MEDS2003 Mid-semester Exam Revision

2.01 Recall the elements that nucleic acids are componsed of and apply this information experimentally.

(deoxy) ribose sugars - have 1' carbon which changes depending on the sugar and helps to identify its structure phosphates - allows for phosphodiester bond between sugars - identical between RNA and DNA - negatively charged at physiological pH (hydrophilic) --> allows for electrophoresis --> ethanol precipitation (dehydrating backbone to purify nucleic acid and make it drop out of solution) nucleobase (A,G,C,T/U) - connected to ribose by N-glycosidic bond (connecting nitrogen of nucleobase to ribose sugar) - absorbs UV at 260nm due to aromatic ring structure and associated Pi electron clouds --> ratio between absorbance at 260 nm (DNA) and 280 nm (protein) can be used to determine purity - different amino acids have different binding properties (A/G - 3 h-bonds, C/T - 2 h-bonds) --> this results in varying melting points due to varying strength between strands (higher G/C content --> higher Tm)

1.02 Avery, McCarty, McLeod - narrowed down the transforming principle molecule

- boiled cells to produce lysate - treated to destroy verious elements (carbohyrate - enymes, chloroform - fats, proteases - proteins) - only destroyign DNA resulting in no transformation

2.05 Name 3 structural features of DNA that make it a good store of genetic information

- double helix - deoxyribose sugar - thymine base (rather than uracil)

1.02 Griffith - the transforming principle

- mixed smooth pathogenic bacteria, and rough non-pathogenic bacteria - killed smooth bacteria, released cell contents - mixed dead cell contents with live rough bacteria, and smooth bacteria was produced --> showed that something in dead cell contents was giving rise to new pathogenic cells

3.05 Explain positive and negative supercoiling

- negative coiling (in opposite direction to helix) which makes it easier for helicase to separate - positive coiling (in the same direction as helix) which makes it harder for helicase to separate

2.05 Describe how the double helix of DNA makes it a good genetic store

- storing genetic information at the centre of the double helix protects it - negative phosphate, polar sugar, hydrophobic/polar bases, polar sugar, negative phosphate make it very unlikely due to varied charge that free radicals and other molecules will damage genetic material

2.05 Describe how the deoxyribose sugar of DNA makes it a good genetic store

- the deoxyribose sugar in DNA means that it is not as susceptible to a proteolytic attack as RNA is --> this means that DNA is more stable than RNA In a proteolytic attack, the 2'OH on RNA ribose can deprotonate in alkaline conditions - this becomes negatively charged, and is attracted to the delta + partial charge of phosphor, breaking the phosphodiester bond and cleaving the RNA molecule

11.03 Describe how translation is initaited in prokaryotes

1. Initiation factors IF1,2,3 binds to small ribosomal subunit 2. The 30S initiation complex is formed when mRNA, IF2, and initiator tRNA bind to the P site of 30S - this is the only tRNA that can bind to IF2 3. Initiation factors that stop the binding of the small and large ribosomal subunits are ejected, and the 50S and 30S subunits form a 70S complex (this is the rate limiting step)

11.04 Describe how the polypeptide chain is elongated in prokaryotes

1. fMet tRNA binding in P site establishes reading frame 2. Elongation factor Tu (EF-Tu) recruits next tRNA and protects ester bond in activated amino acid and checks accuracy of codon-anticodon binding. 3. fMet and is moved to the tRNA on the A site 4. Peptide bond is catalysed in the 50S subunit by proximity and orientation (lined up correctly in binding sites) 5. AA in site A launches nucleophilic attack on aa/tRNA ester linkage in the P site, which leads to a transition state that collapses to form the peptide bond --> peptide chain comes off the 50S P site, but is connected to tRNA in the 30S A site 6. Translocation occurs, where the 30S unit shifts to the left and the tRNA in its A site moves to its P site to make way for the next tRNA in the sequence

3.02 Descibe the challenge of temporal and spatial coordination of prokaryotic DNA replication and how it is overcome

A bidirectional replication fork is used, originating at the origin of replication (OriC), where the 250bp initiator sequence is read by initiator protein DNAa, and the two loops of the circular DNA of E. Coli split apart creating two replication forks which run simultaneously around the loop and meet again at the bottom.

7.07 Define operon

A cluster of genes under the control of a single promoter/coordinated unit of gene expression

9.03 Understand the importance and features of the 5' cap

A modified guanine nucleotide linked to the mRNA via a 5'-5' linkage between end ribose. This protects the mRNA from phosphatase and nuclease degradation. The process: - linkage is done by RNA triphosphatase removing the outermost phosphate group of mRNA - beta phosphate attacks alpha phosphate of guanine and they bind together - guanine is then methylated

9.09 Explain the process of alternative splicing

Alternative splicing allows one gene sequence to give rise to various protein isoforms to fulfull an alternative function or generate diversity. This is the case in >70% of the human genome. One possible mechanism is the retention of some introns in the mature mRNA.

10.08 Identify the role of aminoacyl-tRNA synthetase and describe where amino acids are placed on their corresponding tRNA

Aminoacyl tRNA synthetase recognises the specific tRNA, binds the correct amino acid to it on the CCA sequence (and is specific to each amino acid), and activates it for incorporation into the polypeptide chain. Amino acid + ATP + tRNA + H2O 🡪 aminoacyl-tRNA + AMP + 2Pi AAs are placed on the end of the CCA arm of their corresponding tRNA.

10.10 Describe how aminoacyl-tRNA synthetases recognise their tRNA substrate

Aminoacyl-tRNA synthetases must also choose the correct tRNA partner, recognising it by - anticodon sequence - unusual/modified base identifiers - structure

7.08 Describe the regulation of the trp operon in the presence and absense of tryptophan

Bacteria can take up tryotophan from the environment, or produce in using genes in the trp operon. In the presence of tryptophan Tryptophan acts as a co-repressor and binds to the trp repressor, both binding to hte operator site as a dimer to prevent transcription as it is not needed. In the absense of tryptophan When not present, tryptophan co-repressor activity is not occuring and the repressor is inactive, so the operon is expressed and tryptophan is produced.

7.07 Describe the regulation of the lac operon in the presence and absense of lactose and glucose

Bacteria use glucose for energy, but can use lactose when needed. Therefore, some mechanism for creating enzymes to break down transport lactose is required - there are three of these inlcuded in the lac operon. In the presence of glucose Lactose is not needed. The promoter for the lac repressor protein is always exposed and producing, and the repressor prevents operon expression. --> because the lac promoter is weak, it has a low affinity for RNA pol In the absence of glucose In the absence of glucose, cAMP concentration increases and binds to CAP protein which binds upstream of lac operon to act as a promoter site, recruiting RNA pol. Then, the lactose derivative allolactose (or its more stable analogue, IPTG) can bind to the repressor protein, removing it and allowing for expression of the lac operon.

2.04 Explain which weak forces work to maintain the dobule helix, and how they function

Base stacking (hydrophobic interactions, electronic interactions) - interactions betwen pi stacking in aromatic rings are actually a stronger force than h-bonds in base pairing - base stacking of hydrophobic and aromatic bases is favourbale, as they are flat - these are buried in the interior of the molecule and shielded Base pairing (h-bonds) - GC base-pairing contains 3 h-bonds, and is therefore stronger than the 2 h-bond T/C base pairing - sinlge ring pyrimidine bases always bind to double ring purine bases Ionic interactions - Mg2+ and other ions interact with the negatively charged phosphate backbone to stabilise DNA structure --> can be exploited in re-annealing applications, such as PCR Van der Waal's forces - electrostatic forces between negatively charged phosphates repel each other - bases attract each other and counter this --> results in the twist in the DNA double helix to accomodate both

2.08 Explain why DNA contains thymine instead of uracil

Because cytosine can oxidatively deaminate to uracil (~100x a day), these need to be identified, removed, and replaced - by using thymine instead of uracil, 'rterm-17eal' uracil is not confused with deaminated cytosine, and can be corrected for --> this results in DNA being a more reliable genetic store

4.05 Identify the unique challenges of eukaryotic replication at initiation

Becuse eukaryotic replication has more than one oriC, there are two challenges that arise: Selecting origins - humans have 20,000 initiation sites ~30-300kb apart, with clusters of 20-80 used at any one time - origin selectin occurs during G1 while pre-replicative complexes assemble Initiating replication/activating origins - phosphorylation of cyclin dependant kinsases enables complexes to recruit DNA polymerases - forks extend in both directions from the origin of replication

5.06 Explain how the properties of DNA polymerases are exploited in anti-cancer and anti-viral drugs

By adding DNA polymerase substrate analogs that take advantage of the fact that DNA polymerase cannot continue polymerisation without a 3'OH to continue from, we can stop replication. This is used in the the antiviral drug AZT - reverse transcriptase inhibtior - alternative DNA pol substrate with azide group in place of the 3'OH This is also used in anti-cancer drugs such as fluoridated nucleotide drugs, targeting the specificity of thymine to DNA replication - 5-fluorodeoxyuridine is added to cells - enzyme attemps to convert to thymine, gets stuck - enzyme commits suicide, and cannot convert false substrate to thymine triphosphate and replication stops --> because TTP is unique to DNA replication, this only affects rapidly dividing cells such as hair, nails, and cancer cells

9.10 Explain the role of alternative splicing in generating protein diversity.

By including/omitting various domains of a protein, different isoforms can be synthesised to generate diversity. E.g. Omitting/including a membrane anchoring domain in an antibody molecule to generate a soluble/membrane bound form of a protein, generating diversity.

12.07 Outline how ferretin translation is regulated in response to cellular iron levels

Cellular iron is tighly controlled - it is required for many proteins such as haemoglobin and in the process of producing ATP, but excess Fe can damage proteins, lipids, and nucleic acids in free radical interactions. Control is as follows 1. transferrin carries Fe in the blood 2. Transferrin receptor binds Fe+transferrin and enables entry into cells when the receptor+Fe+transferring complex is endocytosed 3. Ferretin stores Fe in the liver and kidney In reference to translation, not transcription. --> when low Fe, transferrin receptor increases and ferritin decreases --> when high Fe, transferring receptor decreasese and ferritin increases

6.07 Interpret the numbering system of chromosome gene locations

Chromosome banding is used to give 'addresses' for genes in a given chromosome. Bands can have regions, and sub-regions, giving rise to the following naming system (see image), for example 7q31.2 - chromosome 7 q arm band 31 sub-band 2

6.07 Identify the main structural components of chromosomes

Chromosomes compact the 3 Mbp human genome into one cell. Centromere Central region of the chromosome where spindle fibres attach. More repeated sequences as it is less acessible. Telomeres The ends of the chromosome containing a distinct repeating sequence to ensure genetic material is not degraded/lost during replication. Less accessible. p/q arms the short/long arms of the chromosome resectively

9.07 Identify the role of conserved splice sites in the splicing process

Consensus sequences at the end of introns identify sites where splicing needs to occur to remove introns. These sites are typically GU (start) and AG (stop).

4.02 Describe the role of Cyclins in the cell cycle

Cyclins are cytoplasmic proteins which are switched on and off to regulate the various stages of the cycle. --> these increase/decrease in gene expression (and therefore cellular concentrations) throughout the cell cycle --> Cyclin dependant kinases (CDKs) depend on cyclins to phosphorylate their protein targets so that they can function properly

2.05 Describe how the use of thymine in DNA makes it a good genetic store

Cytosine can spontaneously deaminate to become uracil - this happens 100 times a day - these need to be removed and replaced --> if this occurs in RNA, the sequence is ruined because it is impossible to distinguish actual uracil from deaminated cytosine - in DNA, any uracil can be definitively traced back to a cytosine, and corrected for --> this makes the error rate of the DNA sequence much more robust

7.01 Identify the key components needed for transcription

DNA Template Antisense strand is used so that complementary RNA product is the right way around Building blocks Ribonucleotide triphosphates provide the nucleic acids and energy required for biopolymerisation Enzymes RNA polymerase is required, along with its Mg2+ cofactor

2.02 Identify the main components of DNA and RNA with reference to Chargaff's Rules

DNA and RNA consist of (deoxy)ribose sugars, phosphates, and nucleobases. Chargaff's rule postulates that these should invariably subsist in a 1:1:1 ratio in RNA and DNA.

5.08 Describe how DNA is synthesised in vitro in DNA sequencing

DNA is synthesised in Sanger sequencing by effectively running a PCR where a small number of dideoxynucleotides (ddNTPs) are added to the reaction tube. These lack the required 3'OH group to continute polymerisation, and so the sequence will be stopped at varying points depending on the specificity of the ddNTP substrate. This can also be done with fluorescence tagged ddNTPs (tag added to alpha position, as this is conserved) to determine the sequence by absorption characteristics or by running gel electrophoresis.

3.02 Descibe the challenge of unwinding and rewinding of DNA for replication

DNA is tightly wound and bound together. Helicase unwinds DNA double helix h-bonds, bot doesn't break the phosphodiester bonds in the backbone. - hexameric protein - gets energy by hydrolysing NTPs (NTP -> NDP + Pi) Single stranded binding protein (ssbp) keeps the strands apart while replication occurs. Topoisomerases relive torsional strain on DNA (from +/- supercoiling) by breaking the pd backbone and rebinding

3.02 Descibe the challenge of reconciling limitation of 5'->3' polymerisation with the replication fork producing in two strands running in opposing directions

DNA polymerase only has 5'->3' polymerisation activity. This means that a leading, and lagging strand are formed at each replication fork. This poses no issue for the leading strand, where DNA pol can run down the strand towards the fork. On the lagging strand, several things occur - RNA primers are laid down at intervals - intrastand loop is formed where single stranded binding protein (ssbp) binds to the unzipped lagging strand to stop it reannealing - loop brings the lagging strand around to the fork where DNA pol is bound - DNA pol III synthesises Okazaki fragments - DNA pol I takes away the DNA primers and replaces them with DNA - fragments are joined by DNA ligase

3.06 Relate the unique properties of DNA polymerases to the need for fidelity in the copying process.

DNA polymerases have exonuclease activity, which allows for proofreading of new biopolymers. - DNA pol I (5'->3', 3'->5') - able to take away RNA primers on lagging strand and proofread new strand - DNA pol III (3'->5') These properties of DNA polymerases allow them to hydrolyse incorrect base pairing on the new strand when transcription stalls momentarily. They can also hydrolyse incorrect nucleotides added to the template strand. This allows them to resist systematically incorrect biopolymer synthesis and respond to point mutations by removing the incorrect nucleotides, maintaining the integrity of the genetic material over many courses of replication.

3.01 Outline the general mechanism for nucleic acid synthesis (DNA and RNA), where the supply of energy is coming from, and the orientation of the copying

DNA polymerases, RNA polymerases, and reverse transcriptase operate with largely the same mechanism ot synthesise nucleic acids. - 5' -> 3' orientation of synthesis (and only run in one direction) - (deoxy)nucleotide triphosphate substrate is used and added to the 3'OH of the ribose sugar to grow the chain - lone pair of electrons on 3'OH attract positively charged gamma phosphate in (d)NTPs and a nucleophilic attack occurs, cleaving the alpha and beta phosphates - energy stored in the (d)NTP is used to drive the unfavourable formation of the phosphodiester bond to sustain polymerisation by breaking down the molecule and releasing pyrophosphate (2-Pi)

3.05 What activities promote positive and negative supercoiling and how is positive supercoiling relieved?

DNA replication promotes negative supercoiling as it makes it easier for the strands to be separated and makes it easier for helicase to be released. Positive supercoiling arises from helicase activity (?). --> both can be relieved by making breaks in the phosphodiester backbone, and rapidly repairing the break to relieve pressure - this is done by topoisomerase enzymes

7.05 What are the features of transcriptional elongation

De novo synthesis Ribonucleotides are added within the transcription bubble, where - RNA polymerase unwinds DNA to give access - RNA synthesis starts without a primer - first base is either A or G Directionality - RNA strand grows 5'->3' at a rate of 1000 nt/s Proofreading RNA polymerase has proofreading ability in the presence of accessory proteins. - error rate is 1/10^4 - 1/10^6 nucleoties - far more error prone than DNA pol, but not bad

11.11 Understand how antibiotics target specific components of translation machinery to inhibit protein synthesis

Differences in eukaryotic and prokaryotic translation can be used in development of antibiotic drugs. This may operate by - inhibiting intitiation and causing misreading of bacterial mRNA - binding to ribosomal subunits ot inhibit aminoacyl-tRNA synthetase activity (50S) or binding of aa-tRNAs

2.10 Explain how base pairing is exploited in molecular biology techniques

Different nucleobases have different binding characteristics, determined by molecular structure, most notably the number of possible h-bonds. On a surface level, this is used in - PCR amplification - sanger sequencing This can be used to - determine the G/C:T/A ratio by the melting temperature (Tm) or vice versa - inform the effective pH for nucleic acid synthesis --> pKa (pH of 50% dissociation) of Adenine nitrogen is 4, and so is really only an h-bond acceptor at physiological pH --> pKa of guanine N ~9, and so will act as a proton donator at physiological pH

10.04 Explain the features of genetic code

Directionality Codons can only be read 5'->3' as they encode specific amino acids. Punctuation Codons have no punctuation, no skip/pause etc Non-overlapping Codons have to be read sequentially, and while alternative reading frames do exist unregulated out of frame translation be catastrophic.

8.07 Describe how nuclear hormone receptors are drug targets

E.g. Oestrogen - hydrophobic, diffuses across plb - binds to nuclear hormone recepetor in cytoplasm/nucleoplasm, changing the binding pocket so that coactivators/repressors are recruited - receptor has zinc fingers which bind to specific DNA sequences - receptor ligand complex recruits coactivators/corepressors to modulate gene expression --> oestrogen binding site can be blocked by a drug in some breast cancer treatments

10.09 Describe how aminoacyl-tRNA synthetases recognise their amino acid substrate

Each aaRS has a specific amino acid substrate, and typically incorporate various binding structural properties to bind only the appropriate amino acid. E.g. Threonine tRNA synthetase has to distinguish between three very similar amino acids - threonine (the correct one) - valine (methyl group instead of OH) - serine (very similar, as OH group but missing a methyl group and is smaller) There are therefore 3 mechanisms of elimination - Zinc/asp mechanism to distinguish Val and eliminate it by interacting with OH group - Size of activation site, which eliminates val as it is larger - editing site which accepts ser due to its size and cleaves it (the extra methyl group of thr means that it cannot be accidentally cleaved)

8.04 Identify the role of enhancer regions and their difference to promoter elements.

Enhancer regions stimuate transcription by recruiting transcription factors but have no promoter activity of their own. Properties include - acting over a long distance (1000s of bp) - can be up/down stream - can be on either strand - can cause cancers if they become unregulated

11.08 Describe differences in projaryotic and eukaryotic translation start signals

Eukaryotic AUG start codons recruit an unmodified Met rather than fMet - because of this the first AUG in the sequence is taken as the start codon. In eukaryoties there is no shine-dalgarno sequence which signals initiation of translation. This is required in prokaryotic initiation as the first AUG is not sufficient for starting translation as you have polycistronic mRNAs

11.10 Describe the circularisation of mRNA for translation

Eukaryotic mRNA is circular, bound at one end (5' cap) by IF, and at the other (3' poly(A) tail) by PABP. This is presumably to prevent translation of non-polyadenylated mRNA, and prevent the transcription of defective or non-mature mRNA.

8.02 Understand that eukaryotic promoters are cis-acting elements, and that they combine in different ways in different positions relative to the unregulated genes

Eukaryotic promoters are sequences that attrract RNA polymerase, among other protins in eukaryotes. Properties include: Cis-acting On the same DNA molecule as the one they regulate. Combinatorial control The basis of functional complexity in eukaryotes - gives ability to upregulate gene expression on a cell/tissue basis to allow for greater variety of function.

4.01 Indentify the phases of the cell cycle and describe primary activities in the context of DNA synthesis.

G1 - gap1, growth and preparation of chromosomes for replication and cell division S - synthesis of DNA (and centrosomes) where DNA in thecell is replicated to have '2n' amount of DNA G2 - gap2, preparation for mitosis and preparation to distribute DNA into daughter cells M - mitosis (prophase, metaphase, anaphase, telophase) --> also worth noting - interphase (the time between one mitosis and the next cell cycle)

6.08 + 6.10 Describe how histone modifications can influence gene expression

Histone epigenetics can influence gene expression by modifying the availability of DNA. - heterochromatin is condensed, for transcriptionally inert genes - euchromatin is loosely wrapped, and more accessible for expression

6.04 Describe the packaging of DNA with histones to form nucleosomes, and nucleosome structure

Histones forms a complex to package DNA at a molecular level, forming a 30nm thick fibre. This can - shield negative phoshate charges - allow bending and DNA wrapping - restrict transcription acess Histones have many lysine and arginine residues which are positively charged at pH 7, giving them the required electrostatic attraction to hold DNA. Histones have 5 variants, of which 4 (2H2A + 2H2B + 2H3 + 2H4) form an octamer which the DNA wraps around with 1.7 turns (~147 bp), forming a nucleosome.

12.02 Describe how IRESs can be used for co-expression of genes

IRESs can be used to express several genes in sequence under one promoter. For example, expressing a gene of interest tagged with a reporter gene (such as a fluorescence tag) with its own IRES allows for co-expression of the two genes in a 1:1 ratio. This allows us to determine when and how a gene of interest is expressed.

10.12 Understand that invididual genes within polycistronic mRNA can be translated independently

In bacteria, many mRNAs are polycistronic, containing multiple genes. In this scenario each gene has its own start/stop signals and can be translated independently.

9.12 Descibre the post-translational transport of mRNA from the nucleus to the cytoplasm.

In eukaryotes, mRNA translation are separated temporaly and spatially, with transcription occuring in the nucleus and translation in the cytosol. --> mRNA must therefore be transported out of the nucleus through the nuclear pore complex. - helper proteins transport mRNA and break off once out of the nucleus and return to the nucleus

11.09 Describe how translation is initiated in eukaryotic mRNA with a 5' cap

In eukaryotic transcription, the 40S small ribosomal subunit carrying unmodified Met forms a complex that binds to the 5' cap of the mRNA and searches for the first AUG start codon. -->This complex is moved along by helicases, energised by ATP. When bound, initiation factors IF1 and IF3 are released and the large ribosomal subunit binds and translation proceeds

3.02 Descibe the challenge of proofreading and editing of DNA during replication

In order to maintain the integrity of genetic information stored in DNA, DNA polymerase has exonuclease activity acting from 3'->5' (i.e. reverse, back along the sequence) which is able to 'proofread' genetic information in the background. DNA polymerases have slow acting nuclease activity, so they only cleave a new nucleotide if the enzyme is temporarily stalled by an incorrectly paired nucleobase. This activity is more sporadic and therefore less relialbe in RNA polymerases.

9.08 Describe how incorrect splicing can occur and identify a possible consequence of incorrect splicing

Incorrect splicing can lead to disease and can occur in the pre-mRNA or in splicing factors. Typical examples may include premature/in-frame stop codon, and cause >15% of all genetic diseases. --> the result of this may be - an abnormal mature mRNA that is degraded - a deficient or alternative form of a protein - a non-functional protein

5.02 Explain why increasing the ionic strength will promote base pairing and what effect it will have on melting temperature (Tm)

Increasing the ionic strength of the solution with cations such as Mg2+,Na+ shield the negative charges of the phosphodiester backbone, reducing their repulsive effect on each other. --> increasing ionic strenght of solution will increase Tm, as strands are more strongly held together

10.13 Describe the translational start signals on prokaryotic mRNA

Initiator codon Usually AUG (encodes methionine), and binds to anticodon of first tRNA Shine-dalgarno sequence Purine-rich sequence ~-10 from start site in the untranslated region that helps bind the 16S ribosomal subunit (the small one)

11.02 Understand the role of the prokaryotic initiator tRNA in translation

Initiator tRNA that pairs to the AUG start signal recruits fMet (N-formylmethionine) - this tRNA is distinct from that which inserts Met at internal positions. fMet is removed in about half of all new polypeptide sequences when they are ~10aa long.

1.02 Hershey and chase - T2 phage tagging

Isolated DNA specifically as the molecule responsible for replication/expression of protein. Separated DNA by tagging the protein component of T2 phage with the 35S radioisotope, and the DNA component with 32P (possible because the unique component of DNA is the phosphor). --> phage was added to bacteria such that they would interact - blended the sample to remove the phage - centrifigued to separate supernatant (protein, other cell concents) from cell pellet (DNA) - in the 35S tagged sample, the protein supernatant was radioactive - in the 32P tagged sample the cell pellet was radioactive, indicating that DNA was the molecule responsible for carrying genetic information

3.03 Describe the function of single stranded binding protein in DNA replication

Keeps DNA strands apart during replication, and coats the lagging strand to prevent it from re-annealing.

3.03 Describe the function of primase in DNA replication

Lays down RNA primers on the leading and lagging strand for DNA pol I/III

2.10 Explain how base stacking is exploited in molecular biology techniques

Leads to pi electron clouds above/below aromatic rings, which due to their resonance properties absorb UV light at 260 nm. --> this can be used to determine the purity of a DNA sample with potential protein contaminants, as protein absorbs at 280 nm and this this A260:A280 ratio is used as a purity indicator (>=1.8 is considered pure)

2.06 Explain the importance of the major and minor grooves to gene expression

Major and minor grooves arise because the steps of the double helix are not evenly spaced. Grooves give access for proteins to access the genetic material in order to modulate expression. This is exploited by some drugs.

6.09 Understand how histone acetyl transferases and histone deacetylasese influence the binding interaction between histones and the DNA backbone

Modification can be done by histone remodelling, most commonly using acetylation of lysine residues. - histone acetyl transferases (HATs) add acetyl group to aa side chain of lys, neutralising its positive charge - this packs the DNA less tightly due to reduced charge --> histone deacetylases (HDACs) are associated with transcriptional cofactors and relax chromatain - they also provide binding sites for proteins that stimulate transcription

2.03 Describe the key covalent bonds in DNA and RNA

N-glycosidic bonds - connects nucleobase (A,G,C,T/U) to the ribose sugar molecule - named for what it connects (the nitrogen of nucleobase, to a sugar) Phosphodiester bonds - connects the 3'OH of the ribose sugar to the 5'OH - negatively charged

12.05 Describe the process of no-go mRNA decay

No-go mRNA decay detects and destroys instances where ribosome stalls before the stop codon is reached. The stalled ribosome recruits a factor that promotes ribosome dissociation, mRNA cleavage, and mRNA degradation. This may be due to - rare codons (organisms may have a degree of codon specificity due to varying quantities of a given tRNA). - secondary structure that impedes to forward movement of the ribosome, such as a stem loop

12.04 Describe the process of nonstop mRNA decay

Non-stop mRNA decay detects and destroys transcripts without a stop codon. In this case, translation continues down the poly(A) tail, and encodes lysine (AAA). The stalled ribosome recruits Ski7 protein, which degrades mRNA by recruiting exosome complex which recognises the poly(lysine) tag

12.03 Describe the process of nonsense-mediated mRNA decay

Nonsense mediated mRNA decay detects and destroys transcripts with a premature stop codon. This results in exon junction complexes (which mark the site of each excised intron) being left on the mRNA, as typically the ribosome would transcribe the lenght of an mRNA, removing them on the first pass. In this instance, some of the EJCs are left behind. These recruit proteins which cleave the 5' cap and the mRNA is degraded by phosphatasese and nucleases.

8.06 Describe the features of nuclear hormones and their role in regulating gene expression

Nuclear hormones are able to enter the cell as they are hydrophilic, and form receptor ligand complexes which modify expression of specific genes by recruiting coactivators or corepressors.

10.01 Explain why nucleic acid structure are similar while amino acid structures are vastly different

Nucleic acid structure differ to allow for base pairing and to support the structure of the DNA double helix with its various elements to support requirements of encoding genetic information. Amino acids vary greatly to allow for various conformation and structural characteristics of aa sequences in functional proteins. They may vary in - side chains - acidity - charge - binding ability etc...

11.06 Desribe the translocation process and the need for this process during translation.

Once the peptide bond has been formed, the aa from the tRNA in the A site is attached to the polypeptide chain stemming from the 50S P site, but attached to the tRNA in the 30S A site. This means that elongation cannot continue because subunits are not lined up. Translocation is done by elongation factor G (EF-G), where substrates stay attached to where they are in the 50S unit, and shift one place towards the 5' end of the mRNA to realign and continue elongation. 1. the peptide-tRNA moves from the A site into the P site 2. the empty tRNA moves from the P site to the E site 3. the next codon is positioned into the A site 4. tRNA in the E site is released by the ribosome --> Proteins are made in the N to C terminal direction in the A site, as with the first aa addition

3.03 Describe the function of DNA pols I and III in DNA replication

Performs the bulk of DNA biopolymer synthesis, including synthesising DNA on the leading strand, and continuing synthesis on the lagging strand from the RNA primers to form okazaki fragments.

9.04 Understand how and why the pola(A) tail is added to eukaryotic mRNA

Polyadenylation of hte 3' end increases RNA stability and enhances translation. ~250 Adenylate residues not coded for in the DNA sequence are added after transcription using ATP as the substrate. Polyamine binding protein (PABP) binds to poly(A) tail to protect the 3' end.

7.02 Describe the role of promoters in transcription

Promoters signal the start of transcription, and are DNA sequences that direct RNA polymerase to the proper site to begin transcription.

1.04 Watson and Crick's contribution to understanding nucleic acid structure

Proposed the double helix, deducing this structure from Franklin's images. "It has not escaped out notice that the specific pairing we have postulated immediately suggest a possible copying mechanism for the genetic material" - proposed the correct structure - immediately gave rise to a plausible mechanism of replication - gave rise to all modern molecular biology

8.01 Which eukaryotic RNA transcripts are synthesised by which RNA polymerase?

RNA pol 1 - 18S (small subunit), 5.8S (non-coding), 28S (large subunit) rRNA - RNA pol II - mRNA and snRNA precursors (present in larger quantities as it synthesises mRNA) RNA pol III - tRNA and 5S rRNA

9.01 Outline how eukaryotic rRNA is processed

RNA polymerase 1 transcribes 45S precursor transcript, which is then modified - nucleotides are modified to change conformation/structure - pre-rRNA is assembled with ribosomal proteins - cleaved into 18S (small subunit), 28S, and 5.8S (large subunit) rRNA

5.07 Describe how DNA is synthesised in vitro in PCR

Reaction tube is made up with - template DNA - dNTP substrates - primers (forward and reverse, sequence specific) - taq polymerase - buffer to maintain correct pH and tonicity The process - Denature - heat sample to 95°C - Anneal primers to specific sites on DNA- allow to cool to 60°C (too cold and they'll bind in places they're not meant to) - Taq optimisation - heat to 72°C so that taq polymerase can work its magic most effectively --> Repeat cycle ~30 times so that the new strands can denature, and new primers can be added and more dsDNA can be produced, each flanked by primer sequence

10.07 Describe how tRNAs recognise codons via typical base pairing and the wobble effect.

Recognition of the 3rd base in a codon is less descriminating than the first two, and the tRNA is therefore less picky. This means that tRNAs can recognise more than one codon There is steric freedom around the 3rd base of a codon (also the 1st base of the anticodon). --> the 2st and 3nd bases of the anticodon bind normally with the codon --> the 3rd base of hte codon and the 1st base of the anticodon exhibit the wobble effect, where the base pairing is less precise See image for possible variations, but basically (1st base of anticodon - 3rd base of codon) C = G A = U U = A/G G = U/C I = U/C/A

10.03 Explain what a codon is

Regardless of organism, a set of 3 nucleotides (a triplet) constitutes a codon, and encodes 1 amino acid. With 4 nucleotides, you need at least 3 to code for the 20 amino acids (4*4 = 16, 4*4*4 = 64 combinations). Codons are - non-overlapping (read sequentially) - have directionality (5'->3') - degenerate

6.01 Describe and interpret simple calculations for the size of genome as evidence for non-coding DNA in the genomes of complex organisms

Relative number of protein coding genes in the genome is inversely proportional to the sophistication of the genome.

5.04 Explain how the properties of DNA polymerases are exploited in PCR

Requiring a primer - used to select a specific sequence for amplification Taq polymerase's heat resistance/stability - fast and effective at high temperatures, so can operate at whatever temp. (up to 95C for ~1.6h) is required to keep DNA separated

10.11 Describe the components of ribosomes and how they bind to mRNA and tRNA

Ribosomes coordinate translation using tRNA, amino acids, and cofactors. Their key catalytic sites are composed of RNA and proteins, and have two main subunits (50S and 30S in prokaryotes) with 3 tRNA binding sites. - Aminoacyl (anticodon-codon pairing) - Peptidyl (anticodon-codon pairing) - Exit Template mRNA is bound within the small ribosomal subunit.

5.05 Explain how the properties of DNA polymerases are exploited in Sanger sequencing

Sanger sequencing exploits the fact that DNA polymerases require a 3'OH group for extension. By adding a small number of dideoxyATP in the polymerase reaction tube, you can terminate polymerisation at varying points to identify the sequence.

3.03 Describe the function of ligase in DNA replication

Seals the sugar phosphate backbone betwen the former okazaki fragments in the final stages of replication.

4.01 Describe the states of cells not in the cell cycle

Senescence - cells cannot re-enter the cell cycle - E.g. Adult neuronal cells Quiescence - cells are in G0, but can be induced to re-enter the cell cycle by a mitotic signal - E.g. B cells and T cells which respond to a foreign body by replicating G0 - not dividing (terminally differentiated cells are in G0)

1.04 Rosalind Franklin's contribution to understanding nucleic acid structure

Showed the shape of double helix using x-ray diffraction photos in a humid environment to replicate internal conditions, so structure was more preserved

12.01 Identify the role of internal ribosome entry sites as alternative translation sites

Some mRNAs lack the 5' cap and require an alternative start site in eukaryotes. This is typically found in viral transcripts, but is sometimes found in essential eukaryotic transcripts to provide an alternative method for initiation. In these cases there is an IRES (internal ribosome entry site) on the 5' side (upstream of the start codon). Here the IRES interacts with the 40S (small) ribosomal subunit or the initiation factor elfF4F to start translation. Also interacts with initiation complexes.

9.06 Describe the purpose of the splicing process in eukaryotic mRNA

Splicing of introns from eukaryotic mRNA is important as it removes non-coding regions. If introns are translated they can severely impact the functionality of the protein produced.

11.07 Describe how translation is terminated in projaryotic cells

Stop codons for termination are UAA, UGA, UAG, and there are no tRNAs to accomodate these stop codons. Instead, release factors (RFs) recognise these codons. 1. RFs bind to stop codon on P site to bridge two subunits 2. RFs break the ester linkage between tRNA and the polypeptide chain in the P site and polypeptide is released 3. tRNA, ribosome, and mRNA complex dissociates and can be reused

8.03 Identify the common RNA pol II promoter elements and describe their role in initiating transcription

TATA box Located -25 at 5' end (upstream), and can be significantly impaired by a single bp mutation CAAT and GC box Located -40 to -150 upstream of start site, and can be found on either template or coding strand. These are common in constitutive expression where a gene is always expressed in the cell. Initiator element Centred at +1, can compensate for absent or degenerate TATA box Downstream promoter element Centred at +30, and commonly found in genes that lack the TATA box

4.08 Explain why telomerase is an attractive drug target for cancer treatments

Telomerase is active in highly differentiating cells in the body such as germ-line cells and highly proliferative stem cells to facilitate rapid replication. This also occurs in 80-90% of all cancers, and targeting telomerase in these cells may inhibit their indefinite replication capacity. Strategies include - Targeting the RNA component of telomerase in these tissues with antisense oligodeoxynucleites and RNaseH - reverse transcriptase inhibitors (e.g. AZT) - inhibitors of catalytic subunit

4.07 What is telomerase? Predict the implications of telomerase activity

Telomersase is a ribonucleoprotein with protein and RNA components that recognises the overhang of the two DNA strands after replication. - protein component is reverse transcriptase - RNA component acts as a template for repairing telomeres Implications of telomerase activity may be immortality, indefinite terminal differentiation --> indefinite hayflick limit, which dictates that cells will replicate a finite number of times inducing replication erosion and then suffer cell death --> see HeLa cells - cervical cells that survived Henrietta Lacks for ~50 years --> see Dolly the sheep, who was cloned from an older sheep but exhibited characteristics of an older sheep at ~6 y/o

5.11 Interpret and troubleshoot PCR data

The A260:A280 ratio can be used to check the purity of a DNA sample from a PCR reaction for protein contamination. A260:A280 > 1.8 is considered pure.

4.02 Describe the checkpoints on the cell cycle

The cell cycle is monitored by various checkpoints, which make sure the cell cycle is occuring as intended. - stalling/arresting of the cycle will result in apoptosis --> if the restriction point in G1 is reached cells will either complete the cycle, or undergo apoptosis - DNA damage is checked in G2 - at the end of S check for okazaki fragments is done, as this would halt mitosis - Spindles are checked in M to check mechanism for daughter cell distribution

5.01 Identify which component of nucleic acids give them their UV absorbance and why it is not sequence or source dependant.

The conjugated ring structure of DNA/RNA gives their absorbance properties, and as each base has a similar absorption and bases are conserved across organisms the final absorption is not sequence or source dependant. Worth appreciating that - Absorbance 1 at 260nm gives 50 ug/mL dsDNA - Absorbance 1 at 260nm gives 40 ug/mL ssRNA

1.01 Explain the central dogma and the flow of genetic information in cells

The flow of genetic information in cells is set. DNA <--> RNA --> Protein

11.05 Describe how peptide bonds are formed

The formation of a peptide bond is catalysed by the 50S ribosomal subunit, where proximity and orientation are exploited to line up the amino acids for bond formation. The inherent reactivity of the amine group on amino acid in site A with the ester of the aa in site P is used to induce a nucleophilic attack. This leads to a transition state, which collapses into a peptide bond where the N terminus of the incoming aa is attached to the C terminus of the previous aa/polypeptide chain.

6.03 Account for the large amount of non-coding DNA in the genomes of non-complex organisms

The human genome codes for 30,000 proteins, but this only accounts for 1-2% of the whole genome. This includes - regulatory sequences - introns (25%) - pseudogenes - non-coding RNA (infrastructural RNA, regulatory RNA - tRNA and rRNA (?)) - repetitive DNA of various classes

5.03 Explain the hyperchromic effect

The hyperchromic effect demonstates that single stranded nucleic acids absorb more UV as the h-bonds limit the resonance of the aromatic rings.

7.03 Describe the facotrs that determine promoter efficiency

The more a consensus sequence resembles the promoter, the more frequently it will be transcribed (~2 secs --> 10 mins). Distance between promoter and sequence is also a factor in expression frequency (~17 nt is optimal)

7.02 Identify the common bacterial promoter elements

The most common types of bacterial promoter elements are - -10 (10 nt upstream of start) - -35 (35 nt upstream of start) These are most commonly TTGACA, and TATAAT.

2.11 Explain how backbone charge is exploited in molecular biology techniques

The phosphodiester backbone is negatively charged at physiological pH, and this property can be exploited. - the backbone is higly hydrophilic, meaning that DNA is highly soluble in water - negative charge can be exploited in electrophoresis, where the constant mass:charge ratio is used to separate different length strands of DNA and RNA by drawthing them through a gel towards a cathode (+) terminal - can also be exploited in ethanol precipitation, by dehydrating the backbone to purify DNA/RNA, making it drop out of solution

9.05 Describe how the poly(A) tail can be used for mRNA purification.

The poly(A) tail can be exploited using oligo-dTs that bind to the tails. These beads are magnetic and stick to the walls of a separating column, and can therefore be used to separate mature polyadenylated mRNA from teh rest of a sample. Oligo-dTs can even be used as a primer to make cDNA from mRNA for analysis.

5.10 Interpret and troubleshoot DNA sequencing data

The sequence produced by sanger sequencing is actually the complementary sequence (think about what is being produced relative to the original strand). --> shorter sequences represent sequences at the very end of the template sequence, so the sequence can be read top to bottom as the bigger strands will not move as far on the gel --> need to be converted to complementary to be read

7.04 Describe the role of RNA polymerase in initiating bacterial transcription

The sigma subunit of RNA polymerase recognises the promoter sequence, and even though its affinity for DNA is low it slides along looking for the start site. The sigma subunit is released after transcription start.

10.02 Explain the universality of genetic code

The univeral language of genetic code refers to the nucleic acid sequences encoding different amino acids. This means that under the right conditions, human genes can be expressed in bacteria, for example.

10.05 Descibe the degeneracy of the genetic code and identify the benefits that degeneracy confers

The universal genetic code has degeneracy, i.e. redundancy - multiple codons can code for the same amino acids, and seqeuences always code for the same aa. Without this redundancy, an incorrect nt would certainly result in a truncated or non-functional protein, but because even errors can code for an aa the sequence may retain its function. This reduces the negative impact of mutations.

2.09 Explain why DNA contains deoxyribose instead of ribose

The use of deoxyribose sugars in DNA makes it more stable and degrade less easily, and a more reliable genetic store. This is because - when pH increases, OH- concentration increases and the 2'OH of ribose is deprotonated, forming H20 with the OH- - the remaining oxygen is negatively charged, and forms a bond with the delta-positive phosphor in the phosphate group, breaking the phosphodiester bond and cleaving the RNA molecule

4.03 What are the differences between prokaryotic and eukaryotic replication?

There are a lot more similarities than differences. Similarities - bi-directional - DNA polymerases work 5'->3' - include leading and lagging strands - primers are required Differences - linear chromosomes in eukaryotes - have telomeres - significantly more genetic material which needs to be compacted, which requires appropriate machinery - different, slower enzymes with more proofreading ability (e.g. eukaryotic ssbp has 3 subunits, as opposed to 3 in prokaryotes)

4.02 Explain how cancers arise, and how mutations in the cell cycle control mechanisms can lead to cancer

There are two main causes of oncogenesis - mutations in proto-oncogenes turns them into onco-genes --> these can become damaged and mutate to cause unregulated growth - mutations in tumour supressor genes --> these normally slow down cell division and either repair DNA or tell cells to undergo apoptosis Mutations may include: Myc, a gene which switches on cyclin D which is required for CDK 4 and 6 - these CDKs phosphorylate the tumour supressor gene retinoblastoma (Rb) and assocaited histones - tumour supressing function is inhibited Micro-RNAs involved in regulating cyclin dependant kinases P53, a tumour supressor gene active in nearly every cell and is mutated in more than 50% of human cancers

7.06 Outline the stop signals for prokaryotic transcription

There are two main transcription stop signals Hairpin loop - palindromic GC rich + 4 uracil sequence forms a loop, caused by AAAAA sequence at the end of DNA - structure anneals to itself, and stops RNA pol and U region dissociates from the DNA strand Rho protein - binds to the newly made RNA strand --> RNA goes through a tunnel in Rho using ATP to drive this at high speed, and when it collides with the transcription bubble it breaks it, removing new RNA from the DNA

11.12 Describe how diptheria toxin inhibits protein synthesis

Toxin adds ADP-ribose to dipthamide, an essential aa in EF2 --> this blocks EF2 from carrying out translocation of ribosome and stops translation

11.13 Describe how ricin toxin inhibits protein synthesis

Toxin removes adenine from adenosine on 28S rRNA, inactivating ribosomes by preventing the binding of elongation factors

9.11 Understand that transcription and post-transcriptional processing and closley coupled and are coordinated by RNA polymerase II

Transcription and mRNA processing are closely coupled due to the function of the c-terminal domain of RNA polymerase II. When phosphorylated, the CTD is able to recruit and bind capping, splicing, and polyadenylation enzymes for post-transcriptional processing within the nucleus of the cell. - at 25nt long capping starts - once introns are made, they are cut out - once finished transcription cleavage occurs and the poly(A) tail is added.

8.05 Describe the role of transcription factors in the combinational control of transcription

Transcription factors (TFs) are proteins that bind to promoter/enhancer sequences to help recruit RNA polymerase. These are often trans-acting factors, found on a gene other than the one they regulate. These work in conjunction with other regulatory elements - TF bind to promoter elements - act as site for more TFs to bind - TF complex unwinds DNA and phosphorylates the RNA polymerase to kick off elongation

3.06 Why does copying of DNA need to be more accurate than transcription

Transcription is transient, and can be rapidly degraded. This will typically cause, at worst, non-functional proteins or which will simply be degraded, or the RNA will be degraded itself. If such errors appeared in DNA damage/production of non-functional genes could be ongoing and catastrophic.

2.07 Identify the structural differences between DNA and RNA

Two key structural differences between DNA and RNA are - no hydroxyl group in the 2' position on ribose in DNA --> resists nucleophilic attack that cleaves RNA - methlyated 5-carbon in uracil in DNA, making it thymine --> allows for identification of spontaneous deamination of cytosine

12.06 Describe the process of RNA interference induced by siRNAs

Undesirable mRNA degradation can result from exogenous dsRNA. Exogenous dsRNA is cleaved by and RNase called Dicer into 21 nt-long fragments called siRNAs (small interfering RNAs). siRNAs bind to argonaute proteins to form RNA induced silencing complex (RISC) which binds to its complementary mRNA target and inhibtis protein synthesis.

12.06 Describe the process of RNA interference induced by miRNAs

Undesirable mRNA degradiation can result from endogenous dsRNA, generated by large endogenous sequences cleavaed into 21 nt-long dsDNA. This is processed into ssRNA, which binds to argonaute protein to form RNA induced silencing complex (RISC) to bind to mRNA complements and inhibit protein synthesis.

3.03 Describe the function of topoisomerases in DNA replication

Untangle and relax DNA by preventing supercoiling ahead of the fork by breaking the phosphodiester backbone and reforming to to relieve torsional stress. This works for type I and II topoisomerases - type I is negative coiling (in opposite direction to helix) which makes it easier for helicase to separate - type II is positive coiling (in the same direction as helix) which makes it harder for helicase to separate

3.03 Describe the function of helicase in DNA replication

Unwinds the DNA double helix h-bonds by hydrolysing NTPs (NTP --> NDP + Pi)

4.07 Identify the unqiue challenges of eukaryotic replication at the ends and how the end problem is resolved in the cell

When primers are laid down right to the end of DNA, once these are removed there is an overhanging section of the parent strand, as the daughter strand is slightly shorter at its 5' end. --> the solution to this is telomeres - repeating structures at the ends of eukaryotic chromosomes - in humans the sequence is 5'-TTAGGG-3' - extends for as much as 10kb beyond the end of the daughter strand - these are slowly chewed up with every round of replication

5.09 Descibre how DNA is synthesised in vitro in reverse transcription for RNA analysis

cDNA can be synthesised for use in sequencing and analysis using an RNA template. This is done by effectively running a PCR reaction but with reverse transcriptase instead of taq polymerase. The way this differs is what is added to the reaction tube - RNaseH to digest the RNA template - reverse transcriptase instead of taq polymerase - different primers, often oligo-dTs to hybridise polyA tail --> diffent primers can be used to amplify a specific RNA of interest (specific primer), or the total RNA (random hexamer) of the cell to sequence the lot

11.01 Explain how methionine is modified to fMet in bacterial cells

fMet is synthesised by transformylase, an enzyme that formylates metionine attached to the initiator tRNA. - this enzyme does not formylate free methionine or methionine attached to an internal Met tRNA - this process results in fMet-tRNAf

10.06 Describe the functional role and general structure of tRNAs

tRNA is an adaptor molecule which binds to a codon and brings and amino acid with it to be incorporated into an aa sequence. - these are 73-79 ribonucleotides long - there is at least one tRNA for each aa (which attaches to the 3'CCA group on the tRNA - anticodon binds to a codon (these are codon-specific) --> structure must vary enough to bind specific codons and amino acids, but not so much that it is incompatible with translational machinery

9.02 Outline how eukaryotic tRNA is processed

tRNA is transcribed by RNA pol III, and is processed as follows - nucleotides from both ends are cleaved - CCA added to 3' end (this is the amino acid attachment site) - some nucleotides are modified - intron is removed and products are ligated, then two ends re-bind to form the anticodon

6.01 Describe and interpret reannealing experiments as evidence for non-coding DNA in the genomes of complex organisms

~40% of the human genome is composed of repetitive sequences, and was discovered by studying DNA association. --> simple strands of DNA are easier to re-anneal/match up, such as long repeated sequences. Therefore, reannealing experiments are used to gauge C0t1/2 - the point at which half the DNA is re-annealed, and can be used in conjunction with the hyperchromic effect to measure complexity.