GEN 3000 Study Guide Exam 3
GEN 3000 Study Guide Exam 3 85033 - GEN 3000 - 002
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85033 - GEN 3000 - 002
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This 17 page Study Guide was uploaded by Lisa Blackburn on Sunday March 6, 2016. The Study Guide belongs to 85033 - GEN 3000 - 002 at Clemson University taught by Kate Leanne Willingha Tsai in Fall 2015. Since its upload, it has received 168 views. For similar materials see Fundamental Genetics in Biomedical Sciences at Clemson University.
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Key Terms Imporant People Key Ideas 11 Study Guide Exam 3: Chapters 9-12 Chapter 9: 1. Important People: a. Griffith: Proposed the Transforming Principle i. Transforming Principle: early name for DNA. At the time of this proposed concept, they did not know that DNA is what held the genetic information. However, they knew that something could transform one type of bacterial into another. b. Avery, MacLeod, and McCarty: isolated the transforming bacteria. Determined that DNA is the “Transforming Principle” in bacteria. i. The experiment: Destroyed each component of bacteria that could be the Transforming Principle (RNA, DNA, and proteins) to determine which component was the one that is transforming the bacteria. They destroyed one of the components and left the others alone. If they type of bacteria doesn’t change when the component was destroyed, then this would be the Transforming Principle. They conducted this experiment with a strand of pneumonia. (Explained again and more in depth below in the Key Ideas section.) c. Hershey and Chase: conducted tests to determine the Transforming Principle in viruses, (proteins or DNA). i. The experiment: Resulted in them determining that DNA is the Transforming Principle in viruses. They conducted their experiment with a T2 Bacteriophage and E. coli. They used radioactive labeling to determine which one is the Transforming Principle. (Explained again and more in depth below in the Key Ideas section.) d. Chargaff: isolated DNA and determined the base pairing of DNA, called this Chargaff’s Rule. Disproved the previous belief of Tetranucleotide Theory. i. Chargaff’s Rule: Adenine bonds with Thymine. Guanine bonds with Cytosine by discovering different amounts of these base pairings. ii. Tetranucleotide Theory: DNA has four bases that are in a fixed sequence (base pairing) and of equal amounts. e. Watson and Crick (Wilkins and Franklin): Franklin used X-rays to take photographs of DNA, a process called crystallography. The photo she took was then given to Watson and Crick (by her boss Wilkins). This was the missing piece in Watson and Crick’s work. They were then able to build the first model of DNA. 2. Key Ideas: a. Nucleic Acids: Can be DNA or RNA i. Nucleotide: this is the basic component of DNA. 1. Cons ists of: consists of a sugar, phosphate, and a base. 2. For DNA: the sugar is deoxyribose; the bases are adenine, thymine, guanine, and cytosine 3. For RNA: the sugar is ribose; the bases are adenine, uracil (replaces thymine), guanine, and cytosine. ii. Purine: a nitrogenous base that has two rings and is bigger in size. 1. Consists of: adenine and guanine. iii. Pyrimidine: a nitrogenous base that only has a single ring and is smaller in size. 1. Consists of: Cytosine, Thymine/Uracil iv. Nucleotide Components: consists of a sugar, phosphate group, and a base. 1. Bases: A, T/U, G, C 2. Sugar: deoxyribose or ribose (for DNA or RNA respectively) 3. Phosphate group: gives the negative charge. Consists of a phosphate surrounded by oxygen. a. Monophosphate: is when a single phosphate group is attached to a sugar b. Phosphodiester Bonds: the bond between a phosphate group and a sugar of different nucleotides. Very strong bond b. Transforming Principle: A concept that Griffith proposed. i. Griffith’s experiment: 1. Used a strain of bacteria that causes pneumonia. A virulent strain has a smooth coat (the lethal strain). An avirulent strain has a rough/lack of coat (the nonlethal strain). 2. Griffith noticed that if a mouse gets injected with a lethal smooth strain that has be heat killed, the mouse does not die. 3. Griffith also noticed that if a mouse gets injected with a mixture of lethal smooth strain and nonlethal rough strain, the mouse dies. 4. Why? He proposed that something in the mixture “transforms” the rough strain into a smooth strain, which is what causes the mouse to die. He called this the Transforming Principle. However, he did not know what was doing the transformation. ii. Avery, Macleod, and McCarty’s experiment: Wanted to determine the Transforming Principle of bacteria. 1. They used the same strains of pneumonia that Griffith used (the smooth lethal strain and the rough nonlethal strain) 2. Took the filtrate of the lethal smooth strains (they knew that something within this filtrate is the Transforming Principle) and destroyed individual components in separate experiments to determine what was doing the transformation. (The components were DNA, RNA, and proteins.) 3. In the first experiment, they destroyed RNA within the filtrate through the use of RNase. They then put this filtrate in with the nonlethal rough strain. This resulted in a mixture that they then injected into a mouse to see if the mouse would live or die. The mouse died, meaning that the rough strain transformed into the smooth lethal strain. RNA is NOT the Transformation Principle. 4. In the second experiment, they destroyed proteins within the filtrate through the use of protease. They then put this filtrate in with the nonlethal rough strain. This resulted in a mixture that they then injected into a mouse to see if the mouse would live or die. The mouse died, meaning that the rough strain transformed into the smooth lethal strain. The proteins are NOT the Transformation Principle. 5. In the third experiment, they destroyed DNA within the filtrate though the use of DNase. They then put this filtrate in with the nonlethal rough strain. This resulted in a mixture that they then injected into a mouse to see if the mouse would live or die. The mouse lived, meaning that the rough strain did not transform into the smooth lethal strain. DNA IS the Transformation Principle. 6. Conclusion: When DNA was destroyed, the filtrate of smooth lethal strain did not transform the nonlethal rough strain. Resulting in the mouse living. This means that the DNA is what “transforms” the rough strain, therefore, it is the Transformation Principle. 7. Issues with this study: though the data backed up this claim, many people did not want to believe DNA could be the Transforming Principle. This is due to the belief that DNA was not complex enough to be the Transforming Principle. The team had to replicate the experiment many times to get people to believe the data. iii. Hershey and Chase: Wanted to determine the Transforming Principle of viruses. 1. Knows that something happens within a cell to allow for replication of a virus to take place when a virus infects a cell. Proteins or DNA is what causes this “transformation.” 2. Proteins have sulfur in them and DNA has phosphorus in it. 3. Took the virus and labeled the sulfur/phosphorus of the proteins/DNA as radioactive. Allowed for the viruses to infect an unlabeled nonradioactive E. coli cell. The virus multiplied and then it was blended, and then placed in a centrifuge. The heavy portion will sink to the bottom while the lighter portion will go on top. a. For protein/sulfur: the radioactivity was on top within the protein coats and not in the phage that was produced. b. For DNA/phosphorous: the radioactivity was on the bottom within the phage that was reproduced and not on tope with the coats. 4. Conclusion: DNA is what makes the next generation of viruses, therefore, it contains genetic material (it is the Transformation Principle). iv. Eukaryotic Evidence: this was done to support that DNA contains the genetic information. 1. Was indirect evidence 2. UV light causes mutations in heredity, meaning it can change heredity components. 3. Whatever is absorbing the UV light at a certain wavelength is what is being mutated (is what contains the genetic information). 4. Nucleic acids and proteins both absorb UV light, but nucleic acid absorbs UV light at the same wavelength that UV light causes genetic mutations. 5. Conclusion: DNA contains genetic information/is the Transforming Principle. c. Chargaff’s Rule: Chargaff established that Adenine binds with Thymine and Cytosine binds with Guanine. He also disproved the Tetranucleotide Theory. i. Looked at the ratios of nitrogenous bases of different organisms. Noticed a 1:1 ratio between bases in E. coli. ii. Noticed that this ratio changed as the organism changes. 1. Conclusion: Tetranucleotide Theory cannot be true, since the ratio of the bases is not constant in all organisms. iii. Noticed a common ratio between Adenine to Thymine and Guanine to Cytosine. 1. Conclusion: This means that for every Adenine there is a Thymine and for every Cytosine there is a Guanine. Therefore: A bonds with T and G bonds with C. d. Franklin/Wilkins and Watson/Crick: with the research of both people, the first model of DNA was created. i. Franklin’s photograph showed the required width that DNA would have to be. ii. Watson and Crick then used pyrimidine and purines to determine what pairing is being down to reach this required width. 1. Paired a pyrimidine with a pyrimidine and resulted in the DNA not being thick enough. 2. Paired a purine with a purine and resulted in DNA being too thick. 3. Paired a purine with a pyrimidine and resulted in DNA being the right width. 4. Remember: Chargaff’s Rule states that A bonds with T (Purine bonds with Pyrimidine) and G bonds with C (Purine bonds with a Pyrimidine) iii. Conclusion: they were able to find out the way DNA is structured and built the first model, their model is based on B-form DNA. (Discussed down below). Also, they proved that Chargaff’s Rule is correct. iv. Alpha Helix: the “backbone” (proteins and sugar) wraps around the nitrogenous bases, does not twist. e. DNA Characteristics: i. Bonds of DNA: 1. Phosphodiester Bonds: the bond between a phosphate group and a sugar of different nucleotides on a DNA strand. This is a very strong bond that makes up the “backbone” of DNA. a. These are hydrophilic bonds, meaning that they are water loving. So they will wrap around the bases bonds, since these are hydrophobic. 2. Base Pairing Bonds: the bonds formed between base pairs are hydrogen bonds. These are weak bonds that are hydrophobic, meaning water fearing. a. Cytosine and Guanine: cytosine and guanine bond together forming 3 hydrogen bonds. b. Thymine and Adenine: thymine and adenine bond together forming 2 hydrogen bonds. c. The C-G Bond is stronger than the T-A Bond! i. This is due to the greater amount of hydrogen bonds in C-G. ii. Directionality of DNA: DNA is antiparallel 1. Antiparallel: this means that the strands run parallel to each other but in opposite directions. 2. DNA is read from 5’ to 3’ a. So, one strand is going 5’ to 3’ while the antiparallel strand is going 3’ to 5’. However, DNA is still read in the 5’ to 3’ direction (more on this subject in a later chapter). 3. Re-association: DNA wants to re-associate, meaning that when one strand is by itself, it wants to pair its bases back up. a. Reassociation Kinetics: How quickly can DNA renature? i. Double stranded DNA is denatured through heat, causing the strands to separate and become single stranded DNA 1. Breakage of the hydrogen bonds between base pairs, the phosphodiester bonds remain intact, meaning the that backbone does not break. This results into a single stranded DNA ii. Single stranded DNA is allowed to cool, causing the DNA to want to renature into double stranded DNA. 1. The DNA wants to snap back together (re- associate) iii. The time it takes for DNA to renature into double stranded DNA is measured to determine bases within the DNA iv. Since there are 2 hydrogen bonds between A-T, then these bonds can be broken faster/easier than G- C, since these bases have 3 hydrogen bonds. 1. The time it takes for A-T bonds to break is shorter and can be done at a lower temp. than G-C. v. The graph that is produced from DNA being denatured to being renatured will have three gradual decreasing bumps. 1. First bump: is highly repetitive, meaning it will have a lot of copies of a base, therefore, the renature of this segment can be done quickly. a. Why? Since it is highly repetitive, then there are a lot of areas that will match this, it doesn’t have to be an exact copy. 2. Second bump: moderately repetitive, meaning it will have repeats, but not as much as the first one. Can be still renatured quickly, but not as quick. a. Why? Same as above, but the reason for it being slower is due to needing close to exact copies. 3. Third bump: unique, meaning that it only has one copy (no repeats). This takes time to be renatured. a. Why? There is only one copy that will match with it. 4. DNA Structure Types: a. B-form DNA: The DNA structure that Watson and Crick made a model for. i. Exists in the presence of water. ii. Most stable of the DNAs under physiological conditions iii. Alpha helix (right-handed) 1. Wrapping motion, not twisting iv. The DNA that you commonly think of v. Most important b. A-form DNA: Located outside of the cell. i. When the environment changes, the DNA structure can modify. ii. Shorter/wider than B-form iii. Alpha helix (right-handed) iv. Most likely does not exist in nature c. Z-form DNA: dramatically changed structure i. Found inside the cell ii. Left handed helix iii. Zig-zag backbone iv. Sites of active genes can make Z-DNA 1. Regions of B-DNA can change to Z-form. 2. There are two theories behind this: a. Protein binding to DNA puts tension on the outside, Z-form releases some of this pressure b. B-form changes to Z-form to let a protein attach f. Central Dogma: the general flow of information within a cell i. ii. Generally, the above depiction is the flow of information from within a cell. iii. It starts with DNA, where it can be replicated or transcribed. Once transcribed it goes to being RNA, where it is translated and creates a protein. Chapter 11: Chromosome Structure and Organization 1. Key Topics: a. DNA within Prokaryotes: This DNA is DNA without free ends (circular), allowing for this DNA to be supercoiled to take up less space. i. Overrotate: the DNA adds two extra turns within the DNA itself. 1. Results in a Positive Supercoil: look at the way the overlapping is in the depiction of the positive supercoil. 2. ii. Underrotate: The DNA takes away two turns within the DNA itself. 1. Results in a Negative Supercoil: look at the way the overlapping is in this depiction of negative supercoil. 2. a. Most DNA is negatively supercoiled iii. Twisted loops of DNA are then attached to proteins to further organize the DNA and too keep it from tangling up b. Chromatin of Eukaryotes: i. Chromatin: complex of DNA and proteins in eukaryotic chromosomes. Proteins are always interacting with DNA. During interphase, DNA is called chromatin because of the proteins interacting with the DNA. ii. Heterochromatin: a highly condensed chromatin. Such as an inactive X chromosome (Barr body). Highly condensed and turned “silent/off” iii. Euchromatin: chromatin can be transcribed active or “open” chromatin 1. DNA being uses is called active/open. Certain segments of chromatin can be active/open even when condensed, they need to be available to interact with proteins. This part is called Euchromatin. iv. Order of Packaging: 1. DNA wraps around histones to form nucleosomes a. Protects DNA b. Exists in repetitions, every so many base pairs it happens c. “Beads-on-a-string” d. DNA between each histone is called the “linker DNA” e. Nucleosomes are made of 4 different types of histones with two copies of each (8 histones total) 2. Nucleosomes coil together to form a solenoid 3. Loop domains 4. Chromatid a. End result of DNA compaction, most condensed 5. DNA needs to be accessed even when compacted c. Chromatin Remodeling: structure must change to allow access to DNA, compacted DNA needs to be accessed when needed i. Histone tails: the targets for binding to access DNA segments 1. Histones are positively charged and DNA is negatively charged 2. Acetylation: neutralizes positive charge of histones, which results in the relaxation of histone hold. a. DNA can be accessed 3. Methylation and Phosphorylation: increases the charge of the histones, which tightens the hold the histones have on DNA a. DNA is closed down and cannot be accessed. d. Chromosome Banding: allows quick identification of chromosomes i. Is a “finger print” of chromosomes ii. The band order for a particular chromosome in a species is the same for every member of that species. Example: chromosome 1 in humans has the same banding for every human iii. Allows you to identify specific things within a chromosome, it names the regions. iv. Banding is the same for chromosomes of homologous pairs, allows for matching of pairs e. Specialized Chromosomes: Polytene Chromosome i. Is rare. It is only seen in certain cells of certain tissues, at a certain time. ii. By studying the rare/specialized chromosomes, we can understand the normal ones iii. This chromosome can be seen in interphase (normal chromosomes cannot) iv. The Chromomeres are the individual lines of the Polytene v. The puff is a region of the Polytene that is relaxed and pulled out. 1. Allows for a ton of product to be made quickly 2. High amount of gene activity (transcription) occurs here f. Centromeres: i. Point Centromere: a small centromere ii. Regional Centromere: a large centromere 1. In most plants and animals iii. CEN region: Critical regions. 1. Without this region, the centromere will not work, meaning that spindle fibers will not attach iv. Centromeres are very repetitive and is the same for all other chromosomes within a species. 1. Since centromeres are relatively the same, if you try to stain one, all get stained g. Telomere: the caps at the ends of chromosomes to prevent unraveling i. Replication of ends can be tricky to accomplish 1. Does not occur in somatic cells, or they will be shortened until death. 2. Single celled organisms and germ cells do not have to worry about this ii. Telomeres have to be lengthened to prevent death iii. Problems: 1. Replicative enzymes cannot replicate the ends of chromosomes 2. Chromosomes would get shorter to the point of death 3. Solution: have an enzyme that replaces the ends iv. Telomere repeats can be very big 1. TERRA: (Telomeric repeat containing RNA) v. Telomerase: allows for elongation of telomeres h. Repetitive Sequences: i. Transposable sequences: “jumping” genes. Segments that can move within a genome ii. SINE: short interspersed element 1. Alu family: must common sine within the human genome iii. LINE: long interspersed element iv. Retrotransposons: RNA intermediate using reverse transcriptase 1. Exception of central dogma v. Pseudogenes: risen through duplication, but no longer functional Chapter 10: DNA Replication and Recombination 1. Important People: a. Messelson and Stahl: i. The Experiment: 1. Placed E. coli into N medium and spun 15 a. The N is heavy and the original DNA was located on the bottom 2. Transferred to N medium and allowed the DNA to replicate and then spun a. All of the DNA was in the middle, due to containing N14 15 and N 3. Allowed for replication to occur in N medium again and spun a. ½ of the DNA was in the top (where N would be found) and ½ of the DNA was in the middle. 14 4. Allowed for replication to occur in N medium again and then spun 5. ¾ of the DNA was in the top and ¼ of the DNA was in the middle ii. Conclusion of experiment: 1. Semiconservative was proven correct a. 2. Conservative was disproven in the first generation created a. 3. Dispersive was disproven in the second generation created a. 2. Important Topics: a. Replication: i. Theta Replication: occurs within circular DNA 1. Chromosome has to have an origin of replication 2. Where the chromosome starts to unwind 3. Replication bubble: the area around where the origin opens up a. The bubble expands in both directions (bidirectional) ii. Rolling-Circle Replication: occurs within Circular DNA 1. Has to have an origin, cuts one strand and then the strand is peeled away 2. Replication occurs of the remaining strand 3. The peeled segment reforms into a circle and is replicated 4. There is no replication bubble because the strands just peel away 5. Replicated in one direction (unidirectional) iii. Linear DNA Replication: can have multiple origins 1. Common in Eukaryotes 2. There is a lot more DNA to replicate within Eukaryotes to replicate a. More complex so the machinery is slower 3. Multiple origins can be on one chromosome 4. Multiple replication bubbles a. Bidirectional b. The bubbles continue to open until they meet another bubble b. DNA replication requirements and steps: i. Requirements: 1. Single Stranded DNA template 2. dNTPs: all of the nitrogenous basses 3. Triphosphate: three phosphate groups attached to a sugar molecule a. When the phosphate is cut off, there is a release of energy. This allows for the base to be added ii. Replication fork: where the double strand splits into two single strands 1. DNA is only replicated from 5’ to 3’ 2. DNA strands are antiparallel, therefore bases are added in opposite directions on the strands a. Leading Strand: continuous DNA synthesis, moves as the replication fork moves forward b. Lagging Strand: discontinuous DNA synthesis, has fragments of DNA called Okazaki Fragments i. These fragments are created because of how DNA has to be synthesized from the 5’ to 3’ direction ii. Has to wait for the replication fork to move forward before it can place down more bases 1. iii. Linear and Theta replications have leading and lagging strands. Rolling-circle replication only has a leading strand c. Mechanisms of Replication: Bacterial DNA Replication (Eukaryotic is the same process essentially) i. Requires initiation: needs to be able to find the origin 1. Initiator proteins: bind to origin and twists up the DNA to release tension. This allows for the Replication Bubble to open up. ii. Unwinding: the DNA needs to be unwound 1. Helicase: unzips DNA by breaking the hydrogen bonds of base pairs a. Cannot initiate unwinding, moves as the replication bubble opens, moves with fork and causes tension 2. Single strand Bonding proteins: bind to single strand bases, this prevents the DNA from snapping back into place 3. DNA gyrase: a type of topoisomerase, reduces the tension that builds up as the replication fork moves, prevents the DNA from supercoiling, located outside the replication bubble iii. Priming: 1. DNA polymerase: responsible for DNA synthesis a. Cannot initiate DNA synthesis requires a 3’-OH group to be present 2. Primase: an RNA polymerase a. A primer: that puts down the initial 3’-OH with RNA. b. Lagging: needs primers for each segment c. Leading: needs one primer iv. Synthesizing: DNA polymerase can now synthesize 1. E. coli has 5 DNA polymerases: a. DNA polymerase I: Not critical polymerase i. Also known as Kornberg Polymerase ii. has 5’ to 3’ activity (building DNA) iii. has 3’ to 5’ exonuclease activity (correction of errors) iv. has 5’ to 3’ exonuclease activity (removes primers) b. DNA polymerase III: critical polymerase i. Has 5’ to 3’ activity (building DNA) ii. Has 3’ to 5’ exonuclease activity (correction of errors) c. Holoenzyme: complex, multiple proteins come to create enzyme i. Just know that multiple proteins come to form polymerase III v. General order of replication in bacterial cells: 1. Initiator proteins 2. DNA helicase 3. Single-stranded-binding proteins 4. DNA gyrase 5. DNA primase 6. DNA polymerase III (elongates a new nucleotide strand from 3’- OH group provided by primer) 7. DNA polymerase I (removes RNA primers and replaces them with DNA) 8. DNA ligase (joins Okazaki fragments and seals the breaks) vi. Why does replication lack mistakes? 1. DNA polymerases are choosy: they usually pick the correct nucleotide 2. Insertion of the wrong nucleotide leads to incorrect positioning of 3’-OH and stalls polymerase a. A type of “red flag” if not in the right place 3. Mismatch repair: fixes any errors after replication a. Double checks after polymerase is done and fixes errors d. Termination of DNA synthesis: i. Occurs when two forks meet, this causes termination ii. Sequences in some systems bind a termination protein and blocks helicase 1. The presence of the protein=the termination e. Eukaryotic DNA Replication: i. Overall is the same as for prokaryotes, just more complicated ii. Differences with bacterial replication: 1. Multiple replication events at the same time on the same chromosome 2. More polymerases: large variety of DNA polymerases 3. Nucleosome assembly makes it messy. a. chromatin structure and how machinery gets through 4. Linear chromosomes: replication of the ends of chromosomes iii. Eukaryotic Origins: identical sequences, initiator proteins find the origins 1. ARS: autonomously replicating sequences, enable the DNA to replicate iv. Licensing of DNA replication: 1. How do we know if we started replication already at an origin? a. Replication licensing factor: attaches to the origin then and only this one time will the initiator proteins function. A type of pre-initiator protein v. Unwinding: still uses single stranded DNA binding protein, but does not have DNA gyrase (still has a topoisomerase, just not gyrase) vi. DNA polymerase: function in replication, recombination and repair vii. Issue with Telomeres shortening: 1. Telomerase is a ribonucleoprotein: it binds to the end of chromosomes to add telomere f. Replication Fork on Eukaryotes i. If synthesis occurs on both strands at the same time, then two DNA Pol II are required 1. Replication is a continuous process 2. As the bubble opens, replication occurs on both strands 3. Each Fork requires: a. Helicase, single stranded DNA binding proteins, DNA gyrase, DNA primase, DNA polymerase 4. Two Poly III work together on one fork, they are held together a. Issue: the leading strand is in the forward direction while the lagging strand is in the reverse direction. b. DNA is looped for the lagging strand so that the pol III can move in the same direction Chapter 12 Study Guide: 1. Important Topics: a. RNA vs. DNA i. RNA is usually single stranded 1. Can acquire shape, meaning that it can fold upon itself. You can see base pairing within RNA 2. Has a ribose sugar (the sugar contains two OH groups). This makes RNA less stable than DNA 3. Ribonucleotides contain uracil instead of thymine. Uracil is still a pyrimidine 4. RNA can be catalytic ribozymes a. DNA is only carrying genetic information, it is not acting out any jobs Characteristic RNA DNA Composed of nucleotides Yes Yes Type of Sugar Deoxyribose Ribose Presence of two OH groups No Yes Bases A, G, C, T A, G, C, U Nucleotides joined by phosphodiester bonds Yes Yes Double or single stranded Usually Double Usually Single Secondary Structure Double Helix Many Types Stability Stable Easily degraded b. Types of RNA i. There are multiple types of RNA. We focus on rRNA, mRNA, and tRNA 1. rRNA: Ribosomal RNA. a. Function: structural and functional components of ribosomes 2. mRNA: Messenger RNA a. Function: carries genetic code for proteins 3. tRNA: Transfer RNA a. Function: helps incorporate amino acids into polypeptide chain c. Transcription: very selective process. Only bits of the genome are ever transcribed into RNA. Needs to be able to identify which parts are to be transcribed i. Requirements: 1. DNA Template (needs to be able to access the DNA) 2. Substrates to make RNA (ribonucleotide triphosphates) a. Loses two phosphates and provides enough energy 3. Transcription machinery ii. Directionality: 1. Transcription only occurs on one strand in one section, meaning that there can be many sections being described in different directions, due to it not occurring on both strands. a. But it still occurs in the 5’ to 3’ direction while reading the strands from 3’ to 5’ iii. Template vs. Nontemplate strands 1. Template Strand: this is the DNA strand that is being actively read/transcribed 2. Nontemplate Strand: This is the DNA strand that is not being read by the polymerase, the ignored strand a. Also known as the coding strand: the coding strand will look like the mRNA created, only difference is that where the coding strand has a T, the mRNA has a U. iv. Transcriptional Unit: 1. The stretch of DNA that codes for an RNA molecule and the sequences necessary for transcription 2. Upstream: anything to the left of the transcription site 3. Downstream: anything to the right of the transcription site 4. Key parts: a. Promoter: DNA sequence that the machinery that does transcription recognizes. This indicates which strand of DNA will be described, determines start site, and usually not transcribed i. If the promoter is damaged/messed up, the gene is essentially useless b. RNA coding sequence: this is the template, the area that is to get transcribed c. Terminator region: the spot that transcription will stop at. Says that enough has been transcribed and can be stopped now. v. RNA Polymerase: takes care of everything 1. Bacterial RNA polymerase: usually only one type of polymerase, synthesizes all classes of RNAs (mRNA, tRNA, rRNA, etc.) a. Four key subunits to CORE enzyme: do not need to know the four parts b. Sigma Factor: added to core subunits=holoenzyme i. This will initiate transcription at a promotor ii. Different sigma factors direct initiate at different promotors iii. This gives polymerase specificity, directs it to which genes are to be transcribed iv. If needs to transcribe rRNA, then the sigma factor identifies the rRNA promotor 2. Eukaryotic RNA polymerase: multiple kinds. a. They are the result of large multiprotein complexes b. RNA polymerase II: premRNA become mRNA vi. Bacterial Transcription: three major stages 1. Initiation: machinery assembles on promoter, begins the synthesis of RNA. Identification of promoters a. Promoter recognition/binding: key to determining frequency that a gene is transcribed i. Downstream: positive bases ii. Upstream: negative bases iii. +1: first gene that is placed down iv. 10 (Pribnow box): mostly A and T, few Hbonds, a good place for transcription bubble b. Formation of Transcription Bubble c. Generation of first bonds between ribonucleotides d. Escape of machinery from the promoter i. RNA polymerase does not need primers ii. Sigma subunit will not let go of promoter iii. CORE enzyme has to “escape” promoter by kicking out the sigma subunit 2. Elongation: RNA polymerase reads DNA, adds ribonucleotide to growing RNA, this makes the RNA a. RNA Polymerase: unwinding of DNA ahead of transcription bubble and rewinding behind it b. 5’ to 3’ extensions of RNA c. DNA/RNA duplex within bubble i. RNA that is already transcribed trails off behind the RNA polymerase, it can fold upon itself 3. Termination: end of transcription, separation of RNA from DNA template, the stopping of transcription a. RhoIndependent termination: terminator will be transcribed. Contains inverted repeat sequences, when this is transcribed it halts the process, during this halt, the bonds break, freeing the mRNA b. RhoDependent Termination: requires the protein Rho i. The transcribed segment trails behind the polymerase, Rho slides up as the transcription continues. When the terminator is transcribed, it halts the process, allowing for Rho to catch up. This knocks off the mRNA
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