Pro and Euk

Question 1

Structure and organisation of prokaryotic genome

Answer 1

• generally a single, circular molecules, double helix DNA
• relatively less histone-like proteins
• relatively low level of coiling, DNA double helix – folded into looped domains by protein DNA associations –> undergoes supercoiling with the help of DNA gyrase and topoisomerase

Question 2

structure and organisation of eukaryotic genome

Answer 2

• multiple linear molecules, large amounts of histones, scaffold proteins
• high level of DNA packaging:
  DNA double helix –> negatively charged DNA is wound around 8 histone proteins to form nucleosomes, with linker DNA joining adjacent nucleosomes, forming a 10nm fibre –> coils around itself to form 30nm fibre –> forms looped domains with scaffold proteins, forming 300nm fibre –> which supercoils to form metaphase chromosome

Question 3

How telomerase works

Answer 3

1. A short 3-nucleotide segment of the RNA within telomerase binds to part of the DNA repeat in the 3’ overhang by CBP
2. The adjacent part of the RNA within telomerase is used as a template to synthesise a short complementary 6-nucleotide repeat
3. Telomerase catalyses the formation of phosphodiester bonds between the existing 3’OH group of existing DNA overhang and 5’ phosphate group of incoming deoxyribonucleotide
4. After the repeat is made, telomerase translocates 6 nucleotides to the right in the 5’ to 3’ direction of the DNA overhang and begins to make another repeat.
5. Then primase makes an RNA primer near the end of the telomere. DNA polymerase adds nucleotides to the 3’OH end of the primer and synthesises a complementary strand. The nick is then sealed by ligase. The RNA primer is eventually removed.

Question 4

Introns

Answer 4

• non-coding DNA sequences found within a gene, specifically between exons in a specific segment of DNA
• enables alternative RNA splicing, where introns and different combinations of exons are excised and the remaining exons join together such that different mature mRNAs are produced
• one gene can code for more than one polypeptide

Question 4

promoter

Answer 4

• located just upstream of the transcription start site, proximal control element
• has critical elements such as the TATA box and CAAT and CG box
• TATA box is the precise location of transcription start site
• CAAT and CG boxes improve efficiency of promoter
• recognition and binding site of general transcription factors which recruit RNA polymerase to form transcription initiation complex which initiates transcription

Question 5

enhancer

Answer 5

• distal control element
• recognition and binding site for specific transcription factors called activators which increase transcription frequency by promoting formation of TIC by bending of spacer DNA

Question 6

silencer

Answer 6

• distal control element
• recognition and binding site for specific transcription factors known as repressors. that prevent the formation of TIC

Question 7

telomere

Answer 7

• found at both ends of linear eukaryotic chromosomes
• non coding DNA made up of a series of tandem repeat sequences

Function:
- prevent important genes from being eroded and. vital information is not lost with each round of DNA replication due to the end replication problem
–> DNA pol requires a free 3’OH end to add nucleotides, the last RNA primer on the lagging strand cannot be replaced with DNA, hence DNA molecule shortens with each round of replication at the telomeres, telomeres which are non-coding sequences at the ends will be lost before any vital genetic information is –> can be lost without deleterious effects

• protect and stabilise the terminal ends of chromosomes by forming a loop using the 3’ overhang
  –> prevents ss terminal end of one chromosome from annealing to a complementary ss terminal end of another, preventing fusing of 2 chromosomes
  –> formation of loop prevents the cell’s DNA repair mechanism form detecting the chromosomes as damaged DNA and trigger apoptosis
• telomeres allow for their own extension as they have 3’ overhang which provides an attachment point for the correct positioning of the telomerase

Question 8

centromeres

Answer 8

• constricted region on chromosome where kinetochore microtubules attach during nuclear division
• non-coding DNA made up of a series of tandem repeat sequences
• allow sister chromatids to adhere to each other
• allow kinetochore proteins to attach to allow spindle fibres to attach so that sister chromatids can align along metaphase plate and separate to opposite poles –> proper alignment and segregation of chromosomes

Question 9

regulation at genomic level

Answer 9

Histone acetylation: add acetyl groups to lysine residues by histone acetyltransferase which removes the positive charge of histones, hence reduces electrostatic attraction btw histones and negatively charge DNA, promoter region now more accessible to RNA pol and GTFs –> promotes transcription as it promotes formation of TIC

DNA methylation: add methyl groups to DNA by DNA methylase to selected cytosine residues in CG sequences, prevents formation of TIC by
- blocks binding of specific transcription factors at the promoter and hence preventing the formation of TIC
- recruits DNA binding proteins

chromatin remodelling complex: alter the structure of nucleosomes temporarily –> can cause DNA to be more tightly coiled, and prevents access of RNA pol and GTFs to the promoter –> inhibits formation of TIC and inhibits transcription

Question 10

Regulation at transcription level for eukaryotes

Answer 10

• activators: bind to enhancers, promote assembly of transcription initiation complex as bending of spacer DNA allows interaction of activators with RNA pol and GTFs at the promoter –> transcription frequency increases
• repressors: bind to silencers, to inhibit formation of TIC as bending of spacer DNA allows interaction of repressors with RNA pol and GTFs at the promoter –> transcription frequency decreases

Question 11

Regulation at transcription level of prokaryotes

Answer 11

• activator
  –> CAP which binds to CAP binding site at the promoter of the lac operon and increases affinity of RNA polymerase to the promoter –> transcription frequency increase (positive gene regulation)
• repressor
  –> binds to the operator to prevent RNA polymerase from binding to promoter
  –> transcription frequency decreases –> negative gene regulation

Question 12

Regulation at post transcriptional level

Answer 12

• Addition of the 5’ cap
  –> addition of 7-methylguanosine cap at the 5’ end of pre-mRNA, occurs after transcription begins
  –> helps the cell to recognise mRNA so that subsequent steps can occur
  –> acts as a signal to export mRNA out of nucleus
  –> stabilises and protects the growing pre-mRNA chain from degradation by ribonucleases
• intron splicing
  –> when the introns are excised and exons are joined together by spliceosomes, which recognises the sequences at intron-exon boundaries/ splice sites so that functional proteins can be produced
  –> alternative splicing: introns and different combinations of exons are excised, while remaining exons are joined, so that diffferent mature mRNAs and hence different proteins are produced
• addition of poly A tail
  –> addition of a long sequence of adenine nucleotides at the 3’ end of a pre-mRNA by poly A polymerase, forming a polyA tail
  –> acts as. a. signal to export mature mRNA out of the nucleus through nuclear pores
  –> stabilizes and protects the mature mRNA from degradation by ribonucleases, interacts with eukaryotic initiation factors and the 5’ cap for initiation of transcription

Question 13

Regulation at translational level

Answer 13

• mRNA half life
  –> longer the poly A tail, longer the half life of mRNA to be used as a template to make proteins
  –> the poly A tail is removed by ribonucleases in the 3’ to 5’ direction until a critical length is reached which triggers the removal of the 5’ cap and degradation of the mRNA from the 5’ end too
• formation of translation initiation complex
  1. anti-sense RNAs which are complementary in bases to a few codons of the mRNA to be degraded will bind the mRNA, hence blocking translation of mRNA and will be targetted for degradation by nucleases

2. during translation initiation, small ribosomal subunit, eukaryotic initiation factors and initiator tRNA form a complex which bind to 5’ cap and 3’ poly A tail causing mRNA to circularise
   –> this can be prevented by binding of translational repressor to 5’cap, 5’UTR, 3’UTR which interferes with the interaction needed for translation initiation
3. during translation initiation, euk translation initiation factors facilitate the binding of the small ribosomal subunit to the 5’ cap
   –> the availability of activated eukaryotic translation initiation factors is determined by whether or not they are phosphorylated

Question 14

Regulation at post-translational level

Answer 14

1. formation of functional proteins
   –> covalent modification of polypeptides make them functional proteins
2. phosphorylation of eukaryotic initiation factors can activate the protein and hence upregulate its activity
3. protein degradation
   –> by proteasomes determines how long a protein remains in a cell to carry out its function
   –> proteins targetted for degradation are tagged with ubiquitin so that they can be recognized by proteasomes, the proteins enter the proteasome where they are degraded by enzymes into peptides which can be further hydrolysed into amino acids by enzymes in cytosol, while the ubiquitin is recycled and released