Bacteria

Question 1

Bacterial Structure

Answer 1

1. Chromosome
   - ds circular DNA found in dense nucleoid region
   - Forms loop domains + Supercoiling
   - No introns and no post transcriptional modification
2. 70S ribosomes
   - All freely floating w simultaneous transcription and translation
3. Plasmids
   - Small circular autonomously replicating DNA molecule
   - Usually conferring advantages like fertility or antibiotic resistance
4. Cell Membrane and Cell Wall
   - Phospholipid bilayer + Peptidoglycan Cell Wall cross linked by short peptide chains between alternating NAG and NAM
   - Protects cell from osmotic lysis and confers rigidity
   - Gram positive = thick peptidoglycan, gram negative = thin peptidoglycan with thick outer membrane
5. Capsule
   - Layer of polysaccharides called the glycocalyx that protects bacteria from phagocytosis and desiccation + allows bacteria to adhere to each other to form biofilm
6. Fimbriae, Pili and Flagella
   - Fimbriae attach to bacterial surfaces
   - Pili are for conjugation and motility
   - Flagella are long appendages for motility

Question 2

Binary Fission

Answer 2

Asexual means by which bacteria produces genetically identical cells
1. DNA replication begins at the origin of replication
2. Double helix separates using helicase to form a replication bubble, with DNA polymerase synthesising leading and lagging strand (if asked how genome divides, describe semi conservative replication)
3. The 2 newly formed Ori move to opposite poles and attach to plasma membrane
4. The cell also elongates to prep for cell division
5. But because DNA is circular with no free ends, an interlocking structure is formed, so topoisomerase needed to cut, separate and reseal the 2 DNA molecules
6. Invagination of plasma membrane and deposition of new cell wall (division septum) that divides into 2 identical daughter cells

Question 3

Transformation

Answer 3

1. Uptake of naked foreign DNA fragments from dead lysed neighbouring cells in surrounding medium are taken up by competent bacteria cells with proper surface proteins
2. Foreign DNA incorporated into bacterial DNA via crossing over/homologous recombination
3. If the foreign DNA contains a different allele now expressed, the bacteria cell has transformed (change in genotype and phenotype)
   - Cells can be made artificially competent through electroplating (High CaCl2 and heat shock to E. coli)

Question 4

Transduction

Answer 4

Generalised
1. A phage infects a bacterium, injecting its viral genome into host cell
2. During assembly, a small fragment of host cell’s degraded DNA can be randomly packaged within a capsid instead of the phage’s own genetic material during assembly of new viruses (thus generalised bcos any DNA can be packaged)
3. Following donor cell lysis, defective phage is released and can infect another recipient bacterium, injecting donor DNA fragments into new bacterium (no new phage w no viral DNA)
4. Foreign bacterial DNA can replace homologous DNA if homologous recombination takes place, and different phenotype expressed if different allele expressed

Specialised
1. A temperate phage infects a bacterium, injecting its viral genome into host cell
2. Upon spontaneous induction, the prophage, formed from integration of viral DNA into bacterial chromosome via lysogenic cycle, is improperly excised, such that bacterial DNA adjacent to prophage is also excised
- “specialised” because DNA transferred is restricted to bacterial genes adjacent to prophage
3. Phage-host hybrid DNA is replicated and packaged into capsid head during spontaneous assembly of new virus. All viral progeny are recombinant phages (but can still synthesise functional virus structures)
4. Upon cell lysis, recombinant phage infects recipient cell and injects bacterial DNA from host cell
5. Expression of a different allele occurs when DNA replaces the homologous region:
- Crossing over/homologous recombination
- Integration of phage-bacterial hybrid DNA into recipient cell genome via viral integrase

Question 5

Conjugation

Answer 5

• Direct transfer of genetic material from bacterial cell to one another via a temporary link
• Always one-way from donor F+ to recipient F- cell
• F factor on F plasmid contains genes that codes for proteins necessary for sex pili formation and subsequent cytoplasmic mating bridge, allowing for conjugation to occur, increasing genetic variation
  1. Sex pilus on surface of F+ cell makes contact with F- cell
  2. Sex pilus retracts, pulling both cells closer and forms a mating bridge
  3. One of the 2 strands of plasmid DNA is nicked and transferred from F+ to F- cell through mating bridge and replicated by rolling circle DNA replication as the other DNA a strand is used as a template for elongation
  4. Single strand of plasmid DNA in recipient cell recircularises and serves as a template for synthesis of complementary daughter strand
  5. Through semi-conservative replication, both cells now contain a ds F+ plasmid and are both F+ now

Question 6

Rolling Circle Replication

Answer 6

1. 1 strand of ds F plasmid nicked by a nuclease, breaking phosphodiester bond
2. Free 3’ OH end of nick elongated by DNA polymerase using the intact strand as template
3. Newly synthesised strand displaced nicked strand at the 5’ end, which is transferred concurrently across the mating bridge into recipient cell
4. After 1 round, another nick occurs to release original strand, recircularises and ends replication of newly synthesised strand
5. In the recipient cell, the single strand of F plasmid DNA recircularises and serves as a template for synthesis of a complementary DNA strand (semiconservative) to form a double stranded circular F plasmid

Question 7

lac operon Structure

Answer 7

• An operon is a cluster of structural genes with related functions, controlled as a unit under a common promoter and operator to produce a single polycistronic mRNA coding for functionally related proteins
• Operator is a site on DNA where a repressor protein can bind to prevent initiation of transcription
• Operon can be turned on/off based on demand and circumstances = economic use of resources
• lac Z codes for B-galactosidase, it hydrolyses lactose to glucose and galactose AND converts lactose to allolactose
• lac Y codes for permease, it codes for membrane transport protein enabling cells to take up hydrophilic lactose efficiently
• lac A codes for transacetylase (function not known)
• Regulatory gene lac I codes for lac repressor protein in active form because it’s an inducible operon

Question 8

lac operon - Negative Regulation

Answer 8

• Default: lac I is constitutively transcribed, producing active lac repressor protein binding to lac operator sequence via its DNA-binding site, denying RNA pol. access to promoter

[Scenario 1: Lactose present]
1. Since repression of lac operon is weak and may sometimes dissociate, a basal level of permease and B-galactosidase present within cell
2. A small amount of permease can then transport lactose into the cell and B-galactosidase can convert it to allolactose
3. Binding of allolactose (inducer) to allosteric site inactivates repressor protein by altering the conformation of DNA-binding site to be no longer complementary in shape and charge and can no longer bind to operator
4. RNA polymerase can access and bind to promoter
5. Structural genes transcribed as single polycistronic mRNA coding for several proteins
6. All 3 enzymes translated w increased synthesis, turning on the operon and lactose can now be broken down to glucose + galactose
- Lac operon thus an inducible operon — catabolic pathway

Question 9

lac operon - Positive Regulation

Answer 9

[Scenario 2: Presence of both lactose and glucose]
1. Presence of glucose = low cAMP and CAP NOT activated
2. No upregulation of transcription, lac operon promoter has low affinity for RNA polymerase, relief of negative regulation insufficient
3. B-galactosidase and permease not expressed, lactose remains unchanged
4. Glucose is the preferential respiratory substrate, thus lac operon is effectively off when glucose is present
[include briefly in essays if asked why lactose not broken down]

[Scenario 3: Lactose present, glucose absent]
1. In absence of glucose, cAMP levels rise, binding to allosteric site on catabolite activator protein (CAP) to form a CAP-cAMP complex to activate CAP, which then binds to CAP-binding site within promoter
2. Binding activated CAP increases affinity of promoter region for RNA polymerase, increasing rate of transcription
3. Thus, for B-galactosidase and permease to be produced in appreciable quantity, glucose must be in short supply. Positive regulation thus functions as volume control

Question 10

trp operon

Answer 10

• Repressible operon normally on by default and turned off by end product repression

[Scenario 1: Low tryptophan]
1. Regulatory gene trap R is constitutively transcribed, but repressor protein synthesised in its inactive form and is unable to bind to operator
2. RNA polymerase able to bind to promoter with no repressor, thus operon default ON

[Scenario 2: High tryptophan]
1. As tryptophan accumulates, it binds to allosteric site of trp repressor protein, changing conformation to its active form, thus tryptophan acts as a corepressor
2. Activated repressor protein binds to operator
3. RNA polymerase cannot bind to promoter, no transcription of genes, operon switched off