Biochem Exam 4 Challenge Questions
What would the effect of a mutation in the HIV-1 reverse transcriptase gene that decreased the rate of mismatches?
Any decrease in the error rate of HIVRT would result in an overall decrease in the rate of viral mutations. This would probably not be an advantage for the virus, as the relatively high error rate often helps to thwart the use of antiviral drugs.
Approximately how many ATP are required by helicase for the replication of the E. coli chromosome?
Because ATP requires 1 ATP per two bases, it requires 2.3 x 10^6 molecules of ATP to replicate 4.6 x 10^6 bases
What would happen if a mutation in the lambda phage Xis gene occurred such that the resulting protein was not functional?
Because Xis is required for lambda phage DNA excision, a mutation rendering Xis nonfunctional would prevent excision and prevent initiation of a lytic cycle
The modified base 5-methylcytosine is used in higher eukaryotes as an epigenetic mark of particular chromatin states. But 5-methylcytosine is also known to be a hot spot for mutation (it has a higher rate of mutation from C to T than most C residues. Explain why this is the case.
Cytosine residues are deaminated at a very low rate to form uracil, which will lead to a uracil-guanine (U-G) base pair. Because U is not a normal component of DNA, this is quickly repaired to a C-G base pair by uracil N-glycosylase and subsequent events. Deamination of 5-MeC would lead to formation of a T-G base pair, and the cell would not rapidly repair this, as there is no way to be sure which base is the correct base. If DNA replication occurs before the mismatch is repaired, it fixes a C-to-T transition mutation in one of the progeny duplexes.
What are Okazaki fragments?
DNA segments produced by discontinuous replication of the lagging strand. One is started each time a new beta clamp loads onto the lagging strand template and is completed when DNA polymerase encounters the preceding Okazaki fragment.
A scientist wishes to express a eukaryotic protein in bacterial cells. The gene is cloned along with its promoter region and is inserted into a plasmid. After transforming the plasmid into bacterial cells, protein expression is initiated, but no protein is observed after the cells are lysed. Why? How could this problem be fixed?
Eukaryotic and prokaryotic promoters have different sequences for RNA polymerase, and so the bacterial sigma factor would most likely not bind the eukaryotic promoter sequence, thus RNA polymerase would not associate with the DNA, preventing transcription and thus protein expression. Closing the gene behind a bacterial promoter sequence could solve this problem.
in the galactose regulatory system of yeast, what would be the behavior of mutants that lack one of the following components: GAL3, GAL4, or GAL80? Consider the expression of the target genes with and without the presence of galactose. What would be the behavior of a GAL3-GAL80 double mutant?
GAL3 and GAL4 mutants will not express the target genes in the presence of galactose. A GAL80 mutant and a GAL3-GAL80 double mutant will express the target genes in the absence of galactose
Describe the mechanism by which HIV-1 reverse transcriptase (HIVRT) produces DNA for chromosomal integration
HIVRT uses ssRNA of the HIV genome as a template to produce a DNA-RNA duplex, using the 3' end of the host cell Lys-tRNA as a primer. HIVRT has RNase activity and degrades the RNA strand once DNA is produced. RNA fragments serve to prime synthesis of the complementary strand to generate dsDNA.
In the experiment by Jacob and Meselson, bacteria were grown for several generations in a medium containing heavy isotopes (15NH4Cl and 13C-glucose). The bacteria used these isotopes to synthesize all cellular components, resulting in their incorporation into all of the bacterial nucleic acids and proteins. The bacteria were switched to a normal medium after infection with bacteriophage, and the ribosomes were isolated. Jacob and Meselson discovered that the isolated ribosomes contained the heavy isotopes. Why did this result indicate that ribosomes are not carriers of genetic information?
If ribosomes were the source of genetic material, the bacteriophage would synthesize its own ribosomes soon after infection as a way of propagating itself. This would occur in the normal medium, and so the ribosomes would not contain heavy isotopes. However, if ribosomes were only sites of protein synthesis, then the viral mRNA would be translated by bacterial ribosomes, which were synthesized while the heavy isotopes were present. The presence of isotopes in ribosomes enabled Jacob and Meselson to confirm that ribosomes were not the viral genetic material.
Consider the regulation of the trp operon in E. coli. If this system were present in yeast instead of E. coli, would you expect it to function properly?
It would not work in eukaryotes because this mechanism depends on transcription and translation occurring simultaneously; in eukaryotes, the two processes occur in different cellular compartments.
formation of the 5' m7G cap requires that the 5' nucleotide be a triphosphate. Explain why this nucleotide retains all three phosphoryl groups.
RNA synthesis requires NTPs; the first nucleoside added has three phosphoryl groups. The rxn to add each subsequent nucleotide is initiated by a nucleophilic attack on the alpha phosphate, resulting in cleavage and release of pyrophosphate (PPi)
RNaseP can cleave both pre-tRNA and pre-rRNA sequences. If the primary sequences of these RNAs are not always similar, how does the enzyme recognize and bind its substrates?
RNaseP recognizes the secondary structure of tRNA and rRNA rather than the primary structure. Such structures form co-transcriptionally, so the primary transcript is cleaved on the basis of the presence of these structures and not on any individual sequences
What is the primary difference between strong and weak bacterial promoters? How does transcription differ from these two types of promoters?
Strong bacterial promoters differ slightly from the conserved sequences at the -35 and -10 boxes, whereas weak promoters can contain several base pair changes. These are the sites of sigma-factor binding, so the closer the -35 and -10 boxes are to the conserved sequences, the stronger the association between RNA pol and the promoter. Therefore, strong promoters are more likely to be bound by sigma factor and transcribed than weak promoters.
From the following sequence, locate the sites of potential photoproduct formation. Indicate what photoproducts will most likely form and what the potential effect on the DNA would be if left unrepaired. 5'-ACGTCAGTTACGTACTGACGT
The 'TC' photoproduct is likely to be a (6-4) photoproduct, which can stall rep. forks if not repaired. The 'TT' and 'CT' photoproducts would most likely be pyrimidine dimers and, if unrepaired, would most likely result in adenine on the daughter strand. The TT photoproduct would not result in a mutation, but CT would result in a mismatch, with high potential for mutation, depending upon when repair occurs. If the daughter strand were used to correct the lesion, then the C would be converted to a T.
What is the nucleophile in the reaction catalyzed by DNA polymerase? What is the significance of this in terms of initiating DNA synthesis?
The 3' hydroxyl of the previously added nucleotide is the nucleophile for DNA polymerase. Initiation requires a primer, a short stretch of RNA produced by primase, to provide the nucleophile.
If the sequence 5'-AACGC-3' were damaged by reactive oxygen species, what would be the most prevalent product, and what would be the result of replication (show both strands after replication)?
The G would become 8-hydroxyguanine and tautomer, 8-oxoguanine. These base-pair with adenine, resulting in a G-A mismatch that is not likely to be repaired by DNA repair mechanisms. After a second round of replication, the result is a G->C substitution in one of the daughter strands.
What must the rate of nucleotide incorporation by primase be such that it completes a primer in exactly the same time that an Okazaki fragment is completed?
The avg. Okazaki fragment is 2000 bases and synthesized in 2 seconds (rate = 1000 bases/second). A typical primer is about 10-12 bases; thus, primase would need a rate of 5-6 nucleotides/second to avoid being the rate-limiting step of replication.
Calculate the # of Okazaki fragments produced during the replication of a single E. coli chromosome.
The avg. Okazaki fragment is 2000 bases; the E. coli chromosome is 4638 kb. Therefore, about 2319 Okazaki fragments form per replication.
An Ames test of a suspected mutagen was examined both before and after incubation with rat liver extract, giving the following results. What can you conclude about the suspected mutagen?
The results suggest the suspected mutagen is mutagenic without liver metabolism, and even more mutagenic after metabolism by liver enzymes.
RNAi-based modification of amylopectin content in wheat occurs through decreased expression of starch-branching enzymes. If the siRNA used to knockdown plant starch-branching enzymes was transmitted to humans, what enzyme might it affect, and why would this be harmful?
The structure of amylopectin is similar to that of glycogen, as both contain an alpha(1->4) chain of glucose molecules with alpha(1->6) branches. The spacing between branches varies, so it is reasonable to conclude that starch-branching enzymes and glycogen-branching enzymes that catalyze the formation of the same bond are similar. If siRNA against a starch-branching enzyme was transmitted to mammals, it could result in repression of a glycogen-branching enzyme, and the cell would store only straight-chain glycogen, reducing the rate of glycogen breakdown and the capacity to store glycogen. These effects would negatively affect ATP production and could lead to symptoms similar to those observed for glycogen storage disorders.
You are studying the regulation of a Drosophila gene, and you have available the genome sequences of several related Drosophila species. By comparing these sequences in the vicinity of your gene, you identify several regions that appear to be conserved. You also identify potential binding sites for known regulatory proteins. You speculate that this region is an enhancer. Suggest how you might test this idea. Assume that you are competent at recombinant DNA work and that you have a method to introduce desired constructs into Drosophila.
To validate your hypothesis, you would use recombinant DNA to place the proposed enhancer upstream of a reporter gene in Drosophila embryos. Several constructs should be prepared, with the enhancer in either orientation and at variable distance from the promoter for the reporter. If the DNA region includes an enhancer, you should see expression of the reporter.
Plaques of bacteriophage lambda differ from those of most viruses. Most viruses kill all the cells in the plaque, so there are no surviving cells and the plaques are clear; you can see through them. In contrast, plaques of bacteriophage lambda are cloudy or turbid because lysogens arise in the plaque and grow. a) in the plaque, lysogens are often infected by other lambda viruses present in the plaque. Why don't these newly infected viruses grow lytically and kill the cells? b) mutants of lambda can arise that form clear plaques. Which virus genes are likely to be mutated?
a) a lysogen in a lambda plaque can be superinfected by other lambda virions, but the lysogen makes CI, which will bind to the lytic promoters of the incoming viral DNA, repressing them and preventing them from growing lytically. This is termed "immunity" b) clear plaques have mutations in the cI gene; they cannot make CI and cannot est. or maintain the lysogenic state. Clear plaques led to the discovery of this gene and two others; hence the name of the gene "cI" as clear plaque, class I.
In the trp attenuation system of E. coli, predict the effects of the following mutations on the operation of the regulatory system, and give your reasons. To simplify matters, assume that the Trp repressor is not active. Three broad possibilities exist for whether the operon is transcribed in the absence or presence of charged tTRNA^trp: behavior is normal, transcription always low, or transcription always high a) mutation in region 4 so it cannot pair with region 3 b) mutation in region 3 so it can pair with region 4 but not with region 2 c) change of the 2 trp codons to glycine codons d) change of AUG codon of leader peptide to AUA e) combination of the first mutation with each of the other three
a) constitutive (no terminator is formed) b) noninducible (antiterminator cannot form in absence of trp-tRNA, so terminator would always form) c) noninducible (ribosome would not pause at the gly codons in presence of low trp-tRNA) d) noninducible (ribosome would not be able to stabilize the antiterminator) e) consitutive
Describe the typical major elements of a replication fork.
ahead of the rep. fork, topoisomerase adds neg. supercoils to relieve the positive supercoiling caused by separation of strands. Helicase and primase are coupled at the front of the replication fork to separate the strands and synthesize RNA primers, respectively. Single-stranded binding proteins maintain single-stranded DNA and protect it from damage. DNA pol III, tethered to helicase, consists of a core enzyme and a beta clamp, which increases processivity of the polymerase. DNA pol I and ligase follow the replication fork to remove RNA primers and seal nicks in the newly formed DNA.
your colleague is studying the regulation of an E. coli gene, and she purifies a protein that stimulates expression of the gene in an in vitro transcription system. In this system, the gene is expressed at a very low level in the absence of this protein and at a high level in its presence. She interprets these data to mean that it is an activator protein. Can you suggest another possibility, along with one or more experiments that would distinguish between the two models?
another possible explanation is that the purified protein is an alternate sigma factor. Sigma determines the promoter binding specificity, and there are multiple sigma factors. To distinguish between these models, an in vitro transcription system could use an RNA pol without the usual sigma70 factor, called the "core" RNA pol. The new sigma model predicts that the target gene would be transcribed; the activator model predicts no transcription.
many antibacterial agents work by inhibiting RNA pol. These agents are typically molecules that interact with regions of the catalytic center to prevent DNA binding/transcript synthesis. Why are many of these considered "broad-spectrum antibiotics"; that is, antibacterial agents that act against a large number of different bacterial species? Why are these broad-spectrum antibiotics advantageous for both physicians and their patients? These agents, however, do not affect eukaryotic RNA Pol I, II or III. What does this tell you about the sequence conservation between the bacterial and eukaryotic enzymes?
bacterial RNA pol is highly conserved; therefore, an inhibitor of its activity would likely affect all bacteria, meaning the antibiotic can be used without knowing the species of bacteria causing the infection, an advantage because most bacterial infections require immediate treatment. The structure of RNA pol is conserved from bacteria to eukaryotes, but these enzymes do not share a high degree of sequence conservation. The interaction between the antibacterial molecule and RNA polymerase depends on specific base contacts, and the differences at the primary sequence level means that the eukaryotic RNA pols do not form the same contacts, and therefore the inhibitor has no effect
a nucleosome is positioned with two cis-acting sites lying on the DNA close to the exit point of the DNA. At a low frequency, the two sites are transiently exposed by dissociation of a segment of DNA from the histones. If the second site is exposed, the green factor can bind to it. What effect will have this on the ability of the red trans-acting factor to bind to its site? Can this same effect on binding occur by other mechanisms?
binding of the green factor would enhance the ability of the red factor to bind, as its binding site is now accessible, not bound up in the nucleosome. This is a form of cooperatively, which can also involve protein-protein interactions between the two proteins bound to DNA or in solution.
in bacteria, the same DNA strand is often a template for both replication and transcription. Why are codirectional collisions a common occurrence?
both DNA and RNA pol read 3' to 5', resulting in a 5' to 3' polymerization. When both replication and transcription complexes proceed down the same region of DNA, the DNA pol can overtake RNA pol because transcription is much slower than replication; this results in a co-directional collision
assume that you have identified all the cis-acting sites in a prokaryotic genome for a particular gene regulatory protein. Can you predict the consequences of the regulatory protein binding to these sites?
no you cannot be certain what the effect of protein binding will be. numerous examples indicate that the effect of a bound trans-acting factor depends on its context. Bound bacteriophage lambda CI protein and CRP-cAMP both can act as either activators or repressors. the mammalian Oct4-Sox2 heterodimer activates several ES-specific genes and works to repress developmental genes
What are the primary replicative polymerases in prokaryotic and eukaryotic cells?
prokaryotes: DNA Pol III eukaryotes: DNA pol delta (lagging strand) and DNA pol epsilon (leading strand synthesis)
What is meant by semiconservative replication? How is this different from conservative or dispersive replication?
semiconservative replication creates 2 daughter DNA molecules, each containing one template strand and one newly synthesized strand. Conservative replication would yield one DNA duplex made up of both templates and one made of two newly synthesized strands. Dispersive rep. would result in both DNA molecules made of mixed portions of template and newly synthesized DNA, template paired with template, and new DNA paired with new DNA.
