Midterm Study Guide
Midterm Study Guide BI 320
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This 10 page Study Guide was uploaded by Jared Brandon on Tuesday April 28, 2015. The Study Guide belongs to BI 320 at University of Oregon taught by Jana Prikryl in Spring2015. Since its upload, it has received 328 views. For similar materials see Molecular Genetics in Biology at University of Oregon.
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Date Created: 04/28/15
0 Central dogma of biology DNA is transcribed into RNA which is translated into proteins which determine phenotype 0 One gene can lead to many RNAs and proteins via alternative splicing etc When where and how much of a protein is made can be regulated via 0 non coding DNA cis elements o proteinsRNAs that interact with DNA trans factors Shape and surface properties charge hydrogen bonding ability determine if two molecules will interact Basal eXpression is the level of eXpression when not affected by any other factors This is determined by how well the promoter binds RNA polymerase Sequence conservation suggests functional relevance eg the 35 and 10 regions of the promoter The promoter is the site on DNA recognized by the RNA pol holoenzyme and dictates the direction position and strength of transcription initiation The more similar a promoter is to a consensus sequence the more ef cient it is TYPES OF MUTATIONS null amorph complete LOF Typically recessive hypomorph partial LOF Typically recessive conditional functional for one condition but not another i e temperature dependence hypermorph GOF typically dominant antimorph dominant negative indicates protein acts as a homomultimer A single mutant allele can disrupt function of the whole compleX 6 neomorph typically dominant Heterochronic LIIIkt lvl k LOF mutations are typically recessive to wild type One copy of the gene is usually enough to produce an adequate concentration of protein for a WT phenotype codominance A cross between organisms with different phenotypes produces offspring with a third phenotype where both traits are eXpressed together haploinsuf ciency incomplete dominance One copy of the gene is not enough to produce a wild type phenotype A phenotype in the middle is eXpressed Prokaryotes Haploid circular genome high gene density Eukaryotes Haploid or diploid linear genomes low gene density 0 DNA is packages With proteins These proteins are typically positively charged at biological pH so as to attract the phosphate backbone Operon A group of functionally related genes transcribed from a single promoter An operon Will make a polycistronic mRNA For every geneoperon only one DNA strand is transcribed not necessarily the same one Different genes can be transcribed in different directions The direction of transcription is determined by the orientation of the promoter Transcription rate is determined by 0 Intrinsic promoter strength closer to consensus stronger different sigma factors have different consensus sequences 0 Regulatory proteins repressors activators o Operator repressor binding site 0 Activator binding sites Activators increase transcription rate in three ways 0 Help recruit RNA polymerase via cooperative binding 0 Help the transition from closed to open complex 0 Help bump off sigma Repressors can work by inhibiting these steps RNA polymerase Composed of a core enzyme subunits 2 or 3 339 b There is a carboxy terminal domain CTD on the end of the alpha subunits Which binds to the up element an extra promoter stability region Has no speci city and is very processive Sigma factor Binds to the core enzyme to make the holoenzyme Recognizes promoters Contains 4 domains 1 Domain 4 binds the 35 region via a helix turn helix motif Domain 4 must move for promoter escape 2 Domain 2 binds the 10 region and helps melt DNA by folding out two bases on DNA and stabilizing the open complex 3 Domain 11 must move for the open complex to form blocks DNA 4 34 linker lies in the RNA exit channel and must move for promoter escape There are many different sigma factors each of which confer different promoter speci cities to the core enzyme 0 RNA is made in the 5 to 3 direction 3 steps to transcription 1 initiation o holoenzyme binds to DNA to make closed complex this step is reversible 0 DNA melts to form open complex this step is irreversible o abortive initiation makes 10 nucleotides then starts over 0 promoter escape 2 elongation 3 termination o TRCF dependent in cases of RNA pol arrest i e due to DNA damage TRCF knocks off RNA pol while transducing along DNA 0 rho dependent utilizes rut site C rich G poor Knocks off RNA pol while transducing along RNA Only works on RNA not being translated o Intrinsic terminators rho independent The RNA structure causes termination There is a sequence with dyad symmetry that forms a stem loop followed by a run of Us AU base pairs are the least stable of all base pair possibilities Non template strand aka sense strand or coding strand Identical to the RNA that s formed Template strand aka antisense strand Complimentary to the RNA that s formed Second site suppresor analysis provides evidence of direct interaction between two molecules For example two proteins work as a dimer A mutation in the dimer domain of one eliminates function A different compensatory mutation in the other subunit restores some function This is also referred to as a pseudo revertant These mutations are allele speci c there is only one certain allele that can restore function not just any mutation o A cis mutation can t be saved by addition of a plasmid and cannot effect expression from the plasmid o A recessive trans mutation can be rescued by a plasmid o A dominant trans mutation can t be rescued by a plasmid and imparts the mutant phenotype to the plasmid Genetic redundancy If a mutation in gene causes no phenotype change it doesn t necessarily mean the gene has no function Biology employs a lot of backups TRANSLATION The prokaryotic ribosome is composed of a 50s and 30s subunit 23 of its mass is rRNA For both subunits the catalytic sites are almost entirely RNA The ribosomal proteins stabilize the RNA i e by shielding the charge of the backbone Ribosome binding sites o eXist only in prokaryotes 0 contain a Shine Dalgarno sequence AGGAGG that pairs With 16s rRNA of the 30s subunit start codons eXist in prokaryotes and eukaryotes 3 steps 1 initiation the 30s subunit binds at the RBS This subunit contains E P andA sites 2 elongation a charged tRNA recognizes the A site Peptide bond formation occurs 3 termination once a stop codon is reached release factors aid in disassembling the compleX Translation is not continuous from gene to gene Translation starts before transcription ends and multiple ribosomes can translate the same mRNA at the same time in prokaryotes LAC OPERON composed of three genes lacZ encodes 3 galactosidase Which converts lactose to galactose and gluctose a side reaction makes allolactose lacY encodes permease Which allows transport of lactose into cell lacA encodes acetylase Which breaks down bad compounds that come into the cell With lactose Cis sites promoter poor 35 site no UP element low levels of basal expression CAP site activator binding site that lies upstream of promoter Operator the repressor binding site overlapping and downstream of promoter The lac operon has three operator sites The lacI dimer has the same af nity for any operator as the tetramer but the tetramer represses better Trans factors Lac repressor Negative regulator constitutively expressed at low levels It is induced by allolactose to fall off DNA Functional domains of the lac repressor DNA binding LOF mutation is dominant causes constitutive expression Inducer binding LOF mutation is dominant causes uninducible expression Dimerization LOF mutation results in constitutive expression CAP Catabolite Activator Protein Positive regulator CAP binds DNA When cAMP is bound to it When there are low levels of glucose in the cell adenylate cyclase converts ATP to cAMP so CAP binds When there are low levels of glucose Mutations lacIS mutant can t bind inducer and never leaves DNA Expression is off This is dominant because it blocks functional repressors from binding lacID mutant can t bind DNA but can still form a multimer This is dominant because it inhibits the function of the multimers IPTG is an analog of allolactose that binds the repressor directly Circumvents need for lacZ and lacY Mutations that cause constitutive expression operator lacI lacID Mutations that cause uninducible expression lacIs CAP promoter adenylate cyclase ARABINOSE OPERON Two regulatory proteins AraC Both an activator and repressor Activates PBAD the arabinose operon promoter in the presence of arabinose When concentration of araC is especially high araC can bind ara01 and repress its own expression 01 overlaps the promoter for araC This happens in both the presence and absence of arabinose CAP activates PBAD in the absence of glucose Both CAP and araC are needed for full induction 0 When arabinose is present it binds to araC causing it to have a high af nity for both half sites of araI site slightly upstream of PBAD and recruits polymerase 0 When arabinose is not present araC binds a single half site of araI and araOz ara02 is far upstream from araI so the DNA must fold over This folding of DNA inhibits expression 0 Arabinose effects araC by causing shifts in the way it binds to itself lining up the DNA binding domains in different orientations TRP OPERON This operon encodes enzymes that synthesize the amino acid tryptophan Is repressible instead of inducible Regulated by 0 control of transcription trpR trans factor doesn t repress unless bound to tryptophan and trpO operator cis site 0 attenuation transcription termination is regulated within the operon Controls how long the RNA is The operon contains a long leader sequence which has four sites capable of pairing in a stem loop 1 amp 2 can bind generally pair along with 3 amp 4 2 amp 3 can bind 3 amp 4 can bind When 3 amp 4 bind they make an intrinsic terminator they re followed by a long string of uracils which leads to shorter RNA How does it work While the operon is still being transcribed the ribosome begins translating the RNA There is a small ORF in the leader sequence which contains two tryptophan codons in a row If there are low levels of tryptophan there will also be low levels of tryptophan charged tRNA The ribosome stalls at the trp codons since it can t nd the tRNAs fast enough The trp codons are strategically placed so that the ribosome overlaps site 1 and prevents it from base pairing to site 2 as it is transcribed Instead sites 2 amp 3 base pair as they re transcribed This prevents 3 from binding to 4 and forming the intrinsic terminator so the full mRNA is transcribed and is available to be translated If there are high levels of tryptophan the ribosome falls off the rst ORF but not before blocking 2 amp 3 from binding 3 amp 4 get synthesized and base pair and transcription is terminated early LAMBDA PHAGE Viruses are simple they contain few genes and use the host cell s gene expression machinery A phage is a virus that infects bacteria Prophage Viral DNA integrated into a host cell s chromosome Lambda phage infects E coli and has two different developmental programs it s genome can be interpreted in one of two ways 1 Lytic cycle attaches to cell and injects DNA The immediate early genes are transcribed by host polymerase PL PR PR the only promoters that don t require activators Only N and cro are expressed because of terminators after N and cro and immediately and PR The delayed early genes are activated by N The N protein activates genes beyond N and cro N regulates by stopping transcription termination In presence of N protein terminators after N and cro are not recognized by RNA polymerase N allows RNA polymerase to bypass the terminators tr and tL activating the delayed early genes 0 P and Q are now expressed O and P activate phage DNA replication Q turns on the next batch of genes These encode the phage coat How does N know which terminator to work on There are nut sites which have dyad symmetry A stem loop forms in nut RNA which N binds to There is a late Q positive regulator Like N it acts as an antiterminator This one works just after PR allowing RNA polymerase to read into lysis head and tail genes Then the progeny phage are released cell lysis and infect neighbors form plaque on a lawn of cells This is an example of a regulatorv cascade CI is never expressed at high levels during lytic infection since cro inhibits PRM CI is a regulator needed to establish and maintain lysogeny 2 Lysogenic cycle The phage DNA is injected into the cell Only PL PR PR are active N and cro proteins are made PR terminates before its expressed There is delayed early gene transcription N helps express cIII stablizes CH CH transcriptional activator unstable protein Activates promoter for int gene that promotes integration of lambda DNA 0 P and Q CH and C11 activate PRE promoter for establishment of lysogeny which drives cI expression CH and cIII also activate P1 which is the promoter for integrase Integrase promotes lambda DNA integration into host chromosome at a speci c site cI is both an autorepressor and activator How is the choice between c1 and cro made Stochastic due to random uctuations in synthesis of regulatory molecules Cell s physiology Viral infection count cII is readily degraded by FtsH cIII likely acts as an alternative substrate to FtsH FtsH is very active if E coli growth conditions are good If there s high concentrations of FtsH there will be lower levels of CH and c111 and lytic growth will be favored A plaque is a clearing in a lawn of cells resulting from initial infection of a single cell by lytic phagevirus Clear plaque all infected cells lysed Turbid plaque Live cells remaining in plaque are lysogens The choice for lysis or lysogeny is made soon after the cell is infected The choice is made from scratch in each newly infected cell In happily growing cells 50 of all infections result in lytic cycle Repressors Cro represses CI eXpression by blocking PRM CI lambda repressor blocks P1 and PR represses whole thing Activators cI activates its own transcription via PRM cII activates PRM CI expression PRE repressor establishment P1 integration proteins PQA represses Q via antisense RNA Lysogenic induction DNA damage in E coli elicits an SOS response where RecA is made which stimulates proteolytic cleavage of LeXA which inhibits genes involved in DNA repair proteins cI resembles LeXA so RecA also stimulates autocleavage of cI METHODS Does protein X bind to DNA Y To get the DNA sequence we want we do a polymerase chain reaction PCR 0 Denature make single stranded o Anneal bind primers designed to recognizes sequences anking region of interest 0 Amplify polymerase adds nucleotides after primer 5 to 3 0 Heat and repeat Gel electrophoresis separates DNARNAprotein by length alone 0 Positive charge is applied to cause DNA negatively charged bc of phosphate backbone to migrate down gel 0 DNA can be labeled with radioactive phosphate or ethidium bromide can be used to visualize under UV light 0 The smallest pieces will migrate the farthest down the gel Electrophoretic mobility shift assay gel shift 0 tell us if protein binds to DNA 0 uses a non denaturing gel so protein stays bound to DNA 0 doesn t tell us where the protein binds o A negative result is not necessarily informative the conditions could just unfavorable for binding DNase footprinting 0 gives more speci c info about where protein binds o DNase cuts DNA randomly everywhere except where protein of interest is already bound 0 uses a denaturing gel to stop cleavage remove proteins from DNA 0 there should be gaps in the gel based on where the protein was bound How can we tell if a DNARNA sequence we re interested in is in a sample DNA digestion o DNaseMnase non sequence speci c 0 Restriction enzymes sequence speci c Southern blot Used to identify size present or amounts of DNA sequence Uses a DNA probe to detect speci c sequences DNA is fragmented by restriction enzymes A gel is soaked in alkali to denature the DNA prior to transfer from gel to membrane Non speci c DNA fragments are incubated With the membrane Strands are heat denatured and incubated With radiolabeled DNA probe then the bound DNA bounds are exposed on lm by X rays Northern blot Similar to southern blots but With RNA No need to cleave RNA because relative lengths of mRNA are quite small compared to DNA Membrane is heated along With DNA to get rid of any secondary RNA structure Band migration distance in the gel is dependent only on the length of the target DNARNA not on the length of the probe Western blot Proteins are extracted from cells and denatured SDS is a detergent that causes proteins to unfold BME is a reducing agent that breaks disul de bridges Proteins are run on a polyacrylamide gel then transferred to lter paper Antibodies are applied and allowed to hybridize to protein of interest then detected by HRP or immuno uorescense
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