🎧 New: AI-Generated Podcasts Turn your study notes into engaging audio conversations. Learn more

Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...

Full Transcript

BIOL 366: Molecular Genetics Text sections Include all images found in each text section, do not include the sections in the textbook that say “highlight” as part of the deck - 2.4, 6.1, 6.2, 6.3, 6.4, 3.1 7.3 (DNA microarrays only) 7.11 7.2 When making the anki deck can you make sure these learni...

BIOL 366: Molecular Genetics Text sections Include all images found in each text section, do not include the sections in the textbook that say “highlight” as part of the deck - 2.4, 6.1, 6.2, 6.3, 6.4, 3.1 7.3 (DNA microarrays only) 7.11 7.2 When making the anki deck can you make sure these learning objectives are covered - Know the pathway of genetic information flow o DNA to RNA to Proteins (central dogma) 1. Central dogma focuses on how genetic information flows from one pathway to the next 2. Does not discuss RNA to RNA information flow seen in viruses 3. Also does not discuss how RNA mediates the function of DNA that is not turned into anything 4. Reverse transcriptase makes cDNA which is easier to work with than RNA because RNA is very unstable o DNA Replication 1. DNA to DNA 2. Semiconservative meaning that each parental strand separates and gains a daughter strand which makes half of the original DNA conserved in the 2 new DNA strands 3. Procced in the 5’ to 3’ direction o Transcription 1. RNA polymerase II separates the double helix chooses a template strand which makes the non-template strand the coding strand 2. RNA polymerase begins moving down the template strand the 3’ to 5’ direction creating the mRNA strand in the 5’ to 3' direction 3. Transcription factors help RNA polymerase to add the proper base-pairs o - Structure of nucleotides o Nucleotides have a phosphate group, nitrogenous base and pentose sugar o Nucleotides are different from nucleosides because they have a phosphate group o RNA have a ribose sugar that has a hydroxyl group on the 2’ carbon 1. Prefers a C-3 endo structure which gives it a more compact A-DNA structure o DNA has a deoxyribose sugar that is lacking a hydroxyl group on the 2' carbon 1. Makes it prefer a C-2 endo pucker which contributes to the B-DNA structure o Nucleotides can either be purines like adenine and guanine or pyrimidines like cytosine, thymine and uracil o - 4. Once the mRNA strand is made, the RNA polymerase is cleaved off Translation 1. Happens in the ribosome. 2. Assisted by rRNA and tRNA 3. Ribosome has 3 sites labelled E site, P site, and A site 4. tRNA carrying the anticodon for the mRNA enters through the A site 5. The P site is where the growing peptide chain is assembled and once a tRNA is in the A site the peptide chain is moved over to the A site 6. The tRNA in the P site moves over to the E site and is then expelled from the ribosome Cytosine can undergo deamination to turn into uracil Nucleotide-nucleotide interactions in DNA and RNA o Nucleotides base pair to create DNA and RNA o AT and GC are the base pairs o Methylation happens mostly on GC pairs o DNA and RNA can be denatured from nucleotide-nucleotide interactions 1. One is through high pH which turns dsDNA to ssDNA 2. Another example is heating DNA which can cause denaturation • Tm (melting point) is when 50% of the DNA is denatured • It is easier to denature AT base pairs than GC pairs because GC pairs have a triple bond which requires higher energy for the hydrogen bonds to break • Denaturation can be visualized through the hyperchromic effect o Hyperchromic effect is the increase of UV light absorption as DNA denatures from double-stranded to single-stranded o This is due to the fact that double stranded DNA has base-stacking and H bonds that limit resonance and in turn limit UV absorption so if these bonds are broken there is more resonance o o Denatured DNA can anneal back together rather rapidly 1. The first step is complementary sequences find each other through random collisions 2. Second step is unpaired bases find their base pairs and zip the DNA back together When making DNA and RNA nucleotides stack in the 5’ to 3’ direction 1. So a new nucleotide is added on to the 3’OH group of the pentose sugar and binds to the phosphate group on the 5’ OH of the new nucleotide - Structure of DNA and RNA molecules o DNA is B DNA 1. DNA prefers a C2 endo structure 2. This makes it less compact than A DNA 3. B DNA structure is preferred in aqueous environments 4. Its provides a wide major groove and a narrow minor groove (groves are used for easier access of binding proteins) 5. Pretty stable and has 10.5 base pairs per turn 6. Double stranded helix (which makes DNA generally more stable than RNA) o RNA uses A DNA structure 1. A DNA structure makes it more structurally stable than DNA 2. 11 base pairs per turn 3. 2’OH group allows for more binding of proteins 4. Uracil is smaller than thymine so allows for more compact DNA 5. Structure can allow the binding of of metal ions for more compaction 6. Structures are single stranded so use complementary sequences in a singular strand to form secondary structures • Uses atypical base pairing like A-A and G-U • Can form hairpin and internal loops and bulges - Methylation aspects of DNA o Methylation inhibits DNA expression because methylated DNA is less efficiently turned into RNA 1. Can affect gene transcription by physically blocking the binding of transcriptionfacilitating proteins and can also recruit protein that form compact, inactive regions of chromosomal DNA o Methylation in humans usually happens on the cystidine residues of CpG base paired nucleotides 1. Usually happens on the fifth carbon of cystidine o Methylation is not evenly distributed in a DNA molecule as it will mostly be concentrated in the CpG regions o RNA can also be methylated on the 6th carbon of adenine to make m6A 1. M6A is made by methyltransferase enzymes and M6A is associated with diseases like Cancer and affects mRNA production and stability o Enzymatic methylation in DNA happens after DNA synthesis and adds methyl groups to DNA nucleotides o Importance of DNA methylation 1. Chromosome stability 2. Gene expression regulation o - 3. DNA replication 4. Growth and development Abnormal hypermethylation of gene regulatory regions 1. Can be significant in many cancers because it silences the genes that would otherwise be responsible for controlling cell growth DNA synthesis in a living organism o Proceeds in the 5’ to 3’ direction o Require a primer and a template 1. Only one primer needed for the leading strand but lagging strand needs a primer for each Okazaki fragment o dNTPs are added to the 3’ end of the primer, these dNTPs are complementary to the template but the same as the non-template strand o DNA strands have a phosphate on the 5’ end and a hydroxyl group on the 3’ end Synthesis of DNA can also occur from an RNA template which is known as reverse trascriptase o uses RNA dependent DNA polymerase which can synthesize complementary DNA (cDNA) strand - Chemical synthesis of DNA o Happens in the 3’ to 5’ direction o Starts with a nucleotide protected by DMT on the 5’OH end, cyanoethyl on the phosphate and diisopropylamino on the phosphate group as well o This nucleotide is placed on a solid support like an Si bead o The nucleotide is then deprotected at the 5’ end to remove DMT and expose the nucleotide for base pairing o A new nucleotide is introduced and the disopropylamine on the phosphate of the new nucleotide is removed/deprotected o The new nucleotide then makes a phosphite linkage at the 5’ end of the original nucleotide o The phosphite linkage is then oxidated to become a phosphodiester bond o These steps are repeated to grow the strand o Once the strand is made, the bases are deprotected, the phosphates are deprotected and the chain is cleaved off the support. 1. These reactions are very precise since they are automated and are helpful when making probe sequences for southern and northern blotting and are also helpful in DNA sequencing 2. Note that a solvent wash is done after each step of the reaction is performed - Extraction and quantification of DNA/RNA o Denaturation 1. A denatured double helix has no hydrogen bonds holding base pairs together, so the DNA turns into single stranded DNA but does not break the phosphodiester bonds in the nucleic acid 2. Hyperchromic effect 3. Tm: the point at which half the DNA is denatured • Factors that affect melting point are o Salt concentration: stabilize DNA by interacting with the negatively charged backbone requiring more energy to be o DNA concentration: the more DNA = more heat required o DNA length o modifications o DNA can be purified using Qiagen Plasmid Kits these are commercial kits 1. Extraction of DNA 1. Cells must be broken up to release DNA/RNA which can be done through a. lysis using enzyme and alkaline buffers or b. through homogenization (using strong reagents for animals, mechanical grinding for plants or c. by freezing and drying tissues 2. Contaminating cellular materials are then removed a. Cell membranes and insoluble proteins are centrifuged out b. Soluble proteins are removed by precipitation c. RNA is removed by RNase enzymes 3. DNA is then precipitated out with an alcohol like ethanol or isopropanol and also purified and solubilized in an aqueous solution o Purity can be calculated by measuring the absorbance/ optical density OD of the aqueous at 260nm and 280nm 1. The 260/280 absorption ratio is important. For very pure DNA the ratio should be 1.8 and 2 for RNA. Ratios deviating from this show presence of contaminants o You obtain DNA/RNA concentration by using the formula 1. DNA/RNA = EC x OD260 x DF 2. OD260 is the absorbance of the sample at 260nm 3. EC is the extinction coefficient. • The EC for double stranded DNA is 50 ug • The EC for single stranded DNA is 30 ug • The EC for RNA is 40 ug 4. DF is the dilution factor if the sample was diluted. If not diluted the DF is 1 - Understand how to do microarrays. o Microarrays provide snapshots of all the genes in an organism and allow large scale study of gene expression, the microarrays/chips reveal the expression of thousands of genes present on the slide. o Can be used to see which genes are activated and which ones aren’t o There are two types of microarrays 1. cDNA-based microarrays (probe for cDNA) and oligo-based microarrays (probe for primers o Micro arrays procedure for when look at activated vs non activated genes 1. Grow two populations of the cells that you wish to learn more about in different test tubes (2 test tubes), one can is used as a control. 2. Centrifuge both populations to isolate the mRNA and then treat with an extraction buffer to extract the mRNA 3. Move the mRNA into fresh tubes 4. Using reverse transcriptase the mRNA is converted into cDNA. The cDNA are then labelled green and red using fluorescently labelled dNTPs 5. The two cDNAs (green and red) are then mixed together into one tube and will be used to probe the cDNA based microarray 6. The cDNA are loaded on to a chip which contains DNA coding sequences from numerous genes of the species to learn how the conditions required for the gene’s to be expressed 7. Wash excess cDNA 8. Scan the chip with green and red lasers 9. Analyze the results • Green indicates gene expression/activation under the tested conditions • Red indicates gene silencing or that the gene is not expressed in those conditions • And yellow indicates no change in gene expression - Sanger method of ddNTPs chain termination method Why are ddNTPs used in sanger? Because ddNTPs lack an OH group on the 3’ hydroxyl they cannot be extended to form a phosphodiester bond with another molecule o o o o o Type of DNA sequencing where the precise order of nucleotides in a DNA segment can be determined. Sanger sequencing requires the enzymatic synthesis of a DNA strand complementary to the strand under analysis using radioactive primers and dideoxynucleotides(ddNTPs) which are nucleotide analogs ddNTPs interuct DNA synthesis because they lack the 3’hydroxyl group needed for a new nucleotide to add on, this halter synthesis at a particular nucleotide causing a pool of short fragments If it says sequenced from 5’ end then that is the coding strand (so you first have to make the template strand from the top strand and then make the complementary strand which will be identical to the top strand) and if it says sequenced from the 3’ end then that is the template strand that is used as a sequence fragment Methodology 1. Make four beakers each with • • • 4 dNTPs in large quantities (dATP, dCTP, dTTP, dGTP) DNA primers One ddNTP (these are fluorescently labelled nucleotides that will halt synthesis when added on to the growing chain) o Each beaker gets only one ddNTP • 2. 3. 4. 5. 6. Taq DNA polymerase which will withstand high temperatures without denaturing Start with a primer and a nucleotide bond to the 3’ hydroxyl group of the primer dNTPs start extending the chain when ddNTPs are encountered when making the complementary strand, synthesis will stop resulting in products of various lengths which can be used to identify the overall nucleotide sequence • ddNTPs are incorporated into the strand instead of the matching dNTP so ddATP instead of dATP Polyacrylamide gel is used for gel electrophoresis to see the separation of fragments • Note that shorter fragment migrate through the gel faster so they will represent the nucleotides that are closest to the primer (5’ end) and so the sequence is read from bottom to top which is the 5’ to 3’ end Because each ddNTPs is fluorescently labelled with a different colour, DNA sequencing can be identified by identifying the colour sequences as they pass through a detector - Northern vs Southern Blotting o Southern and Northern Blotting are used for detection of DNA and RNA o Southern is for DNA detection which is important when wanting to look at genetic variation o Northern is for RNA detection which is important when wanting to know the different RNA types present in a specific tissue and also to know more about gene expression in different tissues, If genes are transcriptionally active o Works by - 1. Resolve DNA or RNA on agrose gel 2. DNA/RNA is then transferred to a nitrocellulose mebrane 3. A probe is prepared and labelled for detection 4. The probe is allowed to bind and detect the gene of interest through hybridization 5. The signal produced by the probe is detected Learn the principles and applications of next generation sequencing o 2nd Gen Sequencing 1. Pyrosequencing and illumina sequencing 2. High throughput but is more prone to errors 3. Relies on PCR amplification (make more copies of a given fragment) o Pyrosequencing/454 sequencing 1. Nucleotides are added one at a time and nucleotide incorporation generates light for detection 2. Uses pyrophosphates for light emission, pyrophosphate are the two phosphates that get cleaved off once a nucleotide forms a phosphodiester bond with another nucleotide • Light emission occurs when the pyrophosphate reacts with APS to produce ATP • ATP provides energy to luciferase • Luciferase oxidizes lucefirin to generate light 3. Pyrosequencing/ 454 sequencing steps • DNA to sequenced is fragmented into smaller sizes • Adaptor proteins A and B are added to the fragments • Adaptor B binds to a primer that is fixed on to a bead • Fragments are multiplied on beads by emulsion PCR • Complementary strands are washed away • Primers are added to adaptor A • Beads are placed on a wells/plate for pyrosequencing • Pyrosequencing happens one base at a time, the right base added results in light emission for sequencing. Light intensity corresponds to the amount of bases added so higher intensity means more of the same base added • Sequence is complied by computers An issue with pyrosequencing is that the intensity of light is not always accurately proportional to the number of nucleotides in reads so 20 A nucleotides may be read as 19, 20 or 21A o Illumina Sequencing 1. Sequencing is done on a flow cell 2. Uses fluorescently labelled nucleotides 3. Steps • DNA of interest is fragmented • Adaptor oligos are placed on each side of the fragments • All DNA is denatured and applied to a flow cell making the single strands of DNA bind to complementary oligos on the flow cell o Flow cell contains both types of oligos and when the hybridized fragment binds, polymerase creates a complement strand on the flow cell’s oligo primer • The original fragment is washed away • The strands are then clonally amplified (PCR) through bridge amplification o The strand folds over and hybridizes with the second complementary oligo primer o DNA polymerase make a the complementary strand resulting in a double stranded bridge • The double stranded bridge is denatured resulting in 2 singel stranded molecules tethered to the flow cell • This bridge amplification process is repeated over and over and occurs simultaneously • After bridge amplification the reverse strands are cleaved leaving only the forward strand o 3’ ends are blocked to prevent unwanted priming o Sequencing starts when one of the primers starts to be extended by the addition of nucleotides generating a read ▪ Fluorescently tagged nucleotides randomly collide with the strand to form base-pairs, but only one is added at a time based on the sequence of the template ▪ When tagged nucloetides are added to the chain they emit a flourescent light o One the read of the forward is done, the read product is washed away and bridge amplification occurs again and this time the forward strand is cleaved leaving the reverse strand ▪ The same procedure as forward occurs again to make read 2 o Th reads are then analyzed ▪ Poor reads are removed ▪ Reads are overlapped to generate loner sequences ▪ o Illumina uses sequencing by synthesis

Use Quizgecko on...
Browser
Browser