MBS 344 Exam 2 Study Guide PDF

MBS 344 Exam 2 Study Guide The questions on this study guide are identical to the questions in the study guide that is posted on Canvas. The purpose of having a study guide in UT Box is to allow the class to collaborate and answer all the questions in the study guide so everyone has access to the an...

MBS 344 Exam 2 Study Guide The questions on this study guide are identical to the questions in the study guide that is posted on Canvas. The purpose of having a study guide in UT Box is to allow the class to collaborate and answer all the questions in the study guide so everyone has access to the answers. The instructor and TAs will review the answers on this study guide to make sure they are correct and complete. Before answering any questions on this study guide please review the general guidelines below. General Guidelines for Exam Study Guide 1 on Microsoft Teams This study guide is a new learning material for this course so there will likely be revisions to how this is used throughout the semester. Please feel free to provide constructive criticism to the instructor or TAs for how it can be improved. Answer any question or add on to any answer below. For now, there is no limit on the number of questions one student can answer. Only provide answers for questions when you have some confidence in your answer. All student answers should be in a red font. This will allow the instructor and TAs to review student answers and provide feedback. All comments from the instructor and TAs will be in a blue font. We will indicate if the answer is correct, incorrect, or incomplete and provide additional feedback when appropriate. The goal for this study guide is to have all questions answered and verified before the upcoming quiz or exam. There are over 100 students in this class so if everyone works together this can be a good resource to prepare for the upcoming quiz and exam. To edit this document, click on the “Open” prompt to the left of “Download” and select “Microsoft Word Online”. This will open a new window that will allow you edit the document. Lecture 7: DNA Replication Describe the process of semiconservative DNA replication using the terms template strand and daughter strand. Complimentary strands of DNA are pulled apart and used as template strands for the synthesis of the daughter strands, new strands of DNA. Nucleotides are added to the bases on the template strand using the base paring rule to create the daughter strands. Correct Describe the functions of the following enzymes during DNA replication: DNA helicase, topoisomerase, primase, DNA polymerase. DNA helicase breaks the hydrogen bonds between DNA strands to initiate DNA replication. Topoisomerase removes supercoiling generated by helicase. Primase synthesizes RNA primers to give DNA polymerase a 3’ OH. DNA polymerases synthesize the new DNA strand in the 5’ to 3’ direction by reading the template strand in the 3’ to 5’ direction. DNA polymerase needs a primer present. Correct Describe the synthesis of the leading strand and lagging strand during DNA replication. How is the synthesis between the two strands similar? How is it different? You should be able to draw leading and lagging strands if given a replication fork. The leading and lagging strand are present in the replication fork that are synthesized in opposite directions. For both strands, DNA polymerase adds nucleotides in the 5’ to 3’ direction. The leading strand of DNA can be synthesized in the same direction as the replication fork and is continuous. The lagging strand is synthesized in the opposite direction as the replication fork. The lagging is discontinuous since the strand is synthesized in Okazaki fragments. Correct What is an origin of replication? An origin of replication is a specific location on a chromosome where DNA replication initiates. Correct Describe the prokaryotic chromosome. How many origins of replication are observed on one prokaryotic chromosome? Prokaryotic cells have one circular chromosome in their cytoplasm. Prokaryotic chromosomes have a single circular origin of replication. Correct What is bidirectional replication? When replication is initiated at an origin of replication, how many leading and lagging strands are observed at each replication fork? Bidirectional replication is when DNA is replicated in both directions at the origin of replication. There are two replication forks, and each fork has a leading strand and lagging strand (two leading and two lagging strands at each origin of replication) Correct How many different types of DNA polymerases are used by prokaryotic cells? There are 5 different DNA polymerases used by prokaryotic cells for DNA replication. Correct Describe the function of DNA polymerase III in prokaryotic cells. How is it different from DNA polymerase I? DNA polymerase III assists in chromosomal replication by synthesizing both leading and lagging strands in 5’-3’ direction and proofreading mechanism (also 3’-5’ exonuclease activity only). On the other hand, DNA polymerase I removes RNA primers and replaces them with DNA nucleotides. Correct What is a nuclease? Describe the difference between an endonuclease and an exonuclease. A DNA nuclease is a type of enzyme that degrades DNA by removing the phosphodiester linkages. Endonuclease is a nuclease that degrades DNA at an internal location, and exonuclease degrades DNA from a 3’ or 5’ end. Correct Describe 3’ to 5’ proofreading. How does it occur? What is the purpose of 3’ to 5’ proofreading? DNA polymerase III adds the wrong nucleotide once every 100,000 bases. In order to reduce this error rate and fix the mistakes it uses a 3’-5’ proofreading mechanism. While synthesizing the daughter strand in the 5’ to 3’ direction, if it comes across a mismatched base pair (between template and daughter strands) then the polymerase stops. It then moves the 3’ end of the daughter strand into a pocket called the exonuclease site. There the daughter strand is digested in the 3’-5’ end until the mismatched base is removed. Afterward, it adds the correct nucleotides in the 5’-3’ direction. Correct Describe the function of DNA polymerase I in prokaryotic cells. What is nick translation? DNA polymerase 1 removes RNA primers and replaces them with DNA nucleotides using a process called nick translation. Nick translation is the concurrent 5’-3’ excision of RNA nucleotides and DNA polymerization. Correct What is the function of DNA ligase in DNA replication? DNA ligase uses ATP to remove the “nick” in the DNA backbone by joining adjacent DNA nucleotides in a strand together. Correct How many different types of DNA polymerases are used by eukaryotic cells? Eukaryotic cells have around 15 DNA polymerases. Correct Describe the functions of DNA pol e and DNA pol d in eukaryotic DNA replication. How are they similar to DNA polymerase III in prokaryotic cells? How are they different? DNA poly e synthesizes the leading strand and DNA poly d synthesizes the lagging strand. Both of these types of DNA polymerase are only in eukaryotic cells and use only 3’-5’ exonuclease activity (no nick translation mechanism). DNA polymerase III does both functions (creating leading and lagging strands at the same time), only in prokaryotic cells, and also has 3’-5’ exonuclease activity. Correct How many origins of replication are observed on one eukaryotic chromosome? Do eukaryotic cells use bidirectional replication? Eukaryotic chromosomes have thousands of origins of replication and also are bidirectional. Correct Describe the initiation of DNA replication in eukaryotic cells using the following terms: kinases, prereplication complex (preRC), S phase, replication complex (RC), G1 phase, helicase, DNA polymerases. During the G1 phase, the preRC begins to form. This process involves ORC recognizing and binding the origin of replication followed by Cdt1 and Cdc6 recruiting Mcm2-7 (inactive helicase) to the origin of replication. When S phase begins, the kinases CDK and DDK phosphorylated Cdc6 and ORC to activate the preRC, activate helicase, and form the replication complex. Helicase then begins the process of strand separation so that DNA polymerases can synthesize the new strands. What is the function of DNA pol a? DNA pol a is the eukaryotic primase that creates RNA primers on both the leading and lagging strands. Correct What is polymerase switching? Does polymerase switching occur on leading strands, lagging strands, both, or neither? Does polymerase switching occur in prokaryotic and eukaryotic cells? Polymerase switching is the transfer of the DNA template between different DNA polymerases. Specifically in eukaryotic cells for the synthesis of the lagging strand between DNA pol a and DNA pol d. Correct What is the function of the flap endonuclease (Fen1) in eukaryotic cells? Describe the association between the flap endonuclease and DNA pol d. Flap endonuclease (Fen1) removes RNA primers on the lagging strand. When DNA pol d elongates Ozaki fragments, a flap of RNA nucleotides is created. Fen1 recognizes these flaps and removes them. Next, DNA pol d replaces the RNA with DNA nucleotides. Correct What are telomeres? Where are they located? What is their purpose? Telomeres are repeat sequences that are at the end of linear chromosomes and help stabilize the ends from unraveling. Correct Are telomeres observed in eukaryotic and prokaryotic chromosomes? Telomeres only exist in eukaryotic chromosomes. Correct What is the end replication problem? Why does the end replication problem exist? What is the contribution of RNase H to the end replication problem? It describes the inability of DNA polymerases to replicate the ends of eukaryotic chromosomes. Telomeres are shortened on both daughter strands after each round of DNA replication – the 3’ overhang occurs because there is no primer for DNA Poly to fill in this gap. Correct What is cell senescence? Describe the relationship between cell senescence and telomeres. Almost all cells in a multicellular organism undergo cell senescence. Identify one type of cells that do not undergo cell senescence. How do these cells avoid cell senescence? Cell senescence is when a cell stops replicating despite having optimal conditions for growth. When telomeres get too short, cell senescence occurs because shortened telomeres cannot stabilize ends of chromosomes so DNA replication cannot occur. One type of cell that does not undergo cell senescence are STEM cells as they have telomerase active that helps to prevent telomere shortening. Correct What is telomerase? Describe the function of telomerase during DNA replication. Telomerase extends 3’ overhang by synthesizing additional telomeric repeats using RNA as a template which helps prevent telomere shortening. In DNA replication, telomerase first extends the 3’ overhang by synthesizing additional repeats from the template of telomerase RNA. Then, primase makes a new RNA primer on this DNA made by telomerase. Lastly, DNA pol makes new DNA strand and the primer is removed. This extension of the 3’ overhang is what prevents telomere shortening. Correct Lecture 8: DNA Repair What are DNA mutations? Describe the difference between point mutations and chromosome mutations. Permanent changes in the DNA sequence of an organism. A point mutation is a one base change, whereas a chromosome mutation is when a segment of a chromosome is lost, deleted, or moved. The scale of a chromosome mutation is much greater than that of a point mutation. Correct Describe the difference between base substitutions, insertions, and deletions. Which of these mutations could be considered a frameshift mutation? Base substitution is one base is changed; one codon changed. Insertion is one base inserted; many codons are changed. Deletion is one base is deleted; many codons are changed. Both insertions and deletions can be considered frameshift because they can shift the reading frame of the ribosome. Correct Need to know the differences between silent, missense, nonsense, and frameshift mutations. Which of these mutations is associated with disease? Why? Silent mutations are base substitutions that changes the DNA sequence but does not alter the amino acid sequence, so protein function is not altered. Missense mutations are base substitutions that changes the DNA sequence which results in altering one amino acid. Protein function by be altered if protein shape is changed. Nonsense mutation is when there is base substitution that changes DNA sequence and generates a premature stop codon. Protein function is altered because many amino acids are changed. Frameshift mutation is an insertion or deletion that changes the reading frame of a gene sequence. Protein function is altered because many amino acids are changed. Point mutations must change protein function to cause disease; protein function changes only if an amino acid is altered, so silent mutations do not cause disease (missense). Correct Describe how missense and nonsense mutations are indicated in scientific papers and clinical reports. Missense mutations are denoted by letters and a number (ex. T214P = at position 214, a threonine was changed to proline). Nonsense mutations are denoted by the second letter (ex. X = L217X = at position 217, a lysine was changed to a stop codon). Correct Describe the difference between loss-of-function (LOF) and gain-of-function (GOF) mutations. Describe some causes of LOF and GOF mutations. Loss-of-function (LOF) mutations are changes in the DNA sequence that destroys protein function. It is typically recessive in diploid organism. Loss-of-function mutations can be caused by changing protein shape or decreasing gene expression. Gain-of-function (GOF) mutations are changes in the DNA sequence makes protein hyperactive, which is rare. It is typically dominant in diploid organism. Gain-of- function mutations can be caused by changing protein shape or increasing gene expression. Correct What are spontaneous mutations? What are induced mutations? Describe some similarities and differences between these two causes of mutations. Spontaneous mutations are point mutations that occur naturally. There is no mutagen involved and They have a low mutation rate as long as DNA repair mechanisms are functional. Induced mutations are also point mutations. They are induced by exposure to a mutagen, either chemical, drug, or radiation. They have a much higher mutation rate compared to spontaneous mutation. Correct Describe the causes of the following mutations: errors in DNA replication, oxidative damage, depurination, deamination, DNA adduct, UV light, exposure to radiation. For each description, indicate if the cause generates a spontaneous or induced mutation, and they type of mutation that is produced (base substitution, indel, etc.). - errors in DNA replication (spontaneous): mistakes by DNA polymerase can generate DNA mutations if mismatched bases are not corrected by proofreading or DNA repair. Errors in DNA replication create base substitutions (silent, missense, or nonsense mutations) - oxidative damage (spontaneous): reactive oxygen species (ROS) made by cell promotes oxidation of DNA bases (G to 8-oxoG) (base substitution) - depurination (spontaneous): covalent bond between deoxyribose sugar and purine base (A or G) is spontaneously broken by hydrolysis. Depurination creates an abasic/apurinic site: position in DNA backbone that has sugar and phosphate group, but no base (basic substitution). - deamination (spontaneous): removes amino group (NH2) from base in nucleotide (base substitution). - DNA adducts (induced): covalent modification of base in nucleotide that is caused by exposure to a chemical; chemical attaches to base. DNA adducts create base substitutions by altering base pairing of nucleotide (G with DNA adduct pairs with A instead of C during DNA replication) - UV light (induced): promote formation of pyrimidine dimers: covalent bond between adjacent bases in same strand of DNA. Pyrimidine dimers cause a significant distortion in DNA dimer recognized as one base during replication = frameshift mutation Correct Describe the two-step process that produces a point mutation in the chromosome of an organism. Does this process apply for spontaneous mutations, induced mutations, neither, or both? First, the incorrect nucleotide is incorporated into DNA strand and then the mismatch is not corrected before replication, so a point mutation is produced. This is the process for both spontaneous and induced! Some indels cause frameshift mutations, some indels do not cause frameshift mutations. How is this possible? If a frameshift consists of 1 or 2 added/deleted bases it will shift the reading frame. This will cause a frameshift mutation as every encoded amino acid after the addition/deletion will be affected. However, if the mutation consists of 3 additions/deletions it is not considered a frameshift since the reading frame did not change. What causes triplet expansion diseases? Are triple expansion diseases caused by spontaneous or induced mutations? Repetitive DNA is unstable, so inserting three bases is common at repetitive DNA because of template slippage. This leads to triplet expansion diseases as triplet sequences are inserted during replication, so it is a spontaneous mutation. Explain how PCR and agarose gel electrophoresis can be used to diagnose triplet expansion diseases in human patients. Triplet expansion disease occur when a gene sequence is increased in length by a multiple of three bases due to template slippage. As a result, the gene’s length will differ in individuals depending on if they have the disease or not, and the length is typically proportional to disease severity. To diagnose these diseases, you can use PCR to amplify the gene of interest isolated from the individual. Then, determine the length of the gene in the individual by gel electrophoresis. The larger the gene, the more repeats, the shorter is travels on the gel, and the more likely it is for them to have the disease. For each of the DNA repair mechanisms listed below, you need to know the type of damage and the basic steps used to correct the damage. Base excision repair (BER) – What type of DNA damage is correct by BER? What are the roles of DNA glycosylase, AP endonuclease, DNA polymerase, and DNA ligase in BER? BER is used to fix spontaneous single strand DNA damage such as depurination or deamination. DNA glycosylase is used to cut out the damaged base, and an AP endonuclease can come in and cut the backbone. DNA polymerase replaces the removed base with the correct one and is sealed by DNA ligase. It can be used to fix endogenous (oxidation, depurination, and deamination) damage to small distortions. Nucleotide excision repair (NER) – What type of damage is corrected by NER? What are the roles of excinuclease, helicase, DNA polymerase, and DNA ligase in NER? NER is used to fix larger single strand DNA breaks. It can be used to fix induced mutations like thymine dimers. XPC detects an abnormal base and signals for exinucleases to come in and remove the base. DNA polymerase is then able to install the correct base and DNA ligase can reseal the cut. NER repairs large distortions with exogenous damage (DNA adducts, thymine dimers from UV light). Mismatch repair (MMR) – What type of damage is corrected by MMR? What is the difference between MutS and MSH2? What are the functions of MutS/MSH2, nuclease, DNA pol, and DNA ligase in MMR? Mismatch repair is used to fix base pairs that were incorrectly matched by DNA polymerase. MutS/MSH2 can detect the mistake using methylation. MutS is present in prokaryotes and MSH2 is used in eukaryotes. The MutH nuclease can bind and remove a section of the DNA strand with the mismatched base. DNA polymerase then is able to replace the bases and can be sealed by DNA ligase. Nonhomologous end joining (NHEJ) – What type of damage is corrected by NHEJ? What is end bridging and end processing? What is a common consequence of using NHEJ to repair DNA damage? NHEJ repairs double strand breaks. End bridging refers to proteins recognizing and binding to the ends of the broken strands. The proteins then digest bases from the broken portions. This is called end processing. By removing bases, NHEJ creates a self-inflicted deletion mutation. Homologous recombination repair (HRR) – What type of damage is corrected by HRR? Describe the similarities and differences between NHEJ and HRR. HRR repairs double strand breaks in DNA. It can only be used right after DNA replication since it uses a sister chromatid as the template for repairing the DNA. It uses end processing similar to nonhomologous end joining, but the biggest difference is that HRR does not generate a deletion mutation after it has repaired the strand. Lecture 9: Prokaryotic Transcription What is gene expression? What are two steps in gene expression that are used for many genes? Gene expression is the two-step process of using the genetic information in DNA to make proteins. First part is transcription which creates an mRNA strand from a DNA template using RNA polymerase. Second part is a translation which uses the mRNA to make a protein using ribosomes. Correct Define transcription and describe the function of RNA polymerase. How is it similar to DNA polymerase? How is it different? Transcription is the first process in gene expression which forms an mRNA strand from a DNA template strand to help later form proteins. The enzyme RNA polymerase makes an RNA strand. RNA polymerase binds to naked DNA only at one strand, does not need a primer, uses ribonucleotides instead of deoxyribonucleotides, and only makes a leading strand. Both RNA and DNA polymerases synthesize in the 5’ to 3’ direction by making phosphodiester bonds between nucleotides. Correct What are the roles of promoters, regulatory sequences, and terminator sequences in transcription? A promoter is a specific sequence that is located before the actual gene sequence and is where the RNA polymerase binds to the template DNA. Regulatory sequences are sequences where transcription factors bind in order to control the rate of expression. The terminator sequence is a sequence at the end of the gene sequence that makes the polymerase eject off the DNA template and stop transcription. Correct What are the three stages of transcription? What is the RNA polymerase doing in each stage? First is initiation, when the RNA polymerase binds the promotor and DNA denatures. Second, elongation is when the RNA polymerase leaves the promotor and starts creating mRNA in the 5’-3’ direction when far enough. Third is termination; when the RNA polymerase reaches the terminator sequence, it is then ejected off the DNA template strand. Correct What is the transcriptional start site (TSS)? What is the difference between the TSS and the promoter? The transcriptional start site is the one base after the promoter sequence and actually starts transcription when read by RNA polymerase. The promoter is a separate sequence that is not added to the mRNA, but the TSS is part of the gene sequence (first base). Correct What does upstream and downstream from the TSS mean? Upstream means bases away from the TSS site, in opposite direction of RNA polymerase, and are in the promotor sequence (negative values). Downstream means bases after the TSS, in same direction of RNA polymerase, and are in the gene sequence (positive values). Correct What does +1, +10, and -10 mean on a gene sequence? Which are transcribed and which are not transcribed? Which sequences are upstream from the TSS? Which sequences are downstream from the TSS? +1 refers to the TSS site, +10 means 9 bases downstream of TSS, and –10 means 9 bases upstream of TSS (in promotor sequence). Sequences that are downstream (+1 and +10) are only transcribed. Correct What is the difference between the template and coding strands? What is the association between the template and coding strands with the RNA strand during transcription? The template strand is the DNA strand that is read by RNA polymerase to make mRNA and its sequence is complementary to the RNA strand. The coding strand (non-template strand) is not transcribed by RNA polymerase but does have the same 5’ and 3’ orientation and a similar sequence to mRNA except that the T’s are replaced with U’ in RNA. Correct What is the difference between the RNA polymerase core and RNA polymerase holoenzyme in prokaryotic transcription? RNA polymerase core is made up of five subunits in a prokaryotic RNA polymerase. RNA polymerase holoenzyme is the active form of the enzyme (can bind to DNA) and it is formed when a sixth subunit (sigma factor) binds to the polymerase core. Correct What is the function of the sigma factor in prokaryotic transcription? The sigma factor is part of the RNA polymerase that helps find the promotor sequence on the template strand of DNA. Correct What are functions of the -35 and -10 sequences in prokaryotic transcription? The –35 and –10 sequences are what the sigma factor specifically scans for when it finds the promotor. The –10 sequence is an AT rich region with weak hydrogen bonds that can be broken to form a open complex to start transcription at the transcriptional start site. Correct Describe the initiation stage of prokaryotic transcription using the following terms: open complex, closed complex, sigma factor and RNA polymerase core, RNA polymerase holoenzyme, and -10 sequence. The RNA polymerase core is a heteropentamer (i.e., consisting of five subunits that are not all the same). The sigma factor binds to the core to create the RNA polymerase holoenzyme, and it scans for and binds with RNA polymerase to the –35 and –10 (AT-rich) sequences to form the closed complex. Once DNA strand separation occurs at the –10 sequence, the open complex forms. Why do different genes within a prokaryotic chromosome have different -35 and -10 sequences? Prokaryotic cells use different sigma factors to bind to different -35 and -10 sequences in promoter RNA polymerase can bind to only one sigma factor at a time, which recruits it to specific promoter sequences. Why do point mutations at the lac promoter tend to increase gene expression? Need to be able to explain experimental results presented in class. When the mutated promotor sequence is similar to the consensus sequence (lac operon especially) there is a trend where gene expression increases in this case. Correct Describe the transition from initiation to elongation using the following terms: promoter clearance, open complex, and elongation complex. Six protein subunits on RNA polymerase combine to form holoenzyme which binds loosely so it can slide down the DNA. Then, the sigma factor subunit on the RNA polymerase scans DNA for -35 and -10 sequences in the promoter. After sigma factor recognizes promoter, RNA polymerase binds tightly so that there is no more sliding. To begin synthesizing RNA, DNA strands are separated at the -10 sequence (AT-rich region) to create open complex. Once RNA synthesis begins, RNA polymerase leaves promoter to start elongation phase and sigma factor is released. What is abortive initiation? What is the cause of abortive initiation? Abortive initiation is the process of the RNA polymerase binding and unbinding to the template. Since there are no primers used, RNA polymerase is not stable on the template and can “fall off.” The sigma factor remains at the promoter so RNA polymerase can rebind and continue transcription until it gets far enough away from the promoter region. At this point the sigma factor will leave and be recycled. Identify and describe the two strategies used by prokaryotic cells to terminate transcription. 1. Rho independent termination: this mechanism of transcription termination relies on the RNA’s secondary structure and adenine repeats to form a hairpin loop. The RNA polymerase is paused and A-U base pairs can be easily disrupted to release the RNA from the template DNA strand. 2. Rho dependent termination: The terminator region contains a Rho binding site. The rho protein can bind to the region and separate the RNA from the template strand to end transcription. Rho is a helicase. Lecture 10: Eukaryotic Transcription Describe the similarities and differences between prokaryotic and eukaryotic transcription. The polymerases have similar structure and synthesize RNA from a DNA template in the 5’ to 3’ direction. However, eukaryotes have more RNA polymerases (three) than prokaryotes (one), more regulatory sequences, more nucleosomes that may influence transcription, and more proteins required for initiation (in the form of TFs). How many RNA polymerases are used by eukaryotic cells? Describe the function of each eukaryotic RNA polymerase. Do they recognize the same promoter sequences or different promoter sequences? Eukaryotic cells have three RNA polymerases. RNA polymerase I produces most of the ribosomal RNA (rRNA) in the cell; accounts for 80% of all transcription. RNA pol I produces large rRNA strands but produces low number. RNA polymerase III produces all the tRNA and some rRNA in the cell. RNA pol III produces small tRNA strands but produces high number. RNA polymerase II produces all the mRNA, microRNAs, and some noncoding RNA. What is a transcription factor? Describe the differences between general transcription factors and regulatory transcription factors? A transcription factor is a molecule that can bind to the promoter and affect the rates of transcription. A general transcription factor (GTF) is one that is required for transcription and cannot be left out. The 5 GTFs in eukaryotes are TFIID, TFIIB, TFIIF, TFIIE, and TFIIH. A regulatory transcription factor is one that can influence the rate of transcription. These include activators and repressors. Need to know the function of the following proteins during transcription: TFIID, TBP, TAFs, TFIIB, TFIIF, TFIIE, and TFIIH. TFIID contains TBP (binds the TATA box) and TAF (TBP-associated-factor) subunits to recruit it to the promoter where it bends DNA (to allow TFIIB to bind) and binds the Inr, TATA box (if there is one present), DPE, and histone marks. TFIIB helps to recruit pol II to the promoter and binds TBP and BRE (TFIIB recognition element). TFIIF binds pol II when it is not bound to the promoter and also helps to stabilize TFIIB-pol II interactions. TFIIE helps to recruit TFIIH to the promoter. TFIIH facilitates the initiation to elongation transition through its helicase activity, where it separates the template and coding strands, and kinase activity, where it phosphorylates the CTD of pol II to mediate TFIIB, TFIIE, and TFIIH release from the core promoter. What is the preinitiation complex? What is the difference between the preinitiation complex and the initiation complex? The preinitiation complex is formed by TFIIE and consists of the core promoter + five GTFs. The initiation complex forms once TFIIH has performed its actions as described above. Which general transcription factor is responsible for the transition from the preinitiation complex to the initiation complex? What does it do that promotes the transition? TFIIH facilitates the initiation to elongation transition through its helicase activity, where it separates the template and coding strands, and kinase activity, where it phosphorylates the CTD of pol II to mediate TFIIB, TFIIE, and TFIIH release from the core promoter. What triggers RNA polymerase II to begin the transition from the initiation stage to the elongation stage during transcription? Phosphorylation of CTD of RNA Polymerase II promotes release of TFIIB, TFIIE, TFIIH from core promoter. So the phosphorylation of RNA Pol II triggers the transition from initiation to elongation. What is mediator? Describe two different functions of mediator during eukaryotic transcription. A mediator is a protein that facilitates long range interactions between regulatory sequences and the promoter region during transcription. It can be used to bring activators close to the promoter. Mediators regulate transcription by bringing transcription factors closer to regulatory sequences and by regulating RNA polymerase phosphorylation by TFIIH. What happens to RNA polymerase II during the termination stage of transcription? At termination, RNA polymerase II is dephosphorylated and recycled so it can transcribe another gene. Lecture 11: RNA Processing, Part I All cells make mRNA, tRNA, and rRNA. What is the most abundant type of RNA in cells? Least abundant? 80% rRNA and 2% mRNA. Correct What does it mean to say that transcription and translation are coupled in prokaryotic cells? Are translation and transcription coupled in eukaryotic cells? Why or why not? Coupling in prokaryotic cells refer to when transcription is not even finished that ribosomes are making proteins (transcription and translation occur at the same time). Coupling does not occur in eukaryotic cells due to compartmentalization (eukaryotic cells have a nucleus). Transcription occurs in the nucleus and translation in the cytosol. mRNAs also need to be processed before they are translated into protein. Correct What is RNA processing? Which type or types of RNAs are processed? RNA processing is the modification process RNAs have to go through during transcription in Eukaryotic cells. Mostly occurs to mRNA, rRNA and tRNAs in eukaryotic cells. Prokaryotic cells do not process most mRNAs but do process tRNAs and rRNAs. Correct What is a primary RNA transcript (aka pre-mRNA? What type of modifications occur on primary RNA transcripts? Primary RNA is the initial RNA made in transcription that needs to go through RNA processing before it is functional. Modification to mRNAs include 5’ capping, polyadenylation, and RNA splicing. Modifications to tRNAs and rRNAs include base modifications and cutting the RNA. Correct Why is mRNA processing important in eukaryotic cells? Processing helps regulate expression and nuclear export. Before mRNA can leave the nucleus and start translation, it is checked for the chemical modification of 5’ capping. Correct What is the association between mRNA processing and exoribonucleases? RNA processing helps protect mRNA from being degraded by exoribonucleases because RNA processing adds proteins on 5’ and 3’ ends that protect the RNA from digestion. Exoribonucleases are ribonucleases that degrade (unprocessed) RNA from the end. Correct RNA processing requires several enzymes. Where are these enzymes located at the initiation of transcription? Enzymes located on tail of RNA polymerase II after polymerase is phosphorylated by TFIIH at the end of the initiation phase of transcription. Correct What is the 5’ cap? Why is the 5’ cap often depicted as an upside-down nucleotide? 5’ cap is a modified guanine base at the 5’ end of RNA that serves as a binding site for RNA-binding proteins. It is added after 25 nucleotides are made on the RNA during transcription. It is upside down because it is attached to an mRNA strand using 5’-5’ triphosphate bond that acts like a phosphodiester bond. Correct Describe how the 5’ cap is added to mRNA. In your description, include the three enzymes that were discussed in class and describe the function of each enzyme. First, RTPase removes a phosphate from the base at the 5’ end to create a nucleotide that has two phosphate groups instead of three. Then GTase adds GMP at the 5’ end after removing two more phosphates from GTP. This allows formation of a 5’-5’ triphosphate bond between GMP and the 5’ end of RNA. Next, guanylyl-7-methyltransferase adds a methyl group to create 7- methyl-guanosine. Lastly, additional methyl groups are added to the 2’ OH ends on the first and second nucleotides. Correct What are the functions of the 5’ cap? 5’ cap helps protect RNA degraded by exoribonucleases by serving as a binding site for proteins. Helps initiate translation, nuclear transport, and increase RNA splicing efficiency. Correct What is 3’ polyadenylation? What is the function of 3’ polyadenylation? 3’ polyadenylation is added extra adenosine nucleotides to RNA to make a 3’ poly-A tail that helps protect it from degradation. It also creates a binding site for proteins. Correct Describe how 3’ polyadenylation is added to an RNA strand using the following terms: endonuclease, PolyA binding proteins, polyadenylation signal, polyadenylated polymerase. 1. A polyadenylation signal (AAUAAA) is made on the mRNA 2. PolyA binding proteins bind to the poly-A signal on the mRNA 3. An endonuclease comes in and cleaves after the signal sequence 4. Polyadenylated polymerase (PAP) adds repeated adenines to the end of the cleaved poly-A signal sequence 5. PolyA binding proteins can bind to the adenine repeats and protect the 3’ end from being degraded. These proteins are also required for nuclear export. Processing of mRNAs typically requires several steps. Describe the sequence of three processing steps during transcription. What type of RNA processing happens after transcription? 5’ cap processing happens at start of transcription, 3’ polyadenylation happens just before termination. The majority of RNA splicing occurs between the addition of the 5’ cap and addition of 3’ poly-A tail. Correct What is RNA splicing? What is the difference between exons and introns? RNA splicing is the removal on introns and exons off the primary RNA transcript using the spliceosome. Exons are DNA part of genes that make proteins while introns are typically spliced out during transcription. Correct What are some important facts about introns? Are introns “junk DNA”? Why or why not? Prokaryotes do not have introns, but only one long exons. Introns are important since they contain regulatory sequences and are sometimes used as protein encoding sequences. Correct What are small nuclear ribonucleoproteins (snRNPs)? What are snRNAs? What is a spliceosome? snRNPs are RNA-protein complexes that are involved in splicing and combine to form the spliceosome. Their RNA component is called snRNA and helps to identify splice site sequences. What are splice sites? How many splice sites are there for one intron? Splice sites are regions within a pre-RNA transcript where the transcript is cut to produce mature RNA. They are most often located within introns, though there are exceptions to this. One intron typically has two splice sites: 5’ and 3’ splice sites. Identify the consensus sequences that are important for splicing. Why are consensus sequences in an intron important for RNA splicing? The 5’ splice site is GU, the branch point is an A located close to the 3’ end, and the 3’ splice site is AG. The consensus sequences are important because they are recognized by five snRNPs (such as U1 and U2) for the purpose of forming the intron lariat and allowing for exons to be joined together. Describe the process of RNA splicing. How is the lariat formed? How is the lariat removed? How are the two exons attached? Your description should include the following terms: branch point, 5’ splice site, snRNPs, snRNA, 3’ splice site, 2’ OH on adenosine at branch point, lariat, intron, exon 1, and exon 2. 1. The snRNPS recognize the 5’ splice site and branch point consensus sequences using snRNA and bring them into close contact with one another. 2. The 2’ -OH on the A of the branch point attaches to the phosphate at the boundary between the 5’ splice site and exon 1. The lariat is now formed. 3. The 3’ -OH of exon 1 make a phosphodiester bond with exon 2 by attacking the phosphate between the 3’ splice site and exon 2 while releasing the intron lariat. Lecture 12: RNA Processing, Part II What is alternative splicing? Is alternative splicing common for human genes? Alternate splicing is the process of processing a premature RNA transcript differently by removing different combinations of exons/introns. Alternative splicing occurs in around 70% of human genes. Describe the different types of alternative splicing discussed in class: exon skipping, intron retention, alternative 3’ splice site, and alternative 5’ splice site. Exon skipping occurs when exons are spliced out. Intron retention occurs when introns fail to be spliced out. Alternative 5’ and 3’ splice sites occur when there are several “options” of where the spliceosome will splice at these ends, increasing the total number of isoforms of a protein. What is the difference between constitutive and alternative exons? Constitutive exons are in all variants and are required for the protein to be functional. Alternative exons are only in some variants and are not required for the protein to be functional. How do regulatory proteins regulate alternative splicing? Explain how two different cell types can make two different proteins from the same gene using alternative splicing. Positive and negative regulatory proteins regulate alternative splicing by binding to splicing enhancers or silencers, respectively, to facilitate (recruit snRNPs) or prevent (mask splice sites from snRNPs) splicing. Different cell types may utilize different splice sites from the same protein such the mature RNA transcript differs between them and, therefore, a different sequence of amino acids will be translated. What are splice site mutations? Why do splice site mutations cause disease? What are some examples of splice site mutations? Splice site mutations are changes to the DNA that alter the function of the spliceosome during splicing. One example could be the point mutation of CT to GT on the coding strand of DNA, resulting in a new GU 5’ splice site being introduced in the mRNA. These mutations cause disease because they lead to alternative splicing of mRNA which may create nonfunctional protein, especially when introns are retained. What is RNA editing? RNA editing is the process of modifying an RNA sequence without altering the DNA that corresponds to the RNA. It is rare in mammals and may occur before or after splicing takes place. What is the function of adenosine deaminase acting on RNA (ADAR)? ADAR modifies A and changes it to an I (inosine), which acts as a G by base pairing with C. How does ADAR perform RNA editing? ADAR removes an amino group from A through oxidation with water. One example of its effect involves the Glu receptor where it changes a CAG codon to a CIG codon, resulting in a missense mutation (Gln to Arg). What is the function of cytidine deaminase? Describe how cytidine deaminase performs RNA editing in the ApoB gene. Cytidine deaminase removes an amino group on C to convert it to U. In liver cells, full-length ApoB100 is produced, but in intestinal cells cytidine deaminase introduces a premature STOP codon to instead produce a truncated ApoB48 protein. Identify at least one similarity and one difference between alternative splicing and RNA editing. They both alter the structure of a premature mRNA transcript, but alternate splicing does so by how it cuts out portions of the transcript while RNA editing does so by altering individual bases. Identify at least four different reasons for why eukaryotic cells edit mRNA. mRNA editing may introduce new start codons, new STOP codons, alternate splice sites, or missense mutations. Does RNA editing influence transcription, translation, or both? Explain your rationale. It influences both because it alters the resultant RNA produced by transcription but also modifies translation because it may change the resulting protein that is translated. Identify and describe the proteins that are bound to an mRNA that has undergone RNA processing but has not left the nucleus. What role do these proteins have in nuclear export? CBC (cap binding complex) binds the 5’ methylated guanine cap. PABP (poly-A binding protein) binds the poly-A tail. EJC (exon junction complex) proteins bind exon-exon junctions to protect the mRNA. All of these proteins must be bound for the mRNA to be allowed to transport through nuclear pores into the cytosol. What is RNA degradation? Describe how RNA degradation occurs to mRNAs in eukaryotic cells. RNA degradation occurs when RNases break RNA into its component nucleotides. RBPs (RNA-binding proteins) located in the cytosol induce RNA degradation when they bind to the 3’ UTR. The process of degradation involves removal of the 5’ cap and poly-A tail when the exosome degrades the RNA from the 3’ end. What is the exosome? How is the activity of the exosome regulated? The exosome is an exoribonuclease that degrades RNA from the 3’ to the 5’ end. Different proteins up- or downregulated exosome binding to the 3’ UTR to degrade the RNA. Are tRNAs and rRNAs processed in both prokaryotic and eukaryotic cells? Yes. If tRNAs are processed, describe some of the processing that occurs to tRNAs. RNase P cuts the 5’ end and RNase D cuts the 3’ end of the premature tRNA. Introns are also removed and CCA may be added to the 3’ end. If rRNAs are processed, describe some of the processing that occurs to rRNAs. What are small nucleolar ribonucleoproteins (snoRNPs)? What are snoRNAs? rRNA is initially transcribed as a large RNA molecule that must be spliced into small RNAs to produce the large and small ribosomal subunits. The snoRNAs of snoRNPs (RNA-protein complexes) recognize splice sites, and the snoRNPs facilitate modification and processing of the pre-rRNA into smaller subunits. For example, the 45S pre-rRNA in eukaryotes is spliced into the 18S, 5.8S, and 28S rRNAs and bases are modified. Paper 1: Inhibition of telomerase limits growth of human cancer cells Define the terms telomeres and replicative senescence. What is the relationship between telomeres and replicative senescence? Telomeres – repetitive sequences located at the end of eukaryotic chromosomes that help stabilize and protect the chromosomes from shortening. Replicative senescence – is the state of cells that results in the loss of the ability to divide and proliferate. This limits cellular division and mechanisms against uncontrolled cell growth. At a specific length, telomere shorting can result in DNA unraveling and kills the organism. Telomere shortening triggers replicative senescence. Correct What is the function of hTERT? What is the relationship between hTERT and human cancers? hTERT stands for human telomerase reverse transcriptase creates telomeric DNA sequences using an RNA template. It its part of telomerase that helps maintain telomere length against shortening. In cancer cells, there is an increase of hTERT expression and it results in the bypass of replicative senescence (cancer cells divide uncontrollably). Correct What is the overall goal for this paper? In other words, what is the scientific question this paper aims to answer with the figures in the results section? Incorporate and explain the importance of DN-hTERT in your answer. To examine ways to treat cancer cells by targeting mechanisms of telomerase. The goal was to see if DN- hTERT could effectively inhibit telomerase mechanisms and thus if it could shorten the lifespan of malignant cells. Need to be able to analyze and interpret Figure 1C. See discussion activity 4 for questions. GAPDH is a control. This is showing that insertion of plasmids was successful. GAPDH is here to show the DN-hTERT is from plasmid DNA and not from chromosomal DNA. V is a control vector and cells can’t express it, but GAPDH has chromosomal DNA so it does express it. The lab technique used here is RT- PCR followed by gel electrophoresis to measure expression of hTERT in different conditions. Need to be able to analyze and interpret Figures 3A and 3B. See discussion activity 4 for questions. The lab technique used here is Southern blot, which analyzes the size/amount of DNA molecules. DN- hTERT prevented telomerase from lengthening telomeres in 36M. GM487 doesn’t have telomerase function so DN-hTERT had no effect on telomere length. Need to be able to analyze and interpret Figures 4D and 4E. See discussion activity 4 for questions. Cell growth was determined for the different samples in different cell lines. The 36M cell line uses telomerase so cell growth eventually decreases in rate upon the addition of DN-hTERT. The GM847 cell line maintains telomere length via a telomerase-independent mechanism and therefore did not show decreased cell growth upon treatment with DN-hTERT.