencing to determine the site of HIV-1 integration in a clone of infected cells. (i.e. You have a culture of infected cells, all of which derived from a single integration event in a single host cell.) In your explanation, include:

• how you would design your sequencing primer--what sequences it would be complementary to, and what its 5' to 3' orientation would be
• how you would set up your sequencing reactions--how many reactions you'd need, what their essential ingredients would be, and how they would differ from one another
• how you would determine the results of your analysis

I. Create a primer that has homology to a unique region at the 3' end of the virus's genome, probably part of rev or tat. (Want it to only be found once in the HIV proviral sequence.) The 3' end of the primer should be oriented toward the 3' end of the viral genome (i.e. toward the insertion point).

II. Set up four vials for sequencing reactions. In all vials include all four dNTPs, DNA polymerase, and denatured host cell DNA. Also buffer and the primer. In one vial put ddATP, in another put ddCTP, in another put ddGTP, and in another put ddTTP. [The dideoxynucleotide in each vial is at approximately 1/100th the concentration of each of the dNTPs].

III. As the reaction proceeds, DNA polymerase will begin adding dNTPs to the selected primer. As it does, it will sometimes randomly incorporate a ddNTP. When this happens, chain growth will stop because the polymerase can't add to ddNTPs and the polymerase falls off. Since each vial only contains one type of ddNTP, this point where the polymerase falls off corresponds to a part where the DNA is complementary to the given ddNTP.

IV. Thus, if we run these fragments out on an acrylamide gel in four lanes, which allows us to distinguish between fragment lengths 1 bp apart, we should be able to read the sequence right off the gel when looking at it with something that will pick up the radiolabeled primer (e.g. Xray film).

V. What we are looking for is where the sequence of the HIV-1 insert ends and where the genome begins. If we get a large enough part of the genome sequenced, then we should be able to consult a database that tells us which part of the genome we have sequenced and therefore, where the HIV has inserted itself.

Created by: bkbaxter@lclark.edu
Updated: 20 Dec 00