BIOMI 4850 Prelim- HWs

What mechanism is used by all E.coli DNA polymerases to associate with the DNA?

Binding to the β sliding-clamp

Dominant mutation

Describes a mutation that masks the phenotype of another mutant or wild type allele of the same gene.

Epistatic mutation

Describes a mutation that masks the phenotype of another mutation in a different gene.

How do other proteins use this property to carry out other replication associated activities?

Ligase and DNA Pol I associate with β to "find" the lagging strand. Hda associates with β as a signal of DNA replication and, in turn, stimulates ATPase activity of DnaA

Why might a bacterial strain need more than one polymerase?

Like life in all domains, one polymerase will not have attributes for all situations. Given our current understanding, it is impossible to have a DNA polymerase that replicates very accurately and still be able to replicate damaged DNA.

What is one chromosomal feature that could account for a forbidden chromosomal inversion?

Polarity of some DNA sequences, for example ter sites as one-way stop for DNA replication. To a lesser extent Chi sites recognized by RecBCD, or polar DNA sequences around dif that are recognized by FtsK, etc.

What are three gene products that were discussed in class that deal directly with reactive oxygen species that could conceivably cause hypersensitivity to the drug in question when they are lost and be identified in your screen?

SoxR regulated: superoxide dismutase (sod genes) OxyR regulated: catalase (kat genes) and alkyl hydroperoxide reductase (ahp genes)

How does Topo IV activity facilitate the segregation of circular bacterial chromosomes?

The primary function of Topo IV ( a type two topoisomerasae) is to decatenate chromosomes (catenaens are linked chromosomes like links on a chain). If chromosomes are linked in this way they cannot be segregated to individual cells.

Why would a bacterial geneticist use a temperature sensitive mutation?

This type of mutation is especially useful with essential genes because it allows you to still grow and maintain the mutant strain at the permissive temperature (e.g. at 30C), while still being able to analyze the effects of the mutation under another condition (i.e. 42C).

For the genes vsr and dcm discussed in class. What is the phenotype expected for a vsr mutant, a dcm mutant, and a vsr dcm mutant?

vsr mutant - Higher level of mutations in the sequence 5'-CCWGG/GGWTCC - 3' where the second C-G pair would mutate to T-A dcm mutant - no different than wild type for mutations vsr dcm double mutant - no different than wild type for mutations The Dcm enzyme will make the second "C" a 5-methyl cytosine in the sequence above. If the 5-methyl cytosine deaminates it becomes thymine which will not be recognized as atypical in the DNA and will not be removed. Mismatch repair could fix the problem, but will not "know" which one is wrong as it does during replication.

You just identified mutations in two genes that cause an otherwise green bacterium to be the color white. Make up a name for these genes in keeping with the "Demerec convention" for naming genes.

whtA and whtB (italics)

Using the correct format, how would you name six different mutations in one of these genes?

whtA1, whtA2, whtA3, whtA4, whtA5, whtA6 (italics)

Scientists are modifying the genome of E. coli so that it only encodes 57 codons instead of the normal 64 codons. One goal of this work is to make some codons available for amino acids normally not found in natural proteins. You are a genetic engineer that wants to use this strain to make a protein with a new amino acid. What are the steps you must do beyond making a new tRNA to get the new amino acid into a protein of interest?

- Remove endogenous tRNA genes that would recognize your new codon (if not already removed in the strain). - Provide (or engineer) an aminoacyl-tRNA synthetase that recognizes your new tRNA but not any endogenous tRNAs. This enzyme must also specifically recognize your new amino acid and not any other amino acids. - Your new amino acid needs to be produced in the cell or have some way to enter the cell if provided in the growth media.

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DnaK is involved in dealing with denatured proteins as occurs at high temperature. When DnaK is not "busy" dealing with denatured proteins it helps shut off the response to high temperature. Briefly explain two ways that free DnaK contributes to shutting off the sigma H (RpoH) system.

1. DnaK prevents physical binding of Sigma H to the promoter of heat shock genes. 2. DnaK binds to Sigma H, thereby tagging it for degradation.

What would be expected to happen to expression of the genes in the sigma S regulon in each of the three genetic backgrounds? 1. iraDMP 2. rssB 3. iraDMP, rssB

1. heat shock genes never expressed 2. always expressed 3. always expressed

What codons can be use as start codons in bacteria?

83% of start codons are AUG, 14% GUG, 3% UUG, <1% CUG

Describe a situation where a bypass suppressor could account for this suppression (explain what a bypass suppressor is and give an example of the chromosomal mutation and what the mutA allele would be).

A bypass suppressor is a mutation that activates a pathway that allows for the lack of mutA to be compensated for. Many different correct answers could be imagined, but for one if a mutation activated a export pump in the cell that allowed it to be removed from the cell it could work as a bypass mutation. In another example is if the suppressor mutation activated a protein that could break-down or otherwise deactivate Compound X.

lacZ

A gene whose product is only active in the cytoplasm and when used in gene fusions it can indicate domains of a protein that reside in the cytoplasm.

Amber mutation

A mutation from a sense codon to a UAG nonsense codon.

For each of the "mut" genes in the genotype of XL1-Red, give an example of another gene where the mutation would show an epistatic relationship? If there is not one explain why.

Epistatic = mutH, mutL, and mutS; No epistatic relationship mutM, mutY, and mutT;

Why do you think IPTG is called a "gratuitous inducer" of the lac operon? How is it used in the laboratory setting?

IPTG is called a gratuitous inducer because it binds to the LacI repressor and causes it to fall off the DNA, thereby activating expression of the lac operon. However, IPTG itself is not a substrate for LacZ, meaning it is not degraded and IPTG levels stay constant as it induces further expression of the lac operon and of LacZ in particular. Thus, in the laboratory setting, if scientists fuse a gene of interest downstream of the lac operon promoter, PLac, IPTG can be used to upregulate expression of that gene of interest.

You are attempting to clone a gene into one of the many vectors that allow blue / white screening on plates containing Xgal. After plating the cells containing your vector ligated with the insert you find only white colonies. How might a lack of intracistronic complementation explain your inability to find the successful clone and what would you look for in the genotype of your strain of E. coli to ensure that you carried out the experiment in the correct way?

The strain used in the Blue/White screen must have a certain Lac deletion in trans (elsewhere in the cell) to allow intracistronic complementation. One needs to make sure the w-fragment is either on the chromosome on a plasmid in the cell (encoded by the DlacZ(M15) allele).

Would the three genes described above show an "epistatic" relationship if mutated (explain)?

These repair pathways would not be epistatic. While they are all responsible for limiting the effect of 8-oxo-G, they work independently of one another.

Could any of the E. coli DNA polymerases fix the damage caused by the 3'exonuclease in the above example? Why or Why not?

Yes. DNA polymerase I and DNA polymerase III could both fix the damage. This is because they both a 5'-3' polymerase activity that allows them to synthesize new DNA from the now exposed 3' ends of the DNA strand, down the 3' direction. (Actually, any polymerase in E. coli would be able to do this as they all have 5' to 3' polymerase activity)

Do all proteins start with an AUG codon in bacteria?

no

Explain how tmRNA can harness the Clp protease and why this is important?

tmRNA will load into a stalled ribosome on a truncated mRNA. The amino acid charged onto the tmRNA will be added to the protein attached to the ribosome and then the tmRNA itself will be translated like a normal mRNA. However, the amino acid sequence that is encoded in the tmRNA is special, it has evolved to be specifically recognized by the Clp protease machine. This will target the truncated protein that results from the truncated mRNA for destruction. Presumable this is beneficial in that it could prevent any rouge behavior of the partial protein product. Degrading the partial protein will additionally allow the product to be recycled.

phoA

A gene whose product is only active in the periplasm and when used in gene fusions it can indicate domains of a protein that reside in the periplasm.

Indicate where a nonsense mutation could occur in one gene to cause loss of expression of the other gene via an effect of Rho

A mutation changing a "sense" codon to a "nonsense" codon (i.e. one of the three stop codons) can cause a Rho-dependent polar effect on downstream genes. Cryptic RUT sites are found all over the genome. The term RUT "site" is somewhat misleading because the sequence of RUT sites is highly degenerate. These RUT sites are cryptic, because as long as ribosomes are actively translating an mRNA they are not actually recognized by Rho. However, if a nonsense mutation occurs in an open reading frame, translation of a region will stop and if a RUT site is present Rho can load and terminate transcription at a position downstream. This happens when the Rho protein translocates down the mRNA (in the 3' direction) and catches-up to RNA polymerase when it pauses at various sites along the DNA.

Explain why a transcriptional (operon) fusion would not allow you to gather the same information.

A phoA transcriptional fusion would result in the free PhoA protein always residing in the cytoplasm and never dephosphorylating X-phosphate.

Brute force screening

A time consuming screening method where individual mutant strains are analyzed by PCR or some form of biochemical assay.

You discover that your drug also induces transition mutations. How does this differ from a transversion? Which is it more likely to affect the function of the cell a transition or transversion mutation and why?

A transition mutation changes a purine to a purine (A to G or vice versa) or a pyrimidine to a pyrimidine (C to T or vice versa). A transversion is when a purine is mutated to a pyrimidine or vice versa. A transition mutation is less likely to affect protein function, as changes at the third position in a codon are more tolerant for transitions (i.e., transitions are more likely to result in the same amino acid being incorporated into the polypeptide product).

Dominant negative mutation

A type of mutation that allows the isolation of loss-of-function proteins that are not due to simple nonsense mutations.

Deletion mutation

A type of mutation that cannot revert.

The last part of the last question describes a "polar" mutation. Polar mutations can complicate genetic analysis and require specific tests to ensure that one is ascribing a certain phenotype to the correct gene. Explain an experiment that lets you rule out polar mutations in a genetic analysis.

A typical experiment to rule out polar mutations is check to make sure that the phenotype is reversed when you complement with a wild type copy of the genes. If a mutation in a given gene is responsible for a mutant phenotype you should be able to return to the wild type phenotype by providing a wild type copy of the gene in trans (for example by cloning it into a plasmid and expressing it in the mutant cell). If your mutant strain phenotype is due to a polar effect, expressing the gene from a plasmid would not allow suppression of the phenotype. To rule out a dominant negative phenotype, typically one will test a complete deletion of the gene.

Amber suppressor

A tyrosine inserting tRNA with a mutated anti-codon that recognizes the UAG nonsense codon.

In the course of your experiments explained in question 4 you identify a chromosomal mutation that suppresses one of your mutant mutA alleles, mutA304. The mutA304 allele is sensitive to Compound X with mutB provided, but with the suppressor mutation that maps to the chromosome your strain is resistant to Compound X. Describe a situation where an informational suppressor would account for this suppression (explain what an information suppressor is and give an example of the chromosomal mutation and what the mutA allele would be).

An informational suppressor is a type of suppressor that alters the genetic apparatus that cells use to express genes. This typically refers to a tRNA suppressor mutation that recognizes a premature stop codon in mutant allele. An example of how this could be happening in this case is that mutA304 is an amber mutation that changes a formerly sense codon to the UAG amber stop codon. This would result in a truncated polypeptide. A chromosomal mutation that could account for the suppression of this nonsense mutation is a mutation in a tRNA sequence (we'll say a leucine tRNA) that causes it to now recognize the UAG stop codon. When the protein is translated with this suppressor, some will make a full length polypeptide with a leucine in the position that it was previously truncated at.

If you examined proteins that were produced in bacteria what amino acid(s) would you expect to find at the N-terminal end of the protein?

Any amino acid- the first amino acid, the formyl methionine is removed

Below is the genotype for the E. coli strain XL1-Red. As discussed in class this strain will accumulate mutations. Briefly describe why and how you would use this strain to study a hypothetical protein.

Because the genotype of this strain will cause accumulation of base substitutions in DNA, cloning the gene of interest in this strain will result in random mutations in the gene. It is very useful especially when you want to find the functionally crucial amino acids in a hypothetical protein, of which you do not have any idea about its structure and conservation. 1.Clone the gene encoding the hypothetical protein into a plasmid and make a deletion mutation of the original copy on the chromosome of your strain; 2.You should be able to complement the deletion mutant with your plasmid; 3.Introduce the plasmid into the mutator strain XL1-red and subculture the transformants for several generation; 4.Make a plasmid prep; 5.Introduce you mutated plasmid back into the mutant strain and look for increase or decrease of the corresponding phenotype; 6.Sequence your plasmids to figure out which amino acids are important for function.

What is the expected difference between a constitutive mutation in a repressor verses an activator?

Constitutive mutation in a repressor means that repressor shuts off gene expression even under inducing conditions, meaning that no matter what environmental condition, the genes are not expressed. In the case of a repressor any number of mutation that inactivate the protein will cause this constitutive phonotype. Constitutive mutation in an activator means that activator is always promoting transcription, these are likely to be very specific kinds of mutations (for example that cause it to be in the conformation normally found when binding the effector, even when it is not really present).

Briefly explain the difference between a DNA lesion and a mutation and how they are related.

DNA lesions are a type of damage or modification to the string of molecules, typically at the base. Lesions are often quickly dealt with by repair mechanisms in the cell, but if they are not, the change could lead to a base change or a change in its base pairing properties that can be inherited by future daughter cells, thus converting the original lesion into a mutation.

You are comparing the relationship between MutM and MutY, two genes involved in DNA repair. You determine how well the wild type (wt) and the single and double mutants survive exposure to increasing concentrations of mutagen.Do the results suggest that MutM and MutY are in the same or different pathways? Effects are additive.

Different pathways, because the effects are additive. If they were in the same pathway, the double mutant would likely have the same phenotype as either single mutant.

SoxR does not directly deal with reactive oxygen species, but in the regulation of the response to reactive oxygen species. One of the hypersensitive mutations isolated in her screen was due to a mutation in SoxR that resulted in a "dominant negative" phenotype. Explain a specific biochemical function that could be lost in the SoxR protein that could lead to a dominant negative phenotype (as part of this question you should explain what a dominant negative mutation is).

Dominant negative mutation a mutation whose gene product has an adverse effect on the WT gene product. That is, the mutant allele is able to affect the WT allele and cause the mutant phenotype to be dominant. SoxR forms a homodimer structure that binds to DNA in its oxidized form in order to activate expression of SoxS, which in turn activates expression of sod genes. It is possible that the DNA binding ability could be abrogated in one of the subunits, preventing the mutated subunit from binding to DNA, but still interacting with the WT dimer. Since the mutation is dominant, by still interacting with the WT subunit, it can prevent the WT subunit from functioning properly. Thus the entire protein complex becomes non-functional, even though only one of the two subunits is mutated.

You are investigating the interaction of two (fictitious) proteins involved in DNA repair, MutA and MutB. These proteins must interact to allow resistance to the mutagen "Compound X." To understand the interaction between MutA and MutB you want to use reversion analysis via intergenic suppression. Assume you have the mutA and mutB genes cloned into individual plasmids, a strain that has these genes deleted in the chromosome, and a way to make random mutations in the clones. The strain is resistant to Compound X when the plasmid encoded mutA and mutB are expressed from the two plasmids. Explain how you would use reversion analysis via intergenic suppression to suggest regions of the protein that interact (Hint: part of your explanation should include information about what "allele specificity" is and why it is important.)

Firstly, one of the two genes should be mutagenized, and a single amino acid substitution mutant that abolishes resistance to compound Z should be found (in this case we will say mutA was mutagenized and named mutA1). mutB could then be mutagenized and expressed alongside mutA1, and mutants that restore compound X resistance isolated. In order to suggest protein interaction, the mutA1 allele should only be able to be compensated for by certain mutations in mutB; that is the mutB mutation needs to be allele specific where each allele is mutant but specifically two alleles will show a wild type phenotype when together. Additionally, the mutB mutant should not be functional with wild-type mutA.

Name the amino acid(s) that can be inserted first into bacterial proteins?

Formyl-Methionine, f-met

You are interested in a protein that is situated in the inner membrane of a strain of Yersinia pestis. By examining the distribution of hydrophobic and hydrophilic amino acids in the protein sequence you are able to figure out which regions are transmembrane spanning regions. However, to determine how this protein senses the environment in the periplasm you need to know which domains reside in the periplasm and not the cytoplasm. How can you use translational (protein) fusions to the phoA gene and the indicator X-phosphate to find regions of the protein that are likely to reside in the periplasmic space.

Fuse phoA to various positions within the inner membrane spanning protein, express these variants and treat cells with X-phosphate. If PhoA is positioned in the periplasm, it will dephosphorylate X-phosphate and produce a blue color. This indicates that the fusion position is located in the periplasm.

How does FtsK contribute to the ability of XerCD to regulate the resolution of dimer chromosomes temporally and spatially?

Given the orientation of the KOPS sequences in the chromosome and the ability of the FtsK protein to pump DNA only in one direction relative to these sequences the dif sites will be aligned in dimer chromosomes (chromosomes that are not dimers will simply be "pumped" into the correct cells). The dif sites are the sequences used for resolution of dimer chromosomes by the XerC and XerD proteins. The full resolution activity of XerCD depends on an interaction with FtsK. These two processes govern the activity of XerCD only to dimer chromosome and controls the time and place the reaction occurs because the FtsK pump is only assembled at the division zone and only prior to division.

What base is especially sensitive to oxidative stress, how is it changed, and why is it bad for the cell?

Guanine is especially sensitive to oxidative stress, which is altered to 8-oxoguanine, which has a double bonded "O" added to the 5-member ring. This oxidized base is especially dangerous because it can base pair well with adenine, resulting in potential G>T mutations.

When certain genes involved in DNA repair are inactivated the phenotype is a "sensitivity" to certain reagents or conditions. Briefly explain one example of how a gene from E. coli discussed in class would lead to a SENSITIVITY phenotype when inactivated.

If any of uvrABCD genes were inactivated, it would cause the cell to be more sensitive to DNA damage via UV light. The UvrABCD proteins are involved in nucleotide excision repair, which can effectively remove thymine dimers (in addition to other types of damage) that are formed as a result of photochemical reactions. Without this major type of repair, many UV-induced thymine dimers could not be detected and removed, thereby allowing an obstruction that normal DNA polyermases cannot copy and thus the cell would be more sensitive to UV-light and the corresponding DNA damage.

What features would you look for in the genome sequence if you believed that these genes were regulated though an activator or repressor?

If plsA, plsB, plsC, and plsD were regulated via an activator or repressor, you would have reason to believe that the activator/repressor binds to an operator sequence just upstream of the promoter. You would look for this same operator sequence elsewhere in the genome sequence. Generally repressor binding sites tend to be overlapping the sigma binding region and activators tend to be upstream of the sigma factor binding region.

You are working with a new pathogenic bacterium where the genome was just sequenced. You want to work with a subset of genes involved in pathogenesis that you know are stimulated in the presence of blood plasma in the growth media. You know of four monocisstonic genes that are induced by blood plasma (plsA, plsB, plsC, and plsD) and want to use the genome sequence to identify more genes.a) What features would you look for in the sequence if you believed that the plsA, plsB, plsC, and plsD genes were regulated through the use of the same sigma factor?

If plsA, plsB, plsC, and plsD were regulated via the same sigma factor, then the promoters of these four genes would have shared sequence elements a "consensus sequences".

Explain how the experiment would have to differ in "a" if the mutation was a simple null mutation verses a dominant negative mutation.

If the hypersensitivity was due to a null mutation, you could use a chromosomal library from the WT strain and transform it into the null mutant strain. Then, expose the cells to the same antibiotic and see which ones have their WT phenotype restored. These should contain the gene that was absent in the null mutant. If the hypersensitivity is due to a dominant negative mutation, make a chromosomal library from the same hypersensitive strain and transform it into a WT strain. The cells that become hypersensitive to the antibiotic contain the mutated gene that exerts a dominant effect over the WT copy of the same gene.

Briefly explain one example of how a gene from E. coli discussed in class would increase the frequency of mutation when inactivated.

If the mutS gene were to be inactivated, that would increase the frequency of mutation in the cell, leading to a "mutator" phenotype. The MutS protein is involved in mismatch recognition that normally repairs single base-pair errors. The binding of the MutS protein to a site-specific DNA mismatch causes the recruitment of various other proteins that ultimately repair the base-pair error. Therefore, if mutS is inactivated, single base-pair errors would not be repaired due to absence of MutS protein, thereby leading to higher frequency of mutation (mutations in mutL or mutH would have the same effect).

You cloned your gene of interest into the pBAD-TOPO vector to control expression. Explain at the molecular level how Glucose can repress expression of your gene and how Arabinose will induce expression of your gene.

If there are high levels of glucose, there will be an inversely low level of cylic AMP (cAMP) in the cell. Cyclic AMP normally binds to the Catabolite Activator Protein (CAP), and the cAMP-CAP complex will bind to promoters, interacting with RNAPs and ultimately enhancing transcription. Thus, in high levels of glucose, the inversely low level of cAMP prevents the cAMP-CAP complex from adequately forming and binding to the araBADpromoter. The araBAD promoter thereby remains unactivated, and genes will not be expressed. This is how glucose can help reduce leaky gene expression in the system. If there are high levels of arabinose, the arabinose binds to the AraC activator protein, and the arabinose-AraC activator binds to the DNA upstream of the araBAD promoter in the expression vector, interacting favorably with RNAP and thereby upregulating expression of the gene cloned downstream of the arabinose (araBAD) promoter that is used in this vector. This is how arabinose can induce gene expression.

What is a temperature sensitive mutation?

In E. coli (a mesophile) temperature sensitive mutations are mutant alleles that show a wild type phenotype at 30C, but a null phenotype at higher temperatures, like 42C. It is a form of conditional mutation that is especially useful for studying essential genes.

Sigma S is made constitutively only to be quickly degraded. How is degradation signaled at the molecular level and why might the cell carry out this seemingly wasteful process?

In normal conditions, as soon as Sigma S is expressed, it is bound to the RssB protein, which recruits the ClpXP protease that ultimately degrades Sigma S. This expression-then-degradation cycle occurs constantly in non-stress conditions, but it allows the cell to respond to a variety of different stresses via the same pathway. That is, there can be multiple inputs that would each independently activate the same stress response.

The DNA replication proteins were divided into two groups using temperature sensitive mutations. What were the two "classes" of temperature sensitive mutations and give an example of each.

In the study of genes involved in chromosomal replication, temperature sensitive (ts) mutations were particularly informative because they allowed the genes to be separated into ones likely involved in initiation of DNA replication and ones involved in the elongation process of DNA replication based on their ability to incorporate radiolabeled nucleotides during DNA replication when the strain was shifted to the non-permissive temperature.If the gene with the ts mutation stopped incorporating radiolabel essentially immediately (quick stop), it was likely in a gene product involved in the elongation step of DNA replication (e.g. dnaB). If the gene with the ts mutation stopped incorporating radiolabel more gradually (slow stop), it was likely in a gene product involved in the initiating step of DNA replication (e.g. dnaA) because replication events that were ongoing would continue and only new initiation events would be prevented. The gradual drop in incorporation would occur because the cells were at various points in the replication cycle and individual cells would only stop incorporating when each finished a complete chromosome.

Describe how non-coding (ncRNA) can decrease and increase the expression of a protein at the level of mRNA stability and at the level of translation (i.e. you will be describing four different things).

Increase mRNA stability: sRNA is expressed, binds to target mRNA (often with the help of Hfq), mRNA shielded from specific ribonucleases. Decrease mRNA stability: sRNA is expressed, binds to the target mRNA (often with the help of Hfq), target mRNA changes conformation in a way that it is more vulnerable to specific ribonucleases in the cell. Increased translation: sRNA is expressed, binds to the target mRNA (often with the help of Hfq), sRNA binding exposes ribosomal binding site Decreased translation: sRNA is expressed, binds to the target mRNA (often with the help of Hfq), sRNA binding occludes ribosomal binding site

One of the benefits for the synthetic genome with 57 codons is that it is argued that it will be less sensitive to mobile DNA elements like bacteriophage and plasmids. Explain why this might be the case?

Mobile elements found naturally are expected to utilize all 64 possible codons. In the strain where the six "sense" codons and the UAG stop codon were removed the tRNAs and Release Factor 1 were also likely removed. If any of these codons are present in genes essential to the mobile elements function, these genes will not be translated as there will be no machinery (tRNAs and/or aminoacyl-tRNA synthetases) necessary to recognize the removed codons. Also, UAG stop codons utilized by the mobile element will not be recognized.

Which of the two examples display an "epistatic" relationship, MutM and MutY, or MutL and MutH? Explain why.

MutL and MutH. A mutant in one gene masks the effect of the other.

Could any of the E. coli DNA polymerases fix the damage caused by a 5' exonuclease that degraded four nucleotides from the DNA molecule above? Why or Why not?

No. This is because none of the DNA polymerases in E. coli have a 3'-5' polymerase activity.

What is the expected difference between a null mutation in a repressor verses an activator?

Null mutation in a repressor means that the repressor is fully incapacitated. This would mean that whatever genes the repressor normally repressed are now always on. Conversely, a null mutation in an activator means that the activator is fully incapacitated, meaning that whatever genes the activator normally activates are now always off even in the presence of the compound that normally induces the operon.

Explain why truncated RNAs can pose a problem to cells.

Once initiated, translation will normally continue until a stop/termination codon (nonsense codon) is reached (UAG, UAA, UGA). In the normal situation, the stop codon will be recognized by one of the cell's release factors and signal the ribosome to dissociate terminating translation allowing the ribosome to work on a new protein. In the case of a truncated mRNA the ribosome will progress until the end, but then become stalled because there is no information to recruit then next charged tRNA or to signal release. Because many ribosomes will load and translate each mRNA, one truncated mRNA can trap many ribosomes. As ribosome are very energy intensive to make, this is a serious issue for the cell to lose a population of ribosomes.

Why might there be ten times more ribosomes in the cell than RNA Polymerase molecules?

One mRNA (made by a single RNA polymerase) will have many ribosomes translating it at one time.

In class we talked about two "places" in the cell where Topo IV functions. What are the places, why are they important and how is the activity of Topo IV controlled?

One of the two subunit proteins of TopoIV interacts with FtsK. Therefore, Topo IV is active for decatenation at the position where DNA is actively being transported across the septum just prior to division. This is likely important because catenated chromosomes cannot otherwise be segregated. In addition, this is unlikely to be a place where chromosomes could get re- tangled by Topo IV. TopoIV also appears to interact with MukB. This may allow it to unlink chromosomes just before they are condensed into a more compact state (at which point they may be less likely to become linked as catenaens again)

You are doing an experiment where you need to radiolabel a DNA fragment that you amplify using PCR. Which of the three phosphates must be radioactive to make the growing chain of DNA radioactive (, , or )? Why? (Drawing the nucleotide structure as is found on the inside cover of your text book will help)

Only the alpha phosphate will be integrated into the growing chain of DNA from the incoming triphosphate nucleotide. Therefore, the alpha phosphate would need to be radiolabel in the experiment.

Briefly explain a mutation in the lac operon that you could screen for specifically using Phenyl-galactoside, a compound that is a substrate for LacZ but one that does not activate the operon that you would not find using the other galactosides discussed in class.

Phenyl-galactoside (PG) is cleaved for metabolism by LacZ but does not activate the operon, meaning it does not bind to LacI repressor and cause it to unbind the operon as lactose normally would. Wild type cells normally produce very little LacZ in absence of lactose. If cells are grown on media with PG as only carbon/energy source, then the cells cannot survive on what little available LacZ can cleave PG for metabolism, but adequate levels of LacZ expression will not be further induced. However, mutants lacking the LacI repressor, or mutants that don't allow the LacI repressor to bind to the operator will cause constitutive expression of LacZ, which would allow the mutant cells to metabolize more PG and survive on PG-containing media.

How could allele-specific suppressor mutations in the trpL leader region be used to confirm the existence of a stem-loop structure that is important for regulation of the operon by attenuation?

Point mutations in one of these stem-loop structures could preclude the ability of the stem-loop to form, inactivating the ability of tryptophan to regulate gene expression via attenuation. Presumably one could isolate or introduce a complementary mutation that would re-establish the stem-loop structure to allow allele-specific suppression of the original mutation that would restore attenuation, thus confirming that the stem-loop structure is important for regulation.

DNA Pol III is highly accurate due to two mechanisms; one is "Pre-synthetic" and the other "Post-synthetic. What do these terms mean?

Pre-synthetic (Before DNA synthesis): Check of incoming nucleotide complementarily before the chemistry takes place to extend the DNA polymerPost-synthetic (After DNA synthesis): Proofreading of base complementarily with the template strand activity after synthesis

The tree of life has expanded greatly for prokaryotes over the last few years. Which domains of life are included in the term "prokaryote"? Many of the prokaryotic representatives cannot currently be cultured. How were they discovered if they cannot be grown in the laboratory and how do we know where they fit in the tree of life?

Prokaryote is a word with limited utility, it refers to bacteria and archaea. Directly sequencing DNA from a given environment allows investigators to discover new organisms that do not match previously analyzed organisms based on genome sequences. Detailed phylogenic trees can be produced from 16S RNA sequences and DNA sequences from other slowly evolving conserved genes to identify whole new branches of organisms.

In class it was mentioned that codon bias is related to the A+T vs G+C ratio of the strain of bacteria. For Proline and Alanine what are the codons that are more likely to be used in a high G+C organism than a low G+C organism?

Proline - CCC and CCG, Alanine - GCC and GCG

Why might cells that are mutant for DNA Pol I absolutely require the normally non-essential RNaseH protein?

RNaseH can remove the RNA primer found in Okazaki fragments. If Pol I is defective and unable to remove RNA primers, RNase H can play a backup role to allow viable (but sick) cells.

What is (are) the functions of DNA Pol I in E. coli?

Removes RNA primers of Okazaki fragments, while making the complementary DNA strand (also involved in DNA repair events)

Give the overlapping and distinct biochemical activities of DNA polymerase I and DNA polymerase III.

SAME: 5' to 3' polymerase activity 3' to 5' exonuclease for error correction DIFFERENT: Pol1- Not essential for elongation (although processing of the lagging-strand is badly impacted); 3' to 5' exonuclease for primer removal Pol3- Essential for elongation of both strands

You are comparing the relationship between MutH and MutL, two genes involved in DNA repair. You determine how well the wild type (wt) and the single and double mutants survive exposure to increasing concentrations of mutagen.Do the results suggest that MutM and MutY are in the same or different pathways? Effects are not additive.

Same pathway

Briefly explain how SoxR or OxyR is activated by reactive oxygen species to express genes that can combat oxidative stress.

SoxR has a domain that acts as a redox sensor, activating the protein when it is oxidized by ROS. Activated SoxR causes increased transcription of soxS, which upregulates genes that can respond to oxidative stress. OxyR can be oxidized at cysteine residues by hydrogen peroxide, resulting in a disulfide bond between these residues. This bond activates OxyR as a transcriptional regulator, upregulating oxidative stress response genes.

What four nucleotides would be degraded first by a 3' exonuclease from the linear DNA molecule shown below? 5' - A - T - C - G - T - T - C - G - G - A - T - C - G - A - T - 3' 3' - T - A - G - C - A - A - G - C - C - T - A - G - C - T - A - 5'

TAGC

How does FtsK "know" which way to pump DNA?

The FtsK protein can "read" KOPS sequences in the chromosome and pump DNA in only one direction relative to this information. The KOPS sequences are polar, and are situated in one direction from oriC to dif.

When lactose is present in the media, lacZYA expression is upregulated. Explain how the LacI repressor can "upregulate" expression.

The LacI repressor can "upregulate" expression when the regulator unbinds the DNA in the presence of lactose. Under normal conditions where there is no lactose around, the LacI repressor is bound to the lac operon, thereby repressing expression of operonic genes involved in lactose metabolism. However, in conditions where the cell has access to lactose, lactose can bind to the LacI repressor, causing it to unbind the DNA and thereby allowing RNAP to successfully transcribe the lac operon genes.

You have discovered a new highly effective antibiotic. For approval of the drug you have to understand how it works at the molecular level. You determine that E. coli stains that are recA are hypersensitive to the drug (Strains that are recA minus are much more sensitive to the drug than wild type cells). You suspect that your drug damages DNA, but you want to understand the nature of the DNA damage.a) What null mutants could you test to determine if the drug was likely affecting the progression of DNA replication forks to make gaps in the DNA? Explain your experiment, the specific genes you could test, and the normal function of the genes.

The RecFOR pathway is responsible for repairing single-stranded gaps in the DNA. The experiment would involve knocking out the genes involved, which are recF, recO, recR, recQ, and recJ. Deleting these genes and then testing for increased drug sensitivity would tell you if this pathway is involved in repair of DNA damage caused by the drug.Normally, RecQ and RecJ chew back the DNA at the gap, allowing for SSB to bind. The RecFOR complex then binds, displacing SSB and allowing for RecA to bind and initiate recombination.

What null mutants could you test to determine if the drug was likely making flush (i.e. no overhangs) DNA double strand breaks as its mode of action? Explain your experiment, the genes you would test, and the function of the genes.

The experiment would be the same, but the RecBCD recombination pathway is the pathway that would need to be tested. Knock out these genes and then test for drug sensitivity. RecBCD normally functions by binding to the end of linear dsDNA, and chewing away the DNA until a chi site is reached. When the chi site is reached, a 3' overhang is created that is loaded with RecA.

The process of DNA replication is tightly regulated in bacteria like E. coli. Give two mechanisms that prevent re-initiation of DNA replication in this organism.

The origin must be fully methylated at the "A's" in the GATC/CTAG sites in this region to be function. SeqA helps control this process. The fully methylated GATC sites are found normally where the A's on each strand are methylated. The GATC/CTAG sites are hemimethylated right after replication because the new "A" enters the DNA sequence in the unmethylated form and it takes a bit of time for the Dam enzyme to identify the site and methylate the A. SeqA binds to the hemimethylated GATC sites in the origin slowing methylation and thereby delaying a new initiation event for a period of time (~20 minutes).After replication initiation, Hda interacts with the beta clamp and stimulates the ATPase activity of DnaA, coaxing it to the inactive ADP-bound form. DnaA must be in the ATP bound form to initiate DNA replication. Therefore, initiation of a new round of replication will require the DnaA-ADP protein to exchange nucleotide to the DnaA-ATP bound form or sufficient new DnaA must be made to provide enough for initiation (given the high concentration of ATP in cells, new DnaA will essentially always be loaded with ATP.

What is another situation found with transcription/translation that might require tmRNA other than a truncated mRNA?

The tmRNA system has also been shown to be important for freeing ribosomes from mRNAs that encode numerous rare codons that can stall ribosomes for a protracted period of time. Presumably this could play an important role during horizontal transfer of genetic elements from other bacteria with drastically different codon biases than the host. Otherwise a horizontally acquired element could put a serious burden on the ribosome pool in a cell.

Inactivation of the umuD and/or umuC gene leads to an ANTIMUTAGENIC phenotype. Briefly explain why this is the case.

The umuD and umuC genes are activated by the SOS response pathway and can be considered a last resort mechanism for DNA damage that leads to ssDNA. The umuDC gene products form a very error prone polymerase (Pol V mutosome), but this polymerase is only made fully functional after the cell experiences high levels of DNA damage. The umuDC gene are induced late in the SOS induction process, and there are considerable modifications that need to occur to the complex to make it functional as the Pol V mutosome. The PolV mutasome can replicate over highly damaged DNA that would otherwise halt replication with another DNA polymerases. Therefore, evolution seems to tolerate the loss of fidelity because the cell is able to survive. Inactivation of either of the umuDC genes will lead to more cell death, but the overall mutation frequency will be lower in its absence.

Allele specific suppression

This describes a behavior between two mutant alleles that could suggest that amino acids at these positions interact.

There are six different codons that are used for Arginine in the universal genetic code. However, E. coli is said to "prefer" to use CGC and very rarely uses AGG or AGA. Explain what is meant by this and how it can affect the expression or heterologous proteins in E. coli?

This means that E. coli genes have more CGC codons than AGG or AGA and will produce more CGC-recognizing tRNA than AGG or AGA-recognizing tRNAs. If you are using E. coli to express heterologous proteins that have AGG and/or AGA codons, the cells may have trouble producing large amounts of these proteins due to a lack of sufficient quantities of the appropriate tRNAs (or charged tRNAs).

During your experiments, you isolate a null allele of mutA that cannot be complemented by a wild type copy of mutA. What kind of mutation is this (as discussed in class) and why is it useful for molecular genetics?

This type of mutation is called a dominant negative mutation. It implies that the mutant allele somehow interacts with the wild-type allele, in the case of homodimer forming proteins, or somehow interferes with a substrate or another interacting protein (e.g. the mutant binds a substrate but does not cause a reaction, sequestering it). These types of mutations can be useful in determining the function of unknown genes because the protein is made and generally folded correctly, but lacking a specific function.

Could you have used a translational (protein) fusions to the lacZ gene and the indicator X-Gal to find regions of the protein that are likely to reside in the periplasmic space? Explain.

Translational fusions to lacZ would not directly report which regions reside in the periplasm and would instead tell you regions that remain in the cytoplasm where LacZ is active. This information could then be used to infer domains that reside in the periplasm. Ideally both phoA and lacZ fusions would be used to confirm cytoplasmic and periplasmic domains.

How would you write the protein products encoded by these genes?

WhtA and WhtB (no italics)

Most bacterial maintain a single chromosome, but in about 10% of the cases bacteria are found with more than one chromosome. Pick four of the enzymes discussed in class from the following list and explain how they would need to behave differently if the cell maintained multiple distinct chromosomes; XerCD, Topo IV, FtsK, Tus, DnaA and SMC/MukB.

XerCD - XerCD resolves dimer chromosomes at the dif site in circular chromosomes. Each circular chromosome would need a mechanism to resolve dimer chromosomes. Presumably a mechanism would be needed to prevent recombination between the two circles (like having different systems). TopoIV - TopoIV would be required to decatenate any individual circular chromosomes. Including links between the different circular chromosomes and sister circular chromosomes. FtsK - FtsK would likely need to form different complexes for each set of crhomosomes. Presumably the same KOPS would need to be used between the chromosomes to prevent the "wrong" FtsK-like protein from pumping a chromosome that it did not recognize by its polarity sequences. Tus - Active termination systems are not essential in all bacterial species, so each circular chromosome could have a system or not. Presumably there would be little chance that they would recognize random sites for termination on the "other" chromosome given the length of the recognition sites (ter sites). DnaA - All bacteria are known to have a DnaA-like protein. However, there appears to always be one "main" chromosome that is controlled for initiation using DnaA, while the one or more other chromosomes coordinate initiation based on signals from this main chromosome. Replication must be coordinated in some way to prevent loss of the accessory chromosomes. SMC/MukB - SMC/MukB is expected to be involved in compaction in all bacteria. Presumably there must be a mechanism to allow the multiple DNA elements to segregate independently, this may be because only local sides are condensed with the SMC/MukB enzymes.

Most bacterial chromosomes are circular, but there are many examples of bacteria where the chromosome is linear. Pick four of the enzymes discussed in class from the following list and explain why you think they would or would not be needed if a bacterial chromosome was linear; XerCD, TopoIV, FtsK, Tus, DnaA and SMC/MukB.

XerCD - XerCD resolves dimer chromosomes at the dif site. Would not be necessary with a linear chromosome as dimer chromosomes will not form. TopoIV - Theoretically, TopoIV would not be required as catenenes cannot form with linear chromosomes. However, an E. coli strain engineered to maintain a linear chromosome still required TopoIV for growth and bacteria with linear chromosomes have TopoIV-encoding genes in their genomes. An apt metaphor is attempting to untangle two very long cords - a TopoIV-like function would be extremely helpful in separating these two. FtsK - FtsK would likely still be useful in order to properly distribute the two linear chromosomes to daughter cells at the division septum (If there is an example of a bacterium with linear chromosomes then a SpoIIIE protein would likely be essential here too). The accessory role FtsK plays in allowing XerCD recombination to complete, would not be needed. Tus - In standard DNA replication started at the origin, Tus would not be needed as replication will need to proceed to the linear end. DnaA - Replication initiation must still occur so DnaA is still needed. SMC/MukB - SMC/MukB would be just as important with a linear chromosome as a circular.

Your friend is investigating a new antibacterial drug that causes oxidative damage. To learn more about how it functions she isolated a series of mutants in E. coli that are hypersensitive to this drug (Dead at much lower concentrations of the drug than the wild type strain). a) Explain how you could use cloning to directly select for genes that complement the mutant phenotype using a plasmid vector?

You could make a library of randomly generated DNA fragments from the wild type strain and look for clones that now allow robust growth in the presence of the drug that causes oxidative damage. This would find "recessive" mutations (these are more likely), but the library could also be made using the mutant strains and the fragments transformed into the wildtype strain and sensitivity to the drug tested.

How could you genetically identify genes that were similarly regulated if you could randomly introduce gene fusions into the strain

You could randomly fuse a reporter gene (e.g. lacZ) across the entire genome in a population of the organism (i.e. make a lacZ fusion "library"). You would then screen for isolates where the lacZ allele was induced only in the presence of blood plasma. Sequencing these fusions would suggests candidate genes that may be upregulated by the presence of blood plasma.

Using the information from "a" and "b" above, how might you look for more genes induced by blood plasma simply from the DNA sequence?

You would look for these same consensus sequences in the promoter region of other open reading frames elsewhere in the genome sequence for a hint of other genes that could be expressed by the same sigma factor.

For each of the "mut" genes in the genotype (D, S, T) explain what type of mutations you might expect from a strain background with this allele only.

mutD5 is mutation of the gene encoding the proof-reading subunit of DNA polymerase III. This mutation will result in increase of misincorporation in DNA replication. You would expect mostly missense mutations and sometimes nonsense mutations. MutS is an essential part of the mismatch repair complex. Mutation in the mutS gene will leave the errors that occurred during DNA replication. MutT is involved in "cleaning up" the pool of nucleoside triphosphates, inactivating 8-oxoguanine (GO) by converting it to a monophosphate. This mutation will increase the likelihood that GO will be integrated during DNA replication. If the GO is not repaired it will lead to a C-G to T-A transition mutation.

In class, we talked about three gene products that specifically helped prevent damage to one of the four nucleotides. What are the three genes involved in the process and how do they function?

mutT - hydrolyzes oxo-dGTP to oxo-dGMP, so the oxidized triphosphate base can no longer be inserted into DNA. mutM - removes 8-oxoguanine from DNA, leaving an abasic site. mutY - if an A is paired with the 8-oxoguanine, MutY can remove the A.