404 Chapter Questions 10, 11, and 12

Question 12. Compare and contrast the cut-and-paste mechanism of transposition with the replicative mechanism of transposition.

Replicative mode of transposition the transposon gets duplicated because part of the original is broken off and goes somewhere else, and they both become full transposons again. Cut and paste mode of transposition the sequence of the transposon is fully excised and integrated into a different genomic location.

Question 2. Explain why serine and tyrosine recombinases do not require an external source of energy such as ATP hydrolysis for catalysis.

Serine and tyrosine recombinases don't require an external source of energy because they utilize the energy of the broken phosphodiester bond by creating a broken DNA intermediate.

Question 10. Explain why scientists use transposons like Tn5 as a genetic engineering tool to screen a population of bacterial or yeast cells for mutants for a given phenotype. Why does the presence of the transposon in the mutant provide an experimental advantage in a genetic screen, compared to mutations generated by chemical mutagenesis?

Transposons like Tn5 are used to screen a population of cells for mutants of a phenotype because the random insertion of Tn5 can cause a loss of function phenotype. Insertional mutagenesis is advantageous to chemical mutagenesis because it's easier to locate the position of the mutation.

Question 7. Explain why RecA-dependent strand exchange cannot occur between two homologous, double-stranded, covalently closed, circular DNAs (cccDNAs) but can occur between the two double-standed DNA molecules pictured in Figure 11-8c

Two cccDNAs have complementary sequences, but there isn't any ssDNA region that RecA can assemble on. There also aren't any ends that would allow strand exchange.

Calculate the number of mismatches that could occur in one human cell during one round of replication in the absence of mismatch repair.

(3.2 billion bp)/ 10^7 = 320

Calculate the number of mismatches that could occur in one human cell during one round of replication. Assume the size of the human genome is 3.2 billion base pairs.

(3.2 billion bp)/10^10 = .32

Question 2. You are considering two alleles of one specific gene. Describe what feature with respect to the DNA distinguishes one allele from the other. Are the two alleles homologous?

Alleles differ from each other by minor sequence variation. The minor variations in sequence are recognizable by DNA, and it can tell the alleles apart by this. The alleles would be considered homologous since such a large portion of the DNA sequences would be the same.

Question 1. Explain two ways in which DNA replication can introduce doubled-strand breaks (DSBs) in a DNA template.

An unrepaired nick in a DNA strand can lead to collapse of a passing replication fork A lesion in a strand of DNA that makes the strand unable to be used as a template will stop at the replication fork.

DNA polymerase mistakenly inserts a C across from a T during replication. Assuming that proofreading and mismatch repair do not correct the mismatch, is the resulting mutation a transition or transversion after the next round of replication? Explain your choice.

It inserted a C instead of an A. The resulting mutation is a transversion since it is purine to pyrimidine.

The following terms describe the general steps of a DNA repair pathway. Place the steps in the correct order. Name the protein(s) that complete each of the steps in E. coli for the mismatch repair, base excision repair, and nucleotide excision repair pathways. Ligation, DNA synthesis, Recognition, Excision

Recognition, excision, DNA synthesis, ligation The issue is recognized by MutS, which recruits MutL, who recruits MutH. MutH puts an incision in the strand near the mismatch. UVRD unwinds the DNA toward the mismatch. Excision occurs by either RecJ (5' end) or Exonuclease I (3' end). DNA synthesis occurs by DNA pol III, litigation occurs by DNA ligase

Question 12. Compare and contrast synthesis-dependent strand annealing (SDSA) used in mating-type switching with DSBrepair homologous recombination.

SDSA and DSB repair are similar because they both begin with a DSB at the recombination site.

Question 9. Explain the most significant role of homologous recombination in eukaryotes that is not found in prokaryotes.

The role of homologous recombination in eukaryotes that doesn't exist in prokaryotes are crossing over events during meiosis.

Question 5. Describe advantages of using Cre recombinase for genetic engineering in eukaryotic cells.

There is a genetic advantage to using Cre recombinases because they can be used for any cell type, they are site-specific, and can be used to create insertions, deletions, inversions, and translocations.

Given the structure of the damaged base below, circle the modification(s) present relative to the base normally found in DNA. Name the process that produces this type of modification. Name the DNA repair pathway that you expect would recognize and correct this type of DNA damage.

This is a photo of a thymine that got turned into a thymine glycol. This is the result of ROS coming in contact with the thymine and would need base excision repair to fix it.

Given a loss-of-function mutant for dam (the gene encoding the Dam methylase) in E. coli, predict the phenotype one would observe with respect to spontaneous mutagenesis. Briefly explain your answer

Without Dam methylase, the parent strand wouldn't be methylated during replication, meaning MutH can't determine what strand is the parent strand and which is the daughter strand. This would lead to an increase in spontaneous mutagenesis because MutH can't tell which strand to nick to replace the wrongly inserted bp.

14B You are initially surprised to see revertants in the absence of any chemical that you are testing, but you realize that this is normal. Give a specific example of how a revertant can arise in the absence of an added mutagen

Free radicals naturally present can cause mutations, also mutations can be from processes in cells like replication, or you can have hydrolytic attack of the bases. All of these can alter DNA.

Question 11. Define gene conversion and give an example of a mechanism explaining how gene conversion occurs

Gene conversion is when an allele of a gene is lost and replaced by another allele. This can occur during mating-type switching or by DNA repair during recombination.

Aside from DNA damage tolerance, name the repair pathway that potentially introduces mutations. Describe how this pathway introduces mutations.

The repair pathway that potentially introduces mutations is translesion DNA synthesis. This is because it allows DNA pol to synthesize DNA without reading the template strand. The nucleotides incorporated by this are independent of base pairing, meaning there are more opportunities for it to potentially introduce incorrect nucleotides into the strand.

Question 3. Following formation of a DSB, enzymes process the double-stranded DNA to form single-stranded DNA as shown in Figure 11-4b. Does it matter if resection occurs in the 50 to 30 direction or the 30 to 50 direction? Explain why or why not?

Yes, it matters. The resection must occur in the 5' to 3' direction since invading strands end with 3' termini. The 3' termini must serve as primers for new DNA synthesis.

15 Setup You perform a follow-up experiment to that discussed in the Questions for Chapter 9. Here is a review of the setup for the experiment. You have just discovered two new eukaryotic DNA polymerases and want to learn more about their properties. To begin, you obtain purified protein of each DNA polymerase and perform polymerase processivity assays. You use a short primer that is 32P-labeled at the 50 end and binds to a circular single-stranded (ssDNA) DNA template (shown below). You complete the following steps to obtain the processivity assay results for DNA polymerase #1 (Pol #1) and DNA polymerase #2 (Pol #2) 1. In the appropriate buffer conditions, you preincubate the primed, circular ssDNA with Pol #1 or Pol #2 for 5 min at 378C. 2. You add dNTPs and a large excess of ssDNA to initiate the reaction. 3. You allow the reaction to proceed for 10 min, quench the reactions with the addition of SDS, and run the samples on a polyacrylamide gel (which resolves single nucleotides) under denaturing conditions.

You perform a follow-up experiment, and the results are shown below. In this experiment, you separately incubate each DNA polymerase with the accessory proteins (PCNA, RFC, and RPA) and a DNA template that now includes an intrastrand crosslink of two guanines (induced by cisplatin). The cross-link is located immediately adjacent to the 30 end of the primer as indicated by a star (pictured below to the right).

Question 5. What is meant by the term heteroduplex DNA?

DNA duplex resulting after strand invasion occurs, and it usually contains some mismatched base pairs. Mismatched sequences form heteroduplex DNA.

Explain the difference between DNA repair and DNA damage tolerance.

DNA repair recognizes and corrects incorrect bases. DNA damage tolerance bypasses modified bases so the presence of modified bases may not stop replication.

Question 7. Describe the biological relevance to the Hin recombinase catalyzing DNA inversion in the Salmonella typhimurium genome.

Hin recombinase inverts a DNA fragment in the salmonella genome to alternate the expression of flagellar genes. This is advantageous for the bacteria to evade detection by the host.

14A Given these data, propose a function for Protein X.

Protein X cleaves the DNA substrate.

14A What medium must be used in the selective plate as part of the Ames test? Explain how a mutation gives rise to a revertant in this experiment. Be specific

The medium must not have histidine in it. The mutagen can cause a point mutation at the original location of the point mutation in the His gene which would allow the cells to grow in the absence of histidine.

Question 10.Briefly describe how cells generate the DSB required to initiate meiotic homologous recombination in eukaryotes.

Cells generate the double strand break required to initiate meiotic homologous recombination by mediation by Spo11. A tyrosine in Spo11 attacks the phosphodiester backbone to cut the DNA. Spo11 stores the energy from breaking this bond by forming a high energy intermediate with the broken DNA.

Question 11. Researchers have found that treating human tumor cell lines with an inhibitor of reverse transcriptase reduces the rate of tumor-cell proliferation. Predict why reverse transcriptase is expressed in human cells. Hypothesize a general reason why reverse transcriptase activity could be associated with tumor cells.

Inhibiting reverse transcriptase stops retrotransposition events and further genomic instability. Reverse transcriptase is used in humans to mediate the transposition of retrotransposons. Reverse transcriptase further activates the transposition process making the host genome really unstable.

14C Which chemical(s) would you identify as containing a mutagen? Explain your reasoning.

Chemical A because it has a lot of revertants, 100% survival and more CFUs than the control, meaning it has a higher frequency of mutations.

14D Which chemical(s) would you identify as possibly antimutagenic? Explain your reasoning.

Chemical C because there are like no revertants and a lot of the CFUs died.

14B Based on your answer in part A, this protein functions similarly to what E. coli protein?

Protein X performs the same function as RuvC in E. coli.

Question 8. Explain the major feature in the cycle of recombination that distinguishes DNA transposons from retrotransposons

A major difference in the cycle of recombination for transposons and retrotransposons is that retrotransposons are formed by a RNA intermediate during transposition. DNA transposons remain DNA throughout the transposition process.

Consider a loss-of-function mutant in the nucleotide excision repair and translesion synthesis pathway. Predict the level of DNA damage, percent survival, and level of mutagenesis relative to wild type for each mutant after exposure to UV light. In the table below, fill each blank with increase, decrease, or stay the same.

DNA Damage For both NER and Translesion Synthesis, DNA damage increases when a loss of function mutation increases. Percent Survival NER: Decreases when a loss of function mutation increases Translesion Synthesis: Decreases when a loss of function mutation increases Mutagenesis NER: Increases when loss of function mutation increases Translesion Synthesis: Stays about the same when a loss of function mutation increases.

Question 13. Explain why DSB-repair homologous recombination can occur between any two DNA molecules that share homology rather than only between two DNA molecules that carry a specific sequence.

DSB repair homologous recombination can occur between any two DNA molecules that share homology because DSBs can occur anywhere in the genome. The proteins used in homologous recombination are not sequence specific, they don't have to both carry a specific sequence.

Explain why the deamination of 5-methylcytosine leads to hot spots for spontaneous mutations more than the deamination of cytosine in DNA does?

Deaminated 5'methylcytosine results in the formation of thymine, and thymine won't be recognized by DNA pol as the incorrect base since it is a natural base. The deamination of cytosine results in the formation of uracil, which is not a natural base in DNA so that is easily recognized and fixed by DNA pol.

14 Setup There are many claims that certain chemicals that you encounter in daily life are mutagenic. You are interested in learning if chemicals that you commonly use are mutagenic. To do so, you choose to use the Ames test to test for reversion of a point mutation in the HisG gene in Salmonella typhimurium. You added a chemical into the growth medium for the bacteria. Assume you plated an equal number of cells for each mutagenesis plate. You also calculated percent survival for chemically treated cells relative to untreated cells. Remember that the plate media for survival is not selective. A summary of the results is shown below.

No chemical added: 100% survival, 28 His+ colonies per selective plate Chemical A: 50%, 1400 Chemical B: 70%, 20 Chemical C: 100%, 7

Question 6. List the different enzymatic activities that RecBCD catalyzes anddescribe the significance of each activity in the steps of homologous recombination (via the DSB-repair pathway).

RecB does 3' to 5' DNA helicase and nuclease functions RecD does 5' to 3' helicase function RecBCD produces ssDNA required for strand invasion RecC promotes activities of RecB and RecD

Question 4. List the similarities and differences between the mechanisms of tyrosine and serine recombinases during conservative site-specific recombination.

Serine recombinases cleave 2 strands of DNA at a time while tyrosine recombinases cleave all 4 at the same time. Serine recombinases turn the DNA complex 180 degrees, but tyrosine recombinases do not. Both recombinases form a covalent-recombinase-DNA intermediate. They both catalyze reversible reactions, and they both don't require ATP.

Question 3. Explain a major difference between site-specific recombination and transposition.

Site-specific recombination is mediated by site-specific recombinase, and DNA strands are also broken and rejoined at a determined location. For transposition, it is mediated by transposase. For transposition, genetic elements move randomly between different genomic locations.

Question 8. Using the Holliday junction DNA substrate pictured below (50 - 32P end labeled), propose an assay that researchers may have used to determine RuvA protein binding to the DNA substrate. Propose a modification to the DNA substrate to use as a negative control that demonstrates an aspect of the specificity of RuvA binding. (A, B, C, and D are labels for the unique ssDNA fragments that were used to assemble this substrate.)

The DNA can be analyzed by an EMSA. We can incubate the RuvA protein with the DNA substrate and run a nondenaturing gel. We will get two bands on the gel. The faster band will be the DNA substrate and the slower is the DNA bound to RuvA.

Question 1. Given a linear piece of double-stranded DNA that includes two separate crossover regions surrounded by recombinase recognition sites, describe the alignment of the recombination sites that determines if the recombination outcome is a deletion or inversion. Explain why this arrangement dictates the outcome of the reaction.

The alignment of the recombination sites that determines if the recombination outcome is a deletion or an inversion is the alignment of the CSSR complex and the alignment of the DNA.

Exposure of DNA to the chemotherapy drug, cisplatin, causes the formation of intrastrand cross-links between two adjacent guanines in DNA. Explain why the instrastrand crosslink between two adjacent guanines is a better candidate for nucleotide excision repair rather than for base excision repair

The guanine instrastrand crosslink causes distortion in the DNA helix like thymine dimers. NER proteins can recognize this distortion and excise the stretch of DNA including the crosslink. BER can only excise one nucleotide.

Question 9. Provide an explanation for how the human genome can contain greater than 50% transposon-related sequence but does not experience major genetic instability as a consequence of transposon movement.

The human genome doesn't experience major genetic instability from transposon movement because most of the transposable elements are inactive. Some are silenced by DNA methylation or encode siRNAs that inhibit their own transposition.

Question 6. Infection of E. coli with bacteriophage lambda involves integrative recombination for the phage to enter the lysogenic state and excisive recombination for it to enter lytic growth. Does (lambda)Int have a role in integrative recombination, excisive recombination, or both? Explain your reasoning.

The integrase lambda is important for both excisive recombination and integrative recombination.

7. Describe a possible advantage and disadvantage of repairing 3-methyladenine through base excision repair relative to repairing O6 -methylguanine through direct reversal by a methyltransferase.

When base excision repair is done, proteins and enzymes are used, which can be used again. With direct reversal, the methyltransferase enzyme is used which cannot be used again. A disadvantage of base excision repair is that we need multiple different proteins and enzymes to fix it, which means we are counting on multiple different variables to show up and do their job correctly. There's more room for error.

Predict the immediate consequences to a cell in which the system of transcription-coupled nucleotide excision repair stopped functioning properly.

Without NER a cell wouldn't be able to repair pyrimidine dimers before they become mutations. This means there would be many more mutations in the cell.