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186 Cards in this Set
- Front
- Back
Differences between prokaryotes and eukaryotes and viruses
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Bacteria are smaller than eukaryotes, move via flagella, have a circular dbl stranded DNA chrmsm
Eukaryotes have mitochondrian and/or chloroplasts, a nuclear membrane, a cytoskeleton, and a nucleus. Viruses are acellular, need cell machinery to make proteins and get energy. Virus genetic material can be ssRNA, dsRNA, ssDNA, or dsDNA. A bacteriophage is a head and tail virus. One phage = one plaque. Viral life cycles can by lysogenic or lytic. A lysogenic life cycle is one in which the virus incorporated its genome into the bacterial genome where it dvides passively. A lytic life cycle consists production of more viral DNA, construction and cell lysis. |
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What kind of genetic material can viruses have?
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Virus genetic material can be ssRNA, dsRNA, ssDNA, or dsDNA
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What is a bacteriophage?
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A bacteriophage is a head and tail virus. One phage = one plaque.
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diff life cycles of virus?
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Viral life cycles can by lysogenic or lytic. A lysogenic life cycle is one in which the virus incorporated its genome into the bacterial genome where it dvides passively. A lytic life cycle consists production of more viral DNA, construction and cell lysis.
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Why are bacteria and bacteriophage good experimental organisms?
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They are simple, fast and easy to grow, easy to radioactively lable, and WT is prototrophic (can grow on simple medium)
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Describe what happends at the bacterial replication fork
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Autoradiography shows that bacterial circular chromosomes form theta structures when replicating. Viewing a bacterial chromosome requires growing bacteria dor several generations on radioactive thiamine. The gorwing it for ½ generation on regular thiamine. The bacteria is put in liquid, gently lysed and allowed to settle on filter paper. The filter paper is exposed to X-ray film and the radioactivity os detected.
Replication starts at the ORI and is bi-directional forming a theta structure. |
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How do plasmids replicate?
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Follows a rolling circle, or sigma form.
A break occurs in one strand and it begins to “roll off.” A New strand is synthesized behind the old ones. |
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In eukaryotic replcation where does it start?
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There are many ORI.
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What direction is DNA synth?
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5’--> 3’
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WHat is necessary for DNA synthesis?
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needs primer- short length of nuclotides
needs template DNA polymerase |
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Difference btw DNA and RNA
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DNA has thiamine and dideoxyribose as a sugar. (Dideoxyribose makes the DNA more stable than RNA) DNA does NOT hydrolyze in alkali.
RNA has a uracyl instead of a thiamine and ribose as a sugar. RNA has a tendency to undergo oxidative deaminization (a change form a C to a U). The cell can detect this change and fix the RNA. |
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Characteristics of the ORI in prokaryotes
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The origin of replication in Prokaryotes begins from a fixed origin called oriC and proceeds in both directions. There is a dnaA box to which DnaA binds. This box is repeated 5 times in the oriC and in respons to binding AT rich regions begin to unwind. More DnaA proteins coat the oriC as it is opened. Then units of then helicase DnaB bind and slide in a 5’ à 3’ direction to unzip the helix at the replication fork. Primase and DNA poly III are recruited and DNA synthesis begins.
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Characteristics of the ORI in eukaryotes
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In Eukaryotes there are many oriC and replication proceeds in both directions from these points. ORC first binds to the oriC and then recruits Cdc6 and Cdt1. These proteins ensure that DNA replication will take place only in S phase. They in turn recruit the MCM complext which licences the ori and allows for replication.
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what are some diff types of mutations
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Suppression: restores reading frame and function
Synonymous or same sense mutation: maintatins same amino acid sequence. (because of the degeneracy of the code) Nonsense mutiation: A nonsense mutation is always profound. It results in an early stop codon Missense mutation: Can be profound or cannot. It changes one amino acid. Tautameric shift- all 4 bases can undergo a tautameric shift. In a tautameric shift one base pair takes on a different form and may be able to pair with a nonmatching base. This may cause a mustation if it is not fixed with a proofreading enzyme |
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Suppression mutation
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restores reading frame and function
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Synonymous or same sense mutation
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maintatins same amino acid sequence. (because of the degeneracy of the code)
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Nonsense mutiation
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A nonsense mutation is always profound. It results in an early stop codon
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Missense mutation
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Can be profound or cannot. It changes one amino acid.
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Tautameric shift
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all 4 bases can undergo a tautameric shift. In a tautameric shift one base pair takes on a different form and may be able to pair with a nonmatching base. This may cause a mustation if it is not fixed with a proofreading enzyme
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What is the start codon?
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5’ AUG 3’
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What happends when AUG is in a ribosome and a shine delgorno sequence preceeded it?
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This is a start codon:
The start codon does not cause the insertion of a methyanine by tRNAMet but it is recognized by a special tRNA called an initiator. In bacteria, a formyl group is added to the met while it is attatched to the tRNA forming N-formylmethionine. |
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Griffin experiment: what is it? what does it prove?
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Discovered transformation in bacteria: “transforming principle”
S live strain à mouse dies S heat killed à mouse lives R strain à mouse lives R + heat killed S à mouse dies * |
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Avery Experiment: What is it and what does it porve?
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Same experiment as Griffin. Heat killed virulent bacteria + R strain but with different things disabled(protease, DNAse, etc...)
Mouse lives only when DNA is destroyed Dies under other conditions (DNA is intact) futher pointed to DNA as hereditary material. |
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Hershey and Chase Experiment: What is it and what does it prove?
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Used T2 bacteriophage
Labeled DNA with radioactive Phosphorus Labeled Proteins with radioactive sulfur (There is no sulfur in DNA and no phosphorus in protein) Infected Ecoli Sheared phage ghosts off of bacteria with kitchen blender Separated ghosts from bacteria in centrifuge Found: labeled phosphorus INSIDE bacterial cells Labeled sulfur in phage ghosts THEREFORE: transforming principle is DNA |
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Chargaffs Rules
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1. A = T
2. G = C 3. A + G = T + C 4. A + T ≠ always G + C |
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When is melting point of DNA high>
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Melting point is higher at higher G + C content
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describe Molecular Density Gradient centrifugation
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Use cesium chloride. If it is spun at high speeds for many hours ions tend to be pushed down to the bottom of the tube and an ion gradient is established with hisgest ion density at bottom. DNA centrifuges with the cesium chloride forms a band at the position identical with its density in the gradient. DNA of different densities form bands at different places.
If asked about what percentage if heavy/light, do a punnet sq |
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where do you find Phosperous atoms?
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In DNA not protein.
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Meselson-Stahl experiment: what is it and what did it prove?
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replication method of DNA is semiconservative
cells are grown in 15N (heavy) medium as a label for several generations and then transferred into a normal light 14N form of medium for one or two cell divisions. Molecules were separated by density with cesium chloride gradient centrifugation. found: parental: one heavy band fisrt gen: band in middle (1/2 heavy, 1/2 normal) second gen: two bands one all light, one 1/2 heavy 1/2 light |
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Replication fork
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zone where the dbl helix is unwinding
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Leading strand
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New strand synthesized towards the replication fork 5’--> 3’
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Lagging strand
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New strand synthesized away from replication fork with Okazaki fragments.
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Okazaki fragment
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short (1000-2000) nt stretched of newley synthesized DNA on the lagging strand
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DNA polymerase I
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Removes RNA primer and fills in with DNA. Has 3’ à 5’ exonuclease activity and 5’ à 3’ exonuclease activity (the Kornberg enzyme) A distributive enzyme.
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DNA polymerase II
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back up role in DNA repair
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DNA polymerase III
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The main DNA polymerse in bacteria. Uses RNA as a primer. Has 3’ à 5’ exonuclease activity as a proofreading function. A processive enzyme with an associated sliding clamp.
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DNA polymerase III holoenzyme
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The main prokaryotic DNA polymerase. It has10 subunits, there are 10 polIII holoenzymes per cell, it is highly processive (synthesizes 1000 nucleotides per second) and has NO 5’ to 3’ exonuclease activity. The most important individual subunits are:
Alpha- This is the actual polymerase. There are 2 alpha subunits per holoenzyme, one for the leading and one for the lagging strand. Epsilon: the 3’ to 5’ exonuclease (proofreading or editing activity) Beta: sliding clamp or processicity factor. Looks like a doughnut and there is one per each alpha subunit. Gamma: clamp loader, requires and recognizes a 3’ hydroxyl end (the 3’ end of a primer) to use ATP to load the clamp. |
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alpha subunit of DNA polIII holoenzyme
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This is the actual polymerase. There are 2 alpha subunits per holoenzyme, one for the leading and one for the lagging strand.
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Epsilon subunit of DNA polIII holoenzyme
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: the 3’ to 5’ exonuclease (proofreading or editing activity)
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Beta subunit of DNA polIII holoenzyme
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sliding clamp or processicity factor. Looks like a doughnut and there is one per each alpha subunit.
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Gamma subunit of DNA polIII holoenzyme
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clamp loader, requires and recognizes a 3’ hydroxyl end (the 3’ end of a primer) to use ATP to load the clamp.
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How is eukaryotic replication diff from bacteria?
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Eukaryotic replication is not highly processive because there is no sliding clamp
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How is euk replication similar to bacteria?
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One continuous, one discontinuous strand; helicase; topoisomerase
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Integrase
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A site specific recombining enzyme in specialized transduction
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Telomerase
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An enzyme important in synthesizing the telomeres of a lagging strand of linear DNA. Telomerase adds a noncoding sequence to the 3’ end. It carries a small RNA molecule that acts as a template for the polymerization of the telemetric repeat unit
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Telomere
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- the ends of linear DNA
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What is the difference between a telomere and a broken chromosome
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A telomere has end is stable, a broken chromosome (usually sticky ends)? and can be translocated across the chromosome.
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How do telomeres stabilize?
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Because telomeres have a repeating sequence with a 3’ overhang, the overhang can loop arount in a D-loop-T-loop to form a stable structure.
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Base stacking interactions
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Interactions between electrons of bases stacked above each other. They stabilize the double helix. A decrease in base stacking interactions results in an increase of UV absorption (hyperchromicity)
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PCNA
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proliferating cell nuclear antigen acts like a sliding clamp in eukaryotes?
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DNA ligase
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Joins knicks by reforming a phoshodiester bond. (Fills in Gap left by poly 1 and poly 3 on lagging strand)
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DNA topoisomerase I
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relaxes supercoiling of DNA due to unwinding by breaking a single strand of the DNA and then rejoining them.
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DNA topoisomerase II
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Relieves catenation by breaking dbl strand of DNA. Catenation happened at the end of the replication of a circular chromosome.
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DNA helicase
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part of the replisome that breaks DNA hydrogen bonds. It unzips the dbl helix
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DnaA protein
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first step in assembly of the replisome. DnaA binds to a specific 13-bp sequence (called a DnaA box) that is repeated 5 times in the OriC. In reponse the origin encircles the DnaA proteins and is unwound at a cluster of AT rich region. DnaA then reqruits DnaB.
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DnaB
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a helicase that binds after DnaA and slides in a 5’ à 3’ direction to unzip fork.
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Primase
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an RNA polymerase made up of the central protein primosome that synthesizes a short stretch of complementary RNA to a DNA template. No exonuclease activity.
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Single stranded binding protein
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prevents single stranded DNA from reforming a dbl helix by stabilizing the single-stranded form
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Nucleosome
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basic unit of chromatin which consists of DNA wrapped around histone proteins.
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Chromatin Assembly Factor I (CAF-I
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protein that binds histones and targets them to the replication fork, where they can be assembled with new DNA.
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Proliferating cell nuclear antigen (PCNA)
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eukaryotic version of the clamp protein.
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RNA polymerase holeoenzyme
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The bacterial RNA polymerase that scans DNA for a promoter sequence. It is made up of four subunits of the core enzyme plus the sigma factor. It needs ATP, UTP, CTP, GTP, a DNA template. It does not need a primer in vitro.
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Sigma factor (σ)-
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In bacteria the part of the RNA polymerase holeoenzyme that binds to the –35 and –10 regions of the promoter and postitions the holeoenzyme to initiate transcription at the correct start site. It also helps melt the DNA at the –10 region. Once RNA poly is bound sigma disassociates from the holeoenzyme. Different sigma factors recognize different promoter regions.
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Transcription bubble
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Region of single stranded DNA where the template strand is exposed.
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RNA polymerase I
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In eukaryotes, transcribes rRNA genes, 18S and 28S (the larger subunits (except 5S rRNA)
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RNA polymerase II
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In eukaryotes, transcribes all protein-coding genes (mRNA) and some snRNAs
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RNA polymerase III
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In eukaryotes, transcribes small functional RNA genes (tRNA, some snRNA, and 5S rRNA)
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Reverse transcriptase
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RNA dependant DNA polymerase Not all RNA viruses have reverse transcriptase
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Ribozyme
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an RNA that acts as a catalyst
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Types of RNA
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mRNA- intermediate btw DNA and protein.
Functional RNA- RNA that is not intended for translation and serves its own purpose such as: Transfer RNA (tRNA)- molecules responsible for bringing the correct amino acid to the mRNA in translation Ribosomal RNA (rRNA)- the major components of ribosomes Small Nuclear RNAs (snRNAs) in eukaryotes, a part of a system that further processes RNA transcripts. Some snRNAs guide the modification of rRNA’s others unite with other proteins to form the spicosome that removes introns from eukaryotic mRNA. |
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mRNA
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intermediate btw DNA and protein.
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Functional RNA
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RNA that is not intended for translation and serves its own purpose
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Transfer RNA (tRNA)-
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functional RNA
molecules responsible for bringing the correct amino acid to the mRNA in translation |
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Ribosomal RNA (rRNA
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Functional RNA
the major components of ribosomes |
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Small Nuclear RNAs (snRNAs)
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Functional RNA
in eukaryotes, a part of a system that further processes RNA transcripts. Some snRNAs guide the modification of rRNA’s others unite with other proteins to form the spicosome that removes introns from eukaryotic mRNA. |
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Spicosome
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A functional splicing unit made up of small nuclear ribonucleoproteins (snRNPs)
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Coding strand
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strand that is NOT used by RNA polymerase as a template but that resembles the mRNA transcript in content. When giving the DNA sequence for a gene give this strand.
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Template strand
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The strand used as a template to synthesize the mRNA transcript but is opposite as far as content.
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Stages of transcription:
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Initiation-
Prokar: Binding of RNA polymerase to the promoter with the help of sigma factors Euk: Binding of RNA poly to promoter with help of GTFs. First stem is the binding TBP to the TATA box. Then the Preinitiation Complex (PIC) is assembled Elongation- synthesis of mRNA Prok: RNA holoenzyme synthesizes mRNA off of the template strand Euk: RNA poly II becomes disassociated form GTF and synthesizes the primary transcript off the template strand. GTF stay at the promoter and recruit another RNA poly II for multiple transcripts. Elongation id ended when an enzyme recognized a conserved AAUAAA or AUUAAA sequence and cuts of the transcript 20 bases further. Termination- end of transcription. Prok: can be intrinsic or rho dependant. EUK: an enzyme recognizes AAUAAA or AUUAAA and stops transcription 20 bases down. |
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Initiation
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Prokar: Binding of RNA polymerase to the promoter with the help of sigma factors
Euk: Binding of RNA poly to promoter with help of GTFs. First stem is the binding TBP to the TATA box. Then the Preinitiation Complex (PIC) is assembled |
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intiation of transcription in prokaryotes
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Binding of RNA polymerase to the promoter with the help of sigma factors
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initiation of transcription in eukaryotes
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Euk: Binding of RNA poly to promoter with help of GTFs. First stem is the binding TBP to the TATA box. Then the Preinitiation Complex (PIC) is assembled
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Elongation of transcription
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synthesis of mRNA
Prok: RNA holoenzyme synthesizes mRNA off of the template strand Euk: RNA poly II becomes disassociated form GTF and synthesizes the primary transcript off the template strand. GTF stay at the promoter and recruit another RNA poly II for multiple transcripts. Elongation id ended when an enzyme recognized a conserved AAUAAA or AUUAAA sequence and cuts of the transcript 20 bases further. |
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Elongation of transcription in prok
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RNA holoenzyme synthesizes mRNA off of the template strand
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Elongation of transcription in euk
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RNA poly II becomes disassociated form GTF and synthesizes the primary transcript off the template strand. GTF stay at the promoter and recruit another RNA poly II for multiple transcripts. Elongation id ended when an enzyme recognized a conserved AAUAAA or AUUAAA sequence and cuts of the transcript 20 bases further.
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termination of transcription
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end of transcription.
Prok: can be intrinsic or rho dependant. EUK: an enzyme recognizes AAUAAA or AUUAAA and stops transcription 20 bases down. |
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Why is transcription more complicated in eukaryotes?
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1. more genes; more non-coding DNA makes initiation more difficult;
2. Eukaryotes have a nucleus where RNA is synthesized. RNA processing must take place before it is exported outside of the nucleus. 3. chromatin in Eukaryotes |
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Intrinsic termination
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termination where terminator sequence contains about 40 bp ending in a GC rich region followed by a run of six or more A’s. The G and C’s interact with each other and form a hairpin loop in the mRNA transcript. As the RNA poly synthesized the weak bonding U’s from the A rich region it backtracks, comes across the hairpin loop, and falls off.
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Rho dependant termination
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A hexamer called rho binds to the rut (rho utilization site), RNA polymerase pauses and rho mediates the disassociation of RNA poly.
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General transcription factors (GTF’s
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enzymes that aid in transcription
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RNA processing
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modification of RNA in eukaryotes before it is transported outsider the nucleus. These processes include: the addition of a cap at the 5’ end, the addition of a 3’ poly A tail, and splicing to eliminate introns. It is usually contratranscriptional.
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Primary transcript (pre-mRNA):
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mRNA transcripts without any processing or modifications
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Mature RNA (mRNA):
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mRNA with modifications and processing
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Promoter
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a specialized DNA sequence located upstream (5’ of the gene) to the start of transcribes region where RNA polymerase binds. Prokaryotic promoters have –35 and –10 regions. Eukaryotic promoters
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PIC
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preinitiation complex that consists of 6 GTF and RNA polymerase II core in eukaryotes
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Consensus sequence
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a general promoter template for an organism
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5’ UTR
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5’ untranslated region. The segment of mRNA between the start of transcription and the first translation codon.
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3’ UTR
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3’ untranslated region. The segment of DNA that is transcribed into mRNA but does not code for protein.
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Exonuclease
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chews in and destroys DNA or RNA form ends
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Endonuclease
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cuts from within
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Replisome
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The large complex consisting of DNA polymerase that coordinates the activity at the replication fork
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Poly III holoenzyme
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consistis of two catalytic cores (one for leading, one for lagging strand) and many accessory proteins. The accessory proteins form bridges that connect the catalytic cores, coordinating the synthesis of the leading and lagging strands.
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Sliding clamp
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A Poly III holoenzyme accessory protein which encircles DNA like a doughnut and keeps Poly III attatched to the DNA.
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A distributive enzyme
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adds only a few nt before falling off the template. (primase
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A processive enzyme
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stays attatched for a long time and adds many NTs. (poly III)
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ORC
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Origin recognition complex. A protein requires to assemble a replisome in Eukaryotes. It binds to sequences in yeast origins and reqruits other proteins (Cdc6 and Cdt1) Then this combo of proteins recruit the replicative helicase, called the MCM complex.
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Cdc6 and Cdt1
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proteins that help initiate replication in eurkaryotes. They are synthesized during late mitosis and gap 1 and destroys by proteolysis after synthesis have begun. They ensure that the replisome is assembled only before S phase.
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MCM complex
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Eukaryotic helicase that is said to licence the origin
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Kozak sequence
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like the shine delgarno sequence but in vertebrates. It is purine rich (A or G)
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Shine-Delgarno Sequence
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: In prokaryotes a special sequence in the 5’ untranslated region that preceeds a start codon. It pairs with the 3’ end of an rRNA called the 16S rRNA, in the 30S subunit. The pairing correctly positions the initiator codon in the P site.
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Initiation factors
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Proteins that help initate translation:
prok: IF1: initiation factor helps initiator tRNA enter P site IF2: initiation factor helps initiator tRNA enter P site IF3: initiation factor keeps 30 Subunit dissacociated from 50 S. |
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IF1 (prok)
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initiation factor helps initiator tRNA enter P site
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IF2 (prok)
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initiation factor helps initiator tRNA enter P site
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IF3 (prok)
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initiation factor keeps 30 Subunit dissacociated from 50 S.
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Formyl-methianine
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the changed form of methianine that is attatched to the initiator tRNA. It is later removed.
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TATA box
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place in the Eukaryotic promoter (–30) where the first transcription event takes place. This is where the TATA binding protein (TBP), one part of one of the 6 GTF that recruit RNA poly II binds. It attracts other GTF.
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TBP
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TATA binding protein, one part of one of the 6 GTF that recruit RNA poly II binds. It attracts other GTF.
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Carboxyl Tail domain (CTD):
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): A part of RNA poly II that helps RNA poly II disassociate with GTF at the promoter. Initiation of Eukaryotes ends and elongation begins when CTD is phosphorylated by one of the GTFs. Important in coordination transcriptional modifications.
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5’ mRNA capping
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process coordinated by the CTD that helps protect mRNA from degradation and helps initiate translation. A cap consists of a 7-methylguanosine residue linked to the transcript by three phosphate groups.
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Poly A tail
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150- 200 A’s added to the 3’ end of eukaryotic mRNA after an enzyme recognized a polyadenylation signal.
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Polyadenylation signal
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: In euk the conserved AAUAAA or AUUAA region that signals the end of elongation and beginning of termination and where the poly A tail is added.
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Alternative splicing
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different arrangements of exons from same initial mRNA usually in diff cell types or at different stages in development.
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GU- AG rule
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Introns usually have a GU at the 5’ end and AG at the 3’ end. This is where they will be cut.
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How does splicing occur?
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snRNP’s recognize conserved sequences within the pre mRNA introns such as the GU-AG regions and the A residue that is 15 to 45 nucleotides upstream from the 3’ splice site. The snRNP’s make up a sliceosome. Components of the sliceosome interact with the CTD and attach to intron and exon sequences. The sliceosome moves the mRNA in such a way that it performs two cuts, one between the GU (5’) region and the A residue and one btw the GU and AG regions.
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What is the RNA world theory?
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The discovery was made that RNA can sometimes self-splice introns. This suggested that because RNA can act as both genetic material and a reaction catalyst it could have been the genetic material in the first cells.
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Pribnow box
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the –10 sequence in prokaryotic promoter
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CAAT box
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Is an unsteam promoter element in eukaryotes that is –40 to –110 from the transcription site and controls levels of transcription.
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Hayflick limit
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the limit of the number of divisions a cell can undergo due to the deactivation of telomerase and the loss of end chromosome genetic material. Cancer cells have no Hayflick limit and most likely have a reactivation of telomerase.
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What did Yanofsky prove? How?
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Yanofsky proved the colinearity og gene and protein by indicing mutations in the gene for triptophan synthetase alpha subunit, making it inactive. He then, by recombination, mapped the mutant sites on the DNA and biochemically identified changes in the amino acid structure of the protein. He then showed that the mutations in the DNA occurred in the same order as the changes in the protein and the distance between them corresponded.
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Why must the genetic code of DNA code for amino acids in triplets?
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If 1 base = 1 amino acid and there are 4 bases and 20 amino acids
41= 4 possible amino acids If 2 bases = 1 amino acid 42 = 16 possible amino acids If 3 bases = 1 amino acid 43 = 64 possible amino acids (which is more than enough Futhermore, Crick and Brenner showed that the reading frame can be fixed with the addition or deletion of 3 bases. |
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Describe the Crick Brenner Experiment
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Induced a single deletion or addition of a base with the chemical proflavin in phage T4 so that they would not grow on K strain bacteria. Then they found revertants but after analysis found that they were not identical to the wild type strain. So a second suppression mutation must have occurred. The suppression mutation was either a deletion or an addition of a single base that restored the reading frame. The did more exact experiments to prove that 3 deletions would be “fixed” with 3 additions.
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What do you need to make mRNA in vitro? Why would you do this?
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This was a way to figure out the amino acid mRNA code. Went on to prove code degeneracy.
Need mRNA (U, G, C, A) nucleotides, a polynucleotide prosphorylase (the protein factor), ribosomes, and charged tRNAs. The amino acids are assembled at random. |
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Why is the genetic code degenerate?
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1. To protect against mutation
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Wobble
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loose pairing of the third base (5’) in the anticodon with mRNA
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Isoaccepting tRNA
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tRNA that accepts the same amino acid but are transcribed from different tRNA genes.
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What are the stop codons?
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UAA, UAG, UGA, UGG
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What direction is mRNA read? TRNA?
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mRNA is read 5’ --> 3’
tRNA is read 3’ --> 5’ |
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aminoacytl-tRNA synthetase
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attatched amino acids to tRNA. There are at least 20 kinds of this enzyme each specific to an amino acid/tRNA combo.
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Inosine
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a rare base in tRNA that can pair with U, C, or A.
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charged tRNA
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one with an attatched amino acid
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nonsense suppressor
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: a mutated tRNA that had an anitcodon that recognizes a termination codon. Supresses a nonsense mutation. Usually these tRNA mutations happen in the minor tRNA (low codon preference in organism) and bacteria usually have two stop codons. These features protect against the mutation being lethal.
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What are the sizes of prokaryotic and eukaryotic ribosomes?
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Prok 30S and 50S. Together 70S
Euk 40S and 60S. Together 80S |
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What are the positions where tRNA fits into a ribosome and what happends there?
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EPA 3’
The A site is where tRNA enters first. This is called the aminoacyl site. The A site binds an incoming aminoacyl tRNA whose anticodon matched the codon in the A site of the 30S subunit. The P site is called the peptidyl site of the 30 S subunit. The tRNA in the P site contains the growing amino acid chain. The amnio acid chain exits through a tunnel in to 50 S subunit. The E site contains a deacylated tRNA that is ready to be released from the ribosome. |
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The A site
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where tRNA enters first. This is called the aminoacyl site. The A site binds an incoming aminoacyl tRNA whose anticodon matched the codon in the A site of the 30S subunit.
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The P site
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called the peptidyl site of the 30 S subunit. The tRNA in the P site contains the growing amino acid chain. The amnio acid chain exits through a tunnel in to 50 S subunit.
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The E site
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contains a deacylated tRNA that is ready to be released from the ribosome.
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Decoding center
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area in 30 S subunit ribosome that ensures that only tRNAs carrying anticodons that match the codon will be accepted into the A site
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Peptidyl transferase center
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area in the 50 S subunit of the ribosome where cognate tRNAs associate and a peptide bond is catalyzed
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How is translation initiated in Prokaryotes? Eukaryotes?
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In prokaryotes Initiation factors in the 30 S subunit help move mRNA into the ribosome and lead the initator tRNA bind to the P site. Then initiation factors attatch the large subunit to the small subunit. This can happen as transcription is taking place.
In eukaryotes, Eukaryotic initiation factor E4 (elF4E) binds to the 5’ cap of the mRNA. Then elF4G and elF4A bind to elF4E forming the complex elF4F. The small subunit the initiator tRNA and elF2 combo recognize elF4F and the mRNA moves into the small subunit, until the AUG codon is in. Then elF2 releases and the large subunit binds. |
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How is translation initiated in Prokaryotes?
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In prokaryotes Initiation factors in the 30 S subunit help move mRNA into the ribosome and lead the initator tRNA bind to the P site. Then initiation factors attatch the large subunit to the small subunit. This can happen as transcription is taking place.
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How is translation initiated in Eukaryotes?
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In eukaryotes, Eukaryotic initiation factor E4 (elF4E) binds to the 5’ cap of the mRNA. Then elF4G and elF4A bind to elF4E forming the complex elF4F. The small subunit the initiator tRNA and elF2 combo recognize elF4F and the mRNA moves into the small subunit, until the AUG codon is in. Then elF2 releases and the large subunit binds.
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EF-Tu
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Elongation factor Tu is a protein factor that attaches to an aminoacyl-tRNA to form a ternary complex. It is removed once the tRNA enters the A site. It can then form a peptide bond.
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EF-G
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Elongation factor G is a protein factor that is GTP driven. It moves into the A site and moves the tRNA down into the P and E sites. It then leaves and the A site is open again.
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What is the role of GTP in protein synthesis?
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GTP can bind to protein initatiation factors and change their state so that they can no longer bind to the ribosome. Specifically IF2.
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Polyribosome
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In eukaryotes, many ribosomes translate one sequence of mRNA. Interactions between the poly A tail and the 5’ end of the mRNA form a circle that makes it easy for ribosomes to attatch again to the mRNA once they are done.
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Release factors
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proteins that recognize stop codons by use of tripeptides not anticodons. The do not participate in forming a peptide bond.
RF1: recognizes UAA or UAG RF2: regognizes UAA or UGA RF3: aids RF1 and RF2 |
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Native conformation
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a correctly folded protein
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Nascent
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newly synthesized protein
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Chaperones
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: a class of proteins that help other proteins fold
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What are some common port-translational modifications?
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Phosphatation (adding of a phosphate via phosphatase)
Removal of phosphates (kinase) Removal of beg MET |
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Signal sequence
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at amino-term end for secretion
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Nuclear localization sequence
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signal seguence for movement out of nuclear pore.
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Interin
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segment of a protein that is able to excise itself and rejoin the remaining portions (the exteins) with a peptide bond.
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Prototrophic
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can grow on defined (minimal) medium of inorganic salts and organic carbon. Can synth all amino acids, vitamins, nucleotides etc..
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Auxotrophic
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cannot synthesize an essencial growth factor
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What are the three different classes of mutants that can be used for selection?
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Auxotrophs, mutants with the inability to utilize a certain sugar source, mutants that are antibiotic resistant/sensitive, mutants that are phage resistant.
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Transformation
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When a cell takes up naked DNA. A cell must competent in order to take up DNA. Cells become competent just before growth stops in B subtilis Recombination must occur between the donor DNA and the recipient chromosome. For two markers to be transferred via transformation, an even number of cross-over events must occur. Not all cells become competent and not all calls can be transformed. Bacillus subtilis can be transformed and is prototrophic and is favored for experiments. Streptococcud pneumoniae can be transformed but requires complex media.
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How do you determine linkage between two markers by transformation?
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Two markers are said to be linked if they are close enough to be on the same transforming fragment of DNA. You would do a cross:
Donor: is protorophic Recipient: trp- his- After mixing DNA with recipience cells and allowing time, cells are plated onto min medium + trp. Since there is NO his in the medium only cells that are transformed his+ will grow. This selects for his+ tansformants. The resulting transformants are replica plated onto min media WITHOUT trp. Only trp+ will grow. These are double transformants (trp+ and his+) If 20% are trp+, there is a 20% linkage between trp and his. Linkage is a function of gene closeness and molecular weight of donor DNA. |
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How do you determine order of linked markers by transformation?
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3 point cross!
Donor: a+ b+ c+ Recipient: a- b- c- Select for transformants and find linkage numbers. Lowest numbers found indicate a quadruple crossover. |
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Conjugation
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Transfer of chromosomal markers with the help of an F plasmid or Hfr integrated plasmid. Cell to cell contact is required. The F plasmid codes for genes that allow for sex pilli to be attatched to another cell. The other cell is drawn in close, the F plasmid nicks and one strand is transferred to the recipient cell 5’ first. Chromosomal markers are transferred also. Hrf cells are ones in which the F factor is integrated into the chromosome.
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What are the differences btw F+ and Hfr cell?
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F+ cells don’t tranferer chromosomal genes as well as Hfr but they do transfer themselves at a high freq while Hfr don’t transfer the F factor frequently. F+ are sensitive to acrine orange because they are outside of the chromosome while Hfr are not.
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Counter selection
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selection against donor cells so that it does not grow on selective medium.
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How do you determine the order of chromosomal marker transfer in Hfr cells?
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You need selectable markers for recipient and counter selection for donor. Determine the frequency of unselected markers in the recipient. Order follows order of transfer.
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Why does Hfr seem linked to the last marker?
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Transfer starts with a nick in the middle of the oriT of the F factor. It will only be complete again in the recipient cell if the whole chromosome is transferred, which is rare. Only if you select for the last maker will you see Hfr recipients.
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Episome
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: a genetic element that can exsist on its own or be integrated into the chromosome.
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F’
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An Hfr that, when it excises itself from the chromosome, has chromosomal material attatched. Is used to produce partial diploids and to test dominance of markers. Transfer can occur as a result of crossover btw F’ and chrmsm.
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Generalized Transduction
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The transfer of any gene from donor to recipient by a phage vector. Any chromosomal marker can be transferred. A virus head can only handle about 1% of the bacterial chromosome in size. If two bacterial genes are within 1% of the chrmsms length they will be co-tranducible.
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Specialized transduction
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Phage lysogenic lamda is an episome that always integrates at a specific site between gal and biotin gene. This is because of the enzyme integrase. If lamda takes parts of other genes it is defective and cant replicate
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How do you determine if two makers are co-transducible?
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Two and three factor crosses.
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Horizontal transmission
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markers passes along diff types of bacteria with self replication plasmids. NOT DNA or chrms transfer
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Constitutive enzyme synthesis
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enzyme is continually produced in fixed amounts
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Adaptive enzyme synthesis
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Inducible: An enzyme is produced (and gene turned on) in the presence of an inducer. The inducer is usually the enzyme’s subratrate. (Beta- galactosidase is iducible)
Repressible: An enzyme’s production is turned off, repressed, by the co-repressor which is usually the end product. The enzymes for argentine synthesis are repressed by argentine. |
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how does the Lac operon work?
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Lac Operon ( an inducibe operon):
The lac Operon controls the production of three enzymes from three genes: The lac Z gene codes for the protein Beta-galactosidase which cleaves lactose. The lac Y gene codes for the protein Permease which transports lactose into the cell. Lac A codes for an enzyme that is needed in lactose metabolism. The Lac I gene which codes for the Lac repressor is upstream of the Lac promoter and operator. The operator is downstream of the promoter and is where the repressor binds. The loc Operon is induced by lactose. Lactose binds with the repressor and allows transcription. ---- I repress gene ---….----P—Operator—LacZ—LacY—LacA— When glucose levels are low, cAMP level high, lactose level high, and cap high the repressor is turned off by lactose so it cannot bind to the operator. The cAMP made as glucose levels are low binds with CAP then binds to the promoter and increases transcription rate. CAP, catabolite activator protein, when bound with cAMP interacts with RNA poly to increase RNA poly’s affinity for the lac.promoter. The lac repressor can work in cis, meaning that in a heterozygote with one functioning Lac I gene (and a second lac-), the Lac operon is still regulated correctly. If the operon region is mutated so that it cannot bind the repressor than the genes will be expressed even in the presence of the repressor and absence of the inducer. If the Lac I gene is an Is gene (which is dominant to I) it produces a repressor that cannot bind with the inducer. The repressor always binds with the operator region in this case. |
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can the lac repressor work is cis?
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The lac repressor can work in cis, meaning that in a heterozygote with one functioning Lac I gene (and a second lac-), the Lac operon is still regulated correctly. If the operon region is mutated so that it cannot bind the repressor than the genes will be expressed even in the presence of the repressor and absence of the inducer.
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What is the Lac IS mutant?
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If the Lac I gene is an Is gene (which is dominant to I) it produces a repressor that cannot bind with the inducer. The repressor always binds with the operator region in this case.
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How does the ara operon work
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In the absence of arabanose, araC bind with O2 and I1, to bend the DNA so that the RNA polymerase cannot bind.
With Arabanose, arabanose binds with araC so it now binds with I1 and I2 and not to O2. The pomotor can now bind RNA poly. If glucose levels are low CAP- cAMP can also bind and initiate transcription. ---araC---…---O2--------O1------P cap site and promotor ----I1—I2---- PROMOTOR---- No arabanose: ________________O2______ | araC | araC |___O1___Pcap_____I1__I2____PR_ß RNA poly can’t bind Arabanose: Cap Cap araCaraC RNA polyà ---araC---…---O2--------O1------P cap site and promotor ----I1—I2---- PROMOTOR---- |