17 - Genome structure, replication and expression* Flashcards
Shine Dalgarno sequence
region where ribosomes bind to initiate translation of protein
How is chromosome organised
So that ori and ter sites are found at poles of cell
Isomerase
Unwinds supercoil
Catenated
Two circular chromosomes intertwined
FtsK-XerCD recombination machinery
Resolves dimer and cleaves holliday junction
Size of prokaryotic ribsome
70S
subunits:
- 30S (16S rRNA and 21 polypeptides)
- 50S (23S rRNA and 5S, 31 polypeptides)
Describe the basics of DNA replication in prokaryotes
Isomerase unwinds supercoil
Helicase unzips DNA double helix exposing
nitrogenous bases, forming two replication forks
Strands are kept apart by single-stranded DNA binding proteins (SSB)
DNA is replicated 5’-3’
DNA polymerase binds single-stranded DNA, recruits deoxyribonucleoside
triphosphates and matches them to the template strand by complementary
base pairing and catalyses the phosphodiester bond
Antiparallel strands: Okazaki fragments are synthesised by DNA primase using the template strand, extended from the 3’ end by DNA polymerase III
Describe the basics of DNA transcription
Initiation
* RNA polymerase binds promoter and initiates transcription at the start point
Elongation
* Successive addition of ribonucleosides to RNA strand
Termination
* Completed mRNA transcript released
Describe the basics of DNA translation
Initiation
* Ribosome binds Shine Dalgarno sequence and facilitates tRNAformyl-Met alignment with the start codon
Elongation
* Ribosome moves along mRNA, new tRNAs align and peptide bonds
formed between amino acids
Termination
* Ribosome reaches stop codon, releases new polypeptide and
dissociates
Draw and label a diagram of a bacterial gene
Features:
1. (-35) and (-10) box = promoter - region
where RNA polymerase binds to initiate
transcription
2. +1 sight the first point of translation
3. 5’ untranslated region
4. Shine Dalgardo sequence - region where ribosomes bind to initiate translation of protein
5. Start codon - ATG
6. Stop codon - TAA, TGA, TAG
7. 3’ untranslated region
8. Transcription termination signal
Explain how the structure of the nucleoid changes with growth phase and how this affects transcription/translation
Nucleoid is tightly wound to allow for maximised transcription and translation in exponential/growth phase
In the stationary phase the chromosome is relaxed with little transcription and translation occurring
Explain how the two copies of the bacterial chromosome are terminated and resolved during bacterial fission
Circular chromosomes are interlinked or catenated
A double-stranded break is introduced into the dsDNA by topoisomerase on one chromosome
The strands are joined to create a dimer at the two ter sites
Homologous recombination occurs at the ter to form Holliday junctions (strand exchange between each dsDNA pair)
The dimer is resolved by the FtsK-XerCD
recombination machinery which cleaves the Holliday junction
DNA ligase repairs the phosphodiester bonds in the backbones to create an uninterrupted backbone