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BIOL 308
Dragana Miskovic

BIOL 308 - Study Questions Lectures 1-4 1. What information(s) could you obtain from a genetic approach of studying mutants defective in a particular process? The potential function and importance of the gene/protein can be determined through identification, isolation and survival of mutants. The number of genes, the order of the genes (using complementation analysis), and the interactions between different genes (using genetic suppression) can also be determined. 2. How would you define permissive conditions in respect to temperature sensitive mutants? An environment where the mutant can survive, since it displays the wild type phenotype. Under nonpermissive conditions (too cold) mutant phenotype is displayed and mutants can’t survive. 3. Define (or compare and contrast): a) gene expression; transcription; replication; translation; b) gene; allele a) transcription: an RNA copy of the DNA is made, translation: RNA copy is used to create protein, gene expression: gene has undergone transcription and translation for form a product, replication: DNA makes a copy of itself before undergoing mitosis b) gene: contains code to produce RNA or a protein, allele: variations of the gene 4. Explain by using your own words the meaning/significance of gene expression. Gene expression occurs when a gene is “turned on,” and forms a product such as a protein. The gene will be expressed depending on the environment, developmental stage, or tissue type. 5. What are the roles of model organisms in molecular biology studies? Choose two model organisms and explain your reasoning. Model organisms can be easily studied and are used in place of species that can’t be study in the same manor. Ex. Arabidopsis and Drosophila are simple, easy to come by in bulk, cheap, easy to maintain, lots of data is already available, and reproduce quickly allowing scientists to observe future generations. 6. What are three main functions of DNA? Explain the importance of each of them. 1. Stores information: coding for proteins and RNA, and contains regulatory signals 2. Replicates faithfully (preservation of information): semi-conservative, complementary base pairing, preserves genes important to organism’s survival 3. Has ability to mutate (variability of information): allows for adaptation, may be harmful or have no effect 7. What is (are) the role(s) of phospho-diester bonds in DNA structure? What is (are) the role(s) of hydrogen bonds in DNA structure? What is (are) the role(s) of hydrophobic interactions in DNA structure? Phosphodiester bonds hold the sugar to the phosphate backbone. Hydrogen bonds assist with complementary base pairing. Hydrophobic interactions occur due to the highly negative phosphate backbone on the outside and the nonpolar bases inside. 8. What noncovalent interactions are involved in maintaining the double-helical conformation of DNA? Van der Waals (stacking interactions), hydrophobic, ionic (salts stabilize phosphate backbone), and hydrogen bonding 9. Learn to recognize nitrogenous bases (A,T,G,C,U). Purine Pyrimidine Adenine Guanine Cytosine Thyamine Uracil 1 BIOL 308 - Study Questions 10. Describe Meselson-Stahl experiment and explain how it showed that DNA replication is semiconservative? E.coli was grown in N15, and then moved to N14. In the first replication semiconservative material contained equal amounts of N15 and N14 in each strand. In the second replication two of the strands contained N15 and N14, while the other two contained N14 only. 11. What is meant by saying that a DNA strand has polarity? That two strands of DNA are antiparallel? That the strands are complementary to one another? The DNA strand is synthesized in a particular direction, 5’ to 3’, 5’ end is negatively charged. The two strands of DNA go in opposite directions of one another (there 3’ and 5’ ends are opposite). The DNA strands have matching pairs (AT and CG), that always line up together. 12. If a C content of a preparation of double-stranded DNA is 20%, what is the T content? 30% 13. What is the difference between nucleoside and nucleotide? What does dNTP stand for? Nucleotides contain a phosphate group, along with the nitrogenous base and the sugar. Nucleosides do not include a phosphate group. Deoxynucleotide Triphosphate. 14. Describe the conformational characteristics of B DNA (or A DNA, Z DNA, triple-helical DNA). When does this (any of the above) form of DNA occur? B DNA is a right handed, double helix, and has a diameter of 80A, a major groove of 22A, a minor groove of 12A, is 34A per helical turn, and represents most in vivo DNA. A DNA is also a right-handed double helix with 28A per helical turn, and is found in dormant spores of bacillus. Z DNA is a left handed, double helix with 45A per helical turn, and is found in chromosomal breakages and Alzheimer’s disease. Triple helical DNA is formed when purines make one strand and pyrimidines the other, the third strand can be accommodated, and they are found in vitro and likely in vivo during DNA recombination or repair. 1A = 0.1nm, 1nm = 1.0X10-9m 15. How does high salt concentration influence denaturation kinetics of DNA? Explain your reasoning. Salt stabilizes DNA since it neutralizes the negatively charged backbone, and therefore slows down denatuation kinetics of DNA. 16. What are the classes of DNA sequences in genomic DNA (based on renaturation kinetics)? Nonrepeating, where the complexity = number of nucleotides, and repetitive sequences where complexity = the number of unique nucleotides and the total number of nucleotides from one copy of each repetitive sequence. 17. What is C toanalysis? Used to determine the rate of renaturation or a measure of complexity of DNA. C is ohe starting concentration (nucleotides/L) and t is the reaction time (s). Units of complexity are measured in terms of nucleotides. 18. Who received a Nobel Prize for 3D DNA structure? Watson and Crick. 19. If you had two solutions of DNA, one single-stranded and one double-stranded, with equivalent absorbance at 260 nm, how would the concentrations of DNA compare in these two solutions? (You can use a diagram if it makes it easier for you to explain.) There would be more double-stranded DNA because single stranded DNA takes up light easier, since purines and pyrimidines exposed. 2 BIOL 308 - Study Questions Lectures 5-7 1. Thinking question re DNA/RNA structure: Certain chemical agents acting on DNA could convert cytosine to uracil through the process of deamination (chopping off the amino group). This mutation is routinely repaired by the existing repair mechanism (uracil is removed and it gets replaced by cytosine). Knowing this, how would you explain why DNA contains thymine and NOT uracil. If the DNA contained uracil instead of thymine the enzyme would remove all uracils, including those intended to be present. The DNA would end up with adenine-cytosine pairs and be greatly mutated. 2. How does complexity of bacterial genome differ from that of eukaryotic (calf) genome? Bacterial genome is smaller, but contains few/no repeats. In eukaryotic genomes there are many repeats, but the length of the gene is substantially longer, and even without the repeats would be greater in length than the bacterial genome. Eukaryotic genes contain multiple isoforms of a protein, which can be produced through alternative slicing of mRNA. 3. Explain C value paradox. There is no correlation between the amount of DNA (size of genome) and the apparent complexity of the organism. 4. List and briefly explain factors that influence DNA renaturation kinetics. DNA concentration: complementary single strands have a better chance at finding another single strand in greater concentrations, Salt concentration: ionic conditions mask repulsion forces of phosphate backbones, Temperature: needs to be 20-25C below Tm, time: enough time need for reaction to occur, size: larger fragments have more difficulty lining up to anneal, complexity: simple sequences re-nature faster than complex. 5. You have found a new species of insects. To evaluate the complexity of the genome of this species, you isolate genomic DNA from, fragment the DNA to uniform 500 base pair pieces, denature the DNA and measure the rate of reassociation. Your data is represented in the curve below (sorry for the bad drawing): (a) How many classes of DNA (in respect to sequence complexity) are found in this organism? Nonrepetative, moderately repetitive, and repetitive. (b) What can you say about the relative complexity of each class? What fraction of the genome falls into each class? Those with repeats will be less complex. 6. List three (3) differences between prokaryotic Topoisomerase I and Gyrase. Topoisomerase 1 relaxes negative supercoiling by adding positive supercoil, but gyrase introduces negative supercoils by reducing positive supercoiling. Topoisomerase 1 makes transient cut in one strand, but gyrase being a toposiomerase 2 makes a double-stranded cut and passes a duplex DNA through it and reseals it. Gyrase is ATP hydrolysiszed. 7. What are topological isomers of DNA? Topological isomers or isomers are used in DNA to reduce stress (in the form of supercoiling) in DNA specifically during replication. 8. Explain the importance of DNA supercoiling for the cell survival? Supercoiling allows the DNA to be store in the cell 9. What are the differences between primary (or secondary, or tertiary) structures of RNA and DNA? Primary RNA is single stranded and contains uracil, and primary DNA contains thyamine instead of uracil. Secondary structures of RNA involve the folding over of RNA on itself causing hairpins and stem loops., while DNA takes on a double helix structure. Tertiary structures of RNA involve possible formation of U:A:U base triple and pseudoknots can form between nonadjacent sequences, and DNA forms supercoils/complexes with proteins. 3 BIOL 308 - Study Questions 10. You have a small 4800 bp long circular DNA. It has a linking number of 450 (L=450). What are the twist (T) and the writhe (W) of this DNA? What assumptions about the structure of the DNA have you made in your answer? (Hint: what is the definition of the twist #?) Assuming the DNA is double stranded: Twist = total # of base pairs/# of base pairs or turns = 4800/450 = 10.7 Writhe = sign of link number = positive 11. Thinking question 1: Many different mutations have been observed in almost all genes. However, only few have been isolated in histones. How would you explain this finding? Histones have a critical function in the cell, and those with mutations would not survive. 12. Thinking question 2: would it be an easy task to produce polyclonal antibodies to histones? Explain your reasoning. It would be difficult because histones are wrapped in DNA, therefore difficult to access. 13. What are some of the distinctive features of eukaryotic chromosomes? (note: I expect you to first define chromosomes and after that you have to briefly explain nucleosomes/histone proteins/octet +H1/wrapped DNA, different levels of chromosome condensation, centromere and telomere regions) Chromosomes are organized structures of DNA. DNA is wrapped onto histone octects or cores, this is referred to as a nucleosome. The octet + H1 is a 25 to 100 fold compression, H1 associates with DNA and octet in linker region binding two distinct regions of the DNA duplex. The nucelosomes can be condensed into a 10nm fiber where H1 is not required or a 30nm fiber. Centromeres are specific sequences where sister chromatids meet. Telomeres occur at the end of the chromosome. 14. What is unusual about the amino acid composition of histones? How is the function of histones related to their amino acid composition? Histones contain large amounts of arginine and lysine, which contain amine groups. Histones can use the positively charged amine groups to bind to negatively charged phosphate backbone of DNA. 15. Name few nonhistone proteins which are a part of chromatin structure and explain why you would expect them to be found there. Matrix attachment regions (MARs) or scaffold attachment regions (SARs) help bind DNA to histone core. DNA replication proteins, transcription factors and chaperone proteins would also be found to help with DNA replication which occurs before DNA is bound to histone. Topoisomerase determines if coiling is too tight. 16. What are heterochromatin and euchromatin? What is their importance in DNA replication and transcription? Heterochromatin is tightly packed DNA and is less susceptible to DNase digestion and transcriptionally inactive. Euchromatin is lightly packed since it contains transcriptionally active DNA, but is susceptible to DNase digestion. 17. Does the degree of chromosomal condensation play a role in controlling replication and transcription? How (explain briefly; use your own words)? More condensed DNA is less available for replication. 18. Would you expect there to be more histones per kilobase in euchromatin or heterochromatin? Explain your reasoning. There would be more histones per kilobase in heterochromatin because the DNA needs to be tightly wound. 19. Thinking question: the sequences of a particular set of genes are found by in situ hybridization (note: methodology is irrelevant for this question, so do not think about it) to be heterochromatic in some cells and euchromatic in cells at different stages of development. How would these sequences be categorized (what would be your conclusion about these sequences in respect to gene expression)? These genes can be turned on and off, and it is better to have them turned off, if not required, and tightly wound to prevent DNase digestion. However, the genes are being translated at certain points of development due to the cells requirments. Lectures 8-9 1. The human gametes have about 3 billion bp of DNA in their chromosomes. a. If the entire DNA was in relaxed B-DNA form, what would be the average length of a chromosome in the cell? (3 billion/10)*3.4nm = 1,020,000,000nm or 1.02m b. On average, how many complete turns would be in each chromosome? 3 billion/10 = 300,000,000 4 BIOL 308 - Study Questions c. If there are around 30-40,000 genes in a human gamete, how many genes are there in an average chromosome? 40,000 genes/gamete = 40,000 genes/46chromosome = 870 genes/chromosome 80,000genes/somatic cell = 80,000 gene/46 chromosomes = 1749 genes/chromosome 2. Define homologous chromosomes. Pairs of chromosomes, one maternal and one paternal, that are the same length, contain the same genes, and have the centromere in the same position. 3. Define non-homologous chromosomes. Chromosomes that do not match. 4. How many homologous chromosomes are there in a germ cell of a woman? 23, men only have 22 because XY is not homologous. 5. Distinguish between homologous chromosomes and sister chromatides. Sister chromatids make up a chromosome and come from the same parent. One of each homologous chromosome comes from a parent. 6. What is the purpose of cell division in Prokaryotes? In Eukaryotes? The purpose of cell division in prokaryotes is to produce “offspring.” In eukaryotes the purpose is to increase mass, or replace dead cells. 7. Distinguish between DNA replication and cell division. DNA replication duplicates the DNA in preparation for cell division, where the cell splits in two, one copy of each DNA going to each daughter cell. 8. Distinguish between reason/purpose for/of mitosis and reason/purpose for/of meiosis? Mitosis increases mass and replaces dead cells, where meiosis creates haploid cells to be used in reproduction. 9. How is variability of genetic information attained by meiosis and fertilization? Eggs and sperm are haploid and when eggs are fertilized the DNA from the father and mother come together and crossing over can occur. 10. What is a cell cycle? What are the stages of cell cycle? The cell cycle is the process the cell undergoes consistently. Interphase – G1: growth occurs as organelles double and the cell carries out normal metabolism, S (synthesis): DNA replication occurs as chromosomes duplicate, and G2: growth occurs as cell prepares to divide. Mitosis: cell division, however some cells never leave interphase (G0). 11. List and briefly describe the checkpoints in cell cycle. What is their purpose? Checkpoints determine if the cell is ready to move to the next stage – if not error is corrected or apoptosis occurs. Start 1: G1 checkpoint, determines if cell is large enough and is the environment is favourable. S phase: Determines if replication is proceeding Enter Metaphase: G2 checkpoint, determines if cell is large enough, if the environment is favourable and if all DNA is replicated. Exit from Metaphase: metaphase checkpoint, determines if chromosomes are aligned on spindle. 12. What can trigger arrest during the cell cycle? Errors such as DNA damage, and cell size must be adequate and the environment must be favourable. 13. How many chromosomes are there in a somatic cell of a person with Down syndrome (trisomy of chromosome 21) 47 a) How many autosomes does this person have in a somatic cell? 45 b) How many sex chromosomes does this person have in a somatic cell? 2 In a germ cell? 2 In a gamete? 1 In a spermatozoid? 1 In an ovum? 1 In a zygote? 2 c) How many DNA molecules does this person have in mitotic metaphase? 47 chromosomes each containing 2 sister chromatids or DNA molecules, 94 In G1 phase? Chromosomes are unipartite, or one of the sister chromotids, 47 d) How many telomeres are there in a person’s somatic cell during G2 phase? 4 per chromosome, 4x47=188 e) There are 4 alleles for a certain gene carried by chromosome 21 in human population. How many alleles does a person with Down syndrome have in a somatic cell in G1 phase? Each chromatid has 1 allele for the gene, are 6 chromatids for individuals with Trisomy 21, 6. How many different alleles for this gene could the same person have? 4 5 BIOL 308 - Study Questions f) There are 13 alleles for a certain gene carried by chromosome 21 in human population. How many alleles does a person with Down syndrome have in a somatic cell in G1 phase? 6 How many different alleles for this gene could the same person have? 6, since only contains 6 alleles. There are 13 alleles for a certain gene carried by chromosome 21 in human population. g) How many alleles does a person with Down syndrome have in a cell which is in meiosis I anaphase? 6 h) How many different alleles for this gene could the same person have in a cell which is in meiosis I anaphase? 6 14. When does chromosome segregation happen in mitosis? In meiosis? When does chromatide segregation happen in mitosis? In meiosis? Chromosome segregation does not occur in mitosis, but it occurs in anaphase 1 in meiosis. Chromatid segregation occurs in anaphase in mitosis, and in anaphase 2 in meiosis. Lectures 10-11 1. List proteins (enzymes and other factors) involved in the process of DNA replication in E. coli. Explain the role of each of these proteins in replication. DnaA: initiates replication by recognition of DnaA boxes DnaB: helicase, separates the strands DnaC: escort to DnaA forming pre-priming complex SSB, single-strand binding proteins: binds to ssDNA, prevents double helix from reforming Primase, DnaG: elongates existing primer strands RNase H: removes RNA primers except last nucleotide, in lagging strand DNA polymerase I: removes ribonucleotides (last nucleotide), fills in deoxyribonucleotide gaps DNA polymerase III: adds nucleotides in 5’-3’ direction DNA ligase: links fragments on lagging strand Topoisomerase: recognize and regulate supercoiling 2. What is meant by replication being bidirectional? Semiconservative? Continuous and discontinuous? There are two replication fork that move in bother directions from the origin. Replication is semiconservative because one strand is the original and the other is produced from the original. Since DNA is anitparrallel and DNA polymerase III can only move 5’ to 3’ a continuous or leading strand and a discontinuous or lagging strand are produced. 3. Contrast the role of DNA polymerase I and III in E. coli DNA replication. DNA polymerase I removes the leftover ribonucelotide from each gap in the lagging strand, and fills in the gap with deoxyribonucleotides in the 5’ to 3’ direction. DNA pol I also proofreads in the 3’ to 5’ direction. DNA polymerase III adds nucleotides in the 5’ to 3’ direction on both the leading (continuous) and lagging (discontinuous) strands. 4. Which subunit of DNA polymerase III provides processivity? Which protein complex loads this subunit onto the DNA? The beta subunit functions as a clamp and increases the processiviy of polymerase. The gamma complex loads and unloads the beta complex. 5. How can discontinuos synthesis of the lagging strand keep up with continuous synthesis of the leading strand? The leading strand’s replication speed is slowed down by the presence of the leagging strand. 6. Why is decatenation required after replication of circular DNAs? Decatenation is required to unloop the two pieces of circular DNA from one another. 7. Why do eukaryotes need telomeres but prokaryotes do not? Eukaryotes have linear DNA, where after replication a 3’end overhang exists from lagging strand since not enough room to fit primer. This overhang is generally cleaved, therefore can’t have it contain important information. 8. What would be the components necessary to make DNA in vitro by using DNA polymerase I. RNA primers, DNA polymerase III, DNA, ligase, topoisomerase, helicase/DnaB, SSBP, primase, RNase H, DnaA, DnaC, and dNTPs. 6 BIOL 308 - Study Questions 9. What properties would you expect an E. coli cell to have if it had a temperature – sensitive mutation in the gene for DNA ligase? Unsealed fragments on lagging strand. Only half the DNA would be viable in harmful temperatures. At correct temperature no problems should occur. 10. What properties would you expect an E. coli cell to have if it had a temperature – sensitive mutation in the gene for DNA polymerase I? Gaps in lagging strand at extreme temperatures, DNA not viable with lagging strand. At correct temperatures no problems should occur. 11. Compare and contrast major eukaryotic and prokaryotic DNA polymerases. Eukaryotes: Pol alpha makes the RNA primer and does some elongation, Pol sigma is responsible for elongation and maturation of Okazaki fragments and proofreads Pol alpha, and Pol epsilon is responsible for assembly of replisome (DNA replicator machinery). Prokaryotes: DNase I, III, and II which is involved in repair. 12. What is telomerase and why is it important? Telomerase carries an RNA template that is complementary to the telomeric repeats and can therefore cause extension of telomeres. This helps preserve the DNA. 13. What is the major difference between bacterial and eukaryotic replication that allows a eukaryotic cell to replicate its DNA in a reasonable amount of time? Eukaryotic cells have multiple sites of replication. 14. Describe the events that occur at an origin of replication during initiation of replication in E. colii? Specific proteins recognize the origin of replication, contains DnaA boxes that DnaA recognizes. The area is rich in A-T base pairs allowing for easier denaturation. Initiation depends on methylation of oriC, on both strands. Dam methylase methylates DNA at GATC sequence. This prevents the DNA from continuous replication since both strands must be methylated ad after one round of replication on one of each strands is methylated. 16. What are cis- elements? What are trans- factors? Cis-elements: sites or sequences on DNA like DnaA box and OriC. Trans- factors: proteins that diffuse through cell/nucleous and recognize cis-elements and bind to them, like DnaA, DnaB and DnaC. 17 Is the following statement true or false: Regardless of whether a gene is expressed in a given cell type, it will replicate at the same, characteristic time during S phase. Explain your reasoning. True, all genes need to be present and are required in the replicated DNA, whether or not expressed. 18. Thinking question: Is making of RNA primers (by primase), which have to be subsequently removed and replaced with dNTPs (by DNA polymerase I) actually wasteful and energetically inefficient process? (Hint: think about fidelity of primer-making vs. proofreading capability of DNA pol I, in other words what is the only purpose of RNA primers – is the accuracy of this process very important at this point?) DNA polymerase can only elongate chains and therefore requires a primer. Need RNA primer to build off of. Also, RNA primer doesn’t need to be prefect, and therefore less energy required, since it will be removed. 19. You preformed Cot analysis using genomic DNA samples obtained from a 2 year-old child and a 76 year-old individual. Results of this analysis show that one of the most rapidly reassociating classes of DNA is substantially reduced in the older individual with respect to the 2 year-old. How can you explain this finding? (hint: think about termination of linear DNA replication.) Telomeres contain repeats, which anneal quickly together in Cot analysis. As we grow older and out cells replicate, our telomeres reduce, and therefore reduce the number of repeats. Lectures 12-14 1. Distinguish between the terms “mutation”, “DNA repair” and “recombination”. Mutation: errors in the DNA, may be lethal, silent, or beneficial, caused by replication errors, spontaneous changes in DNA, and external factors like radiation. DNA repair: mechanisms like DNAP and mismatch repair, repair mutations. Recombination: occurs during replication, repairs lesions the fork encounters to continue replication, fills in one strand by retrieving corresponding single strand from another homologous duplex. 7 BIOL 308 - Study Questions 2. List and briefly explain three major causes for mutation in DNA. Replication errors not fixed by proofreader mechanisms, spontaneous changes in DNA like deaminations and depurinations, and external factors like radiation, temperature, and mutagens. 3. Explain how errors in DNA replication can lead to mutations. Wrong base may be put into place, and go unnoticed by repair mechanisms. 4. Distinguish between the effects of mutations on the somatic and germ cells of multicellular organism. Somatic mutations may effect organism. Mutations in germ cells may effect offspring. 5. List the various types of DNA repair mechanism (we have mentioned seven). 1. Proof-reading or Editing by DNA Polymeras 2. Direct Reversal of Damage 3. Base Excision Repair 4. Nucleotide Excision Repair 5. Mismatch Repair 6. Recombination Repair 7. Error-Prone Repairs 6. Give the detailed description of the base excision (or nucleotide excision) repair process in bacteria. The incorrect base flips to the other side of the backbone. Glycosylase recognizes and removes base (ie. Uracil in DNA) by hydrolyzing glycosidic bond. AP endonuclease and phosphodiesterase removes sugar phosphate backbone. The nucleotide is replaced by DNA pol and DNA ligase seals the nick. 7. Describe the mismatch repair process of bacteria (pay attention at the ways in which the daughter and parent strand are recognized by repair system). MutS scans DNA for distortion and binds to mismatch ie. loop. MutL is recruited by MutS-mismatch-DNA complex. MutS ttranslocates along DNA until a GATC sequence is reached (used to determine which strand is parent). MutSL activates MutH (endouclease), which recognizes GATC and binds MutSL. MutH nicks the unmethylated or daughter strand of DNA. Strand is then progressively degraded from GATC to mismatch and a new DNA strand is synthesized by DNaseIII and sealed by ligase so the mismatch is removed and replaced with correctly base-paired nucleotide. 8. Describe briefly the mechanism of direct reversal of damage in bacteria. Direct reversal repairs thymine dimmers through photoreactivation by photolyase. The thymine dimmer is broken by light-dependent activity. 9. Distinguish between the two DSB repair mechanisms we talked about in class. Double stranded breaks can be repaired by nonhomologous end joining (NHEJ) and recombination repair. NHEJ can directly ligate the two ends of broken DNA by cleaving overhangs, which is error prone. Recombination repair references the sister chromosome to repair the other and is less prone to error. *Recombination repair may also be used when one strand is broken. 10. What is SOS repair mechanism, when is it used and why is it important? SOS translesion repair is higly error-prone because polymerases add nucleotides randomly, without proper base pairing. It introduces mutations by still enables complete chromosome replication. It’s used as a last attempt and is important because it introduces mutations – variability. 11. Describe the term “non-homologous end joining” and explain how this process results in the repair of double strand breaks in DNA molecule. Cleaving of overhangs in a damaged piece of DNA allows for joining of non-homologous ends to join. 12. What are biological roles of DNA recombination? DNA repair, formation of new genes and integration of a specific DNA element. 13. List and briefly describe the ways genomic DNA can be rearranged (there are three of them). General or homologous recombination: genetic exchange between pairs of homologous DNA sequences. Site specific recombination: occurs between sequences with a limited stretch of similarity, involves recombination sites. Transposition: mobile DNA element moves from one site to another, usually little sequence similarity involved. 14. What are the key steps in single stranded model of homologous DNA recombination? The two homologous DNA molecules align. A break is introduced into one strand of each homologous DNA molecule. A single strand region from one strand invades and pairs with the complementary strand from the homologous molecule. The two molecules become connected through crossing DNA strands, forming a Holliday junction. Branch migration occurs, moving the Holliday junction. Cleavage of the Holliday junction results in new combinations of DNA. 15. Explain the relationship between hybrid duplex and heteroduplex. (You can use diagram.) A homoduplex is the strand without homologous chromosomes DNA. A heteroduplex contains a hybrid of DNA. 8 BIOL 308 - Study Questions 16. Describe the function of proteins involved in homologous recombination of E. coli. Used in repair since E.coli is haploid. RecBCD is a nuclease/helicase that binds to duplex DNA and processes it into a substrate for recombination by generating single strands for invasion. RecA starts recombination by facilitating pairing of homologous DNAs and is involved in strand invasion. RuvAB is a complex with helicase activity and recognizes Holliday junctions and catylases branch branch migration. RuvC is an endonuclease that catalyses the resolution of Holliday junction. RecBCD complex enters the DNA at the site of the double strand break and unwinds and degrades both strands of DNA in the presence of ATP. Upon reaching a chi sequence activity changes and the chi sequences control its activity. Degradation of 3’ end stops and degradation of 5’end increase creating 3’ overhang. RecA binds to the overhang and initiates strand exchange. Branch migration is caused by RuvAB and the Holliday junction is resolved by RuvC. 17. Describe the role of chi sequences in homologous recombination of E. coli. Chi sequences enhance recombination frequency. 18. What is the major role of homologous recombination in prokaryotes? Repair 19. What are the roles of homologous recombination in eukaryotes? Repair and variability (in meiosis) 20. When does the programmed creation of DSBs occur in eukaryotes? In which type of cells? Briefly describe the process. Programmed double stranded breaks occur between homologous chromosomes in germ cells during prophase 1 of meiosis 1. Double stranded breaks are formed at recombination hotspots. Spo11 generates double stranded breaks. Mre11 is a nuclease that creates 3’ overhangs. Dmc and Rad1 cause strand invasion. End up with 2 Holliday junctions. 21. Define gene conversion (use your own words). What is the significance of gene conversion? Gene conversion is where DNA from one chromatid or chromosome is exchanged with a homologous pair. It increases variability in the genes. 22. What is the role of mismatch repair mechanism in gene conversion? If the new DNA is incorrectly placed into the DNA ie. Forms a loop, doesn’t match, the mismatch repair mechanism will recognize this and replace the new DNA to match the parent DNA. 23. Explain the role of site-specific recombination in infection of E. coli genome by lambda phage. Lambda phage can insert it’s DNA into the E.coli genome at a recombination site, entering latent prophage state. 24. What are potential effects of transposons on the genome? Transposons can cause genetic changes by inserting into genes/coding sequences and into regulatory sequences, possibly disrupting the cell. 25. List and briefly describe three mechanisms by which genetic elements are able to move from one site in the genome to another. DNA-only transposons: move as DNA either by cut and past or replicative pathways. Retroviral-like retrotransposons: moves via RNA intermediate produced by a promoter in the direct long terminal repeats vertical movement (between cells). Nonretroviral retrotransposon: moves via an RNA intermediate that is often produced from a neighboring promotor, horizontal movement (within a cell). 26. Draw a fully annotated diagram illustrating the transposition mechanism of a simple IS transposon (or retroviral-like retrotransposon, or non-retroviral retrotransposon). 9 BIOL 308 - Study Questions 27. Who was Barbara McClintock and what was her major scientific contribution [no more then two (2) sentences]. Nobel prize winner for her discovery of mobile genetic elements – transposons! Lecture 15 1. What kind of information can we obtain from a northern blot? RNA detection 2. What kind of information can we obtain from a Southern blot? DNA desction 3. Compare and contrast Southern and northern blotting technique. In Southern blotting DNA is denatured and broken into fragments that are separated by agarose gel electrophoresis, then blotted onto nitrocellulose paper, and a probe is added to label DNA to be visualized by autoradiography, In northern blotting mRNA, tRNA, and rRNA are isolated from cells and elecrophoresed, very similar to Southern blotting except RNA does not need to be denatured since already single stranded. 4. Compare the information obtained by northern analysis with the information obtained by microarray experiments. A northern blot determines the steady-state level of a specific transcript in a certain RNA mixture (the abundance of specific mRNA at a certain time and under certain conditions). A microarray experiment is similar to running hundreds of northern blots, since you can compare different mRNAs, or mRNAs from different environments, to see if they are expressed. 5. How do we treat a DNA gel prior to Southern blotting? Explain why. 6. Thinking question: you have made a short probe (50 nucleotides) from a certain genomic DNA (note: genomic DNAs include both exons and introns). Are you sure that you would be able to use this probe for northern hybridization? Explain your reasoning. The DNA is cut with restriction nucleases because then we can determine if the gene sequence occurs more then once, and we can estimate the positions o the gene copies. The same probe used in DNA could not by used in RNA because all the information found in DNA is not transcribed to RNA, and RNA removes introns. The probe used for DNA may contain part of this intron and therefore not bind to the RNA. 7. Distinguish between a template, primer and a probe. Would you be able to use primer as a probe? If yes, explain when/how. A template is a known sequence that is similar to the target, that we can base our probe off of. A probe binds to the unknown DNA if there is a match. A primer binds to DNA to start transcription. A primer could be used as a probe as long as you do not want the primer to act as a primer, because the primer is meant to fall off the DNA strand after replication.. 8. Why is it important to know the exact start site of transcription? It is important in order to insert a gene afterwards, so it will be transcribed. 10 BIOL 308 - Study Questions 9. What does SDS-PAGE stand for? Explain the roles of SDS in SDS-PAGE (keep in mind - two major roles). Sodium deoceyl sulfate polyacrylamide gel. SDS is a detergent that denatures proteins, and gives them a nigative charge proportional to their mass, allowing for size seperation.. 10. Compare the information you could obtain by SDS-PAGE with the information you could obtain by using Western blotting. SDS-PAGE seperates proteins based on size. These separated proteins can be transferred to a membrane and incubabted with a specific antibody that will bind to a specific protein. If that protein is present, a secondary antibody will bind to the first and allow it to be visible. 11. General knowledge: distinguish between antibody and antibiotic. An antibody (immunoglobin) is found in the body and is used to identify and neutralize foreign objects like bacteria and viruses. An antibiotic is a compound or substance that kills or slows down the growth of bacteria. 12. How does salt concentration influence hybridization process between target DNA or mRNA and probe during nucleic acid hybridization step performed in northern (or Southern) blotting method? Higher salt concentration neutralizes charged backbones allowing strands to come together, so probe more easily binds. 13. Change “salt concentration” in the above question to: a) temperature at too high of a temp the probe won’t anneal b) size of the probe* small probe more likely/takes less time to anneal c) probe’s % of identity with the target sequence* The closer in similarity the probe is the more likely of have a correct match, reduces mismatch. Lectures 16-19 1. You have discovered base changes in the promoter region of the operon in a bacterial chromosome. Would you expect these changes to act in trans on another copy of the operon? Explain your reasoning. No, since the change is in the sequence or cis element and not the protein or
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