Module 5: Bacterial Genetics (Mutants and Techniques) Flashcards
Strain
Group of organisms that share a common species BUT differ genetically from other members of the species
(Genetically distinct cells)
(Subdivisions within species)
All strains can be traced back to a…
Wild Type Strain
Wild Type Strain
Organisms that (usually) possess the TYPICAL or representative characteristics of a species
A wild type strain can refer to… (2)
1) Original isolate from nature
2) Common laboratory strain from which all mutants of a study were derived
Examples of a WT strains (2)
E.coli K12
–> Isolated from human feces (1922) and is the WT strain from which most E.coli strains have been derived
Sinorhizobium melioti
–> Nitrogen fixing bacteria derived from an alfalfa root nodule isolate
Mutation
Change in DNA sequence
(relative to the comparable sequence in WT variant)
Mutant
A cell or strain possessing mutation/s
Allele
An alternate form of a gene
Genotype
Description of alleles an organisms possesses (ALL alleles)
Mutations can change the ________ of a gene which can alter ________
Mutations can change alleles which can alter gene product/function
Mutations can cause: (3)
1) Loss of gene function
2) Modification of gene function
3) Regaining of gene function (of a previously mutated gene)
What is the source of mutation?
RANDOM errors in DNA replication that go unresolved
–> Arise spontaneously due to DNA rep. errors that are not repaired successfully
Mutagens
Factors that can increase RATE of mutation
Examples of mutagens:
1) UV Light
2) DNA damaging chemicals
Although mutations are random/spontaneous what can we change about them?
The RATE at which they occur
Genes vs Protein Naming Convention
GENES =
–> 3 letter name, lowercase, italicized
–> If multiple genes in same pathway = denote differences using one CAPITAL letter at the end of the name
Proteins =
–> Gene name (3 letters), 1st letter uppercase, NOT italicized
Ex: lacY (gene) –> LacY (protein)
If multiple genes are involved in the same biochemical pathway, how are they named?
Root gene names are the SAME
BUT each is differentiated by a capital letter at the end
Mutants are most commonly identified and isolated via…
PHENOTYPIC DIFFERENCES
Commonly used phenotypes for mutant detection: (4)
1) Antibiotic resistance
2) Exopolysaccharide Capsules (involved in virulence)
3) Pigment production (color of colonies)
4) Motility (from action of flagella or not)
Main methods used for detection of phenotypic mutants (names):
1) Phenotypic Selection
2) Phenotypic Screening
a) Single Plate
b) Screening by comparison (REPLICA PLATING)
Phenotypic Selection
A method for isolation of phenotypic mutants in which bacteria are plated on a SELECTIVE growth medium that promotes growth of ONLY desired mutants
–> Allows only strains with particular combination of phenotypic characteristics to grow and form colonies
Procedure for Phenotypic Selection:
Isolating antibiotic resistant mutants
1) Media containing antibiotic is placed in growth plate
2) 109 CFUs of culture are plated
3) Incubate
= Colonies that grow are antibiotic resistant mutants! –> Can be extracted to isolate
(everything else dies)
What is a limitation of phenotypic selection method?
Restricted use:
Can only be used to retrieve/isolate mutants with phenotypes that lends themselves to being distinguished by differential survival/growth under different culture conditions
(Ex: antibiotic resistance, high temperature, phage resistance)
Phenotypic Screening
A method for isolating phenotypic mutants in which phenotypic mutants are distinguished and determined VISUALLY
–> ALL cells grow under the test conditions (NOT based on selective growth)
Phenotypic Screening vs Selection
Selection = Determination by selective growth
Screening = Visual determination
What are the two methods of phenotypic screening?
(Why would you use one over the other?)
1) Single plate method (for mutants that LOOK different)
2) Screening by comparison (2+ plates)
(for mutants that DO NOT look different)
Single Plate Phenotypic Screening: Process
1) Culture is plated
2) Incubation
3) Plate analyzed for colonies with visual differences
–> Mutants determined based upon their appearance
4) Mutant colonies retrieved from the screening plate
Screening by Comparison (General Process)
1) Identical colonies are plated onto two separate plates:
1st plate = Supports mutant growth
2nd plate = Does NOT support mutant growth
2) Plates are incubated and then growth is compared between the two
3) Colonies missing on 2nd plate = mutants –> Go over to 1st plate and collect the colonies in the positions that correspond to the missing colonies on the 2nd plate
4) = isolation of mutants from plate in which mutant growth was supported
Replica Plating (Definition)
Technique for producing DUPLICATE plates for use in phenotypic screening
Replica Plating Process
1) Master plate is prepared with growth medium that supports growth of all cells
2) Master plate is “stamped” onto sterile velvet cloth to transfer the colonies
3) 2 fresh plates are prepared
1st plate = same medium as master plate (supports all growth)
2nd plate = selective medium; does NOT support mutant growth
4) Fresh plates are stamped onto velvet cloth with imprinted colonies = two plates with identical colonies plated
5) Duplicate plates undergo screening analysis
What is an alternative process to Replica Plating?
(Explain the process)
PATCHING
1) Master plate prepared
2) Sterile toothpick used to sample from individual colonies
3) 2 fresh plates produced (one identical to master plate and a second with medium selecting against mutant growth)
4) Numbered grid markers placed on the 2 new plates
5) From sterile toothpick, one colony is placed in ONE cell of grid on one plate and then placed on second plate in the same exact cell
6) Incubate
7) Compare plates
–> Cells with no growth under conditions that select against mutant growth = mutant colonies
Benefits of patching over replica plating:
1) Better tracking of individual colonies (due to numbered grid)
2) = Greater precision
3) = Greater reproducibility of results
Changes in _________ provide the __________ upon which environmental ___________ act
Changes in genetic material provide the raw material upon which environmental pressures act
Mutation drives _______, ensuring ___________
Evolution, ensuring the vitality of life
Bacterial colonies are made up of millions of….
CLONES
Clones
Genetically identical cells which originated from a single cell
If a mutation confers an ___________ to a cell, the progeny of that cell will be able to…
If a mutation confers an advantage to a cell, the progeny of that cell will be able to…
OUTCOMPETE cells that lack the mutation
Cells with an advantageous mutation will (2)
1) Outcompete cells lacking that mutation
2) Increase their presence in a population
What did Richard Lenski do?
Started and still leads the “Long Term Evolution Experiment (LTEE)” which examines “evolution in a test tube”
–> Looks at evolution in progresss in a microbial system over time
Process of Lenski’s LTEE
1) Established 12 parallel cultures of E.coli
2) Initial cultures were frozen
3) Subcultured each culture into new growth medium once a day for 75 days
4) After 75 days: Aliquots were removed from each culture and FROZEN
5) Repeated 3 + 4 for 1,500 days (10,000 generations)
6) Fitness (cell size and viability) was analyzed in each frozen sample and compared over time
Analysis of Fitness Process
(Ancestral and evolved cultures thawed)
1) Neutral marker added to one of the cultures (creates different color colonies)
2) 1:1 mix of ancestral:evolved cultures
3) Mix diluted into two flasks
4A) 1 dilution = immediately plated
4B) 1 dilution = incubated (to grow) –> then plated
5A) Initial ancestral:evolved cell ratio calculated (red:white cells)
5B) Final (after growth) ancestral:evolved cell ratio calculated (red:white cells)
6) Calculate relative growth rate (Initial ratio –> Final ratio) = RELATIVE FITNESS
What did Lenski observe (find)?
Fitness INCREASED over generations of evolved cultures
–> Population of cells came to consist of cells increasingly better suited for survival
What was a potential explanation for the fitness gains Lenski observed over time?
Mutations in genes involved in the cell stress response
–> Allowed cells to better survive the stressors of growing in lab conditions
A population that has evolved under certain conditions will…
exhibit enhanced fitness under THOSE specific conditions BUT NOT under other conditions
What was a major mystery surrounding mutations?
Do mutations occur in RESPONSE to environmental conditions or are they spontaneous (occur even in absence of selective conditions)?
(Essentially, do mutations occur because they NEED to occur or do they just simply happen whenever?)
What did Esther Lederberg do?
Conducted experiment to investigate whether Streptomycin resistant mutations arise spontaneously (in the absence of selective pressure; WITHOUT streptomycin present)
What was Esther Lederberg’s previous work that led to her famous experiment?
In her previous work she observed a few resistant variants appear when E.coli was grown in the presence of Streptomycin
–> Led to the questions of whether streptomycin presence was a requirement for mutations for its resistance to occur
Esther Lederberg’s Experiment: Process
1) E.coli culture was spread over an agar master plate (no streptomycin)
2) Replica plating using sterile velvet cloth
3) Some of the plates had NO streptomycin while others were treated with streptomycin
4) Colonies observed growing in streptomycin plate were identified to be resistant mutants
5) The colonies on the NON-streptomycin plate in the corresponding positions as the identified resistant mutants were collected
6) The NON-streptomycin collected colonies (corresponding to resistant colonies) were then transferred to a NEW plate WITH streptomycin
7) Plates were analyzed for growth
Esther Lederberg’s Experiment: Results
Cells grown in NON-streptomycin conditions that were transferred to Streptomycin plates GREW on the streptomycin plates!
== Mutation for streptomycin resistance occurred WITHOUT exposure to streptomycin
(cells were already resistant when transferred onto strepto plates)
What conclusions were drawn as a result of Esther Lederberg’s experiment?
Conclusion: Mutations CAN occur WITHOUT selective pressure
–> Mutation is spontaneous; it does not rely on any environmental condition
What experiment corroborated Esther Lederberg’s findings?
Experiment of Salvador Luria and Max Delbruch:
“Fluctuation Test”
What was the experiment of Salvador Luria and Max Delbruch?
Investigated the generation of resistance to T1 bacteriophage in T1 sensitive E.coli
What was the thought process behind the Luria and Delbruch experiment?
If T1 resistance mutation randomly occurred in cells…
it would be passed down to its progeny…
and as such the # of T1 resistant cells in culture would be representative of WHEN the mutation spontaneously occurred
Fluctuation Test: Process
1) T1 sensitive E.coli were separated into two types of culture
a. Independent cultures (in tubes)
b. Singular culture (in one flask)
2) Independent and singular cultures were incubated
3) After incubation, cells were plated:
a. Independent cultures were each inoculated on their plate
b. singular culture was inoculated onto a # of plates
4) T1 was added to the plates
5) Colonies were counted
What were the results of the fluctuation test?
Independent Culture Plates:
= Observed a variable # of colonies grown per plate (not the same across plates; some had more growth, some had less, some had none)
Singular Culture Plates:
= Observed ~ uniform # of colonies grown on each plate!
Fluctuation Test:
How does results of independent cultures display randomness of mutations?
Because the observed variability in resistance to T1 between independent cultures demonstrates that the mutations all occurred at different times (not standard) if they even occurred at all
Fluctuation Test:
Why did the singular culture plates all have the same colony growth?
Because in the larger culture, progeny were able to disperse throughout the total volume = each plate was therefore more uniform in composition
Fluctuation Test:
The independent cultures produced varying # of resistant progeny based upon what factors (2)?
1) IF any T1 resistance mutation occurred
2) WHEN in the culturing process the T1 resistance mutation occurred
Fluctuation Test:
For the independent cultures, what does the amount of progeny colonies on T1 plates determine about WHEN a given culture mutated?
> # of progeny = > time to proliferate on T1 plate = EARLIER mutation event
< # of progent = < time to proliferate on T1 plate = LATER mutation event
Recombinant DNA
DNA molecule consisting of DNA from two (or more) distinct sources that get linked together
(producing a unique DNA molecule)
What is needed to form recombinant DNA (in a general sense)?
Need to be able to:
1) CLEAVE (cut)
2) STITCH (paste)
pieces of DNA very precisely
Restriction Enzymes
Enzymes that recognize and CLEAVE DNA at specific site/s
Where are all restriction enzymes derived from?
What is their purpose in original source?
All restriction enzymes are derived from bacteria
–> Purpose in bacteria is to protect against phage infection (by cleaving viral DNA)
Restriction Site
A short (usually 4, 6, or 8 BP) DNA sequence recognized by a restriction enzyme
–> Site at which enzyme binds and then makes a cut at a point within this site
Most restriction sites are what kind of a special sequence?
PALINDROMIC
Palindromic (meaning in DNA)
Both strands of dsDNA have an identical sequence if both are read 5’ to 3’ or 3’ to 5’
–> Essentially, when looking at a DNA molecule, the two strands will be identical if read in opposite directions (L+ R)
How/where do restriction enzymes cut DNA?
(What types of cuts?)
They make double stranded cuts in DNA at a point in the restriction site
–> Can make one of two cuts:
1) Sticky End Cuts
2) Blunt End Cuts
Sticky Ends
A cut made by restriction enzymes in which the resulting two fragments have single stranded overhangs
(these overhangs are complementary to each other)
Blunt Ends
Cut made by restriction enzymes in which the two fragments have no overhang
DNA cut by the same restriction enzyme (producing sticky ends)…
DNA cut by the same restriction enzyme (producing sticky ends) will be fragmented into molecules with complementary overhangs that can ANNEAL to each other
What is a major difference in annealing between sticky and blunt end DNA fragments?
Sticky Ends = Can anneal ONLY to other complementary sticky end fragments (made by the same restriction enzyme)
Blunt Ends = Can anneal to ANY other blunt end fragments (does NOT have to be made by same restriction enzyme)
–> Because there is no overhang, their end sequence doesn’t really matter) and thus
DNA Ligase
Enzyme that anneals DNA
What bond does DNA ligase form?
(Forms COVALENT bonds!)
Forms the phosphodiester bond between free 3’ OH grp and free 5’ phosphate grp on nucleotides to be annealed
= Forms the backbone connection between nucleotides
If two fragments of DNA are cut by the same restriction enzyme (that produces sticky ends), what are the potential products of annealing?
4 potential products:
Two molecules the same as original starting molecules
Two recombinant molecules (mix of the two starting ones)
Common restriction enzyme examples (4):
1) EcoRI
2) BamHI
3) HindIII
4) SmaI
EcoRI
Source = E.coli
Res. Site + Cut = 5’-G*AATTC-3’
Cut Type = Sticky
BamHI
Source = Bacillus amylo-liquefaciens
Res. Site + Cut = 5’-G*GATCC-3’
Cut type = Sticky
HindIII
Source = Haemophilus influenzae
Res. Site + Cut = 5’-A*AGCTT-3’
Cut Type = Sticky
H = haemophilus, I = influenzae
SmaI
Source = Serratia marcescens
Res. Site + Cut = 5’-CCC*GGG-3’
Cut Type = BLUNT
S = serratia, ma = marcescens
Restriction Enzyme Naming Convention
(Name reflects origin organism!)
1st Letter = Genus
2nd/3rd Letters = Species
4th Letter = Strain
(+ Roman numeral = order of RE discovery from a given strain)
DNA Cloning
Replication of recombinant DNA
(AKA molecular cloning)
DNA Replication does NOT equal_______
(Why)
DNA cloning!
DNA replication = Natural process
DNA Cloning = LAB TECHNIQUE
General overview process of DNA cloning:
1) Aquire DNA fragment of interest
2) Insert fragment into a VECTOR = Recombinant vector
3) Insert recombinant vector into host cell
4) Recombinant vector replicated by host cell
Cloning Vector
A DNA molecule that can be genetically manipulated AND that can replicate within cells
What are the main types of cloning vectors?
1) Plasmid
2) Phages
3) Cosmids
Plasmid Vectors
Recombinant plasmids
–> made using plasmids with specific restriction sites that can be used to cut the plasmid and insert foreign DNA also cut by the same restriction enzyme
What did Stanley Cohen do?
First person to use a plasmid vector!
–> Made the pSC105 recombinant vector
What is the pSC105 vector made from?
Made from two plasmids:
1) pSC101
2) pSC102
–> Both cut by EcoRI and then ligate to each other
Characteristics of pSC101
1) Tetracycline resistance gene
2) OriV (origin of rep. for E.coli)
3) ONE EcoRI restriction site
Characteristics of pSC102
1) Kanamycin resistance gene
2) THREE EcoRI restriction site
In pSC105 production, which plasmid serves as the vector and which one donates a DNA fragment?
pSC101 (tetracycline res) is the VECTOR
pSC102 (kanamycin res) is the fragment donor
What is the difference in fragments produced by ECORI digestion of pSC101 and pSC102?
pSC101 = ONE linear fragment (cut at one re site)
pSC102 = THREE linear fragments (cut at three re sites)
Process of pSC105 Production:
1) pSC101 and pSC102 are digested with ECORI
2) Complementary fragments anneal by base pairing and then ligation (by ligase) = Mix of ligated and non-ligated fragments
–> SOME of the products will be the pSC105!
(= linear pSC101 fragment + linear pSC102 fragment with the kanamycin res. gene –> anneal at both ends of fragments to circularize)
3) Products of ligation are transformed into E.coli
4) E.coli is plated on kanamycin and tetracycline –> cells that survive = cells successfully transformed with pSC105 and therefore carry it!
Characteristics of pSC105:
1) Tetracycline res. gene
2) Kanamycin res. gene
3) TWO EcoRI restriction sites
4) OriV
What was the FIRST widely used cloning vector?
pBR322
pBR322 size
Small plasmid ~4.36 kb
Characteristics of pBR322
TWO antibiotic res. genes (1. Ampicillin, 2. Tetracycline)
MANY restriction sites throughout AND has restriction sites WITHIN the antibiotic resistance genes!
a-amylase gene
Encodes for a-amylase enzyme which hydrolyzes starch
(Found in bacillus species)
Cloning of a-amylase gene using pBR322:
What restriction enzyme is used for digestion of bacillus and pBR322 plasmid?
BamHI
Cloning of a-amylase gene:
What vector is used?
pBR322
Cloning of a-amylase gene:
Production of cloning vector process
1) pBR322 and bacillus species DNA is digested with BamHI
2A) pBR322 becomes linear (one BamHI site) AND LOSES tetracycline resistance (BamHI site = in Tc res gene –> BamHI digestion = cleaves Tc res gene)
2B) bacillus DNA (many BamHI sites) = many linear fragments
3) Mix linear pBR322 fragment with other bacillus DNA fragments and ligase
= 4 potential product types, one of which is recombinant pBR322 with inserted a-amylase gene
4) ALL ligated products transformed into E.coli (only circular molecules transform)
What are the potential product types from digestion of pBR322 digestion by BamHI?
- pBR322 plasmid reforms (ends anneal to themselves)
- Linear recombinant molecules
- Amp. resistant, Tc sensitive CIRCULAR recomb. molecules
- Amp. resistant, Tc sensitive CIRCULAR recomb. molecules WITH a-amylase gene
In transformation of ligated products, what can and CANT transform?
WHY?
Circular molecules = transform
LINEAR molecules = DONT transform
–> Linear molecules are degraded by bacterial enzymes when trying to enter!
Cloning of a-amylase gene:
Selecting for plasmid with a-amylase gene process
I. Isolation of cells with a functional plasmid
1) All transformed cells plated on ampicillin
a. What grows = got a functional plasmid
b. What dies = did not get a functional plasmid
II. Isolation of cells with a recombinant plasmid
2) Replica plate onto plate with ampicillin AND tetracycline
a. What grows = cells w/ recircularized pBR322
b. What dies = cells w/ a recombinant plasmid (some fragment inserted at the res. site in the Tc res. gene)
3) Find colonies on ampicillin only plate that correspond to the dead ones on the Amp/Tc plate
III. Isolation of cells with a-amylase containing plasmid
–> FUNCTIONAL SCREEN:
4) Screen each collected Amp. resistant/ Tc. sensitive colony for the ability to hydrolyze starch (a-amylase function)
–> Plate colonies on plates of starch and iodine
= Colonies w/ clearing around them = contain a-amylase plasmid!
What functional screen can be used to test for a-amylase presence?
Plate cells with iodine and starch
–> Any “clearing” around colonies that occurs = that colony has a-amylase
Clearing = Areas on plate that are clear in color (compared to other regions with dark starch-iodine complex
Cloning of a-amylase gene:
What feature of pBR322 plasmid is changed by recombination of the vector using BamHI?
Tetracycline resistance is lost!
–> Plasmid is Ampicillin resistant BUT tetracycline sensitive
–> Because BamHI cuts at a site IN the Tc resistance gene!
What is a limitation of pBR322
Not all restriction sites are within the antibiotic resistance genes
= Not all restriction sites allow for easy screening to detect clones w/ desired recombinant plasmids
Multiple Cloning Site
Short DNA segment in a vector that contains a CLUSTER of different restriction sites that only appear ONCE in a plasmid
Methods for screening for presence of recombinant plasmid of interest in host cells: (2)
1) Replica plating –> Screening Method
2) Blue-white screening
Blue-White Screening
A method that allows for the detection of a desired cloned DNA fragment in a SINGLE screening step using coloration differences
What plasmid is commonly used in blue-white screening?
Why?
pUC18 plasmid
–> Has BOTH an antibiotic resistance gene (ampicillin) AND part of the LacZ gene (encoding for alpha fragment of B-galactosidase)
What does the LacZ gene produce? (what does the product do?)
B-galactosidase
= Involved in breakdown of lactose BUT also breaks down a chromogenic substrate: X-GAL
When using the pUC18 plasmid, the LacZ gene within the host cell is only expressed if…
LacZ’ gene from pUC18 is intact (lack of gene insertion)
–> and then transcribed and translated –> producing alpha-fragment of B-galactosidase
–> Alpha fragment (N-terminus) combines with host cell produced omega-fragment (C-terminus) = Functional B-galactosidase
What component of pUC18 makes it good for blue-white screening?
It has the LacZ’ gene with an MCS inserted into it
–> Restriction at one of the sites in the MCS = insertion of DNA into the LacZ gene = nonfunctional gene
–> Allows us to determine if recombination happened depending on whether LacZ gene is functional or not
How is a functional B-galactosidase produced?
Combination of alpha and omega fragments
–> The fragments on their own are not functional, must come together to work!
What does the LacZ’ gene produce?
Alpha-fragment (N-terminus) of B-galactosidase
What fragments of B-galactosidase correspond to the N and C terminals of the functional protein?
N-terminus = alpha-fragment (LacZ’ gene)
C-terminus = omega-fragment (host cell genome)
Blue White Screening:
Results for non-recombinant plasmid (recircularized pUC18)
LacZ’ gene = intact –> No cleavage at MCS site
–> Transcription = alpha fragment produced!
–> alpha fragment combines with omega fragment = functional B-galactosidase
–> X-GAL is broken down = releases blue substance
COLONIES TURN BLUE
Blue White Screening:
Results for successfully recombined pUC18 plasmid
LacZ’ gene = NOT intact –> Cleavage and insertion at MCS site
–> Transcription = no alpha-fragment produced
–> No combining with omega fragment = NO functional B-galactosidase
–> X-GAL is NOT broken down = no color
COLONIES ARE WHITE
What do the colors mean in blue-white screening?
Blue= plasmids unsuccessfully recombined (no gene insert, regular function –> X-gal breakdown)
–> What we DONT want
White = plasmids successfully recombined (has gene insert, dysfunction –> NO X-gal breakdown)
–> What we want!
What are the features looked for in the determination of a desirable plasmid vector? (5)
1) Has an origin of replication that is functional in a host cell
2) Has some selectable marker = helps for screening
3) Has an MCS = helps facilitate cloning and screening
4) Small size = helps facilitate easy transfer into cells
5) High Copy # = allows for many copies to be produced
Selectable Marker
A component of a vector that imparts a phenotype so that only cells containing the vector are able to grow under specific conditions
= allows for selection of cells with desired plasmids!
–> Usually antibiotic resistance
Most plasmids are derived from______ which has the commonly found _________ in most plasmids
E.coli plasmid –> ColE1
–> has the OriV replication origin!
What is the issue with OriV?
Only functions in E.coli = plasmids with OriV have limited host range
Shuttle Vectors
Plasmids that can replicate in >1 host due to presence of >1 origin of replication!
Phage Vectors
the use of bacteriophages as vectors
Phage vectors “take advantage” of__________
The natural ability of viruses to infect cells and efficiently deliver their genomes into them!
Process of Recombinant Phage Vector Production:
1) DNA of interest digested by RE
2) Phage DNA digested by the same restriction enzyme to cut out 20kb region from MIDDLE of phage DNA
= leaves two END fragments
3) Mix and ligate the L + R arm phage DNA fragments with the DNA of interest fragments
= Some products will be recombinant phage vector!
4) in vitro phage packaging: phage vectors incubated with head and tail assemblies
–> Cos sites at the terminals of L + R arms are cleaved upon packaging into phage = linear sticky-ended molecule inside phage
5) Phage infects E.coli
6) Phage DNA circularizes in the E.coli host cells (sticky ends come together)
7) Phage vector replication occurs in the host E.coli
–> LYTIC CYCLE OCCURS = cell bursts releasing many phages (some with the desired recombinant DNA)
8) Recombinant phage DNA extracted from plaque on plates
How big is lambda phage DNA?
50 kb
What does the 20kb of removed DNA in phages (when making vectors) consist of (3)?
Genes encoding for:
1) Machinery for integration of phage DNA into host cell (going into lysogenic phase)
2) Machinery for excision of phage DNA from host genome (exiting lysogenic/entering lytic phase)
3) Other genes unnecessary for viral replication
For lambda phage packaging to occur, what are requirements of the DNA being packaged? (2)
1) Must be ~50kb
2) Must have terminal COS sites!
In lambda phage vector generation, why are genes for integration and excision removed from the lambda phage genome?
To force the phages into LYTIC cycle and prevent integration of phage DNA into host cell genome
–> If integration occurs, no progeny will be formed due to entrance into a quiescent phase
WE WANT PROGENY because progeny = cloning of our DNA we want
What happens to cos sites upon phage packaging?
Endonucleases associated with the phage heads cleave the cos sites when the phage DNA enters the head = phage DNA with dual sticky ends
Lambda Bacteriophage
Temperate (lysogenic) phage that infects E.coli which typically enters lysogenic phase (integrates into host cell genome)
What is the DNA size limitation for lambda phage vectors?
What is this useful for?
DNA must be ~50kb for packaging into phage to occur
= Prevents packaging of recombinant phage DNA that has OTHER fragments (not = to 20kb like our DNA of interest is designed to have)
–> Allows us to select for clones with the appropriate and desired 20kb DNA fragment!
Cosmid Vector
Hybrid vector made up of phage AND plasmid components
Cos- (cos site) -mid (plasmid)
What is the structure and components of cosmids?
Circular DNA molecule with the following components:
1) Selective Marker (Antibiotic res. gene)
2) Internal COS site (within the circle)!
3) MCS (where DNA fragment of interest gets inserted)
4) Origin of rep; doesn’t have to be E.coli specific
What is/are the phage component/s of cosmids?
What is/are their function/s?
COS SITE = Allows for packaging of recombinant cosmid DNA into a phage!
–> Allows for transfer of cosmid into a host cell (via phage injection)
What is/are the plasmid component/s of cosmids?
+ their functions:
1) Selective marker
2) Origin of replication
=Maintains and replicates the cosmid in host cell as a PLASMID
3) Multiple Cloning Site (MCS)
= allows for insertion of DNA of interest
Process for cosmid recombination and cloning:
1) Cosmid vector and DNA of interest digested by same restriction enzyme
–> RE used targets a site within the MCS of the cosmid
2) DNA of interest ligated with linearized cosmid
3) Produced cosmids (all) are incubated with phage head and tail assemblies
–> 3.1) COS site of cosmid is cleaved by endonucleases of phage head (= linearized cosmid)
–> 3.2) Phage is produced containing linearized recombinant cosmid!
4) Produced phages go and attach to host cells and inject DNA
5) DNA injected recircularizes (cos site fragments rejoin) and maintains in the cell as a plasmid (due to the host cell specific origin of replication)
In increasing order, what is the size of DNA that different vectors can accommodate the insertion of?
SMALLEST = Plasmid (up to 15kb inserted DNA)
Middle = Phage Vector (up to 24 kb inserted DNA)
LARGEST = COSMID (35-45 kb inserted DNA)
What plasmid is this?
pUC18