Tuesday's Notes; Chapter 13 and 14
Tuesday's Notes; Chapter 13 and 14 30156
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This 7 page Class Notes was uploaded by Hannah Kennedy on Sunday August 7, 2016. The Class Notes belongs to 30156 at Kent State University taught by Dr. Helen Piontkivska in Spring 2016. Since its upload, it has received 9 views. For similar materials see ELEMENTS OF GENETICS in Biological Sciences at Kent State University.
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Date Created: 08/07/16
8/2/16 Lecture—Ch 13 and 14 Ch. 13—DNA replication and recombination 1. Models of replication a. Conservative b. Dispersive c. Semiconservative replication i. Each new synthesized double stranded DNA molecule has 1 original parental strand and 1 newly synthesized strand 2. Circular DNA replication—e.g. circular chromosome of E.coli a. ORI = site on the bacterial chromosome where DNA synthesis begins; synthesis of new daughter strands proceeds bidirectionally (~245 nucleotides long) i. 3 types of DNA sequences are found within this 1. AT-rich region—2 strands separate here 2. DnaA box sequences—serve as recognition sites for the binding of the DnaA proteins a. Bind DnaA proteins to initiate the actual replication 3. GATC methylation sites—help regulate the replication process ii. 3 tandem repeats required for the formations of the replication bubble iii. more genes are located on the leading strand rather than the lagging strand to avoid detrimental mutation errors b. replication fork = the site where the parental DNA strands have separated and new daughter strands are being made (2 forks move in opposite directions outward from the origin) 3. Major players of replication a. Double stranded DNA b. Replication bubble and ssBPs i. as soon as the ds DNA is unwound, it forms segments of ss DNA so to avoid the ss DNA from being digested, single-stranded binding proteins bind to keep bubble open and intact and kept from being taken up ii. Single-stranded binding proteins (i.e. accessory proteins) are needed to keep the bubble open c. Helicase—unwinds the ds DNA molecule that sits at the top of the replication fork, breaks H bonds d. DNA polymerase—catalyzes the formation of covalent bonds between adjacent nucleotides to make new daughter strands i. DNA polymerase as an enzyme has a limitation because it can only prime things in the 5’ to 3’ direction and it needs something to start off of (i.e. RNA polymerase offers a 3’ end so that DNA polymerase can begin to add nucleotides) 1. RNA polymerase has a 5’ to 3’ limitation but, unlike DNA polymerase, it can start from scratch and doesn’t need a starting point for itself ii. Different DNA polymerases 1. DNA polymerase I excises RNA primers and fills in with DNA with deoxyribonucleoside triphosphates dNTPs 2. DNA polymerases II, IV, and V are involve in DNA repair 3. DNA polymerase III is responsible for most of DNA replication a. Catalyzes the attachment of nucleotides to the 3’ end of each RNA primer b. Processive enzyme = characterizes DNA polymerase III because it doesn’t dissociate from the growing strand after it has caralyzed the covalent joining of 2 nucleotides e. RNA primase—synthesizes an RNA primer for RNA polymerase i. RNA primers are short strands of RNA that are made by the linkage of ribonucleotides via primase 1. 1 primer is made for the leading strand 2. multiple primers are made for the lagging strand f. Origin of replication—ORI i. This is AT rich because there are less H bonds between them and they are easier to reach g. Replication fork h. Gyrase aka topoisomerase—prevents from supercoiling i. Travels in front of helicase i. Ligase—covalently links okasaki fragments together 4. Antiparallel replication—leading and lagging strands a. Occurs in an in antiparallel fashion because 1 strand runs 5’ to 3’ and the other runs 3’ to 5’ i. Because of this and because DNA polymerase can only synthesize in the 5’ to 3’direction, we have a leading strand and a lagging strand 1. Leading strand a. RNA primer is made at origin and DNA polymerase III adds nucleotides in the 5’ to 3’ direction sliding towards the fork b. Continuous replication 2. Lagging strand a. Multiple RNA primers are made so that DNA polymerase III can add nucleotides in the 5’ to 3’ direction away from the fork b. To complete okasaki fragments, RNA primers must be removed via DNA polymerase I, DNA must be made where the primers have been removed, and covalent attachement of adjacent fragments of DNA must happen c. Made discontinuously in okasaki fragments d. Fragments needs to be ligated therefore ligase tapes them together e. Accumulates more mutations than the leading strand due to an increased risk of mismatch b. Process of the synthesis of DNA at the replication fork: i. DNA strands separate at the origin and create 2 replication forks ii. Synthesis of the leading strand occurs in the same direction as the movement of the replication fork iii. The first okasaki fragment of the lagging strand is made in the opposite direction iv. The leading strand elongates v. 2 ndokasaki fragment is made vi. lrdding strand elongates st nd vii. 3 okasaki fragment is made and the 1 and 2 are connected together 5. DNA replication: proteins a. Eukaryotic replication i. ORI 1. Have multiple origins of replication in which groups of them are activated at the same time (euchromatin site start early and heterochromatin regions replicate later) a. Allows DNA to be replicated in a reasonable length of time ii. Linear chromosomes 1. DNA replication occurs bi-directionally from many ORI during S phase iii. Genomes are composed of multiple chromosomes iv. Larger DNA content of the cells when compared to bacteria v. Multiple replication origins vi. Many diff types of polymerase enzymes 1. Lesion-replicating polymerases = DNA polymerases that are attracted to damaged DNA and are able to synthesize the complementary strand over the abnormal region 2. Flap endonuclease = removes RNA primers vii. Special process to replicate the chromosome ends via telomerases 1. Telomere = the telomeric sequences within the DNA and the special proteins that are bound to these sequences (found at the end of linear chromosomes) 2. Telomerase = prevents chromosome shortening by recognizing the sequences at the ends of eukaryotic chromosomes and synthesizing additional repeats of telomeric sequences a. Challenge of replicating telomeres i. Problem: we have a 5’ to 3’ restriction. Can’t link together the first 2 individual nucleotides and it can only elongate only pre-existing strands. 3’ end of the linear chromosome can’t be replicated via DNA polymerase bc a primer can’t be made upstream therefore chromosome would become shorter with each round of DNA replication if the problem wasn’t solved ii. Solution: telomerase: part protein part RNA molecule w built in template that allows the generation of the chromosome ends 1. RNA part has a sequence complementary to DNA sequence in telomeric repeat to allow telomerase to bind to 3” overhang of telomere and synthesize nucleotides at end of DNA strand viii. Assembly of newly replicated DNA into nucleosomes 1. Chromatin is tightly packed into nucleosomes ix. Replicates with high fidelity (i.e. accuracy) 1. H bonding between the appropriate bases are more stable than between mismatched pairs 2. Active site of DNA polymerase preferentially catalyzes attachment of nucleotides when the appropriate bases are across from each other 3. DNA polymerase enzymatically removes mismatched nucleotides a. Proofreading = the mechanism by which DNA polymerase is able to remove mismatched bases Ch. 14—Gene transcription and modification 1. Central dogma of molecular biology a. DNA to mRNA to protein via replication, transcription, translation i. DNA replication makes DNA copies that are transmitted from cell to cell and parent to offspring ii. Transcription produces an RNA copy of a gene 1. mRNA is a temporary copy of a gene that contains information to make a polypeptide iii. translation produces a polypeptide using the information in mRNA iv. the polypeptide becomes part of a functional protein that contributes to an organism’s traits 2. Transcription a. Organization of sequences of a gene and its mRNA transcript i. DNA 1. Regulatory sequences = site for the binding of regulatory proteins a. Regulatory proteins = influence the rate of transcription 2. Promoter = site for RNA polymerase binding; signals the beginning of transcription 3. Terminator = signals the end of transcription ii. mRNA 1. ribosome-binding site = the site for ribosome binding; translation starts near this site (ribosome scans mRNA for a start codon) 2. start codon = specifies the first AA in a polypeptide sequence, usually a Met in eukaryotes 3. stop codon = specifies the end of polypeptide synthesis b. Major players i. RNA polymerase—moves along the DNA strand and causes it to unwind ii. DNA template (i.e. anti-sense strand)—used as a template for RNA synthesis iii. Ribonucleotides—RNA polymerase always connects them in the 5’ to 3’ direction c. 3 steps i. initiation—the promoter functions as a recognition site for transcription factors which enable the RNA polymerase to bind to the promoter; following binding, the DNA is denatured into a bubble (i.e. the open complex) 1. gives RNA polymerase a starting point 2. promoters = the minimum DNA sequences needed for basal transcription; promotes gene expression (3 features found in most promoters) a. transcriptional start site (core promotor) = short DNA sequence necessary for transcription to take place (3 categories of proteins needed for basal transcription) i. RNA polymerase II—initiates transcription ii. General transcription factors—needed for RNA polymerase II to initiate transcription of structural genes (e.g. mRNA) iii. Mediator—mediates interactions between RNA polymerase II and regulatory transcription factors b. regulatory elements = short DNA sequences that affect the ability of RNA polymerase to recognize the core promoter and begin transcription; recognized by transcription factors (2 categories) i. enhancers = activating DNA sequences involved with the alteration of the basal transcription rate (distal promoter element) ii. silencers = DNA sequences that inhibit transcription c. TATA box = proximal promoter element that determines the precise starting point for transcription ii. elongation—RNA polymerase slides along the DNA in an open complex to synthesize the RNA 1. RNA polymerase synthesizes mRNA in the 5’ to 3’ direction from the DNA template 2. Synthesis of the RNA transcript a. RNA polymerase slides along the DNA and creates an open complex as it moves b. The DNA strand known as the template is used to make a complementary copy of RNA c. RNA polymerase moves along the template strand in a 3’ to 5’ direction and RNA is made in the 5’ to 3’ direction using nucleoside triphosphates as precursors d. U is substituted for T iii. termination—a terminator is reached that causes RNA polymerase and the RNA transcript to dissociate from the DNA 1. Terminator = specifies the end of transcription 3. RNA modifications a. Exons = the coding sequences that are contained within mature RNA b. Introns = intervening sequences = sequences found between exons that are removed from the genetic code i. are of different lengths and different densities per gene ii. accounts for the enormous genetic diversity in humans iii. splicing = process that removes introns and stitches together exons in a mature RNA molecule 1. self-splicing = splicing that doesn’t require the aid of other catalysts 2. spliceosome = multicomponent structure that recognizes the intron boundaries and properly removes it; splices pre-mRNA a. made of snRNPSs = snurps = several subunits that contain snRNA and a set of proteins b. process of spliceosome: spliceosome subunits bind to intron sequence and recognize boundaries between introns and exons. Spliceosome holds pre-mRNA in right position for correct splicing, spliceosome catalyzes the removal of introns and the linkage of exons 3. alternative splicing = a possible advantage for a species to contain introns; states that when a pre-mRNA has multiple introns, variation may occur in the pattern of splicing to allow the resulting mRNAs to have alternative combos of exons a. biological importance: 2 or more diff proteins can be made from a single gene to allow organism to carry fewer genes in genome Modification Description Occurrence Processing The cleavage of a large RNA Occurs in both prokaryotes transcript into smaller pieces; and eukaryotes and happens 1+ smaller pieces becomes a for rRNA and tRNA functional RNA molecule Splicing Involves cleavage and joining Common among eukaryotic of the RNA molecule; pre-mRNAs, sometimes in removes introns and stitches rRNAs, sometimes in tRNAs, together exons and rarely in bacterial RNAs 5’ capping The attachment of a Eukaryotic mRNAs; occurs methylguanosine cap to the 5’ while the pre-mRNA is still end of an mRNA; plays key being made by RNA role in splicing of introns, exit polymerase II of mRNA from nucleus, and binding of mRNA to ribosome. Functions: allows mRNAs to exit nucleus, allows for early translation, allows for efficient splicing of introns 3’ poly-A-tailing The attachment of a string of Eukaryotic mRNAs and some adenine-containing bacterial RNAs nucleotides to the 3’ end of mRNA at a site where the mRNA is cleaved; important for stability and translation; important for stability so the mRNA can leave the nucleus RNA editing—a change in the The change of the base Some eukaryotic RNAs nucleotide sequence of an sequence of an RNA after it RNA molecule that involves has been transcribed; can additions or deletions of generate start codons, stop certain bases or conversions codons, and change the of 1 base into another coding sequence for polypeptide Base modification Covalent modification of a Common in tRNAs base within an RNA molecule 4. Different kinds of RNA molecules are transcribed from DNA a. Messenger RNA—mRNA i. Encodes the sequence of AA within a polypeptide ii. Transcribed by RNA polymerase II iii. Modified by polyadenylation = cleavage near their 3’ end and attachment of a string of adenine nucleotides iv. Pre-mRNA = the long transcript produced by the transcription of structural genes within nucleus; usually altered by splicing and other modifications before it can exit the nucleus and start to become mature b. Ribosomal RNA—rRNA i. Necessary for the translation of mRNA; make up ribosomes and protein subunits ii. Transcribed by RNA polymerase I c. Transfer RNA—tRNA i. Necessary for the translation of mRNA ii. Transcribed by RNA polymerase III iii. Requires processing via endonucleases and exonucleases 1. Exonuclease = cleaves a covalent bond between 2 nucleotides at one end of a strand 2. Endonuclease = cleaves the bond between 2 adjacent nucleotides 3. Made as large pre-cursors that have to be cleaved to produce mature and functional tRNAs d. microRNA i. short RNA molecules involved in gene regulation e. scRNA—small cytoplasmic RNA i. found in the cytoplasms and is necessary to target proteins to the ER ii. component of the signal recognition particle (SRP) f. RNA of RNaseP i. Catalytic component that is necessary in the processing of bacterial tRNA ii. Ribozyme = RNA molecule with catalytic activity g. snRNA—small nuclear RNA i. necessary in the splicing of eukaryotic pre-mRNA ii. snRNAs and proteins combine to form small nuclear riboproteins (snRNPs) that are components of the spliceosome iii. transcribed by RNA polymerase II h. telomerase RNA i. enzyme telomerase is made of an RNA molecule and protein subunits i. snoRNA—small nucleolar RNA i. necessary in the processing of eukaryotic rRNA transcripts 5. Important transcriptional differences between prokaryotes and eukaryotes Prokaryotes Eukaryotes Simple transcription process Complex transcription process Smaller cells Larger cells with many organelles Have less genes that encode for cellular Have more genes that encode for cellular proteins proteins Genes don’t need specific environment or Genes need to be transcribed in the correct development stage cell and during the correct development stage Genetic material is transcribed by RNA Genetic material is transcribed by RNA polymerase holoenzyme polymerase I, II, and III to transcribe diff categories of genes Promoter sequence is more simple and less Promoter sequence is more variable and variable more complex 6. Coding—figuring what kind of signal this molecule codes a. The diamond code—coding theory i. Certain combinations of nucleotides are specific for certain functions b. Triple nature of the genetic code i. A triplet code was proposed bc… 1. If the DNA is a 2 letter code, you can have 16 possible combinations that can be coded. There are 20 possible AA therefore this is too little and nonsufficient 2. If the DNA is a 3 letter code, there are 64 possible combinations. Because this is larger than 20, it’s possible 3. If the DNA is a 4 letter code, there are 4 placeholders and 264 possible combinations. a. Looked at insertions of a single nucleotide to look at 2 letter/3 letter/etc. i. Any time 1 or 2 nucleotides were added the protein would appear broken 1. This leads to a frameshift ii. Any time 3 nucleotides were added the protein would appear the same 7. Characteristics of the genetic code a. 3-letter code referred to as a codon i. 3 possible placeholders so there are 64 possible codons b. linear c. co-linear—describes the 1 to 1 correspondence between the sequence of codons in the DNA-codoing strand and the AA sequence of the polypeptide d. universal e. redundant or degenerate i. 1 aa can be coded for several codons ii. degeneracy goes as far as 6 codons (i.e. 1 aa can be encoded by 6 codons) f. unambiguous i. 1 codon 1 aa 8. Exceptions a. mitochondrial genome** 9. Wobble hypothesis a. States that the 1 and 2 ndcodon positions are more important in attracting correct tRNA than the 3 position i. The anticodon of a single form of tRNA can pair with more than 1 codon in mRNA b. Occasionally we can have mismatches btwn tRNA and codon
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