Explain what happened to the Ade2+ gene near the telomere in the red cells and in the white cells. What is the term for this phenomenon?
Telomeres are the physical ends of chromosomes that are characterized by repetitive sequences. In eukaryotic cells, they usually get lost during successive DNA replications as the replication enzymes cannot replicate these end sequences. In yeast, if a gene is placed near telomere, its transcription is reversibly repressed. This phenomenon is known as telomere silencing or telomere position effect. Genes under this effect switch between repressed and active states each being stable for many cell generations. Telomere position effect is a kind of heterochromatin silencing phenomenon. Heterochromatin is a collection of repetitive sequences with low gene density and hypoacetylation of histone proteins that silences the genes transferred to its region.
In Ade2- mutants (red), the gene Ade2+ (white) was inserted near the telomere. On incubation, they should have resulted in entirely white colonies if the gene was inserted and replicated successfully. But the appearance of red as well as white colonies shows that both the silent and the expressed states were quite stable through multiple cell cycles. The appearance of red colonies in white sector and vice versa demonstrates that the cells have the ability to switch expression states.
You take single cells from the white sectors and allow them to form new colonies. You do the same with cells from the red sectors. Predict the color of the colonies. Explain.
As the white sector have cells with the inserted gene at the telomeric end, it is quite possible that even if they had the gene in active state for some generations, it might get silenced in future generations. So, if the cells from the white sector are replicated they may form only white or only red colonies or a mix of white and red colonies.
Whereas, if cells are taken from red sector for replication, the resulting colonies will always be red as they had only the repressed Ade+2gene at initial replication cycles and it is evident that they will result in only red colonies.
You measure acetylation of the N-termini of histones H3 and H4 that are located near the telomeric Ade2+. Do you expect more or less acetylation in the red cells than in the white cells?
Histones H3 and H4 will be hypoacetylated at the telomeric ends as they are present around the heterochromatin sequences. The telomeric sequences alter between the active and repressed forms for many cell generations. Thus, on observing acetylation of the histone proteins near the telomeric Ade+2, white colonies will have fewer acetylation compared to red colonies as they are getting transcribed.
You look for mutants that give only white colonies. You find mutations in three genes: Rap1, which encodes a protein that binds specific short DNA sequences near the telomere; Sir2, which encodes a histone deacetlyase; and Sir4, which encodes a protein that binds Rap1, Sir2, and deacetylated N-termini of histones H3 and H4. Propose a series of events in which these proteins act together to turn a white cell into a red cell.
TPE is under the control of several factors of which Rap1 protein binds it and establishes the telomere position effect. In yeast, telomeric DNA is assembled into a non-nucleosomal chromatin structure called telosome and Rap1 is the sequence specific DNA binding major protein in telosomes. Along with Rap1, Sir2, Sir3 and Sir4 proteins are also required to maintain TPE. Due to interactions between the Sir proteins and subtelomeric nucleosomes, chromatin there continuously propagates inwards. These proteins together play an important role in silencing genes at the telomeric ends.
The subtelomeric nucleosomal DNA is made less accessible to the DNA modifying enzymes such as methytransferases. Sir proteins bind with N terminal tails of H3 & H4 histones forming a complex. Thus, making these histones hypoacetylated compared to their counterparts in the genome. This leads to transcriptional silencing of genes at telomeric ends due to their uncommon chromatin structure. This way these proteins interact together turning a white cell into red.
3. Explain how cells maintain the fidelity of DNA replication.
DNA replication is a process carried out by cellular enzymes and protein factors to makes two double stranded copies of a single double stranded DNA. A base strand forms the complementary strand resulting in the newly formed DNA whereby one strand is parental and the other one is new (semi-conservative). The DNA copies are transferred from parents to their progeny forming the basis of inheritance. DNA contains the essential genetic information that is required for all the functionalities of the cell from its structure to metabolism. The identical copies should be formed during replication which otherwise will lead to loss of genome stability that might result in abnormalities like mutations in future generations.
Nearly perfect copies of DNA are created to maintain its fidelity by keeping a check on the process of replication through proofreading and several error checking mechanisms. A cell replicates nearly six billion nucleotides in just few hours (1000 nucleotides per second) whereby the error rate is so low that it may only cause a single mutation per genome for one cell division. It all depends on the correct base pairing of nucleotides corresponding to the base template. This level of accuracy is mainly maintained by three major factors during DNA replication, one being the high selectivity of DNA polymerases, the second being the exonucleolytic proofreading and the third one being the strand directed repair (Alberts B, 2002).
Selectivity of DNA polymerases: three DNA polymerases are basically involved in catalyzing the replication mechanism each playing a distinct role. DNA polymerase I initiate polymerisation by catalyzing the attachment of a nucleotide as per Watson-Crick base pairing to the primer on a base template. They basically avoid the mismatches between incorrect nucleotide and the DNA template that might occur due to changes in helical geometry or occurrence of rare tautomeric forms (due to which C might pair with A instead of G, etc).
Along with this, polymerases exhibit the proofreading of the DNA template to rectify any impairing. Before any nucleotide is incorporated in the growing DNA chain, a double check is already performed by the polymerases. Their active sites have hydrogen bonded water molecules that get displaced for correct complementary base pairing promoting fidelity as incorrect dNTPs are avoided. This high affinity of correct nucleotide for moving polymerases on the DNA strand in 5’- 3’direction is the first move of proofreading. Second proofreading move by polymerases is its conformational change that binds only the correct nucleotide efficiently after its binding but before the addition to the nucleotide chain. Here, any of the incorrect nucleotides get displaced.
Exonucleolytic proofreading: In cases, even after the initial proofreading by polymerases there might be instances where incorrect bases incorporates in the growing DNA chain. Polymerase can catalyze the polymerization reaction of DNA chain but it is not capable of initiating it. For initiation, it requires a primer with a free 3’OH end. If there is an incorrect base at the primer’s 3’end or at the growing 3’end, the self catalytic 3’-5’exonuclease activity of polymerase comes into action. Figure 1 demonstrates the exonuclease activity of polymerases by removing incorrect and incorporating correct base. It removes the mismatched pairs till the correct nucleotides generate to prime DNA synthesis. This way, polymerases keeps on correcting its mistakes along the polymerized DNA chain.
The DNA replication in 5’-3’direction is also essential for maintaining replication fidelity as the growing 3’OH end can be easily hydrolyzed compared to the free 5’triphosphate. Therefore, any mismatch at 3’end can easily be removed compared to the 5’end as high energy bond is cleaved providing energy for polymerization.
Strand–directed mismatch repair: It rectifies the errors that are further missed by the exonuclease activity. It identifies the distortion in the DNA helix due to non-complementary base pairing. It occurs after replication completion and to repair the mismatch at the newly formed strand is compulsory which otherwise being done at the parent strand can double the error rate. This distinction is carried out via methylation in prokaryotes and nicks/ single strand breaks in eukaryotes. Prokaryotes methylate their adenine bases present in the sequence GATC but after replication completion, it takes a while to do so in the new strand. This delay is detected and the mismatch repaired in the new strand. Similarly, eukaryotic newly synthesized DNA strand has nicks that are detected by the repair system as new strand and mismatches are rectified.
4. Describe the process of double strand DNA break by homologous recombination.
Double stranded DNA breaks can result from many reasons. It can be a result of broken replication fork or due to any damages. Homologous recombination is prominent in every life form as it provides fidelity to DNA by maintaining replication and telomere maintenance. It serves as an error-free saviour from damages caused by DNA gaps, DNA double stranded breaks (DSBs) or by DNA interstrand cross-links (ICLs). This recombination is the exchange of homologous DNA nucleotide sequences thereby generating the genomic variation during meiosis in the gametes. This can either take place between the two copies of same chromosome or between the DNA sections sharing similar nucleotide sequences.
Recombination in this way forms the basis of evolution. Firstly, the two different chromosomes comes close to each other and cross over occurs between the homologous strands by breaking of the two DNA strands and joining them with the opposite partner strands to form two recombinant helices. The site of cross over is not fixed and it can be located on any part of the DNA strand and is known as the heteroduplex joint. This joint is too precise to lose or gain any DNA sequence but still it results in formation of new sequence DNA molecules on either sides of the joint. Homologous recombination occurs via synapsis where the complementary strands form the two DNA molecules undergo base pairing.
Formation of synapsis (Cooper, 2000): For synapsis to happen, the two DNA strands for cross over need to be separated and freed from hydrogen bonds. A unique exonuclease activity creates a nick in both the DNA strands creating breaks. The 5’ends of both the strands are chewed off by exonuclease to create 3’hangings that are single stranded and search for complementary strands for base pairing as shown n figure 2. The single stranded overhangs are stabilized by single strand binding proteins that bind to the 3’ends and keep the bases exposed. In prokaryotes, the unwinding and nicking are carried by RecBCD enzyme. It unwinds the dsDNA by acting as helicase and creates a nick in the DNA strand through the nuclease activity. It is then it finds the complementary strand in the homologous chromosome and pairs with it.
Holliday Junction formation: Rec A protein with two active DNA binding sites then comes in the scene. RecA is a prokaryotic protein and its homologues like Rad51 are present in yeast and further in more complex eukaryotes. It is specialized as it holds the single stranded DNA strand as well as the double stranded DNA helix together to initiate the exchange. This strand exchange is then initiated by RecA protein leading to the formation of holliday junction or cross strand exchange. Holliday junction is characterized by the reciprocal exchange between the two out of four strands of homologues. It can isomerize in different resultants that are described in figure 3.
Resolution of the helices: After forming an open structure, utilizing energy from ATP hydrolysis, several proteins (Ruv A & B in prokaryotes) moves the crossover points and extends the heteroduplex. After the exchange, resolution occurs by cutting the connecting strands on two helices. Ruv C in prokaryotes, Rad1 & Rad10 in yeast carries out the cleavage and ligation at the end completes the process. The resultants can be of two kinds here depending whether the cut is at the crossover strands or the non crossover strands. In first case, helices separate with the single strand exchanged and in the second case, the two helices have been exchanged through crossing over event generating two recombinant chromosomes.
5. Explain the process of transcription of eukaryotic mRNA with emphasis on protective post-transcriptional processing.
Cells need proteins to perform the functions that comprises of structural stability, energy production for survival and a lot more. The information is all coded in the DNA of the organism. Thus genes need to be expressed to fulfil the cellular necessities. Transcription is the fundamental process whereby DNA is copied to RNA forms like mRNA, tRNA and rRNA through specialized RNA polymerase enzymes. Multicellular organisms are made of distinct cell types and their complexity demands regulatory mechanisms. To maintain the gene expression regulation, eukaryotic transcription employs different RNA polymerases to transcribe different genes whereas in prokaryotes single polymerase is used. Eukaryotic RNA polymerases cannot recognize the promoter sequence but after interactions with several proteins (initiation factors) they can initiate the transcription process.
Three RNA polymerase function in eukaryotic cells namely RNAPI, RNAPII and RNAPIII. RNAPI transcribes 5.8S, 18S and 28S subunits of rRNA that forms structural component of ribosome (required in translation of proteins from mRNA). RNAPII transcribes mRNA that carries the protein information and is later translated to proteins and also regulatory RNA’s like micro RNAs and long coding RNAs. RNAPIII transcribes 5s subunit if rRNA and tRNA that is involved in translation for carrying amino acids to the translation site. rRNA and tRNA are non-protein coding RNA. There is a large structural similarity between all the RNA polymerases i.e. they are composed of 8 to 14 subunits whereby five are common in all. Transcription by RNAPII is the focus for study as it copies mRNA which is a protein coding RNA. The transcription machinery here needs to penetrate the chromatin layers to be able to reach the DNA.
Initiation requires special proteins called transcription factors in eukaryotes. General and specialized transcription factors have been characterized to play role. Five general factors have been identified through researches that recognize all the promoters for RNAPII and forms a part of basic transcription machinery. The later ones are for specific genes and regulate the gene expression. TATA box (sequence is TATAA) is the characteristic promoter sequence for RNAPII that is located ~25 sequences upstream from the start site. TFIID is the initial general transcription factor that binds to the TATA box through TATA binding protein (TBP) subunit. It also binds to polypeptides called TBP associated factors (TAF). Second factor, TFIIB binds with TBP followed by RNAPII along with TFIIF. Helicase activity is shown by TFIIH that binds next to the initiation complex unwinding the double stranded DNA. The unwinded DNA structure with the initiation complex is termed as transcription bubble that is stabilized by TFIIB. The RNA polymerase II complex with TFIIB, TFIIE, TFIIF, TFIIH is known as polymerase II holoenzyme.
Elongation is the next step of transcription. TFIIH also comprises of a protein kinase as well that phosphorylates the C-terminal of largest RNAPII subunit and hence allows it to slide along the DNA for transcription. RNA synthesis does not require primer to proceed. Thus, before leaving the initiation complex, polII generates around 10-14 nucleotides of RNA strand and a part of it remains bound to the complex and another to the growing RNA chain. This stage is known as promoter escape followed by promoter-proximal pausing. Eukaryotic DNA is wrapped around histone proteins that need to loosen up before the complex reaches for transcription. FACT protein does this by disassembling the nucleosome and removing two histones temporarily that are reassembled once the genes are transcribed. RNAPII at times pauses for instance or even for long durations before proceeding to the productive elongation so as to check for any errors. Finally, the enzyme precedes the RNA chain in 5’to 3’direction. Here a unique RNA:DNA hybrid structure is formed along the length. RNA so formed is complementary to the DNA template. It then reaches the end and termination occurs.
Termination signals are characteristic for termination of the mRNA chain. There are two signals classified that guide the process. One being the poly A sequence and the stop sites that are located downstream. Rather, RNAPII terminates at random sites passing the gene (sometimes few bps and at times thousands of bps) being transcribed. The newly formed RNA has an upstream AAUAAA sequence and a downstream GU rich sequence. After they get transcribed, CPSF & CstF proteins bind AAUAAA & GU sequences respectively. CPSF cleaves the pre-mRNA at a site ~30 sequences downstream to AAUAAA site and releases it prior to termination. The remains are cleaved off by 5’exonuclease till it encounters the growing end by RNAPII and at that point the complex dismantles and termination occurs.
Protective Post-transcriptional processing of mRNA
The pre-mRNA released during termination of transcription undergoes processing before proceeding for translation to prevent from degradation and increasing their half lives. Pre-mRNA are coated with the stabilizing proteins during processing and export from the nucleus that makes them last for several hours inside the cellular nucloelytic environment. It undergoes three modification processes namely 5’capping, 3’tailing and splicing of intervening non-coding sequences.
5’capping has important roles in protection of mRNA from degradation by ribonucleases (having specificity for 3’-5’phosphodiester bonds) as well as the cap helps recognize these mRNAs by initiation factors of protein synthesis for translation. A cap of 7-methylguanosine is added at the 5’end of the pre-mRNA while it is still under synthesis through a 5’-5’phosphate linkage with the help of phosphatase and guanosyl transferase enzymes. The diphosphate end adds gamma phosphorous atom from GTP molecule to add guanine at 5’-5’end. A methyl group from S-adenosyl methionine is then added to the guanine ring by guanine-N7-methyltransferase. It is the cap 0 structure and caps 1, 2, 3 and so on are formed if the adjacent nucleotides also get methylated at 2’OH groups on ribose sugar. Figure 5 demonstrates the capping process.
3’tailing is the addition of a poly A tail by poly A polymerase enzyme. This PAP enzyme is a part of the cleaving complex described during termination. While the pre-mRNA is cleaved between the AAUAAA and GC rich sequences, this enzyme adds ~250 adenosine residues (poly A tail) at the 3’end as shown in figure 6. Along with savage from degradation and recognition for translation initiation, it also aids in transport of active RNA into the cytoplasm.
Pre-mRNA splicing frees the pre-RNA from the non-coding gene sequences called introns. They are present in the DNA and get transcribed onto RNA but need to be removed before translation because the reading frame will not be effective then resulting into a dysfunctional protein. Splicing take place while the pre-mRNA is still in the nucleus via spliceosomes. Small nuclear RNA along with a complex of proteins form snRNPs and such five units form a spliceosome. Introns are characterized with a GU sequence at 5’end and AG at 3’end. These are recognized by spliceosome complex that cleaves phosphate backbone at G and joins the free G so created with A within the intron as shown in figure 7. It is followed by connecting 3’end of first exon to 5’end of next exon removing 3’intron end in a lariat form.
6. How the eukaryotes maintain the fidelity of the transcribed mRNA.
Transcribed mRNA is significant as their genes codes are important that translates into proteins. Thus, a perfect sequence is must for the correct proteins to be manufactured from RNA templates. It is known that the error rate in translation is much higher compared to transcription but little is known about fidelity maintenance in transcription process. It is also a fact that replication is the most accurate of the three processes and maintains highest fidelity. The error rate in transcription is ~1X10-4.
RNA polymerases are the first key points in maintaining fidelity of the process through their conformational structure and functions. They restrict the non-complementary ribose NTP substrates as well as the complementary dNTPs from adding to the newly synthesizing RNA chain. The trigger loop is a flexible domain of the RNAP active site (Yuzenkova et al, 2010) involved in interactions with the incoming nucleotides. The incorrect nucleotides do not align with loop and hence discarded (phosphorolytic editing). It enforces the formation of phosphodiester bond with the matching rNTP and the 2’OH. Trigger loop functioning is further assisted by Rpb9 protein that ensures further fidelity.
During the transcription elongation, RNA polymerase gradually precedes via three steps namely promoter escape, promoter-proximal pausing and finally productive elongation. This mechanism in itself keeps a check on the incorporation of only the correct ribonucleotides. The polymerase keeps on forming and degrading the RNA segments and does not move forward leaving the promoter until a correct small RNA segment of around 14 nucleotides is created. On moving further, it pauses if any mismatch occurs and rectifies it by degrading a small segment through the 3’-5’exonuclease activity (hydrolytic editing) and then again extending the chain. This ensures the fidelity of RNA transcription.
7. Cells have several mechanisms to prevent the production and accumulation of truncated protein fragments. To illustrate why these fragments can be deleterious, consider a transcriptional activator protein called Groovy that binds to an enhancer element upstream of the Zippy gene. Groovy has a DNA-binding domain at its N-terminus and a domain that binds a histone-modifying enzyme at its C-terminus.
If a cell makes substantial amounts of an N-terminal fragment of Groovy, containing the DNA-binding domain, what is likely to happen to transcription of the Zippy gene?
Truncated proteins are the non-functional and incomplete proteins that are resultant of the nonsense point mutations that results in premature stop codon in a transcribed RNA. These are shortened protein than the actual. When such mutations occur, the mRNA is degraded before they are translated to truncated products through the process of nonsense-mediated mRNA decay (NMD). In cases of translation of such mRNAs, the abnormal proteins exhibit dominant negative effect on the wild type protein products often by combining it and dimerizing it. It is deleterious than the condition when cell does not produce the gene product at all.
In the case of N-terminal production of the Groovy protein, the Zippy gene will not be transcribed as the protein requires both the N and C domains to be functional. In the absence of C domain, Groovy will not be able to unwind the histones around Zippy gene and hence no expression. This condition is similar to the one where no Groovy protein is being produced.
If, in addition to the N-terminal fragment of Groovy, a cell also makes an equal amount of the full-length protein, what is likely to happen to transcription of the Zippy gene?
In such cases, where a wild type protein and its truncated counterpart co-exist in a cell, the truncated protein exerts a dominant-negative effect on the functional protein. In this case also the N-terminal fragment of Groovy will exert a dominant-negative effect on the full-length protein and inhibit it to transcribe Zippy gene. This case is even worse as the proteins tend to accumulate inside the cells that can prove to be deleterious.
Consider two cell lines or strains that each have a fully functional allele of the Groovy gene on one copy of chromosome 3. On the second copy of chromosome 3, strain 1 lacks a second Groovy allele whereas strain 2 contains the Groovy allele that codes for the N-terminal fragment described in part A. If strain 1 and strain 2 have different amounts of Zippy mRNA, which do you expect to have more Zippy mRNA? Explain.
In such case, strain 1 is expected to have more Zippy mRNA than strain 2. Strain will have sufficient Zippy mRNA as and when required by the cell because there is no repression of the expressed Groovy protein. Whereas, in strain 2 the N-terminal fragment of Groovy will inhibit the functional Groovy protein from exhibiting its functionality and when no transcription of Zippy gene will be there then no Zippy mRNA will be observed in these cells.
Strain 3 encodes the same N-terminal Groovy fragment as strain 2, except that the gene in strain 2 contains a deletion of some Groovy coding sequence and strain 3 has a single base-pair change that converts a lysine codon (AAA) to a stop codon (UAA). You find that strain 3 produces the same amount of Zippy mRNA as strain 1. Can you explain why?
When strain 2 carries a deletion of a part of groovy coding sequence, then on translation it will produce a non-functional protein that will subsequently be degraded as it will not comply to proper three dimensional conformations because for appropriate folding and processing of protein, a perfect amino acid sequence is mandatory. Thus, the second copy of allele in strain 2 will not result in any Groovy protein.
In strain 3, the single base pair change in the stop codon in the gene segment coding for only the N-terminal Groovy fragment causes the non-sense mutation. The non-sense mRNA on transcription is degraded by the nonsense-mediated mRNA decay (NMD). As a resultant, no N-terminal Groovy proteins are expressed in strain 3. The functional Groovy protein is saved from the dominant negative effect of its truncated N terminal fragment. On being saved, Groovy proteins will exhibit their role of transcribing Zippy gene and resulting in the same amount of Zippy mRNA as in strain 1.
8. Prokaryotic and eukaryotic cells protect against translating mRNAs that may produce misfolded proteins. Describe the mechanisms by which misfolded proteins are salvaged or degraded.
Proteins are synthesized for several functions of structure formation that are destined to the cell membrane or for secretion. After translation, they go to the endoplasmic reticulum for proper folding and processing without which they are incapable of functioning. At times, they get misfolded in the ER and thus sent for degradation. The degradation pathway depends on teh type of protein being degraded. Newly formed misfolded protein is degraded via ubiquitin dependent manner through UPS or ERAD pathways. Whereas, through autophagy protein aggregates or oxidatively modified proteins are eliminated.
The Ubiquitin Proteasome system (UPS) tags the protein to be degraded with an ubiquitin by E3 ubiquitin ligase. Prior to this, E1 forms a high energy thioester linkage by hydrolyzing ATP between its active site and ubiquitin residue. Then it is passed to E2 which then aligns with E3 that is bound to the misfolded protein. Ubiquitin binds to lysine of misfolded protein. This ubiquitin is further polymerized by several ubiquitin units forming a polyubiquitin chain and gets recognized by 26S proteasome as shown in figure 8. There degradation is done by deubiquitinating enzymes first and then by the proteolytic active sites in proteasome in the cytoplasm. Proteasome breaks down proteins into amino acids that are reused.
Figure 8: Ubiquitin mediated proteasomal degradation. (Source: Burger & Seth, 2004)
The Endoplasmic Reticulum mediated degradation (ERAD) is only for proteins destined for the secretory pathways. These are synthesized in ER and for them to be degraded; they should be misfolded and released from the ER via translocon in the cytosol where they are degraded via ubiquitin mediated proteasomal degradation. This process involves ER chaperons and E3 ligases.
Autophagy degrades large aggregates of proteins and other cellular wastes by forming a phagophore on inception of signals. A vesicle is formed first that gradually increases in size. They convert to autophagosome by contacting with phagosome and carries the degradation destined protein and cellular components. The lysosome then fuses with and degrades the contents.
At times, ER has been playing a critical role in properly folding the misfolded proteins. It is a large secretory compartment in which proteins undergo continuous changes with help of chaperons and folding enzymes.
9. What do consensus sequences in promoter regions of DNA and untranslated regions of RNA signifies?
Consensus sequences are the set of repetitive sequences occurring throughout the genome. They have significance in several aspects. They are thought to play similar role even if present at different locations. The consensus sequences in the promoter regions provide specificity to the transcription factors for binding and regulation. These kind of sequences are calculated by considering all the known examples and a common sequence pattern is derived that can be considered universal for action in different cases. They are conserved through evolution thus offering the universal pathways like replication, transcription, translation, splicing, etc a definite pattern for mechanisms like binding of transcription factors on promoters for initiation or regulation, specific sites for binding of factors and more. At times, the relatedness between two species is calculated by similarity of the consensus sequences.
Consensus sequences in the promoter (ex: TATA box, GC rich sequences) and untranslated regions of mRNA play an important role in providing a unique location for the initiation of transcription and translation respectively. Through these sequences, it becomes definite for the initiation factors to recognize the start site and begin the process. These act as cis-acting elements to which the trans- acting elements bind.
There are many processes taking place in a cell at a time. Many of them are dependent on each other and several have to take place together. For this, a common consensus sequence at regulatory sites ensures co-ordination and enhances the speed of their expression.
Alberts, B., Johnson, A., Lewis, J., Raff, M., Roberts, K., Walter, P. Molecular biology of the cell. 2002, 4th edition. Accessed at http://www.ncbi.nlm.nih.gov/books/NBK26850/ on 18 April 2015.
Berkeley, 2007. Accessed at https://mcb.berkeley.edu/courses/mcb110spring/2007L11to13.pdf on 15 April 2015.
Biolabs Inc., 2015. Accessed at https://www.neb.com/tools-and-resources/feature-articles/polymerase-fidelity-what-is-it-and-what-does-it-mean-for-your-pcr on 19 April 2015.
Boundless. “Initiation of Transcription in Eukaryotes.” Boundless Biology. Boundless, 25 Nov. 2014. Accessed from https://www.boundless.com/biology/textbooks/boundless-biology-textbook/genes-and-proteins-15/eukaryotic-transcription-108/initiation-of-transcription-in-eukaryotes-445-11670/ on 20 April 2015.
Burger, A, M., Seth, A, K. The ubiquitin-mediated protein degradation pathway in cancer: therapeutic implications. European Journal of Cancer, 2004. 40; 2217-2229.
Cooper, M, G. The Cell: a Molecular Approach.2000, 2nd edition. Accessed at http://www.ncbi.nlm.nih.gov/books/NBK9859/ on 19 April 2015.
Smaloy. 2003. Accessed at http://www.sci.sdsu.edu/~smaloy/MicrobialGenetics/topics/genetic-analysis/recombination/rec-molecular.html on 12 April 2015
Yuzenkova, Y., Bochkareva, A., Tadigotla, R, V., Roghanian, M., Zorov, S., Severinov, K., Zenkin, N. Stepwise mechanism for transcription fidelity. BMC Biology 2010, 8:54.