Lecture 1: Nucleic acids and chromosomes 1) Draw the structure of a nucleotide labelling the sugar, base and phosphate(s).
NH2
Deoxyadenosine 5'triphosphate
O
C
N
HC
Phosphate O
C
N
C
N
O
CH N
Adenine (Base) O-
P
O
O-
P
O
O-
P
O
CH2
OH
O
H
H
OH
H
H
Deoxyribose (Sugar) • •
nucleic acids are linear polymers of nucleotides (nucleotides are monomers) phosphate groups are attached to the 5’ carbon of the deoxyribose, hydrolysis of the phosphate groups is very exothermic and used to drive many chemical reactions 2) List the bases found in DNA, indicate which ones are purines and which ones are pyrimidines. • Purines – Adenine and Guanine Pyridimines – Thymine and Cytosine 3) Describe a single DNA chain and explain the difference between the 5’ and 3’ ends. • The phosphate is on the 5’ carbon, which is linked to the OH group on the 3’ carbon on the adjacent nucleotide; each bond formation results in the loss of a diphosphate group and is called a phosphodiester link. • DNA is not symmetrical- it has 5’ and 3’ ends. As it is not symmetrical other molecules can sense the orientation of DNA 4) Draw the structure of the double-stranded helix of DNA (not atomic structure), showing base pairing, the major and minor grooves, and the directionality of the chains.
•
'5
Minor Groove
Major Groove
'3
3'
5'
deoxyribose and phosphate chains run along the outside with negative charges on the outside, the 2 two chains are held together by H bonds • the chains run in opposite polarities to one another 5) Describe melting and reannealing of complementary strands and what is meant by Watson-Crick base pairing. • High temperature/low salt concentration causes the two strands to melt or dissociate; strands with more CG will be more resistant to melting than AT strands as they are joined by 3 not 2 hydrogen bonds; if you lower the temperature/increase the salt, complementary strands will re-anneal (if there is a mixture, strands will find each other) • Watson and crick base pairing: A T; G C • 2 strands are said to be complementry in their sequence, if you know the sequence of one strand then you can predict the sequence of the other- this allows DNA to be copied 6) Compare the genomes of Escherichia coli and Homo sapiens. • Genome = entire DNA content of a single cell ( every cell has the same DNA content) • E. coli – single double helix circular molecule, 4 x 106bp Homo sapiens – several separate double helices called chromosomes, 3.9 x 109bp, 23 chromosomes (haploid cell). 7) Draw a diagram illustrating the packaging of DNA into nucleosomes and relate this to chromosome structure; describe the structure of eukaryotic chromosome. • DNA is packaged to make it more compact (DNA + protein = chromatin), at the lowest level, the DNa is wrapped around proteins called histones to make units called nucleosomes. • Histones are small proteins ~ 100 amino acids in size and are positively charged, due to the high content of basic amino acids. Therefore histones bind well with negatively charged DNA. • There are 8 histones (2 of each H2A, H2B, H3 and H4), Dna is wrapped twice around each nucleosome ~ 200 amino acids of DNA. The H1 histone binds between the nucleosomes. The nucleosomes are joined together by linker DNA. • Nucleosomes cause a 7 fold condensation of DNA, the nucleosome coil again to cause a 40 fold condensation in the DNA.
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In an interphase cell, the 46 chromosomes appear as a diffuse mass called chromatin; in cell division, the replicated chromatin condenses to form two sister chromatids held together at the centromere. Describe the Human Karyotype Somatic cells are diploid and have 2 copies of each chromosome, 23 pairs of chromosome, contain 22 pairs of autosome and 1 pair of sex chromosomes In a standard karyotype the chromosomes are distinguished by size, centromere position and banding pattern Describe what happens to the DNA and chromosomes during the mammalian cell cycle Interphase – contains 3 phases: G1 (10 hours) is the interval between the mitotic phase and the S phase, when the cell is metabolically active and duplicates its organelles and cytosolic components; S phase (9 hours) is between G1 and G2, when DNA replication occurs; G2 phase (4 hours) between S and the mitotic phase, when cell growth continues and enzymes and other proteins are synthesised in preparation for cell division. Mitotic phase (1 hour) Prophase – chromatin fibres condense and shorten, to form the characteristic chromosome with double-stranded chromatids held together by a centromere; later the nucleolus disappears and the nuclear envelope breaks down. Metaphase – the chromosomes align along the metaphase plate. Anaphase – the centromeres split, separating the sister chromatids which move to opposite poles becoming daughter chromosomes; beginnings of cytokinesis. Telophase – daughter chromosomes uncoil to diffuse chromatin; re-formation of nuclear envelope, nucleolus etc.
Lecture 2: DNA replication, the cell cycle and mitosis 1) Starting with a stretch of double stranded DNA draw the 2 new double stranded molecules formed after replication. Take care to show polarity of the strands.
• Each strand of parental DNA acts as a template for the formation of a daughter strand hence – semi-conservative 2) Explain how the structure of DNA is suited to the need to replicate the genetic information of an organism and what is meant by semi-conservative replication. • Each strand of parental DNA acts as template for the formation of a daughter strand, hence it is semi-conservative. • DNA is suited to the need to replicate, as A will only successively bond to T and G to C (i.e. each parental strand can specify the sequence of nucleotidein its complementary strand, therefore DNA can be copied precisely. 3) Describe the reaction catalysed by DNA polymerases. • In E. coli, the principal enzyme in DNA replication is DNA polymerase III, which synthesises DNA by adding deoxynucleotide triphosphates to 3’ end of a DNA molecule; it is an exothermic reaction, driven by the energy that it produces; DNA polymerase needs a primer (it cannot start a new chain) and also a template (to match bases). • DNA is synthesised form 5’ to 3’ 4) Describe the components and function of the replication complex in prokaryotes, including the terms: template, primer, Okasaki fragment, and replication fork. • DNA starts to fork at the DNA origin (fork/unwinding is caused by DNA helicases) • The leading strand replication is continuous and is carried out by DNA polymerase III • The lagging strand replication is dicontinous, because replication must occur from 5’ to 3’. • RNA primer attaches to lagging strand when ~ 1000 based are exposed (the RNA primer is attached by a RNA Pol called primase. • The primer is extended by DNA Pol III until the next RNA primer is reached • DNA stretches which are synthesised in the opposite direction to the movement of the replication fork are called okazaki fragments • DNA Pol I removes the RNA primer using 5’ to 3’ exonuclease activity this then continues the synthesis of okazaki fragments through the RNA primer regions. • DNA ligase then joins the 2 adjacent strands together using ATP. • The replication complex is a single complex containing 2 pol III holoenzymes, which synthesises both leading and lagging strand at the replication fork. Thelagging strand is looped round so that the 2 polymerases can stay together. 5) Describe the function of DNA topoisomerases. • Type1: The unwinding of the replication fork creates tortional stress in the double helix ahead of the replication fork. Type 1 topoisomerase makes a single stranded break ahead of the replication fork, so that the DNA can then rotate freely at the break, relieving any tortional stress. The DNA is then rejoined.
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Type2: E. Coli has a circular chromosome. As the two strands of DNA are wrapped around each other, replication leads to 2 interwined circles. Type 2 topoisomerase causes a double stranded break and allows the other DNA helix to pass through the break. The break is then rejoined. 6) Draw a diagram showing replication of the E. Coli chromosome.
7) Describe the replication of mammalian chromosomes. • In eukaryotes, there are similar mechanisms, but different proteins in use; there are more polymerase enzymes for example • Eukaryotic chromosomes are linear and long, there are multiple origins of replication at about 100kb apart. Each fork is bidirectional and replication is complete when all the forks meet. • In mammals, the DNA is AT rich, but no particular sequence of origin. Proteins recognise the origin and open up the DNA so the 2 forks can be initiated. • Pol α synthesised primers for both leading and lagging strands • Pol β removes RNA primers and synthesises DNA in its gaps • Pol δ synthesises the leading strand • Pol ε synthesises the lagging strand • Pol μ replicates mitochondrial DNA 8) Describe how accuracy is maintained by proof reading and RNA primers. • The RNA primers must be erased so if any mistake is made they are not kept, as would be the case in the use of non-erasable DNA primers; DNA polymerase has built in proof-reading – it will only add new nucleotides if the 3’ end contains an OH group; additionally, mismatched nucleotides at the 3’ end are not as effective as templates. If an incorrect nucleotide is added to a growing strand, the DNA Pol. Will cleave it from the strand and replace it with the correct nucleotide before continuing. 9) Describe how drugs affecting DNA can be used against caner and viruses • Chemotherapy: in cancer, cell cycling is not controlled and cells rapidly divide, we can use a nucleoside, which acts as a stopper in the process of replication as it has no phosphate group, an example of a nucleoside is cytosine aribinoside. • Antiviral drugs: this involves the use of nucleotide analogues e.g. acyclovir which is phosphorylated by a herpes virus encoded thymidine kinase, but not by the host enzyme. Infected cells then incorporate it into their cell which kills them. (AZT is similar, as it incorporated into the DNA by reverse transcriptase in HIV)
Lecture 3: Gene organisation and transcription I 1) Describe the basic differences between DNA and RNA. • RNA, in contrast to DNA, is single-stranded, but has local regions of short base pairing, the sugar is ribose, and it contains uracil instead of thymine, but maintains the same complimentary nature and is less stabile than DNA. • RNA also codes for protein, whereas DNA codes for RNA 2) Describe what is meant by ‘transcription’. • Genetic information represented by the sequence of bases in DNA, serves as a template for copying the information to a complementary sequence in a strand of RNA. • Transcription is a process in which a nucleotide information in DNA is copied into RNA 3) List the major functional classes of RNA and the classes of RNA polymerase involved in synthesising each of these. • Ribosomal RNA (rRNA) – RNA Polymerase I (transcribes r.RNA genes) • Messenger RNA (mRNA) – RNA Polymerase II (DNA into mRNA) Transfer RNA (tRNA) – RNA Polymerase III (mRNA into tRNA) 4) Draw a diagram of the general organisation of a eukaryotic gene.
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5) Describe what is meant by an “intron” and an “exon”. • Intron – non-coding region of a eukaryotic gene that is transcribed to pre-mRNA but removed by RNA splicing when mRNA is produced. Exon – segment of a eukaryotic gene that consists of DNA coding for a sequences of nucleotides in mRNA. 6) Describe what is meant by a “gene promoter”. • The DNA sequence at which the initiate complex assembles in transcription; consists of Transcription Factor Binding Site(s) (which control the rate of transcription) and the TATA sequence which specifies the where RNA Polymerase II should start transcribing. 7) Describe what is meant by a “transcription factor”. • Regulate the level of transcription from a given gene. Can be one of two forms: Transcriptional Activator or Transcriptional Repressor. • They act collectively to bring about cell specific developmental inducible gene expression. 8) Describe with the aid of diagrams the processes involved in transcribing a eukaryotic gene. • The basal transcription complex allows RNA polymerase II to be phosphorylated and start transcription; its assembly requires the binding of specific Transcription Factors (activators), but can also be suppressed by Transcriptional repressors, and additional Transcription Factors can bind to increase the level of transcription, if not, it produces a basal level of transcription.
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TFIID binds to the TATA seq. using TATA binding protein (TBP) and TBP accessory factors (TAFs) On binding TFIID partially unwinds the DNA helix widening the minor groove to allow extensive contact with bases of DNA (unwinding is asymmetrical with respect to TBP and TATA complex, thereby reassuring transcription is unidirectional.
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TFIIA and TFIIB allows the TFIID to bind to the RNA Pol II ( the RNA Pol II binds to TFIID with TFIIF already attached to it)
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TFIIH, E, J futher unwind the helix to facilitate DNA transcription Basal complex allows RNA Pol II to be phosphorylated and engage transcription. In the absence of additional transcription factors, a basal (low) level of transcription takes place • Transcription factors bend DNA on binding, they interact with each other and the basal complex to modulate transcription. • Transcription factor action is regulated by factors outside the cell. E.g. hormones. 9) Distinguish “sense” and “antisense” DNA template strands. • The sense strand is the non-coding strand that has the same appearance as the produced pre-mRNA except for the substitution of thymine for uracil. The anti-sense strand is complementary to the pre-mRNA and is the coding strand. 10) Define what is meant by “pre-mRNA”. • Pre-mRNA is mRNA that has not been processed i.e. it is not ready for translation; it still contains the useless intron sequences for example; also known as the primary transcript or heterogeneous nuclear RNA (hnRNA). • Pre-mRNA is converted to mRNA by the process of splicing
Lecture 4: Gene organisation and transcription II 1) Describe the addition of a “cap” and “poly-A tail” to pre-messenger mRNA. • The cap structure is added to the 5’ end and protects mRNA and also enhances translation; the cap structure Is formed by hydrolysis of the terminal triphosohate of the mRNA to a diphosphate this then reacts with GTP to form a 5’ – 5’ phosphate linkage. • the poly-A tail is added in a process called polyadenylation and is added one base at a time and is found 11-30 bases downstream from the AAUAAA sequence, which is common to all mRNA molecules.
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2) Describe with the aid of diagrams the events that take place in pre-mRNA processing, describing the terms splice donor and acceptor site, the lariat intermediate and the spliceosome. • RNA processing uses small ribonuclear proteins (snRNPs); firstly, U1 binds to the splice donor sequences, which is AGGU (GU is the start of the intron); then U2, and U4 binds to the intron and U5 to the splice acceptor site (U6 binds to the itron also), the end of the intron characterised by the sequence Pyr15NCAG (AG is the end of the intron), and U6 bind completing the formation of the splicing complex or spliceosome • An A residue in the intron is used as a branch point in an intermediate step, the branch results in a phosphodiester bond between the %’ phosphate of the start of the cleaved intron and the OH of the branch point A. This results in the phosphodiester bond between the G at the end of the intron acceptor site and the next extron to be cleaved, therefore the intron is fully cut. The cleaved structure (intron) is called a lariat structure.
3) With examples, describe how mutations in splice sites feature in the human disease. • In β-Thalassemia ( a genetic haematological disease) there is an imbalance in the relative amounts of α and β chains in the haemoglobin molecule (less β chains) and several of the 100 variations of the disease feature splice site mutations in the β-globin gene.
Lecture 5: Protein translation and translational modification 1) Outline the mechanisms by which ribosomes translate an mRNA sequence into an amino acid sequence. • Translation initiation – dissociation of the two ribosomal subunits is followed by assembly of the pre-initiation complex comprising methionine-tRNA, initiation factors and the small ribosomal subunit (40s); the mRNA then binds to the pre-initiation complex (the methionine-tRNA binds to the P site (the start codon AUG); then by the conversion of GTP to GDP, the large subunit (60s) binds also. • Translation elongation – a new tRNA molecule (the anticodon is complementary to the mRNA codon) joins to the A site (adjacent to the P site); peptidyl transferase catalyses the production of a peptide bond between the two amino acids; then the new tRNA molecule is translocated to the P site, and the first tRNA dissociates, and then the cycle continues.
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Translation termination – the stop codon (UAA, UAG, UGA) is recognised, then release factors bind to the vacant A site, which then, using GTP and peptidyl transferase causes the release of the polypeptide chain. Binding of RF causes the peptidyl transferase to transfer the peptidyl group to water rather than to an aminoacyl group. This results in the release of the polypeptide Describe the role of amino-acyl tRNAs in ensuring the fidelity of the genetic code. They have an in-built proof reading function that occurs just prior to the peptide bond formation. State how a ribosome recognises the start and end of a sequence to be translated. Start (AUG) and Stop (UAA, UAG, UGA) codons. Explain why some antibiotics inhibit protein synthesis in prokaryotes but not in eukaryotes. The translational machinery is very complex and is easily disrupted; some antibiotics selectively disrupt to bacterial protein synthesis e.g. inhibit initiation, translocation etc. Identify the features of a newly synthesised protein which are required for it to enter the secretory pathway. The first 20-24 amino acids of the polypeptide chain is the signal sequence; this sequence is detected by the signal-recognition protein on the RER membrane and binds to it; the polypeptide is then translocated through the activated protein channel into the ER lumen; within the ER lumen the signal sequence is cleaved and the protein is folded. Give examples of the ways in which proteins may become post-translationally modified. Addition of carbohydrate (glycosylation – occurs in the Golgi apparatus), phosphate (phosphorylation) or lipid groups (prenylation).
Lecture 6: Analysis of DNA 1) Explain the term hybridisation, used for binding of a probe to a nucleic acid. • A method in which single-stranded DNA or RNA molecules combine to form double-stranded molecules. Standard assay involves a labelled nucleic acid probe to identify related molecules in a mixture of target unlabelled nucleic acids. 2) Explain the concept of stringency of hybridisation, and the factors that contribute to stringency. (=breaking) • Denaturation of a probe DNA is achieved by heating until the hydrogen bonds holding the 2 strands are disrupted. The energy to do this depends on strand length and base composition. Hybridisation stringency increases with: increase in temperature and decrease in Na+ concentration (monovalent cations destabilise the DNA duplex eg Na+). 3) Explain how the polymerase chain reaction (PCR) is used to amplify small amounts of DNA for subsequent analysis. • Brief heat treatment (94 degrees) separates the two strands of DNA. Cooling (55 degrees) of the DNA in the presence of a large excess of the two primer DNA oligonucleotides allows these primers to hybridise to complimentary sequences in the two DNA strands. The mixture is then incubated (72 degrees) with DNA polymerase and the four deoxyribonucleoside triphosphates so that DNA is synthesised starting from the two primers. The cycle is begun again by heat treatment to separate the newly synthesised DNA strands. The cycle is repeated many times and the amount of DNA synthesised doubles with each cycle. 4) Describe in general terms the way in which PCR primers would be selected to amplify a given DNA sequence. • PCR primers or synthetic DNA oligonucleotides are 15bp – 25bp single strands of DNA complimentary to the sequence on one strand of the DNA double helix at opposite ends of the region to be amplified. They determine the region of DNA that is to be amplified because they serve as primers from which DNA polymerase can synthesise a new DNA strand in vitro. The start and end sequence of the gene to be amplified would have to be known and the primers would be synthesised with the same DNA sequences. • When selecting a primer avoid base configurations that have tandem repeats of nucleotides, The G-C ration should be equal in each primer and also avoid sequences that may form hairpins. 5) Describe the reactions carried out by restriction enzymes (restriction endonucleases) and explain their usefulness in analysis of DNA. • Restriction endonucleases cut DNA only at particular sites, determined by a short sequence of nucleotides. They can cut DNA giving either “blunt” ends or “sticky” ends. They can therefore be used to produce a set of specific DNA fragments from a genome. 6) Describe how DNA fragments can be separated on the basis of size. • Gel electrophoresis separates DNA fragments on the basis of their length. The DNA fragments are loaded at one end of a slab of agarose/polyacrylamide gel containing a microscopic network of pores. A voltage is then applied causing the negatively charged fragments to migrate towards the positive electrode. The agarose matrix impedes the migration of larger fragments resulting in the DNA fragments being separated out into a ladder of discrete bands, each composed of fragments of the same length. 7) Describe the principles behind the standard method of obtaining the sequence of a piece of DNA. • DNA polymerase synthesis of single-stranded DNA using a short primer DNA, deoxyribonucleoside triphospates (dATP, dCTP, dGTP, dTTP). Didioxyribonucleoside triphosphates (ddATP, ddCTP etc. or ddNTPs) are also present and incorporated into the growing DNA chain, thus terminating further growth because there is no –OH group for the next nucleotide to bind to. The reaction mixture eventually produces a set of DNAs of different lengths complementary to the template DNA terminating at each of the As, Cs, Gs or Ts depending on which ddNTP was used in the mixture. Four separate DNA synthesis reactions take place, each involving a separate ddNTP with each reaction producing a set of DNA copies that terminate at different points in the sequence. Products are separated by electrophoresis in four parallel lanes of a polyacrylamide gel. The fragments are detected by an incorporated label into the primer or deoxyribonucleoside triphosphates used to extend the DNA chain. In each lane the bands represent fragments that have terminated at a given nucleotide but at different positions in the DNA. By reading off the bands in order the DNA sequence of the newly synthesised strand can be determined.
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