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Chap 15-18 outline

by: Anna Perry

Chap 15-18 outline BIOSC 0160

Anna Perry
GPA 3.5
Foundations of Biology 2
Dr. Swiganova

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An outline of Chapters 15-18. This will be useful for the first test
Foundations of Biology 2
Dr. Swiganova
Test Prep (MCAT, SAT...)
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This 24 page Test Prep (MCAT, SAT...) was uploaded by Anna Perry on Tuesday February 3, 2015. The Test Prep (MCAT, SAT...) belongs to BIOSC 0160 at University of Pittsburgh taught by Dr. Swiganova in Winter2015. Since its upload, it has received 58 views. For similar materials see Foundations of Biology 2 in Biology at University of Pittsburgh.


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Date Created: 02/03/15
Chapter 15 DNA and the Gene Synthesis and Repair I What Are Genes Made Of A The HersheyChase experiment showed that viral genes consist of DNA and not protein and that DNA is the hereditary material B The Secondary Structure of DNA a DNA is typically doublestranded with each strand consisting of a long linear polymer made up of deoxyribonucleotides They link together into a polymer when a phosphodiester bond forms between a hydroxyl group on the 3 carbon of one deoxyribose and the phosphate group attached to the carbon of another The primary structure of DNA has two major features I A backbone made up of sugar and phosphate groups ii A series of bases that project from the backbone Each strand has a directionality or polarity The two strands line up in opposite directions in an antiparallel fashion The strands then twist around each other to t together snugly and form hydrogen bonds to form a double helix Hydrogen bonding of particular base pairs is complementary base pairing i AT ii CG ll Testing Early Hypotheses about DNA Synthesis A Three Hypotheses about how the old and new strands interact during replication 1 Semiconservative replication The old parental strands of DNA separate and each is used as a template for the synthesis of a new daughter strand a Each new daughter molecule would consist of one old strand and one new strand Conservative replication The bases temporarily turn outward so that complementary strands no longer face each other and can serve as a template for the synthesis of an entirely new double helix all at once a This results in an intact parental molecule and a daughter DNA molecule consisting entirely of newly synthesized strands Dispersive replication The parental double helix is cut wherever one strand crossed over another and DNA is synthesized in short sections a Stretches of old DNA are interspersed with new DNA down the length of each daughter strand B The MeselsonStahl Experiment 1 Proved that DNA replication was semiconservative because each newly made DNA molecule comprised of one old strand and one new strand A Model for DNA Synthesis A DNA polymerase polymerizes deoxyribonucleotides into DNA which catalyzes DNA synthesis 1 DNA polymerases can add deoxyribonucleotides only to the 3 end of a growing DNA chain 2 As a result DNA synthesis is always 5 gt 3 3 DNA polymerase I removes the RNA primer and replaces it with DNA 4 DNA polymerase III extends strands or fragments DNA Synthesis is exergonic because the monomers used in the reaction are deoxyribonucleoside triphosphates dNTPs How Does Replication Get Started 1 A quotbubblequot forms at the origin of replication a Bacterial DNA have 1 origin of replication b Eukaryotes have multiple origins 2 DNA synthesis is bidirectional Therefore replication bubbles grow in two directions as DNA replication proceeds 3 Once a replication bubble opens a different set of enzymes takes over at the replication fork the Y shaped region where the parent DNA double helix is split Howls The Helix Opened and Stabilized 1 DNA helicase breaks the hydrogen bonds between the base pairs causing the two strands to separate 2 Singlestrand DNAbinding proteins SSBPs attach to the separated strands and prevent them from snapping back into a double helix 3 The unzipping process creates tension further down the helix DNA should become tightly coiled because of this but does not become of topoisomerases enzymes that cut DNA allow it to unwind and rejoin it ahead of the advancing replication fork How Is The Leading Strand Synthesized 1 A primer a strand a few nucleotides long that is bonded to the template strand provides DNA with a free 3 hydroxyl group that can combine with an incoming deoxyribonucleotide 3 4 a Primase synthesizes a short stretch of RNA that acts as a primer for DNA polymerase b Primase is a type of RNA polymerase an enzyme that catalyzes the polymerization of RNA Once a primer is present DNA polymerase works in the 5 gt 3 direction adding deoxyribonucleotides to complete the strand As DNA polymerase moves along the sliding clamp holds the enzyme in place on the template strand This is called the leading strand because it leads into the replication fork and is synthesized continuously F How Is The Lagging Strand Synthesized 1 4 Only one strand of DNA at the fork can be synthesized 5 gt 3 The other must be synthesized in a direction that runs away from the fork This is the lagging strand The Discontinuous Replication Hypothesis a Says that primase synthesizes new RNA primers for lagging strands as the moving replication fork opens singlestranded regions of DNA and that DNA polymerase uses the primers to synthesize short lagging strand DNA fragments The Discovery of Okazaki Fragments a These short DNAs are Okazaki fragments and DNA ligase catalyzes the formation of a phosphodiester bond between the adjacent fragments All of these enzymes are joined into the replisome a large macromolecular machine IV Replicating the Ends of Linear Chromosomes A Replication of chromosome ends requires a specialized DNA replication enzyme B The End Replication Problem 1 2 The region at the end of a chromosome is called a telomere When the replication fork reaches the end of a linear chromosome DNA polymerase synthesizes the leading strand all the way to the end resulting in a double stranded copy On the lagging strand primase adds an RNA primer close to the end but when it is removed DNA polymerase is unable to add DNA near the tip a The left over DNA stays singlestranded This singlestranded DNA at the end degrades eventually resulting in the shortening of the chromosome C Telomerase Solves the End Replication Problem V Repairing Mistakes and DNA Damage A DNA polymerases work fast but the replication process is also astonishingly accurate B Correcting Mistakes In DNA Synthesis 1 As DNA polymerase goes along a DNA template hydrogen bonding occurs between incoming deoxyribonucleotides and the ones on the template strand DNA polymerases are selective about the bases they add to a growing strand a The correct base pairings are energetically favorable b These correct pairings have a distinct shape DNA Polymerase Proofreads a If a new base is not correctly paired the positioning of it provides a poor substrate for DNA polymerase to extend i The geometry of incorrect base pairs is different ii DNA polymerase s active site can detect these shapes and will add a new base only when the previous base pair is correct b This means DNA polymerase III can proofread i If the wrong base is added the enzyme pauses removes the mismatched base that was just added and proceeds with synthesis Mismatch Repair a Mismatch repair occurs when mismatch bases are corrected after DNA synthesis is complete b Proteins recognize the mismatched base remove the section containing it and ll in the correct bases using the older strand as a template c Mutations in components of mismatch repair are observed in many cancers C Repairing Damaged DNA 1 DNA can be damaged by sunlight Xrays and chemicals If this damage were ignored mutations would accumulate to lethal levels 2 To x problems caused by chemical attack radiation etc organisms have evolved a wide array of DNA damagerepair systems a Nucleotide excision repair works on DNA damage caused by ultraviolet light and some chemicals i UV light can cause a covalent bond to form between adjacent pyrimidine bases within the same DNA strand ii This can create a kink in the structure which stalls DNA polymerase blocking DNA replication b Nucleotide excision repair xes thymine dimers and other damage that distorts the helix i An enzyme recognizes the kink in the helix ii Another enzyme removes a segment of single stranded DNA containing the defective sequence iii The intact DNA strand provides a template for synthesis of a corrected strand and the 3 hydroxyl of the DNA strand next to the gap serves as a primer iv DNA ligase links the newly synthesized DNA to the original undamaged DNA D Xeroderma Pigmentosum A Case Study 1 Xeroderma Pigmentosum XP is a rare autosomal recessive disease in humans a Causes sensitivity to UV light b Skin develops lesions after slight exposure to suanht James Cleaver proposed that people with XP are sensitive to sunlight because they are unable to repair damage induced by UV light Cleaver collected normal skin cells and those from people with XP When grown in culture he exposed them to increasing amounts of UV radiation and recorded how many survived a Cell survival declined with increasing radiation dose in both types of cells but XP cells died off more rapidly The ndings con rmed that nucleotide excision repair is virtually nonexistent in XP individuals If the overall mutation rate in a cell is elevated because of defects in DNA repair then mutations that trigger cancer become more likely Chapter 16 How Genes Work What Do Genes Do A Gene expression is the process of converting archived information into molecules that actually do things in the cell B Alleles that do not function at all are called knockout mutants 1 Creating these and analyzing their effects is one of the most common research strategies in studies of gene funcUon C Beadle amp Tatum developed the onegene oneenzyme hypothesis which stated that each gene contains the information needed to make an enzyme 1 They tested this by mutating genes in the bread mold Neurospora crassa and observed that defects in particular genes resulted in the mold s inability to produce speci c compounds E F Th A C D E 2 The idea was to knock out a gene by damaging it and then infer what the gene does by observing the phenotype of the mutant individual Srb amp Horowitz also tested this They knew that organisms synthesized arginine in a metabolic pathway made of several steps 1 If a colony could grow in the presence of arginine but failed to grow without arginine they concluded that it couldn t make its own arginine A genetic screen is any technique for picking certain types of mutants out of many randomly generated mutants Genes contain instructions for making proteins e Central Dogma of Molecular Biology Genes contain the information for all of the proteins in an organism 1 The genetic code is information coded in the base sequence of DNA 2 The information encoded in the base sequence of DNA is not translated into the amino acid sequence directly Messenger RNA mRNA carries information out of the nucleus from DNA to the site of protein synthesis ribosome 1 RNA molecules act as a link between genes and the proteinmanufacturing centers 2 The enzyme RNA polymerase synthesizes RNA according to the information provided by the sequence of bases in a particular stretch of DNA a It polymerizes ribonucleotides into strands of RNA The Central Dogma summarizes the ow of information in cells It states that DNA codes for RNA which codes for proteins 1 DNA gt RNA gt proteins 2 The sequence of bases in DNA specifies the sequence of bases in RNA which specifies the sequence of amino acids in a protein 3 Transcription is the process of copying hereditary information in DNA to RNA 4 Translation is the process of using the information in nucleic acids to synthesize proteins An organism s genotype is determined by the sequence of bases in its DNA while its phenotype is a product of the proteins it produces Reverse transcriptase synthesizes a DNA version of the RNA genes in viruses The Genetic Code B The genetic code is a triplet code with a three base sequence called a codon specifying a single amino acid 1 The addition or deletion of a base causes a loss of function in a gene because the mutation throws the sequence of codons or the reading frame out of register 2 AUG is the start codon which signals that protein synthesis should begin at that point on the mRNA molecule 3 There are 3 stop codons which signal the end of the protein coding sequence UAA UAG and UGA They do not code for any amino acid The genetic code is redundant because all amino acids except Met and Trp are coded for by more than one codon How Can Mutation Modify Genes and Chromosomes A A mutation is any permanent change in an organism s DNA It is a modi cation in a ce s information archive a change in its genotype 1 Mutations create new aees If a mistake is made during DNA synthesis or repair a change in the sequence of bases in DNA results A single base change is called a point mutation 1 Point mutations that cause changes in the amino acid sequence of a protein are called missense mutations 2 Point mutations that don t change the amino acid sequence are called silent mutations 3 Mutations that alter the meaning of subsequent codons are frameshift mutations 4 Nonsense mutations occur when a codon that speci es an amino acid is changed by mutation to one that speci es a stop codon a This causes early termination of the polypeptide change and often results in a nonfunctional protein Biologists divide mutations into three categories 1 Bene cial Some mutations increase the fitness of the organism in certain environments 2 Neutral If a mutation has no effect on tness it is termed neutral a Silent mutations are neutral 3 Deleterious Mutations that lower tness are termed harmful f point mutations alter DNA sequences that are important for gene expression they can have important effects on phenotype even though they don t change the amino acid sequence of a protein E Changes in chromosome number result from chance mistakes in moving chromosomes during meiosis or mitosis F The composition of individual chromosomes can change as well 1 Chromosome segments can become detached when accidental breaks in chromosomes occur a b The segments may be ipped and rejoined in inversion Or become attached to a different chromosome in translocation c When a segment is lost it s called deletion d When additional copies of a segment are present it s called duplication 2 Chromosomes of cancer cells exhibit deleterious chromosome mutations including aneupoidy inversions translocations deletions and duplications G A karyotype is the complete set of chromosomes in a cell Chapter 17 Transcription RNA processing and translation An Overview of Transcription The rst step in converting genetic information to proteins is to synthesize an RNA version of the instructions archived in DNA A PCP 1 2 RNA polymerase is responsible for synthesizing mRNA Once a NTP nucleotide triphosphate that matches a base on the DNA template is in place RNA polymerase breaks off two phosphates and catalyzes the formation of a phosphodiester linkage between the 3 end of the growing mRNA and the new NMP As the 5 gt 3 matchingcatalysis process continues RNA complementary to the gene is synthesized This is transcription The strand read by the enzyme is the template strand The other strand is called the coding strand matches RNA sequence besides U and T differences RNA polymerase synthesizes in the 5 gt 3 direction but do not require a primer to begin transcription How Does Transcription Begin in Bacteria 1 The initiation phase of transcription helps the RNA polymerase know where and in which direction to start transcription The enzyme cannot do it on its own a A detachable protein subunit called sigma must bind to it rst Bacterial RNA polymerase and sigma form a holoenzyme which consists of a core enzyme RNA polymerase in this case containing the active site for catalysis and other required proteins sigma Promoters are sections of DNA that promote the start of transcription a Sigma proteins bind to promoters The particular section of bases in bacteria similar to TATAAT is the 10 box because it is centered about 10 bases from the point where bacterial RNA polymerase starts transcription a DNA located in the direction that RNA polymerase moves is downstream b DNA located in the opposite direction is upstream c The place where transcription begins is the 1 site ln Bacteria transcription begins when sigma as part of the holoenzyme binds to the 35 and 10 boxes This determines where and in which direction to synthesize 6 Once the holoenzyme is bound the DNA helix is opened by RNA polymerase creating 2 separated strands When an incoming NTP pairs with a base RNA polymerization begins a RNA polymerization is exergonic and spontaneous because NTPs have signi cant potential energy owing to their phosphate groups 7 The initiation phase is complete as RNA polymerase extends the mRNA from the 1 site F Elongation and Termination 1 Once RNA polymerase begins synthesizing the elongation phase is underway 2 RNA polymerase is macromolecular with many parts a In the interior of the enzyme a group of amino acids forms a rudder to help steer the template and coding strands through channels inside the enzyme b The enzyme s active site catalyzes the addition of nucleotides to the 3 end c A group of projecting amino acids forms a region called the zipper to help separate the newly synthesized RNA from the DNA template 3 During the elongation phase DNA goes in and out of one groove NTPs enter another and the growing RNA strand exits to the rear 4 Termination ends transcription ln bacteria this is when RNA polymerase transcribes a DNA sequence that functions as a transcriptiontermination signal G Transcription in Eukaryotes 1 Eukaryotes have 3 polymerases that each transcribe only certain types of RNA 2 Promoters in eukaryotic DNA are more diverse Most include a sequence called the TATA box about 30 base pairs upstream of the start site 3 Instead of a sigma protein eukaryotic RNA polymerases recognize promoters using basal transcription factors which assemble at the promoter followed by the RNA polymerase 4 Termination involves a polyA signal which tells an enzyme to cut the RNA Eventually RNA polymerase falls of the DNA template and terminates transcription ll RNA Processing in Eukaryotes A In bacteria when transcription terminates the result in mRNA that s ready to be translated B When eukaryotic genes are transcribed the initial product is a primary transcript which must still undergo multistep processing before it is functional 1 For proteincoding genes this is premRNA C The Startling Discovery of Split Eukaryotic Genes 1 Eukaryotic genes consist of exons which are parts of the primary transcript that remain in mature RNA a They are expressed They also consist of introns which are removed in forming mature RNA a They are not represented in the nal RNA product b They are intervening c Because of introns eukaryotic genes are much larger than their corresponding mature RNAs D RNA Splicing 1 An transcription proceeds the introns are removed from the growing RNA strand by splicing which is where pieces of the primary transcript are removed and the remaining segments are joined together a Splicing occurs within the nucleus while transcription is still underway b It results in RNA that contains an uninterrupted genetic message Splicing is catalyzed by small nuclear RNAs snRNAs working with a complex of small nuclear ribonucleoproteins or snRNPs Splicing is broken into 4 steps a snRNPs bind to the start of an intron and an A base in the intron b Other snRNPs arrive to form a spliceosome c The intron forms a loop plus a single stranded stem lariot with the adenine at its connecting point d The lariat is cut out and a phosphodiester linkage links the exons on either side producing a continuous coding sequence mRNA E Adding Caps and Tails to Transcripts 1 2 For premRNAs intron splicing is accompanied by other steps a As soon as the 5 end of a eukaryotic premRNA emerges from RNA polymerase enzymes add a 539 cap consisting of a modi ed guanine nucleotide with 3 phosphate groups b An enzyme cleaves the 3 end of the premRNA downstream of the polyA signal and another enzyme adds a long row of adenine nucleotides known as the polyA tail After the addition of the cap and tail the processing is complete 3 RNA processing is the general term for any of the modi cations such as splicing or polyA tail addition An Introduction to Translation A To synthesize a protein the sequence of bases in an mRNA molecule is translated into a sequence of amino acids in a polypeptide Ribosomes Are the Site of Protein Synthesis 1 There is a strong correlation between the number of ribosomes in a given type of cell and the rate at which that cell synthesizes proteins Translation in Bacteria and Eukaryotes 1 In bacteria ribosomes attach to mRNAs and begin synthesizing proteins even before transcription is complete 2 Multiple ribosomes attach to each mRNA forming a polyribosome In this way many copies of a protein can be produced from a single RNA 3 Transcription and translation can occur concurrently in bacteria because there is no nuclear envelope to separate the two processes 4 In eukaryotes transcription and mRNA processing occurs in the nucleus and translation occurs in the cytoplasm a Once mRNAs are outside the nucleus ribosomes can attach to them and begin translation How Does an mRNA Triplet Specify an Amino Acid 1 When an mRNA interacts with a ribosome instructions encoded in nucleic acids are translated into a different chemical language 2 Adapter molecules hold amino acids in place and interact with mRNA codons by hydrogen bonding The Structure and Function of Transfer RNA A Transfer RNA tRNA is the adapter molecule used in translation They transfer amino acids from the RNA to a growing polypeptide What Do tRNAs Look Like 1 Transfer RNA sequences are relatively short and certain parts of the molecules can form secondary structures a Some sequences of bases can form hydrogen bonds with complementary bases sequences elsewhere in the same strand b StemandOloop structures result 2 A CCA sequence at the 3 end of each tRNA molecule offers a site for amino acid attachment while a triplet on the loop at the other end of the structure could serve as an anticodon a set of three ribonucleotides that forms base pairs with the mRNA codon 3 This makes tRNA fold into an upsidedown Lshape the anticodon at one end the CCA at another tertiary structure C How Are Amino Acids Attached to tRNAs 1 An input of energy ATP is required to attach an amino addtotRNA 2 Enzymes called aminoacyltRNA synthetases catalyze the addition of amino acids to tRNAs 3 For each of the 20 major amino acids there is a different aminoacyltRNA synthetase and on or more tRNAs 4 Each aminoacyltRNA synthetase has a binding site for a particular amino acid and a particular tRNA 5 The combination of a tRNA covalently linked to an amino acid is an aminoacyl tRNA D How Many tRNAs Are There 1 According to the genetic code there are 61 different mRNA codons But most cells only contain about 40 tRNA codons 2 This is because of the wobble hypothesis which states that inside the ribosome certain bases in the third position of tRNA anticodons can bind to bases in the third position of a codon in a manner that doesn t match base pairing a This allows for a limited exibility or wobble in the base pairing V The Structure and Function of Ribosomes A The translation of each mRNA codon begins when the anticodon of an aminoacyl tRNA binds to the codon It is complete when a peptide bond forms between the tRNA s amino acid and the polypeptide chain 1 These events happen in the ribosome 2 Ribosomes contain ribosomal RNA rRNA and can be separated into the large subunit and small subunit a The large subunit is where peptidebond formation takes place b The small subunit holds the mRNA in place B During protein synthesis three distinct tRNAs are lined up inside the ribosome 1 The tRNA on the A site carries an amino acid 2 The one in the middle holds the growing polypeptide chain and occupies the P site 3 The E site is for tRNA that no longer has an amino acid attached and is about to leave the ribosome C The ribosome synthesizes proteins in a 3 step sequence 1 An aminoacyl tRNA diffuses into the A site if its anticodon matches the codon in mRNA it stays 2 A peptide bond forms between the amino acid held in the A site and the growing polypeptide held in the P site 3 The ribosome moves down the mRNA by one codon and all three tRNAs move one position within the ribosome a The tRNA in the E site exits b The tRNA in the P site moves to the E site c The tRNA in the A site switches to the P site D Synthesis starts at the Nterminus and proceeds to the C terminus E Initiating Translation 1 A start codon AUG is found near the 5 end of all mRNAs 2 Translation begins when a section of rRNA in the small subunit binds to a complementary sequence on an mRNA a This is the ribosome binding site or ShineDalgarno sequence b About 6 subunits upstream from start codon 3 Proteins called initiation factors mediate the interactions between the small subunit the message and the tRNA a These help in preparing the ribosome for translation b Also prevent small and large subunits from coming together until initiator tRNA is in place at the start codon c Help bind mRNA to small subunit 4 The initiator aminoacyl tRNA binds to the start codon 5 Initiation is complete when the large subunit joins the complex F Elongation Extending the Polypeptide 1 Incoming aminoacyl tRNA New tRNA moves into the A site where its anticodon base pairs with the mRNA codon 2 Peptide bond formation The amino acid attached to the tRNA in the P site is transferred to the amino acid of the tRNA in the A site a When both the P and A sites are occupied by tRNAs the amino acids on the tRNAs are in the ribosome s active site This is where peptidebond formation occurs 3 Translocation The ribosome moves one codon down the mRNA with the help of elongation factors The tRNA attached to the polypeptide chain moves into the P site The A site is now empty a Elongation factors help move the ribosome relative to the mRNA so that translation occurs in the 5 gt 3 direction b Translocation requires GTP c The empty tRNA in the E site is ejected into the cytosol 4 REPEAT 13 until termination G Terminating Translation 1 When the translocating ribosome reaches one of the stop codons a release factor recognizes the stop codon and lls the A site a Release factors do not carry an amino acid b When a release factor occupies the A site the protein s active site catalyzes the hydrolysis of the bond linking the tRNA in the P site to the polypeptide chain freeing it 2 The newly synthesized polypeptide is released from the ribosome the ribosome separates from the mRNA and the subunits dissociate H PostTranslational Modi cation 1 Proteins aren t fully formed and functional by termination 2 Folding a Protein function depends on shape and shape depends on how it folds b Folding is speeded up by molecular chaperones 3 Chemical Modi cations a In the ER and Golgi apparatus small chemical groups may be added to proteins often sugar or lipid groups b Many proteins are altered by enzymes that add or remove a phosphate group Chapter 18 Control of Gene Expression in Bacteria An Overview of Gene Regulation and Information Flow A Gene expression is the process of converting information that is archived in DNA into molecules that actually do things in the cell 1 It occurs when a protein or other gene product is synthesized and active Bacterial cells can be densely packed along intestinal walls competing for space and nutrients 1 A cell has to use resources ef ciently if it s going to be able to survive and reproduce Most gene expression is triggered by speci c signals from the environment Mechanisms of Regulation 1 Gene expression can be controlled at any of the three steps from DNA gt mRNA gt Protein gt Activated Protein a The cell could avoid making the mRNAs for particular enzymes i If there is no mRNA then ribosomes can t make the gene product ii Transcriptional control occurs when regulatory proteins affect RNA polymerase s ability to bind to a promoter and initiate transcription iii It is particularly important due to its efficiency it saves the most energy for the cell because it stops the process of gene expression at the earliest possible point b If the mRNA for an enzyme has been made the cell could prevent the mRNA from being translated into protein i Translational control occurs when regulatory molecules alter the length of time an mRNA survives or affect translation initiation or elongation ii It allows a cell to make rapid changes in the amounts of different proteins because the mRNA is already present and available for translation c Many proteins have to be activated by chemical modi cation such as the addition of a phosphate group i Posttranslational control provides the most rapid response of all three mechanisms because only one step is needed to activate an existing protein d All three forms of control occur in bacteria 2 Among the mechanisms of gene regulation there is a clear tradeoff between the speed of response and the conservation of ATP amino acids and other resources a Transcriptional control is slow but ef cient in resource use b Posttranslational control is fast but energetically expengve 3 Some genes are transcribed all the time or constitutively E Metabolizing Lactose 1 Glucose is E coli s preferred carbon source meaning that is the source of energy and carbon atoms that the organism uses most efficiently a Lactose can also be used but isn t until glucose supplies are depleted 2 To use lactose E coli must rst transport the sugar into the cell Once lactose is inside the cell the enzyme B galactosidase catalyzes a reaction that breaks down the disaccharide into glucose and galactose a E coionly produces high levels of Bgalactosidase when lactose is present in the environment b Lactose itself regulates the gene for Bgalactosidase meaning that lactose acts as an inducer a small molecule that triggers transcription of a speci c gene ll Identifying Regulated Genes A To understand how E coli controls production of B galactosidase and the transport protein that brings lactose into the cell Jacob and Monod rst had to nd the genes that code for these proteins 1 To do this they isolated and analyzed mutants 2 Their goal was to nd E coli cells that couldn t metabolize lactose a These cells would either lack Bgalactosidase or the lactose transporter protein B To nd mutants that are associated with a particular trait a researcher has to complete two steps 1 Generate a large number of individuals with mutations at random locations in their genomes a Monod and colleagues accomplished this by exposing E coli to mutagens Xrays UV light or chemicals that damage DNA and increase mutation rates 2 Screen the treated individuals for mutants with defects in the process or biochemical pathway in question C Replica Plating to Find Lactose Metabolism Mutants D 1 Mutagenized bacteria are spread on a quotmaster platequot lled with a gelatinous growth medium containing glucose but no lactose a It is important that the mutant cells are capable of growing on the master plate The bacteria are then allowed to grow so that each cell produces a colony A block covered with a piece of sterilized velvet is pressed onto the master plate Some cells from each colony are transferred The velvet is pressed onto a replica plate that contains medium that differs by a single component a The second medium had only lactose and no glucose b Cells from the velvet stick to the replica plate s surface producing an exact copy of the locations of colonies on the master plate After the transferred cells grow the colonies on the replica plate are compared with those on the master plate a Colonies that grow on the master plate but are missing on the replica plate are mutants that can t metabolize lactose b By picking cells from these colonies on the master plate researchers build a collection of lactose metabolism mutants Several Genes Are Involved in Lactose Metabolism 1 The initial mutant screen yielded 3 types of mutants a The mutant cells were unable to cleave lactose i Jacob and Monod concluded that they lacked a functioning version of Bgalactosidase and therefore the gene that encoded it is defective ii The gene was lacZ and its mutant allele was lacZ39 b The mutant cells failed to accumulate lactose inside the cell i Jacob and Monod hypothesized that the mutant cells had defective copies of the membrane protein responsible for transporting lactose into the cell ii The protein was identi ed galactoside permease and the gene that encodes the gene that encodes it was lacY c The mutant cells didn t show normal regulation of B galactosidase and galactoside permease expression Instead the mutants made the proteins all the time even if no lactose was present d Cells that are abnormal because they produce a product at all times are constitutive mutants The gene that was mutated to produce constitutive Bgalactosidase and galactoside permease expression was lacl 2 Jacob and Monod identi ed 3 genes involved in lactose metabolism lacZ lacY and lacl a b d e LacZ and lacYcode for proteins required for the metabolism and import of lactose Lacl is responsible for some sort of regulatory funcUon When lactose is absent the lacgene or gene product shuts down the expression of lacZand lacY When lactose is present transcription of lacZ and achs induced The genes are close together which suggests that lacZand lacYare transcribed together lll Negative Control of Transcription A There are two general ways that transcription can be regulated 1 Negative control occurs when a regulatory protein called a repressor binds to DNA and shuts down transcription a The lacZand lacYgenes are under negative control b The lacgene codes for a product that repressed transcription of the lacZand lacYgenes c The repressor binds directly to DNA at or near the promoter for the lacZand lacYgenes 2 Positive control occurs when a regulatory protein called an activator binds to DNA and triggers transcription B Lactose interacts with the repressor in a way that makes the repressor release form its binding site C The Operon Model 1 Operon is a set of coordinately regulated bacterial genes that are transcribed together into one mRNA 2 The group of genes involved in lactose metabolism was termed the lac operon 3 A gene called lacA was found to be adjacent to lacYand lacZand transcribed as part of the same operon a It codes for the enzyme transacetylase which catalyzes reactions that allow certain types of sugars to be exported from the cell when they are too abundant and could harm the cell 4 Three hypotheses are central to the JacobMonod model of lac operon regulation a The lacZ lacY and lacA genes are adjacent and are transcribed into one mRNA initiated from the single promoter of the lac operon i This is cotranscription and it results in the coordinated expression of the 3 genes b The repressor is a protein encoded by lacthat binds to DNA and prevents transcription of the lac operon genes lacZ lacY lacA i Lacl is expressed constitutively and the repressor binds to a section of DNA in the lac operon called the operator c The inducer lactose binds to the repressor When it does the repressor changes shape The shape change causes the repressor to come off the DNA i This is allosteric regulation D How Does Glucose Regulate the lac Operon 1 Transcription of the lac operon is drastically reduced when glucose is present 2 Glucose inhibits the lactose transport activity of galactoside permease through a chain of molecular events a When both glucose and lactose are present the transport of lactose into the cell is inhibited b Because lactose doesn t accumulate in the cytoplasm the repressor remains bound to the operator c Negative control 3 When glucose levels outside the cell are low glactoside permease is active a If lactose is present it is transported into the cell and induces lac operon expression 4 The mechanism of glucose preventing the transport of inducer is known as inducer exclusion which affects the activity of many different sugar transporters E Why Has the lac Operon Model Been So Important 1 Follow up work showed that many bacterial genes and operons are under negative control by repressor genes a This means the ndings are general 2 It introduced the idea that gene expression is regulated by physical contact between regulatory proteins and speci c regulatory sites in DNA 3 It offered an important example of posttranslational control over gene expression a Posttranslational control is best when a rapid response is needed Positive Control of Transcription A In positive control an activator protein binds to a regulatory sequence in DNA when genes are turned on B When bound to DNA the activator interacts with RNA polymerase to increase the rate of initiating transcription C The ara operon provides an important example of positive control 1 It contains 3 genes that allow E coli to use the sugar arabinose which is found in many plant cell walls 2 Without arabinose in the environment the ara operon is not transcribed 3 When arabinose is present transcription of the ara operon is turned on by an activator protein called AraC D The AraC protein is allosterically regulated by arabinose 1 When bound to arabinose two copies of the AraC protein attach to a regulatory sequence of DNA called the ara initiator that lies upstream of the promotor 2 Once AraC is bound to DNA it can also bind to RNA polymerase a The interaction between AraC and the RNA polymerase helps to dock the polymerase to the promoter and accelerate the initiation of transcription E AraC is both an activator and a repressor 1 In the absence of arabinose the two copies of the AraC protein remain together a While one araC copy remains bound to the initiator the other copy now binds to a different regulatory site in DNA b In this con gurator AraC works as a repressor to prevent the transcription of both the ara operon and the araC gene Global Gene Regulation A To compete for resources bacteria must be able to coordinate the expression of large sets of genes 1 An effective way to express sets of genes together is to group them into an operon and transcribe them into a single mRNA 2 But there are limits to the size of operons B Global gene regulation is the coordinated regulation of many genes 1 Genes can be grouped into a regulon a set of separate genes or operons that contain the same regulatory sequences and that are controlled by a single type of regulatory protein a Regulons allow bacteria to respond to challenges that include shortages of nutrients sudden changes in temperature exposure to radiation or shifts in habitat 2 A regulon consists of many genes scattered across the genome a All of these genes are controlled by the same type of repressor protein that binds to the same operator sequences near the promoter of each gene b When an environmental change triggers the removal of the repressor protein from all the operators every gene in the regulon is transcribed 3 Regulons can be under negative control by a repressor protein or positive control by an activator protein a Damaged DNA sets off an SOS signal that induces the transcription of more than 40 genes that code for DNA repair enzymes and for DNA polymerases that can us damaged DNA as a template C Interactions among protein regulators and the DNA sequences they bind produce nely tuned control over gene expression regulating individual genes operons or large sets of genes Chapter 19 Control of Gene Expression in Eukaryotes


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