MCDB 101A: Midterm 3

characteristics of insertion sequence (IS) elements

-0.7-5 kb in size and bounded by terminal inverted complementary repeats (Its) -most IS elements encode transposase (and rarely any other genes) -IS element = transposase gene + Its -upon closer examination of these sequences it was determined that the flanking sequences were inverted repeats that are complementary to each other -they must have the transposase gene -also have genes that confer antibiotic resistance and some regulatory elements -jumping in and out of the genome can be back for the host and result in host death so the regulatory elements make sure that this does not happen too often to cause cell death -transposase gene is essential to allow the DNA to jump

the F factor

-100kb (largest plasmid recorded) -40 genes: DNA metabolism during conjugation genes, plus biogenesis genes, and stabilization -capable of both types of replication but rolling circle replication is the important mode for conjugation -oriT: site of nick (nickase = TraD) gives rise to rolling circle replication

the Holliday model: alignment of homologous regions

-flanking marked locates exterior to homologous regions -site f single base difference between the two homologous regions

an important conclusion of the Holliday model

-recombinant DNA molecules cannot constitute more than 50% of the products of a recombination event -theoretically, every time RecBCD makes the cuts it will either be recombination (50%) or non-recombination (50%) -in reality recombination molecules always make up less than 50% of the expected products (recombination only happens about 1% of the time) this is because the bacterial genome is large and it's a small chance that the two strands of viral DNA inside the bacterial genome will find each other in order for recombination to occur

pre-replication repair: very short patch repair corrects T-G mismatches resulting from methyl cytosine deamination

1) demanination 2) recognition by Vsr 3) cleavage by Vsr 4) removal by 5- to 3' exonuclease of DNA Pol I 5) addition of C by DNA Pol I

2 mechanisms of transposition

1) replicative: the transposon is copied and inserted at the new site while the original copy is retained at the old site. Used by most bacterial transposons -most common mechanism -must replicate in order to jump sites -a piece of DNA that is the entire genome plus an entire plasmid is really bad for the bacteria because it is so big that the bacterial replication machinery cannot replicate it so the bacteria will die if the cointegrate is not resolved (when the donor and target is separated) 2) non-replicative: the transposon moves from the old site to the new. Also known as conservative or cut and paste transposons. -All IS elements use this mechanism -they leave behind a hole in the donor genome location -transposon simply moves from donor to target leaving a hole in the donor

genes regulated by LexA

1) weak SOS Box genes: low affinity binding of LexA; expression of these genes requires a small amount of activated RecA, associated with a low level of DNA damage -genes: RecA, LexA, UvrA, UvrB, UvrC, UvrD 2) strong SOS Box genes: high affinity binding of LexA; expression of these genes requires a large amount of activated RecA, associated with high level of DNA damage -genes: UmuC, UmuD, SulA

pre-replication repair

1. base excision repair 2. nucleotide excision repair 3. very short patch repair corrects T-G mismatches resulting from methyl cytosine deamination

F+ vs HFR strains

F+ strains -efficient transfer of F+ phenotype -low frequency transfer of genomic DNA -F factor replicates as plasmid (independently fro, bacterial genome) HFR strains -inefficient transfer of F+ phenotype -1000x more frequent transfer of genomic DNA than F+ strains -F factor replicates with genome (dependent on genome replication)

which molecule is involved in recombinational bypass?

RecBCD

which molecule is responsible for the initial nick in the two parental molecules that undergo recombination (Holliday model)?

RecBCD

Meselson-radding recombination

an alternatie recombination mechanism

dideoxy sequencing reactions

-32P-labeled primer (radioactive) -heat stable DNA polymerase -dNTPs -one of the four ddNTPs per reactions (4 tubes total for each nucleotide) -sequencing reactions are carried out with thermal cycler to enable denaturation -also need Mg2+ chelator

CRISPR-Cas9 and gene editing

-CRISPR components: Cas 9 + guide RNA (RNA + tracrRNA) -uses: gene knock out and gene replacement -issues: off target effects (small sequences can target other sequences unintentionally), inefficient cutting, brings ethical issues

a specialized transfusing phage lambda with unstable gal operon mutations

-E. coli that is uncle to digest lactose can have a specialized phage inserted into its genome to then have the ability to break down lactose -sometimes when lambda interests itself next to these gal genes where the attb or attp site is, when it excises it can take some of the neighboring gene with it

F+, HFR, and F' strains

-F factor can integrate itself into the genome through homologous recombination -F factor has transposons and IS -another type of strain arises when the F factor is integrated into the genome: HFR (still have oriT) -F factor was already integrated into the genome but then it underwent a nicking event at the oriT site and became excised along with some of the chromosomal DNA: F'

the FLM addiction

-F plasmid encodes a toxin-antitoxin system -as long as the bacteria keeps the plasmid inside of it the bacteria will survive because it has the toxin but also has the antitoxin -the bacteria that does not have the plasmid only has the toxin so it will die -FlnA is a small protein toxin that lyses the cell -FlnB is a FlnA antidote. It is an RNA molecule that is complementary to FlnA mRNA; it leads to formation of a dsRNA molecule that is degraded by RNAse III -FlnA is very stable and persists after the F plasmid is lost -a host cell that loses its F plasmid will die -normally: the plasmid is not wanted by the bacteria so the bacteria tries to outcompete the plasmid by replicating faster than the plasmid so when the bacteria divides then only one of the two cells will have the plasmid; the cell that doesn't have the plasmid will survive, the other that has the plasmid will die

the importance of IRs for transposition

-IS elements containing multiple point mutations within their IRs are typically unable to transpose -the transposase is going to recognize the IR so if there are mutations in the IR then the transposase will not recognize it so the IS is going to be stuck in the genome of that organism (unable to excise) -these transposable elements are part of every genome due to these mutations causing them to be trapped -cut and paste molecular cloning techniques have made it possible to delete the IRs from IS elements. these amputated IS elements cannot transpose -gel shift assays showed that the IRs are the binding sites for transposase (specificity: transposases are going to directly bind to these IR sequences) -add IRs flanking any DNA (size limit) and transposase will act on it and remove it (that piece of DNA will be excised from the genome)

many R factors can undergo conjugate transfer

-R factor: gene that allow a bacteria to survive 1 of the 5 types of antibiotics -conjugation can occur between any bacterial species

proteins involved in recombination

-RecBCD: generates initial nicks (endonuclease), unwinds helix (helices) and loads RecA onto ssDNA -RecA: promotes invasion of a homologous dsDNA region by binding to the invading ssDNA and disrupting base pairing in the target sequence (RecA is good at binding to ssDNA; must bind to the tip of the invading molecule to direct the invasion; causes destabilization of the base pairing) -SSB: binds ssDNA, prevents formation of secondary structures (ensure that the DNA remains straight) -DNA ligase: joins the cut strands by creating a phosphodiester bond -RuvA/B: responsible for branch migration -RuvC: generates nicks in Holliday structure that lead to resolution (endonuclease)

the bacterial DNA damage checkpoint

-SUlA inhibits the GTPase activity of E. coli cell division protein FitsZ -FitsZ needs to hydrolyze GTP to create spring like structure around the bacteria and also a ring that will pinch off two bacterial cells during division -SulA prevents this potentially damaging mechanism -Separation is inhibited -cytokinesis -bind chromosome of the bacteria to the cell wall -this mechanism needs to be inhibited if there is a mutation in the genome because it will break the DNA -filamentous growth occurs

regulation of transposition

-Tn3: resolvase inhibits its own transcription and that of transposase -Tn10: an antisense RNA inhibitor -Tn5 and target immunity: a preexisting transposon can block the insertion of a second transposon of the same type. May promote the spread of transposons to distant sites -mechanism: the transposase gene additionally encodes a repressor of transposase that is translated from a different start codon

transposable elements

-a piece of DNA bounded by inverted terminal repeats that can excise itself and move from one location to another within or between chromosomes, plasmids, or phages -the first examples of prokaryotic transposable elements were associated with E. coli gal operon mutations that reverted to wild type with higher than normal frequency

specialized transduction

-a recombination error during excision can replace a portion of the lambda genome with gal or bio genes -lambda dgal transducing phage have replaced head and tail genes with gal DNA. coinfection with a normal lambda phage is required to allow this phage to undergo the lytic cycle -lambda pbio phage lack the xis and int genes cannot undergo lysogeny without the presence of a normal phage to supply these gene products

for transduction to be successful, the donor DNA must recombine with the host by an even number of crossovers

-a single recombination event causes linearization of the bacterial genome which causes cell death -true transduction requires even number of recombination events to prevent cell death -most commonly 2 or 4 -transductio is extremely rare (even though the pseudo pac site is common) only 1/500 phages undergo transduction; only 1/10000 phages actually carry an entire gene from the host (most transduction phages only carry partial chinks of the genes)

DNA damage repair

-a typical base mutates at a rate of ~ 10^-7 cell/generation -number is bias because we only count the visible mutations (i.e. the mutations that escape mutational repair mechanisms) -in reality the actual rate of spontaneous alternations of DNA sequence is much higher -many of these changes go undetected because they are corrected before they result in an altered phenotype

post-replication repair: the SOS system

-activated when the capacity of DNA repair systems that generate gapped (ssDNA) molecules is exceeded 1) ssDNA binds to RecA, altering its activity 2) RecA promotes auto proteolysis of LexA, a transcriptional repressor of the SOS system (LexA is the repressor, LexA Box is the operator) 3) UmuC and D genes are expressed 4) RecA cleaves UmuD to form UmuD' 5) UmuCD' and Pol III encodes DNA Pol V which can continue to insert bases past a damaged region of DNA; base pairing specificity is lost, resulting in the insertion of incorrect bases and making the SOS system mutagenic -loses ability to proofread -prevents the bacteria from stalling replication and therefore being killed

post-replication repair: DNA damage tolerance

-allows cells to continue replication even though large bulky groups such as pyrimidine dimers and aflatoxin products are present -bacteria has to tolerate these mutations

HFR strains can transfer chromosomal genes between two host bacteria through conjugation

-always transfer oriT first -F+ cells integrates F+ by crossing over -conjugation of the HFR with F- -integrated F factor i nicked, and nicked strand transfers to the recipient cell, bringing bacterial genes with it -transferred strand is copied, and donor bacterial genes are appearing in the recipient

the Holliday model: strand exchange and branch migration

-as the knot is pushed forward in branch migration, the DNA is exchanged in a complex known as a heteroduplex DNA

random chain termination generates a nested series of truncated DNA fragments

-at random the ddNTP will be incorporated into the sequence so they will be randomly terminated (ddNTP competes with the normal dNTP) -this is usually done with 2 primers and one will bind to the 3' end of the top strand and the other will bind to the 3' end of the bottom strand -the gel read out gives the sequence of the complementary strand so if you want to get the template strand you must convert the complementary strand

functions of recombination

-bypass of damaged bases during replication -rescure of collapsed replication forks that result when a DNA polymerase encounters a single strand nick (single strand nick in replication causes DNA polymerase to stall, if it stalls for too long it can kill the cell) -repair of DNA double stranded breaks -exchange of genetic information between different DNA molecules un the same cell and the exchange of genetic information between cells through conjugation and transduction

CRISPR

-clustered regularly space short palindromic repeats -each CIPSR has its own PAM, can genes, and repeats -2 CRISPR classes: -Class 1: each function is for each gene -Class 2: all functions encoded by one gene -primary infection: a phage infects a bacteria -a cas endonuclease cleaves the viral DNA into fragments -a cas integrase inserts a viral DNA fragment into a preexisting CRISPR array (if Class 1 then it's a separate enzyme; if class 2 then it's the same enzyme) -secondary infection: when the same type of phage reinvades the bacterial cell the second RNA, the tracrRNA links effector Cas protein to the crRNA, targeting it to cut the target viral DNA

composite transposons

-consist of a pair of IS elements separated by intervening DNA -contain two complete ISs that each have its own transposase so there is twice the amount of transposase; both ISs must be the same (Ex. both must be IS50 (either R or L) so that they have the same transposase) -the transposase will bind to the outermost IR sequences and excise the fragment of DNA that lies between the two ISs -the intervening DNA often contains antibiotic resistance markers and/or other genes -every time they jump in the genome they can bind a new gene that gives the bacteria a new function (Ex. tetracycline resistance gene)

non-composite transposons

-contain single terminal inverted repeats rather than terminal IS elements -contain a single copy of transposase along with other genes

F' strains

-derived from HFR strains -contain an F plasmid -convert to HFR strains more readily than do F+ strains -always integrate at same location to form HFR strains -contain one or more E. coli genes -can be used to create merodiploids

pre-replication repair: nucleotide excision repair

-detects and removes damaged bases resulting from photodimer formation or the addition of bulky products and covalent inter strand crosslinks 1) UvrA2-UvrB complex scans DNA molecules for damage 2) if damage is detected UvrC replaces UvrA 3) UvrB and C make nicks upstream and downstream from the damaged site 4) UvrD is a DN helices that displaces the cut segment (removes damaged pieces) 5) DNA Pol I replaces the missing sequence and DNA ligase seals the nicks completing the repair

dideoxy sequencing: The Sanger method

-dideoxyNTPs (two H's on the ribose at 2' and 3' so no phosphodiester bond can be formed at the 3' position where OH is normally located) -ddNTPs will be the last nucleotide i the sequence since to the nucleotides can be added after

mechanism of generalized transduction

-donor bacteria has a virulent phage (engages in lysis) that injects its genome into the genome and also breaks down the genome of the bacteria (lytic phages do this but not lambda). Fragments of the host DNA are then packaged into the viruses along with the viral genomes into the viral particles that then go and infect other bacteria, giving new bacteria the genes of the original bacteria host through recombination thus changing the phenotype of the newly infected bacteria -note that recipient bacteria is not lysed because transducing phage lacks viral DNA -P phages (Ex. P21) are the best phages for transduction because their genomes are huge so they can carry a lot of the hot DNA

requirements for conjugation

-donor strain: F+ (contains F plasmid). F factor encodes pilus proteins and endonuclease that initiates rolling circle replication and regulatory factors. F factor is an episome: it can exist as an autonomously replicating plasmid or can integrate into the genome -recipient strain: F-; these strains encode proteins that allow pills to attach to cell wall. When F- strain receives F factor it becomes F+

a three factor cross given linkage and gene order

-donor: d+e+f+ (e is the selected marker; all genotypes must have e+) -recipient: d-e-f- -lowest number of recombinations is the frequency of quadruple recombination events -can then determine the gene order and cotransduction frequencies -you can only determine the distance between a selected marker and an unselected marker

New Sanger sequencing: automated fluorescence sequencing

-each ddNTP is labeled with a different fluorescent dye -can generate ddNTPs with fluorescence on them instead of radioactivity (safer) and instead of separating them into 4 lanes you can run them at once and heat so they will fluoresce with different colors depending on nucleotide -primary technology used to map human genome -multiple nucleotides in a row can be problematic to read -lower case letters come from peaks that are not properly resolved (weaker signal) due to DNA damage

key features of pyrosequencing

-each of the 4 dNTPs is added sequentially to the sequencing reaction -apyrase is added between each nucleotide addition step to degrade unincorporated dNTPs -light emission or luminescence occurs when an added nucleotide is complementary to the template. Luminescence is measured using a sensitive CCD detector -the amount of emitted light associated with each nucleotide addition is proportional to the number of nucleotide copies that can incorporate

the Holliday model: the products of recombination are determined by the orientation of the resolving endonuclease

-endonuclease makes 2 cuts in the strands which can result in 2 outcomes/ the location of the resolving nicks determines the type of resolution 1) resolving nicks that occur on the same strands that were nicked by RecBCD produce patched or parental/non-recombinant products that retain the original flanking markers -if the cuts are made on the original strand that had the nick to begin with then there will be no recombination, parentals restores -the internal DNA strands are the ones that are originally nicked 2) resolving nicks that occur on different strands than the ones that were originally nicked by RecBCD produce spliced or recombinant products that have exchanged flanking markers -if the cuts are made on different strands then there will be recombination -cut the internal DNA strands first and then cut the exterior DNA strands

Pyrosequencing uses a pyrophosphate release to detect the addition of a specific nucleotide

-gels slowed sequencing process so pyrosequencing is based on luciferase -every time you add a new nucleotide to a growing sequence you will produce pyrophosphate as a byproduct -you can add one nucleotide at a time to a growing chain to see if it produces light; if light is emitted then that nucleotide was correct in the sequence

advantages of pyrosequencing

-greater accuracy that dideoxy-based methods -allows for more rapid sequencing in automated systems -no need for ddNTPs, labeled primer or fluorescent-labeled nucleotides -eliminated need for gel electrophoresis (slow step) -sequencing requires a single reaction instead of 4 -one disadvantage: the maximum read length (300-500) is less than dideoxy method (800-1000)

an F plasmid can promote the transfer of non-conjugatable plasmids

-if a bacteria already has a plasmid inside of it and then receives the F plasmid, the F plasmid can have homology with the other plasmid and they can undergo recombination -the recombinant is the F plasmid and the original plasmid already there which can then be passed down together into other recipient cells -recombination can occur again in the recipient cells so that the recombinant plasmid is separated into the F factor and the original plasmid

the Holliday model: resolution

-in order for he two joined DNA molecules to separate, the cutting and ligation steps must occur a second time -the stopping mechanism of the knot moving along the strands is unknown -when the knot stops moving then an endonuclease will cut the DNA -the knot can also slide in either direction along the strands

bacterial immune system

-innate immunity: restriction endonuclease (nonspecific response to invading phage) -adpative immunity: CRISPR (targeted immunity with memory; bacteria remember the phages they are infected with)

replicative transposition mechanism

-key difference: there is only a cut of one single strand DNA per DNA molecule (not a double strand break) -there are still open 3' Oh groups -transposable elements are still attached to the donor DNA -the gap that is formed after the nucleophilic attack is larger to DNA polymerase III is used instead of DNA polymerase I -DNA polymerase III starts replication of the entire honor molecule and then go to the transposable element instead of simply filling the overhang gap; DNA polymerase III stops after it replicates the transposable element (only replicates the donor and transposable element, not the target) -there is now copies of the transposon which is called the cointegrate which consists of the target DNA, donor DNA, and two copies of the transposable element -this structure is so big it can kill the bacteria so must separate the target and donor DNA -2 ways to resolve the cointegrate: site-specific recombination and homologous recombination -site-specific recombination: most common; requires res site (target sequence) and resolvase enzyme; performs recombination between the two res sites of the donor and target DNA which then separates the two elements; resolvase makes this process more safe -homologous recombination: requires host recombination enzymes; RecA dependent (has a high failure rate which is why it is only favored when site-specific recombination is unavailable)

a mispair created during recombination can be changed to a normal base pair by mismatch repair or replication

-mismatch pairs that are created during this process are not observed because DNA has a really goo repair mechanism for mismatches -replication corrects the recombination mismatches -mutants were used to view the mismatch base pairs that are usually corrected through DNA repair mechanism and replication -MutS, MutL, MucH involved in DNA repair mechanism along with DNA Polymerase III -if you use mutants for MutS, MutL, MutH then the mechanism for the mismatch repair no longer works and the entirety of the mismatches that occur through recombination can be observed

the DNA damage response

-monitor the status of DNA which can be done by the DNA physical characteristics/topography (width/diameter) -repair any damage -mechanism: remove damaged fragment and replace it; sometimes the damage is so severe that this mechanism is impossible -tolerate what cannot immediately be repaired (allows for survival but comes at a cost)

important transposon facts

-most transposition is RecA independent and does not involve homologous recombination -most transposons insert into DNA at random locations (Tn7 and Tn10 are exceptions) -transposons can hop into existing genes and inactivate them. reversible transposition occurs when a transposon hops out without removing any of the surrounding host DNA (precise excision). Imprecise excision results in the deletion of surrounding host DNA

replication-coupled transposition of TN10

-non-replicative transposition (triggered by replication) -a double strand break gap is left and DNA polymerase III fills gap -hemimethylation causes the transposon to have higher affinity for the IRs (GATC)

transposons

-occasionally when 2 IS elements insert close to each other they will later excise as a unit carrying with them any DNA that is located in between them -bacterial transposons range in size from 3-50kb -shares many of the same characteristics as an IS on its own

DNA transfer during conjugation occurs through rolling circle replication

-only one copy of the genome through rolling circle replication -only leading strand is passed down and the lagging strand finishes replication inside of the recipient bacteria

Transfer of F plasmid by conjugation

-only the leading strand gets passed down from donor to recipient

lederberg and tatum: recombination between bacteria

-physical contact allows exchange of genetic information: only one bacteria passes own information to another -auxotrophs

PAM

-protospacer adjacent motif -sequence (2-9 bp long) inside the virus -only sequences that have the PAM will be kept: when the viral genome is chopped, only sequences with PAM will be kept by the bacteria; one of the sequences with PAM and is placed in between the palindromes but the PAM is not integrated into the genome (CRISPR Cas needs PAM to be able to recognize it and destroy it) -no PAM = sequence is not cut (safeguard mechanism so CRISPR does not cleave its own DNA) -found in the foreign DNA spaces (located at the 3' end of the crRNA) -gives permission for Cas endonuclease to cut spacer RNA -source is foreign DNA sequence. Only fragments containing PAM are integrated into CRISPR array (PAM is removed during integration) -Canonical sequence for Cas9 is 5' NGG 3' -prevents Cas9 from cutting integrated spacer DNA which would create DNA break in genome

post-replication repair: mismatch repair

-removes mismatched base pair that result from the insertion of a non complementary base during replication -requires that a mechanism must be in place to distinguish between old and newly synthesized strand -prevents mismatches from becoming true mutations where a base pair has been altered mechanism 1) DNA adenine methylase methylates A in the sequence GATC -this occurs a little bit after replication is done -the template strand will have the methyl groups on A, the new strand does not have methyl groups -how common is GATC? (4)(4)(4)(4)=256 2)A-methylated DNA is not found in the newly synthesized strand 3) MuSt recognizes mispair, MutH + L binds MutS (MutS does surveillance) 4) the endonuclease MutH nicks the newly made strand adjacent to the nearest methylated A 5) exonuclease chew away the strand until mismatch is removed 6) DNA Pol III and DNA ligase repair gap

denature DNA fragments, separate on acrylamide gel, and autoradiograph

-run each of the DNA fragments on a gel and then read the gel to get the DNA nucleotide sequence (read the bands from smallest t o largest sequence to get the complementary strand) -gel must be really dense so it can separate fragments by 1 nucleotide and it must be really long to capture the entire sequence (very tedious and hard to interpret; has nucleotide limit of 1000 bps)

denaturation of IS elements in plasmids demonstrates the presence of IRs

-slowly cooling DNA after denaturation causes it to form secondary structures that can be analyzed -discovered that the IR were binding to each other in a loop fashion -the bases that did not base pair together in the loop consisted of the transposase and other genes

differential gene regulation by operator affinity (weak SOS Box genes)

-small amount of damage --> small amount of ssDNA --> small amount of RecA --> small amount of LexA is degraded -this results in a modest derepression of the weak SOS Box genes allowing for the necessary expression of Uvr genes needed to deal with low level DNA damage -the low level expression of LexA and RecA allowed under these conditions assures that the regulation of the SOS operon is determined by the accumulation of ssDNA and not by the availability of either LexA or RecA

pre-replication repair: base excision repair

-specific glycosylase detects and removes abnormal base (uracil, xanthine, and hypoxanthine: these glycosylases cut the base but the phosphodiester bond is still intact) -leaves AP site (site without a base) -BER mechanism also repairs alkylated bases and oxidized bases -AP endonuclease cuts the phosphate backbone -the 5' to 3' exonuclease activity of DNA polymerase I and/or other exonuclease remove downstream nucleotides -DNA Pol I recognizes the cut in the DNA and binds the start removing bases -DNA Pol I in DNA replication removes the primer and replaces it with DNA -DNA Polymerase I fills gap and ligase seals gap

bernard david: the U-tube experiment

-strain A, strain B with fine pored filter between -filter ensures that no bacteria touches physically so there was no growth when plated on minimal media showing that genetic information exchange relies on physical contact

discovery of transposable elements

-studying pest resistant genes in corn -jumping genes -not a mutation because in the same generation there were reversions of the wild type phenotype so she concluded that there were some elements that were disrupting these genes

prediction of mismatches

-the Holliday model predicts that recombination events between non-identical homologous sequences will generate mismatches -mismatches occur because the two pieces of homologous strands only need to be 95% identical so up to 5% of the strands can be different and therefore the bases don't properly pair -observation of these mismatches after recombination is evidence that supports this model -mismatches occur no matter how the second cut is made (regardless if there is recombination or not)

cotransduction frequency

-the frequency with which 2 markers are transmitted from one cell to another by transduction. Cotransduction frequency is measured by selecting for one marker and determining the frequency at which a second marker can be detected. Cptransduction frequency provides a relative measure of the distance between closely linked genes. The cotransduction frequency is given as a decimal or percentage and is inversely proportional to the distance between the genes -extremely rare but happens enough to make a map of relative distances between markers -genes that are closer together are more likely to be packed together than genes that are far apart in the genome -instead of map units, use decimal or percentage: how frequently two genes are packed into the same virus -most of the time auxotrophic genes are used Ex. 50% of the time leu is packed together with azi so they are close together; 0% of the time thr is packed with azi so they are not close together

generalized transduction

-the gene transfer process is carried out by many lytic phages. Transduction takes place when a phage packages host DNA into a new virus particle instead of phage DNA. These genes can then be introduced into another bacterium when it is infected by the phage. An altered phenotype may be seen if the transferred genes are incorporated into the host genome through recombination -called transducing phages

transposition

-the movement of genes between different loci within an organism or between organisms as carried out by transposable genetic elements -a gene is inserted in a certain place and then the gene is excised and moves across the molecule to be inserted at another location in the genome -or a DNA insertion is one organism can move and then insert itself into the genome of a different organism

conjugation

-the transfer of genetic material from a host cell with an F factor (F+) to an F- recipient -pili are used to transfer material from a host cell with F factor to a recipient that is F- (F+ = male, F- = female) -the plasmid that is responsible for the genetic information exchange contains the instructions on how to make the proteins necessary for building the bridge and also for the process of conjugation

packaging of chromosomal DNA is aided by the presence of pseudo Pac sites in the E. coli

-there is a pac site that is recognized by the packing enzymes which push the viral DNA into the reassembled capsid and allow for the rest of the assembly to occur -pseudo pac site is similar to the actual pac site so the pac enzymes will pack the bacteria DNA instead of viral DNA

structure of several insertion sequences

-there is some variation in ISs -the inverted repeats are going to be affected only by the transposase that is produced by that particular IS -Ex. the transposase for IS50 is not going to be able to act on the inverted repeats of another IS (they are specific for each one)

HFR strains are usually formed by homologous recombination between transposable elements

-these events are RecA dependent and don't involve transposase -homologous recombination is dependent on RecA

how to carry out an interrupted mating experiment

-these give bacterial gene order and relative distance between genes 1) mix an HFR Str^s gal+ lac+ ton^r azi^r strain with an F- Str^r gal- lac- ton^s azi^s strain 2) remove samples periodically and run through blender 3) plate on selective media: minimal media + streptomycin and appropriate conditions to select for transfer of individual markers -showed that E. coli genome is circular -yielded the first genetic map of E. coli chromosome

summary of events occurring in replicative transposition

-transposase creates a single nick at each end of the transposon -3' OH groups at transposon ends carry out nucleophilic attack on recipient DNA -DNA polymerase III initiates replication at 3'Oh of recipient DNA creating 2 transposon copies -resolution of the cointegrates occurs through the action of resolvase or homologous recombination -no DSB remains in the donor site

non-replicative transposition mechanism

-transposase is a dimer that binds to both sequences at the same time creating a loop -a double strand break in the backbone of the molecule occurs resulting in the transposon that has excised itself -hydroxyl groups at the 3' ends of the transposon are going to generate a nucleophilic attack on the phosphodiester bond on the target DNA and then integrates itself into the target DNA genome -transposase creates blunt ends but the insertion itself creates a little bit of an overhang due to the hydroxyl groups so there's a little gap (<15 bps) so DNA polymerase I and ligase will come in and fill and seal the strand -now there is duplication of the target sequence at the end of the DNA molecule which is the trademark of non-replicative transposition

summary of events occurring in non-replicative transposition

-transposase makes blunt ended cuts at ends of the IS or transposon, leaving a dsDNA break at the donor site -the exposed 3' OH groups at the transposon ends carry out a nucleophilic attack on the target DNA (phosphodiester bond in backbone) -a gapped composite structure of transposon and recipient DNA is formed -DNA polymerase I fills in the gaps creating direct repeats at the ends of the transposon

features of replication-coupled transposition

-transposition of Tn10 occurs most frequently just after the region containing the transposon is replicated -GATC sequences are located within the transposase promoter and one transposase binding site (inverted repeat) -GATC sequences are hemimethylated immediately following replication -RNA polymerase and transposase bind more tightly to hemimethylated sites than to fully methylated sites -this is a specialized form of cut and paste transposition and is not replicative transposition -allows for repair of DSB resulting from transposition through double strand break repair

gal(-) mutants of bacteriophage lambda form loops with gal(+) bacteriophage

-virus is usually injected into the linear form of bacterial DNA (easier to pack). If you take the linear viruses and then slowly heat them up they will denature then if you cool them the viruses will hybridize (two strains of DNA will bind to one another) one strand is bigger than the other one so a loop will form that is unable to hybridize with the other strand

U-tube experiment revisited

-way to test for conjugation -result: filter failed to block transfer of transfusing agent between strain A and strain B -if conjugation was occurring then it would be blocked by the U-tube filter so no colonies would grow but colonies did grow so it wasn't conjugation (it was a virus)

human diseases associated with inherited mutations in DNA repair genes

-xeroderma pigmentosum: NER defect -trichothiodistrophy-NER defect -hereditary non-polyposis colorectal cancer (HNPCC): defect in mismatch repair -Cockanye's syndrome: defect in cyclobutane dimer repair

how the Meselson-Radding model differs from the Holliday model

1) M-R recombination is initiated by DNA double stranded breaks 2) M-R recombination occurs near Chi sites: 5' GCTGGTGG 3' 3) the M-R model proposes that only one invading strand initiates recombination 4) M-R recombination requires DNA Pol I to fill gaps that result from DNA strand displacement 5) no mismatches result from base pair differences between the recombining molecules

Lederberg and Zinder mixed strain experiment

1) combined 2 strains of Salmonella: strain 1 was met- his- and strain 2 was phe- trp- tyr- 2) plated mix on minimal media 3) observed colonies (bacteria growth) -reversion was out of question because it was so extremely rare (required 3 reversions) -conjugation was not occurring because U-tube experiment still yielded colonies -instead it was a virus that was exchanging the bacterial DNA (virus is small enough to move through the U-tube fine pored filter)

the Meselson-Weigle experiment

1) differentially labeled 2 phage lambda strains with heavy and light isotopes of carbon and nitrogen -viruses that have very heavy DNA and viruses with normal light DNA -had mutational markers on the bacteria in the heavy DNA: c (cloudy) and mi (minute); the light/normal DNA has the wild type phenotypes (+,+) 2) coinfected E. coli with both labeled strains (heavy and light) 3) collected phage from bacterial lysate and centrifuges through CsCl2 (CsCl2 allows things to separate based on density) 4) multiple bands of phage were produced -have the original very heavy band and very light DNA bands along with many hybrid bands -this is evidence that the two original parental strands have exchanged genetic information and have created new genomes for these other viruses; new genomes were combinations of heavy and light DNA 5) removed individual bands, washed phage to remove CsCl2 and infected E. coli with phage 6) some phage from intermediate density bands yielded plaques with recombinant phenotypes -new phenotypes not the same phenotypes as the parents

mutation avoidance mechanisms

1) enzymes that detoxify oxygen radicals a) superoxide dismutase: converts O2- to H2O2 b) catalase: converts peroxide into H2O and O2 2) MutT proteins: converts 8-ocyGTP to 8-oxydGMP (removal of the two phosphates removes potential of the free base being incorporated into the DNA)

the Holliday model of recombination

1) homologous sequences align -sequences must be 95% identical and 50 bp long 2) one strand in each DNA helix is cut -2 cuts, one in each DNA strand 3) srand invasion/exchange occurs -strand of the cut DNA will extend itself and bind to the other strand that was cut 4) ligation of exchanged strands forms Holliday structure -Holliday structure moves through these molecules that are exchanging information 5) branch migration generated heteroduplex DNA 6) resolution: 2 additional cuts occur and cut strands are ligated to produce separate double-stranded DNA molecules -unties the knot by cutting the DNA molecules again which will either result in recombination or not

recombinational bypass

1) in the absence of SOS activation, DNA Pol III skips over a pyrimidine dimer and restarts replication downstream of the damaged region 2) to fill the gap, the RecBCD endonuclease cuts the opposite parent strand and RecA promotes recombination between the two strands such that the missing sequence in the gapped strand is replaced with the corresponding sequence of the intact parental strand 3) DNA Pol I uses the lagging strand template to repair the newly created gap in the donor parent strand (severe DNA damage but can still be managed; start replication downstream from damaged region) -will not repair the dimer but buys time to keep replication going

post-replication repair

1) mismatch repair 2) DNA damage tolerance 3) the SOS system

rules for determining recombinant frequency

1) no odd crosses 2) must start and end cross with the F- strain 3) double crosses are more frequent than quadruplets 4) quadruplet cross allows determination of gene order 5) distances are given in map units 6) recombination frequency is directly proportional to map distance

direct reversal of DNA damage

1) photoreactivation: photolyase bind to cyclobutane photodimers in the presence of visible light and converts thymine dimer back to 2 monomers (can be caused by UV light damage) 2) alkyltransferases: remove alkyl groups from bases (Ex. O-6-methylguanine)

transfer of genetic information by an HFR strain is transient unless accompanied by recombination with recipient genome

an odd number of crossovers results in cell death

Iclicker: for generalized transduction to occur the bacterial genome has to have a sequence similar to the

pac site

transduction

the transfer of bacterial genes between two host cells by a bacteriophage