-single-celled

-prokaryotic (no nucleus or membrane-bound organelles)

-do NOT exhibit meiosis

-circular chromosome

Often alternative phenotypes of prototrophy (self-feeding) and auxotrophy (outside-feeding) are used as markers for genetic analysis.

Class 1

-Ability to synthesize essential macrmolecules such as amino acids or nucelotides

-things it needs to survive

Class 2

-ability to utilize specific energy sources (typically referred to as a carbon source)

Class 3

-Resistance to compounds that normally kill or inhibit bacterial growth

S=sensitive/ R=resistant

Scoring Bacterial Phenotypes

A selection system is used.

-each colony is said to derive from a single bacteria (ie. spread so thin that single cell will be isolated)

-read examples= ch7 p2-3

New genetic material enters the bacterial cell by one of three mechanisms, and then undergoes recombination with homologous regions of the original bacterial chromosome.

1.Conjugation = the exchange of genetic material b/w bacteria involving cell to cell contact

2.Transformation = the uptake of genetic material from the environment by bacteria

3.Transduction = viral (bacteriophage) mediated transfer of genetic material b/w bacteria

One member of the conjugating pair carries a fertility factor (F), conferring donor ability.

-F plasmid - a small, circular, autonomously replicating, extrachromosomal piece of DNA

F⁺ have it

F^- dont

1952- discovered bacterial gene transfer in E. coli

Strain A = bio^-cys^-leu⁺phe⁺thi^-thr⁺

Strain B = opposite

***both strains are auxotrophic

***if strains A and B are combined, a low frequency (1 x 10^-7) of the resulting bacteria are prototrophic

 ..............So how does recombination occur??

Hypothesis 1: Contact required

-placed on either side of a filter (DNA can pass)

-unable to transfer genes through filter

Hypothesis 2: The direction of gene transfer is directed, with one strain acting as a donor and the other as a recipient.

Experiment

-Expose one strain to streptomycin, an antibiotic that allows the transfer of DNA but inhibits the ability to produce progeny.

Strain A: treat with Str and incubate with Strain B... get prototrophs.....(strain A donates to B, which can grow)

Strain B: treat with Str and incubate with strain A... result is no prototrphs

Conclusion: Trasfer is directed; only Strain A may act as the donor and Strain B as the recipient.

**Strain A has F factor**

1. Enables the production of pili, a proteinaceous attachment tube b/w bacteria that facilitates cell to cell contact

2.replication of F plasmid permits F to be maintained (a copy of the F factor moves into F- cells during conjugation)

3.prevents conjugation b/w two F+ cells

**Some F+ strains spontaneously become strains with a high frequency recombination (HFr) of chromosomal genes (1000 times more frequent)

Experiment:

Hfr strain: Str^STon^RLac⁺Gal⁺Azi^R

F^- strain: opposite

-combine the two strains

-tak a sample of cells at specific time intervals and interrupt mating (disrupt conjugation)

-plate the samples on medium containing streptomycin (Str⁺)

-streptomycin kills the original Hfr cells (Str^S) but permits the growth of F^- cells (Str^R)

-determine the genotype, by testing for the Ton^R,Lac⁺,Gal⁺, and Azi^R genetic markers, of the surviving bacteria

Results: The genetic markers are present with different frequencies at different sampling (mating) times

**Conclusions:

1)In Hfr Strains, F factor is integrated into the bacterial chromosome promoting transfer of chromosomal genes

2)There is a fixed point at which transfer begins (origin) and a linear order to the transfer process of the genes

3)The time taken to transfer a specific gene is related to the distance b/w that gene and the origin (use minutes for map distance)

Additional Observations:

-If the mating is permitted to continue, the recipient cells become Hfr

Conclusion:

-The complete F-factor must be transferred, both the origin (first to enter) and the terminus (last to enter) are required to become Hfr.

exconjugate: a bacterium that has undergone conjugation (the recipient cell)

endogenote: the endogenous recipient chromosome

exogenote: the chromosome segment from the donor

merozygote: a partially diploid cell formed through conjugation (it is transient, as the cell destroys linear DNA)

experiment:

Consider 5 different Hfr strains, all derived from the same F+ strain, and determine the sequence of gene transfer.

results

1)For each strain, th gene which enters first is different (integrated in different places)

2)the relative position of each gene is constant (gene order conserved)

ie. gene b always flanked by a and c

Conclusion

The difference between the Hfr strains is the position and orientation of the origin within the circular chromosome

Hypothesis: Using the sequence and time of transfer, a map of the chromosome can be obtained.

1. Based on the order of transfer it is possible to deduce the linear order of genes.

2.based on the first gene transferred it is possible to determine the position and orientation of the F-factor

3.the time taken to transfer the genes, from the donor to the recipient, is equal to the distance b/w the genes

**draw in a circle!!

Hypothesis: Once genes are transferred into the F- strain, they are able to recombine with genes in the host chromosome. The frequency of recombination b/w two genes will depend on the distance b/w those genes

The likelihood of a gene being transferred is directly related to its distance from the origin.

-Genes nearest the origin being transferred at the highest frequency

-genes farthes from the origin have the lowest frequency

gradient of transfer: genes neearest origin are transferred at a higher frequency than those furthest from the origin

Integration: = recombination + transfer

-a gene must be successfully transferred before it can be integrated into the chromosome

-the genes closest to the origin have a higher frequency of being transferred and therefore will be integrated more frequently (than genes further away from the origin)

-**this will distort the recombination frequency as a measure of genetic distance

If mapping by recombination frequency, one must ensure that the transfer is equal for all genes

-if the transfer is equal for all genes, then

...integration = recombination

-and the integration frequency (=recombination frequency) can then be used to determine the distance b/w genes

-when you see the results for frequency of inheritance, there will be one gene that occurs least often.

-we can therefore deduce that this gene is transferred last

-to ensure that the transfer is equal for all genes, we can select for the exconjugate that contains the last gene (furthest from the origin) that we are looking for.

-we do this by killing the others who do not have it.

Once we can ensure equal transfer of genes, we can select for exconjugates that contain this last gene transferred.

-the number of crossover events must be even (otherwise result would be a linear chromosome)

-if there are, say, 3 genes, like in a trihybrid we can determine what is the middle one by crossover frequency. (whichever one deviates from parental)

-then we just find map distance (% recombinants) between both ends and the middle, and add them for the total.

-introduction of DNA fragments from the environment through the bacterial cell wall and membrane (recombination leads to integration)

-the frequency of bacterial transformation is generally low but can be increased in a laboratory setting by:

---making cells "competent" to take up DNA (treatment with Ca++)

---use of a heat shock or an electric shock (electroporation) to increase the uptake of DNA

-transformation can also be induced in plant and animal cells

Linkage Information and Transformation

-DNA is introduced as "small" ragments, no direction of transfer

-the closer together two genes are, the more likely they will be on the same fragment and integrate into the chromosome together

-can deduce the relative distance b/w pairs of genes based on their cotransformation frequency (the introduction of more than one gene simmultaneously)

-if two genes are unlinked, the frequency of cotransformation will be equal to the product of their individual transformation frequencies

1.phage recognizes and attaches to the bacterium (adsorption)

2.viral genome enters the bacterium

3.bacterium makes copies of the viral genome (replication) and capsid (transcription and translation)

-viruses have extremely strong promoters

4.new phages assemble within the host bacterium

5.bacterial cell is lysed and new viral particles are released

-if bacteria are infected and then grown on an agar plate, the lysis of the bacterial cells results inthe formation of a zone of clearing around an initially infected bacterial cell

-the zone of clearing, referred to as a plaque, is the result of neighbouring bacteria becoming infected and lysed as the viral particles are dispersed from the original host cell

-a plaque is assumed to result from the infection of a single host (same idea as how a bacterial colony on a plate is assumed to arise from a single bacterial cell)

-a temperate phage such as lambda is able to ener the lysogenic cycle as opposed to initiating the lytic cycle upon infection. By contrast, a virulent phage is unable to enter the lysogenic cycle and therefore always results in the lysis of its host via the lytic cycle

1.The phage genome becomes integrated into the bacterial chromosome, the phage genome is then referred to as a prophage (and the bacterial host is now lysogenic)

-integration is accomplished through a recombination evet at specific attachment sites (att sites) found in both the phage and bacterial genomes

2.The prophage, through the expression of a specific gene, represses the lytic cycle

-repressors are present within the cytoplasm and maintain the lysogenic state

3.The prophage is replicated each time the bacterial chromosome is duplicated (and is inherited by progeny)

4.Adverse conditions may trigger the prophage to become released from the bacterial chromosome which will subsequently induce lysis (initiation of the lytic cycle)

-excision of the prophage is through reversal of the initial recombination event

-if a lysogenic Hfr bacterium conjugates with a non-lysogenic F- bacterium, the transferred prophage immediately enters the lytic cycle = zygotic induction

-the non lysogenic F- lacks the cytoplasmic repressors and therefore lysis is induced

-bacteriophage mediated transfer of genetic material

-genetic information is transferred b/w two bacteria via a bacteriophage vector

HYPOTHESIS: Physical cotact is necessary for the transfer of genetic material

RESULTS:

-the exclusion limit (point at which genetic material cannot be transferred b/w bacteria) is determined by the size of the phage particle

-if the phage is larger than the pore size of the filter = no transfer of genetic information

-if the phage is smaller than the pore size = transfer of genetic material ccurs

-therefore, the phage is the vector for transfer

GENERALIZED TRANSDUCTION

-random sample of genes are transduced.... occurs via lytic cycle

1.the bacterial chromosome disintegrates as the cell is lysed

2.random genomic fragments may be inforporated into the phage particle = faulty head stuffing

3.subsequent phage infection of a new host cell will introduce these genomic fragments from the previous host

4.recombination leads to intefration of the fragment into the chromosome of the new host

-each gene is equally likely to be transduced

-proximity of genes is implied by....(see figure 7-26)

-can only uptake small pieces of NA - the higher the co-transduction, the closer the genes are

SPECIALIZED TRANSDUCTION

-can only be carried out by a temperate phage (lysogenic cycle)

1.excision ofthe prophage initiates the lytic cycle

2.if reombination is not exact, bacterial genes close to the att site may be incorporated into the phage (faulty outlooping)

-in lambda, the gal gene is adjacent to the att site

-if the phage leaves a viral gene behind, it may become incapable of integrating

3.subsequent infection of a new host can lead to the integration

-recombination integrates the new genetic information

-can also occur by formation of a prophage in the new host cell

note: this may require the presence of an addition "helper phage"

-mapping bacteriophage genomes requires two genetically distinct phages to come together, within the same host cell, to have the opportnity to recombine

-coinfection of a bcaterium, with two distinct phage types, can be accomplished by incubating the bacerial host with a very high concentration of both phages

QUESTION:

What is the map distance b/w the T2 phage genes h and r?

The h gene influences host range where

-h+ alleles has a narrow host range (infect 1 Ecoli strain)

-h- allele has a broad host range (infect two strains)

The r gene influences rate of how lysis where:

-r+ allele = slow lysis (smaller plaques)

-r- allele = fast lysis (larger plaques)

EXPERIMENT

-cross (coinfection) phage h+r- and h-r+

-incubate the progeny of the coinfection with a combination of both E.coli strains

-score the 4 possible genotypes (2 parental and 2 recombinant)

(cloudiness indicates if one, none, or both are infected)

-can then calculate distance using recombination frequency (amount of h+r+ and h-r- over total plaques)

Mendel's Findings:

-one gene, two alleles

-each gene controls a single different characteristic

-one allele is dominant to the other

COMPLEXITY ARISES:

1)a gene may have multiple alleles: each gene consists of many nucleotides, which when mutates result in different alleles

-different alleles may have different phenotypes

2)a gene may be pleiotropic: a single gene acts to control more than one phenotypic characteristic

-ie. gene controlling pollen grain germination also control root hir growth

-suggests a common underlying physiological or developmental mechanism

3)more than one gene may control a single characteristic

-since many biological pathways consist of a series of steps, each controlled by a different protein (coded by a gene), a single characteristic may be the result of multiple gene activity

How do you know if 2 mutant lines with the same phenotype are the result of:

a)mutations in 2 genes within the same pathway

b)2 alleles of the same gene

-to understand genetic control, must break or screwup genome

EXAMPLE

-Consider trp biosynthetic pathway:

(trpA+)---->X---(trpB+)---->Y---(trpC+)---->

---->tryptophan

Two new auxotrophic mutants for tryptophan synthesis are found in the plant Arabidopsis thaliana

Hyp 1: The two mutants are alleles of the same gene, A

Hyp 2: The two mutants are alleles of different  genes, A and B

EXPERIMENT:

-cross the two mutants together and the results will differ depending on which hypothesis is correct

Exp Results for Hyp 1:

-crossing two auxotrophs yields F1 auxotrophs

-the 2 mutations fail to complement, therefore they represent two alleles of same gene

Conclusion: F1 progeny is mutant, therefore mutations do not complement one another, and the mutations are alleles of the same gene.

Exp Results for Hyp 1:

-F1 can be a prototroph (assuming true breeding parents)

 Conc: F1 progeny is wild type, therefore the mutants complement on another, and must be different genes

Rules for a Complementation Test

-can only be done with recessive mutations

-if the mutations are in different genes, the two mutations will complement one another (progeny wil be wild type)

-if the mutations are alleles of the same gene, the two mutations will not complement on another (progeny will be mutant)

-can only be done with recessive mutations

-if the mutations are in different genes, the two mutations will complement one another (progeny wil be wild type)

-if the mutations are alleles of the same gene, the two mutations will not complement on another (progeny will be mutant)

Variation the Complementation Test - The Heterokayon

-in organisms with a haploid life cycle, cannot perform a complementation test

-instead, fuse two cells

-the resulting cells have 2 nuclei in a common cytoplasm (heterokayon)

-gene products from both nuclei act within the common cytoplasm

-mimics the diploid condition

1)epistasis

2)complementary gene action

3)redundancy

4)suppression

-all are modified dihybrid ratios

Recessive

-the dark red colour of wild type Drosophila eyes results from a biochemical pathway with two steps, each controlled by a different gene (v+ and s+).

vermillion----(v+)--->scarlet----(s+)--->dark red

recessive allele v (vv flies = vermillion)

recessive allele s (ss flies = scarlet)

cross: vvs+s+ (vermillion) x v+v+ss (scarlet)

-if no interaction, expect hetero F1 and 9:3:3:1 in F2

***if mutant at v, gene s does not matter

-Vermillion phenotype is epistatic to the scarlet phenotype (epistatic = stands over)

***upstream genes are usually epistatic to downstream genes

***ratio will be 9:3:4(vermillion)

Dominant

-squash colour is determined by 2 genes, W and G

-recessive allele w permits colour expression (ww squash are coloured)

-dominant allele W does not allow colour expression (W_ squash are white)

-recessive allele g (gg squash = green)

-dominant allele (G_ = yellow)

if cross WWgg and wwGG....

F2 ratio wil be 12:3:1

-where mutations in either gene give the same phenotype

-a mutation in any step of the biosynthetic pathway for tryptophan will make the organism auxotrophic for tryptophan

-need both genes!!

-**ratio will be 9:7 (7 being all combinations that don't have both... 3+3+1)

"duplicate genes"

-two genes doing the same thing - loss of one gene can be compensated for by the presence of the other gene

-chemotaxis in Caenorhabditis elegans (a nematode)

-WAN1 and WAN2 - both expressed in sensory neurons

***ratio 15:1.... only need one of the functional genes

-only observe mutant phenotype if the function of both genes are lost

-a mutation in one gene compensates for a mutation in a second gene

-good evidence that the two gene products interact

"two wrongs make a right"

EXAMPLE

-Protein A + Protein B = functional protein (dimer)

heterodimer= protein of 2 different subunits

homodimer= protein of 2 identical subunits

clavata

-mutation in CLV1 causes the division of cells in the apical meristem to become uncontrolled

-get enlarged club-shaped apex

-mutation in CLV3 also causes the division of cells in the apical meristem to become uncontrolled

F2 of CLV1CLV1clv3clv3 x clv1clv1CLV3CLV3:

9 wildtype

3 mutant

3mutant

1 wildtype

***ratio will be 10:6 (one of the supression ratios)

____________________

What if there was an allele of CLV3 (clv3*) which could bind with both wild type CLV1 and mutant clv1?

F2of CLV1CLV1clv3*clv3* x clv1clv1CLV3CLV3:

9wildtype

3wildtype

3 mutant

1wildtype (nearly wildtype.. other changes)

ratio: 13:3

-*****not all mutations alter amino acid sequences in same way (ie. allows them to interact)

1.Inter action b/w alleles (of the same gene)

-ie. dominance/recessive relationships

-can only be defined if a phenotype is expressed in the diploid stage

2.Interaction b/w genes

-ie. epistasis, suppression, redundancy

Genes controlling spore (haploid) pigmentation in Neurospora (fungus)

al=albino

al+=orange

ylo=yellow

ylo+=orange

-expect 1:1:1:1

-however, get 1:1:2.... 2 albino classes

-al must be epistatic to ylo, suggesting the white-yellow-orange biosynthetic pathway

penetrance - the percentage of individuals with a given phenotype who exhibit the phenotype associated with that genotype

B_ = normal vision

bb = blind

-but, of 100 people of enotype bb, only 20 are blind

-therefore, allele b has a penetrance of 20%

expressivity - the extent to which a given phenotype is expressed in an individual

A_ = normal pigmentation

aa=albino

-one individual has genotype aa, but has 30% of normal skin pigmentation

Variation = Polymorphs

-naturally present within populations

Observations of Variation:

1.morphological variation: ie. snail shells, leaf shape inflorescence structure... not just outward expression

2.chromosome polymorphism: ie. inversions, deletions, translocations, extra cromosomes

3.protein polymorphism: changs in amino acid sequence may result in changes to protein charge and physical properties of the protein

-proteins can be positively, negatively, or neutrally charged

4.DNA sequence polymorphism:

-observations of variation in sits recognizd by restriction endonucleases and basepair differences

-detection of DNA sequence polymorphism may be accomplished using:

a)restriction fragment length polymorphisms (RFLPs)

-number of fragments generated following endonuclease digestion .. or.

b)polymerase chain reaction (PCR)

-PCR product size generated (ie. alu)

-defining variation within a population requires that the allele frequency be determined for that population

-requires the conversion of phenotype into genotype

EXAMPLE 1

-if the heterozygous class has a distinct phenotype (ie. incomplete or co-dominance)

What is the frequency of the M allele in MN blood typing?

Let p = frequency of allele M (fMM + 1/2 fMN)

Let q = frequency of allele N (fNN + 1/2 f MN)

***p + q ALWAYS = 1

EXAMPLE 2 - If the heterozygous class is indistinguishable from homzygous

Purple flower colour (A) is completely dominant to white (a). In a population of 100 plants, there are 16 white flowered plants and 84 purple flowered plants. What is the frequency of alleles?

q= frequency of recessive allele a

q= square root of recessive phenotypes

q= 0.4 (sqrt 0.16)

-therefore, p =0.6 (because total is 1)

heterozygosity: a measure of the degree of genetic variation with a population

-indicated by the total frequency of heterozygotes at a locus

-if only one allele is present, no heterozygotes exist and therefore the heterozygoisty =0

**p and q are also a measure of the frequency of gametes within a population

-because p and q are equivalent to the allele frequencies and therefore gamete frequencies, f(alleles) = f(gametes)

-the genotypic frequency of the next generation can be predicted from p and q of the current generation

IF

1)the f(gametes) remain constant and,

2)random mating occurs,

then p and q will remain constant with each generation

=Hardy-Weinberg Equilibrium

*a population will remain in Hardy-Weinberg equilibrium unless outside forces act upon it

1= p² +2pq + q²

Forces Resulting in Variation From Equilibrium:

1)mutation

-production of genetic variants, perhaps during DNA replication or through the actions of mutagenic agents

-mutation events are relatively rare

2)selection

-differential reproductive fitness of a genetic variant, resulting in an increase in frequency of that variant

-ie. thrifty genotype

3)migration

-migration of a population with different genotypes or genotypes in different frequency, ocmpared to the local population will cause a change in allele frequency

-ie. allele frequencies of Native Americans suggests several migratory waves

4)non-random mating

-random mating (independent genotype) results in an equilibrium distribution of genotypes after one generation

-non-random mating occurs when mating is dependent upon genotype

-ie. changes in European allele frequencies map to national border

5)drift

-random changes in gene frequencies in a population

-Techniques involved in the

1)synthesis

2)amplification

3)purification

of specific DNA molecules

Recombinant DNA molecule: a single DNA molecule created from two different DNA sources

1)DNA of interest ... genomic, cDNA(complimentary)

2)Vector (allows for replication by the host cell.. ie. plasmid)

How do we create recombinant DNA molecules?

Early 50s: Luria and Bertani

-certain strains of E.coli able to restrict growth of bacteriophages

1960s: discovered that resistant bacteria have an enzyme system that

1)recognizes foreign DNA and cleaves it

2)protects self DNA from cleavage

-addition of methyle (CH3) groups

1970s: 1st restriction enzyme isolated from Haemophilus influenzae

NOMENCLATURE

1st letter = genus

2nd and 3rd letter = species

4th letter = strain

Roman Numerals = order of discovery

CLASSES
Type I: involved in DNA restriction AND modification. Cutting is random and remote from recognition site

Type II: Most are involved in DNA restriction only. Cutting is usually within the recognition site (most commonly used)

Type III: Like Type I, but require 2 recognition sites. Poor efficiency.

KEYS TO ACTIVITY

-palindromes

-both strands of DNA recognized and cut

-hree types of ends produced: 5' overhangs, 3' overhangs, blunt ends

Three Requirements:

1)Vector: a vehicle to carry our DNA and ensure it will be replicated

2)Host: Provides machinery to replicate DNA. Host of choice is E.coli.

3)Means of Selection: How do we know if the vector contains our DNA? How do we know host has taken up vector?

SUMMARY

1. Isolate DNA

2.Cut /w R.E.

3. Ligate DNA (plasmids as cloning vectors)

4.Transformation

5.Selection (appropriate medium)

Often desirable to make large quantities of protein

-biochemical studies

-commercial applications

Therefore, vector must allow for

-replication DNA
-transcription

-translation

=>expression vector

Native promoters generally do not allow high levels of gene expresion.

Solution is to isolate just the coding sequence

Expression Vectors make use of viral promoters

-extremely high levels of expression

-can be controlled (turned on/off)

Due to "Eurkaryotic Gene Problem", there is a "Eukaryotic Gene Solution" and eukaryotic genes can be xpressed in a prokaryotic host

The application of recombinant DNA technology to a "problem"

WHY?

Research Related:

-DNA sequencing, modifications, structure

-Site-Directed mutagenesis

-Gene products (difficulties in obtaining sufficient quantities, purification, nature of organism, need to modify)

Gene Therapy (medical) is still in infancy

1990- commercial production in China of tobacco resistant to TMV

1994- Calgene developed the Flavr Savr tomato

-affected an enzyme involving ripening

-tomatoes could vine ripen but would maintain a shelf life similar to traditionally harvested tomatoes

-tomatoes pulled from the market in 1997 - bland, bruised easily, consumer concerns

Glyphosate Resistance

N-[phosphonomthyl]glycine

-post emergent, broad spectrum herbicide

-acts by blocking an enzyme (EPSP) central to synthesis of aromatic amino acids

Roundup Ready Canola

-gene encoding for a resistant form of EPSP introduced into canola

-gene encoding for glyphosate exidase from Achromabacter also introduced

-plants devloped include corn, cotton, soybeans, sugarbeets

Fishberry

-urban legend