Bacteria Flashcards

1
Q

Describe the process of binary fission.

A
  1. DNA replication begins at the origin of replication (ori) where DNA is unzipped by breaking hydrogen bonds between bases of the 2 strands to form a replication bubble
  2. DNA replicates by semi-conservative replication where each original strand serves as template for synthesis of daughter strands by complementary base pairing. Replication takes place outward from the origin in both directions forming 2 replication forks.
  3. 2 newly formed ori move to opposite poles of the cell and attach to the plasma membrane
  4. Cell elongates to prepare for division.
  5. Since DNA is circular with no free ends, the 2 daughter DNA molecules will be interlocked with the completion of replication.
  6. Enzyme topoisomerase cuts, separates and reseals the two DNA molecules
  7. Invagination of the plasma membrane and the deposition of new cell wall (division septum) eventually divide the parent cell into two daughter cells -> each inherits a complete genome (genetically identical)
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2
Q

Describe how naked DNA fragments are taken up by competent
bacteria cells in the process of transformation.

A
  1. Fragments of foreign naked DNA from dead lysed bacterial cells
  2. Naturally competent bacteria with cell-surface proteins bind and transport DNA into the cell.
  3. Artificially bacteria can be made competent through immersion in a medium with CaCl2 followed by a heat shock treatment
  4. Foreign DNA incorporated into chromosome through crossing over at 2 homologous regions (i.e. homologous recombination) found on the bacterial chromosome
  5. Result: recombinant cell
  6. If different alleles for a gene were exchanged, the new allele will be expressed –> permanent change in genotype & phenotype
  7. Recombinant genome will be passed on to all subsequent offspring through binary fission
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3
Q

Describe the process of generalised trandusction involving a lytic phage.

A
  1. A phage infects a bacterium, injecting its viral genome(DNA)
    into the host cell
  2. The bacterial DNA is degraded into small fragments, one of
    which may be randomly packaged into a capsid head during
    the spontaneous assembly of new viruses
  3. Upon cell lysis, the defective phage will infect another
    bacterium and inject bacterial DNA from the previous host cell
    into the new bacterium
  4. Foreign bacterial DNA can replace the homologous region in the
    recipient cell’s chromosome if crossing over (i.e. homologous recombination) takes place, allowing the expression of a different allele from the previous host
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4
Q

Describe the process of specialised transduction involving a lysogenic phage.

A
  1. Temperate phage infects a bacterium, injecting its viral genome into the host cell
  2. The viral DNA is integrated into bacterial chromosome forming a prophage
  3. which may be improperly excised to include adjacent segment of bacterial
    DNA and not the entire phage DNA during an induction event
  4. Hence phage-bacterium hybrid DNA may be packaged into a capsid head during
    the spontaneous assembly of new viruses
  5. Upon cell lysis, the defective phage will infect another bacterium and inject
    bacterial DNA from the previous host cell into the new bacterium
  6. New alleles from the previous bacterial cell can be incorporated into the genome of the new host by homologous recombination or integration of phage- bacterium hybrid DNA as defective phage enters the lysogenic cycle
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5
Q

Explain why conjugation can only be initiated by an F+ cell with an F plasmid.

A
  • F+ cells are bacterial cells which posses an F plasmid
    -On the F plasmid, there is a segment of DNA called an F Factor (F=fertility) that carries genes coding for sex pili
    -Due to presence of F factor, the donor cell is able to produce appendages called sex pili to attach itself to the recipient cell
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6
Q

Describe the process of conjugation

A
  1. Sex pilus (coded for by F factor) of F+ bacterial cell makes contact with a F- cell and retracts to bring the 2 cells closer
  2. The hollow pilus then acts as a cytoplasmic mating bridge between the 2 cells
  3. One of the 2 strands of the plasmid DNA is nicked and transferred from the F+ cell to the F- cell through the bridge
  4. The single stranded F plasmid DNA circularizes in F - cell and is used as a template to synthesize a complementary strand for a double-stranded
    plasmid DNA. The F- recipient cell is now a F+ cell
  5. Replication of the plasmid occurs via rolling circle DNA replication
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7
Q

Desribe the process of rolling circle DNA replication

A

a) One strand of ds F plasmid is nicked by a nuclease –> free 3’OH end is then used as a primer for strand elongation by DNA polymerase using the unnicked/intact strand as a template–> elongation process is facilitated by the displacement of the 5’ end of the nicked strand and is transferred across the mating bridge to the recipient bacterium –> Upon completion of a unit length of the plasmid DNA (after 1 round), another nick occurs to release the original strand

b) In the recipient cell, the single strand of F plasmid DNA re-circularises and serves as a template for the synthesis of a complementary daughter stand to form a double stranded circular DNA.

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8
Q

Describe negative gene regulation of the lac operon in the absence of lactose.

A

By default, lac operon is considered repressed i.e. off

  1. The regulatory gene, lacI, is constitutively transcribed resulting in continued production of small amounts of the lac repressor protein
  2. The lac repressor protein is produced in the active form and binds to the lac operator sequence via its DNA-binding site
  3. In the absence of lactose, the repressor binds to the operator site, denying RNA polymerase access to the promoter.
  4. Transcription of the structural genes of the lac operon is hence blocked. This has the effect of switching the lac operon off

Note: binding of repressor to operator mediated by weak interactions, repressor sometimes dissociates from the operator, resulting in a basal level of lac operon products i.e. galactosidase, permease and transacetylase

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9
Q

Describe how the inducer, allolactose, relieves the repression of the lac operon when lactose is present.

A
  1. In presence of lactose, a few molecules of lactose will enter the cell with the help of permease and are converted to allolactose by the few beta-galactosidase molecules present.
  2. Binding of allolactose (inducer) to the allosteric site inactivates the repressor by altering the tertiary structure of the repressor so that its DNA-binding site is no longer complementary to and cannot bind to the operator.
  3. RNA polymerase can access and bind to the promoter to initiate the transcription of the structural genes of the operon (transcribed as a single polycistronic mRNA)
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10
Q

Describe how the inducer, allolactose, relieves the repression of the lac operon in the presence of lactose.

A
  1. In presence of lactose, a few molecules of lactose will enter the cell with the help of permease and are converted to allolactose by the few beta-galactosidase molecules present
  2. Binding of allolactose to the allosteric site inactivates the repressor bu altering the tertiary structure so that its DNA-binding site is no longer complementary to and cannot bind to the operator.
  3. RNA polymerase can access and bind to the promoter to initiate transcription of the structural genes of the operon (transcribed as a single polycistronic mRNA)
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11
Q

Describe positive gene regulation of the lac operon in the absence of glucose when lactose is present.

A

Activator protein: catabolite activator protein (CAP)
-Its DNA-binding site allows it to bind to the CAP-binding site situated within the promoter
-Its allosteric site is specific for binding of cyclic AMP (cAMP), an alternative form of AMP.

  1. In the absence of glucose, cAMP levels will increase
  2. The cAMP binds to the allosteric site of CAP forming CAP-cAMP complex, activating CAP which binds to the CAP-binding site within the promoter
  3. This binding of activated CAP increases the affinity of the promoter region for RNA polymerase. Thus, increasing the rate of transcriptional initiation of the lac operon structural genes.
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12
Q

Describe negative gene regulation of the trp operon in high levels of the corepressor, tryptophan.

A
  1. The trp repressor protein is synthesised in its inactive form with little affinity for the trp operator, which lies within the trp promoter.
  2. As tryptophan accummulates, it binds to the allosteric site of the trp repressor, activating the repressor.
  3. The activated repressor protein binds to the operator at its DNA-binding site.
    -Tryptophan therefore serves as a corepressor which works together with a repressor protein (by activating it) to switch an operon off.
  4. With repressor bound to operator, RNA polymerase cannot bind to promoter and transcription of structural genes cannot proceed, hence turning the operon off.
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13
Q

Describe how the trp operon is regulated in low levels of tryptophan.

A
  1. The regulatory gene, trpR, is constitutively transcribed, resulting in the continued production of trp repressor protein. The trp repressor protein is synthesised in its inactive form with little affinity for the trp operator, which lies within the trp promoter.
  2. The inactive trp repressor protein is unable to bind to the operator at its DNA-binding site.
  3. With no repressor bound to operator, RNA polymerase able to bind to promoter and transcription of structural genes can proceed, hence the operon is default “ON”.
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14
Q

Explain why the trp operon is a repressible operon and produces proteins that are involved in an anabolic pathway.

A

Gene products of the trp operon in E. coli are enzymes involved in the synthesis of the amino acid tryptophan which is essential for protein synthesis hence the operon is usually “on”.

The trp operon is repressible and turned off in high presence of tryptophan.

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