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Let a be an integer and d be a positive integer. Show that

Discrete Mathematics and Its Applications | 7th Edition | ISBN: 9780073383095 | Authors: Kenneth Rosen ISBN: 9780073383095 37

Solution for problem 37E Chapter 5.2

Discrete Mathematics and Its Applications | 7th Edition

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Discrete Mathematics and Its Applications | 7th Edition | ISBN: 9780073383095 | Authors: Kenneth Rosen

Discrete Mathematics and Its Applications | 7th Edition

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Problem 37E

Let a be an integer and d be a positive integer. Show that the integers q and r with a = dq + r and 0 ? r

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Linkage and Chromosome Mapping in Eukaryotes Chapter Five Linkage • Most chromosomes carry a lot of genes • Unit of inheritance through meiosis is a chromosome, not each individual gene • Therefore, sometimes two genes are going to be inherited together because they are so close physically on the same chromosome – Breaking Mendel’s 2Law – Independent assortment of any two genes Complete Linkage • When two genes are always inherited together • Can only produce two gametes: AB (50%) and ab (50%) • Determined by which allele resides on which chromosome in parent • Gametes are known as “parental gametes” because their alleles are exactly the same as the parent’s alleles Complete Linkage YyRr Two Gametes YR (½) yr (½) YR YYRR YyRr Two Phenotypes YyRr (½) ¼ ¼ 3 : 1 yr YyRr yellowgreen (½) roundwrinkled ¼ yyr¼ When genes are transmitted together Crossing Over • Occurs during Prophase I • Homologous chromosomes exchange alleles = Recombination • Crossing over introduces independent assortment between two genes on the same chromosome • However, percentage of crossing over is determined by physical distance between two genes Crossing Over Percent of crossing over is determined by physical distance between two genes: • Closer two genes lay – less likely crossing over can occur – Genes will be linked = inherited together • More distance between two genes – more likely crossing over can occur – Genes may be unlinked – Depends on percent recombination Independent Assortment • With two completely independent genes expect four gametes to be made: AB (25%), Ab (25%), aB (25%), ab (25%) • Two genes on the same chromosome, but that lay far apart will undergo crossing over as often as not – 50% chance of recombination • Thereby producing four gametes in equal amounts = unlinked Independent Assortment YyRr 4 different gametes YR Yr yR yr (¼) (¼) (¼) (¼) Genetic Linkage • Determined by percent recombination • Any recombination less than 50% is considered linked – Only 0% recombination is complete linkage • In reality complete linkage is very rare • Between 0-50% recombination reflects physical distance between genes • Never see more than 50% - why not Genetic Distance • Frequency of crossing over tells us how far apart two genes are on chromosome • Not exactly the same as physical distance because certain sequences lead to more or less crossing over – Recombination hot spots • Instead, recombination measures genetic distance between two genes – Is correlated to physical distance, but not same Why does the frequency change Morgan’s hypothesis: two genes located close together on same chromosome were less likely to have a chiasma between them • Therefore chance of crossing over depends on physical distance between genes • This idea allows mapping of genes • Once you know percent of recombination you can determine genetic distance Genetic Mapping • Mapping is determining the order and distance between genes • Genetic mapping is done through breeding and counting offspring • Now a genetic map can be compared to a physical map determined from sequencing – Correlation has been confirmed • Genetic maps calculate chance genes will nd break Mendel’s 2 Law – inherited together Mapping Genetic Distance Recombination seen between genes: 1. yellow-white 0.5% 2. yellow-miniature 35.4% 3. white-miniature 34.5% Obviously: yellow and white genes are closest together • Yellow and miniature are furthest apart • White lays in the middle Mapping Genetic Distance Recombination seen between genes: 1. yellow-white 0.5% 2. yellow-miniature 35.4% 3. white-miniature 34.5% Mapping Genetic Distance • Basic principle still used today to calculate map distance (cM) between genes Based on a few facts: 1. Limited number of chiasmata form during each meiosis 2. Position of chiasmata is random along the length of each chromosome Therefore likelihood of crossing over is less between loci that are close together Likelihood of Crossing Over Recombinant Gametes Out of four gametes produced: • Only two will reflect crossing over event – Producing recombinant offspring Mapping Genetic Distance • Count number of parental and recombinant offspring Recombinant = % recombination Total Offspring • 1% recombination = 1 cM ex blond hair, blue eyes x Brown Brown 3 = Brown blue 44 = blond blue 5 = blond Brown 48 = Brown Brown Single v. Double Crossovers One long chromosome can contain more than one chiasma: • Single crossover (SCO) = one chiasma between two genes • Double crossover (DCO) = two chiasma between three genes • Still only produce two recombinant gametes Product Rule • Product rule determines the chance of two independent events occurring simultaneously • Combined probability is calculated by multiplying individual probabilities • Probability of double crossover (DCO) is equal to probabilities of each single crossover (SCO) multiplied together Mapping Three Genes To discover the order and distance between three genes at once: • Must account for the presence of all: – Single crossovers (SCO) Between each combination of two genes – Double crossovers (DCO) • Must have heterozygous genotypes – Otherwise can’t see crossing over Mapping Three Genes Let’s work through mapping example in Drosophila with three genes: • Don’t know order of genes but have to assume order just to draw out picture – Do know that these genes exist on X chrom. Mapping Three Genes F1: • Males produce two gametes – Cannot have crossing over because Y chromosome doesn’t synapse to X • Females can produce eight different gametes due to SCOs and DCOs Mapping Three Genes • Which gametes do you expect to be most frequent • Which gametes are least frequent • How will we determine map distance And order of 3 genes Mapping Three Genes Total: 10,000 Mapping Three Genes • Determine order of three genes based on the DCO compared to NCO’s phase – w must be in middle • Determine map distance by % recombinant Mapping Example in Maize • Another thing that needs to be determined in linkage mapping is phase Phase: which alleles are inherited together on the same homologue • Heterozygotes have two different phases: – AB/ab – Ab/aB • Also known as chromosomal alignment Determining Phase • How can we determine phase • Look at results of cross: most common = NCO Determining Order • How do we determine the order of genes • Look at results of cross: least common = DCO Determining Order DCO = pr v bm + + + Possible order (knowing phase): + v bm v + bm v bm + pr + + + pr + + + pr Determining Genetic Distance • How do we calculate cM between genes • Look at results of cross: cM = % Recombinant (161 + 86)/1109= 0.223 or 22.3 cM (395 + 86)/1109= 0.433 or 43.4 cM Mapping Example in Maize • Which genes are 22.3 cM apart and which genes are 43.4 cM apart v + bm + v  pr + pr + v + bm + pr  bm + pr + Undetected Crossovers • Some crossovers are impossible to detect • A small number of offspring might make it impossible to see a rare phenotypic class • Genetic maps are most accurate between genes that are closely linked Interference • A crossover event in one region inhibits a second crossover in nearby regions • Because of physical constraints that prevent chiasmata forming close together • Quantified by comparing expected double crossovers to observed DCOs: Observed DCO Expected DCO Interference • Observed DCO – simply count offspring • Expected DCO = SCO x SCO – Assuming chance of each SCO is independent of other SCO – Use product rule to determine chance of DCO • This ratio gives you an idea of the amount of expected crossovers that were seen • Interference = 1 – Observed DCO Expected DCO Synteny Mapping • Without sequencing entire genomes human genes were mapped to chromosomes • Produce a hybrid cell line that contains some human chromosomes – Know which human chromosomes present • Check to see if cell line produces a protein – If so: then gene must be on remaining chrom’s – If not: then gene must not be on these chrom’s Synteny Mapping • Use data below to map Genes A and B: Gene A must be Gene B must be Gene D must be Mapping in Haploids • Haploid organisms usually replicate through mitosis  exact copies • Some go through a diploid zygote stage where crossing over can be seen • Diploid zygote undergoes meiosis into a spore (acsi) containing four haploid cells – Known as a tetrad • Examining the alleles in each cell (of four) allows genetic mapping Mapping Haploids w/Tetrads Three types of tetrads can be made: 1. Parental type (PD) 2. Non-parental type (NPD) 3. Tetratrype (T) PD = same allele combinations as parents NPD = all genotypes are different than parents (recombination of chromosomes) T = two genotypes are parental, two are recombinant Tetrad Analysis Tetrad Analysis If PD = NPD genes are unlinked If PD >> NPD genes are linked • Calculate map distance: NPD + ½ T RF = Total number of tetrads • RF is a percent, multiply by 100 to get cM Questions Discussion Thomas Hunt Morgan • Columbia University • Drosophila melanogaster • Saw different percentages of offspring phenotypes would occur depending on which two genes he was studying Questions: 1. What is causing different phenotypes 2. Why does the frequency change What is causing different (recombinant) phenotypes • Crossing over between genes • Introduces new allele combination • New phenotypes not seen in parents • Morgan was the first person to describe crossing over Between synapsed homologues during Prophase I LOD score • Log of Odds ratio • Measures the probability that two traits are linked in humans (Can’t set up cross, small number of offspring) • Data taken from multiple, large pedigrees (likelihood of genotype/phenotype data assuming linkage) LOD = lo10 (likelihood of genotype/phenotype data assuming no linkage) • LOD ≥ 3 indicates linkage between genes Physical Mapping • With recombinant DNA technology mapping genes done on a physical map DNA markers: polymorphisms whose location in the genome is known • Use linkage testing to discover a marker that is linked to a gene • Since you know the location of the marker, you now know the location of the gene Physical Mapping • Linked genes that are too close together to follow Mendel’s 2 ndlaw Disease • They will always go into the Marker same gamete in meiosis • Use linkage to known markers to identify disease causative genes Common Markers • RFLP – Restriction Fragment Length Polymorphism – Polymorphic sequence at cut sites • STR – Short tandem repeats (Microsatellites) – Different number of repeating units • SNP – Single nucleotide polymorphism – Different base at one specific location Sequencing a Genome • The ultimate gene mapping experiment • Determine the exact sequence of an entire genome for an organism • Produce a physical map with location and physical distance between all genes – Compare/contrast to genetic map • Sequenced genomes: – Human, Drosophila, mouse, etc Mendel and Linkage • Linkage breaks Mendel’s 2ndLaw – Independent assortment of any two genes • Why didn’t Mendel ever see linkage • Studied 7 traits in an organism with 7 chromosomes – was he lucky • No - physical maps show us that some of his traits were on same chromosome Mendel and Linkage • Genes v and le are physically and genetically linked (Chromosome 4) • Mendel did not report that v-le showed independent assortment – Did not study every gene combination • Do you think: – This is a gene combination he did not study – Or that Mendel saw these two genes didn’t show independence and left out the results Sister Chromatid Crossing Over • Can crossing over happen between sister chromatid – held together by centromere • Experiments have shown yes • Because chromatids have same exact alleles you can’t see the effect in offspring • Does actually matter at all Analysis and Mapping in Bacteria and Phages Chapter Six Bacteria • Single celled organisms • Usually reproduce asexually – Producing billions of genetically identical cells • Single, circular haploid genome – All mutations are expressed in phenotype • Have adapted to exchange and recombine genetic information with unrelated cells – To allow faster evolution • Recombination allows genetic mapping Conjugation • Genetic material from one bacterium cell is transferred to another bacterium • Akin to sexual reproduction, except: – One or a few genes – Replaces genes in one strain with other strain Conjugation This experiment proved conjugation can occur F (fertility) Factor • Some cells can donate their genetic material, but other cells cannot • F+ : cells that can donate DNA • F- : cells that can only receive DNA • F+ cells produce a “sex pilius” which makes physical contact with other cell – A tubular extension of the cell membrane • F factor physically moves into F- cells – Making both cells F+ F factor • Separate, smaller circular chromosome: – Double stranded DNA ~100,000 nucleotides 19 genes – Coding for: Genetic transfer and pilius formation High Frequency Recombination • Hfr = a mutated strain of bacteria that undergoes recombination 1000 X’s faster • Interrupted mating: – Hfr and F- strains mixed – Incubated for varying times – Then sheared apart in a blender • Saw that the genes always recombined in a certain order ex azi then ton then lac+ Chromosome Mapping • Hfr showed an ordered transfer of genes • Suggests that chromosome was transferred linearly • Gene order and genetic distance can be determined by measuring time Chromosome Mapping • Reran interrupted mating with four different Hfr strains • Saw different order of gene transfer – Or did they Look for the pattern. Draw circle. What is the Hfr strain • F factor gets integrated into bacteria’s genome • Site of integration becomes O site – Always the first part to recombine • F factor itself is the last part to recombine • Therefore any genes in the bacteria’s genome can be mapped based on time conjugation and recombination require Integration Conjugation Interrupted Conjugation • F- cell picks up some bacterial genes – Becoming partially diploid • Because mating is interrupted almost never picks up the F factor – Furthest away from O site – Therefore, F- cell does not become F+ • Hfr cell is unchanged after conjugation • Examining genes picked up by F- allows genetic mapping of bacterial genes Chromosome Mapping Example: • In three Hfr x F crosses (all Hfr strains derived from one F strain), the first three markers of transfer from interrupted mating experiments are: Hrf1: QDE... Hfr2: SCE... Hfr3: SFQ... • What is the order of genes on the circular bacterial chromosome (with S placed at both ends to represent circularity) Hfr  F′ • The process of integration can be reversed and the integrated F factor can be excised back out randomly • Excised F factor frequently retains some of bacterial genes with it F′ • F′ cells can go through conjugation with F- cells • In this case the chromosomal genes from the initial bacterium can be transferred through conjugation F- again partially diploid Plasmids • Independent, double stranded, circular DNA that exists in cytoplasm – In bacteria all DNA is cytoplasmic • Contain multiple genes • Classified based on type of genes: ex F factor = plasmid with genes for Fertility • Copied and distributed to daughter cells along with host DNA during mitosis R plasmids • Confer resistance antibiotics that would normally kill bacteria cells • Through conjugation these R plasmids can be transferred to non-resistant bacteria – Producing “super bugs” – Cannot be treated with multiple antibiotics Col Plasmid • Non-transferable plasmid – Not normal plasmid type • Encoding toxins that will kill neighboring bacteria cells • Bacteria that carry the Col plasmid (Col+) are immune to these toxins – “genetic warfare” between cells Note: Most plasmids are used for research – Genes carried are controlled by scientists Tranformation • Process where foreign DNA is stably integrated into the bacterial genome – Remember how Hfr was produced • Difference is that foreign DNA is integrated at homologous sequence in the genome – Not integrated randomly as F factor • Numerous steps, but only two categories: 1. DNA entry into host cell 2. Recombination btwn foreign DNA and genome Transformation Transformation Details • Competence: when a population of bacteria cells can accept foreign DNA – Can occur naturally – Or be induced for genetic studies • DNA strand aligns to homologous sequence and crossing over occurs • Replacing original host sequence – Which is then degraded • Heteroduplex: double stranded DNA that is a mix from two different sources Genetic Linkage • If two (or more) genes are linked they will often be co-transformed simultaneously • Because they are located on the same transforming segment of DNA • If two genes are not linked they will be transformed independently • Therefore, studying frequency of co- transformation can allow gene mapping Bacteriophages • Viruses that infect bacteria cells • Bacteria cells = host Bacteriophages – Lytic Phase All host cells die after lysis >200 live viruses released Viruses can then infect surrounding bacterium Bacteriophages – Lysogenic Phase • Some virus can also be integrated into host genome • As bacteria is replicated, viral DNA will be too • Lysis can be induced under certain conditions Lederberg-Zinder Experiment Prevents conjugation Transduction • When bacterial genetic material is recombined through phage infection • Occurs through the phage’s integration of it’s genome into the host cell’s genome through the lysogeny • If phage carries some bacterial genes in it’s genome along with it’s own genes – Some bacterial DNA picked up by virus Transduction • In step 4 some phages pick up part of the bacterial genome • When this phage infects another cell it carries the first DNA to a new host • Bacterial genes are integrated at homologous sequences into the host cell’s genome Transduction Mapping • If two (or more) genes are linked they will often be co-transduced simultaneously • Because they are located on the same transduced segment of DNA • If two genes are not linked they will be transduced independently • Therefore, studying frequency of co- transduction can allow gene mapping of bacterial genome Bacteriophage Recombination • Phage undergoes recombination too • Which allows for gene mapping of the few genes carried by bacteriophages 1. Mutants were identified in order to determine genes that existed in phage and their functions 2. Measure how often recombinant mutants are seen vs. total plaques Some of the Phage Mutants rapid lysis (r) More rapid and effective lysisected Mixed Infection Infect cells with two different mutant strains: ex h r+ x h+ r Bacteriophage Mapping • By studying mixed infection results – Mutants in two distinct strains of phage – Infecting same bacteria culture at once • Count the number of parental and recombinant plaques • If genes are unlinked parental and recombinant will be equally likely • If genes are linked, see an increase in parental plaques Bacteriophage Mapping • Determine map distance of bacteriophage’s genes by calculating RF • Recombination frequency (RF) Recombinant Plaques = % recombination Total Plaques • 1% recombination = 1 cM • Therefore mapping in phages is exactly the same as eukaryotic gene mapping Complementation Tests Questions Discussion RecA Protein • If recA gene was mutated bacteria could not recombine • Therefore protein encoded by this gene must be necessary for recombination • RecA protein allows recombination between homologous sequences • Other genes and proteins also involved and necessary for recombination as well Map Distance 1.Chapter 7, Q 10 2.Chapter 7, Q 17 • Plus calculate Interference 3.Chapter 7, Q 14 4.Chapter 7, Q 15 • Plus calculate Interference Tetrad Analysis Parents Offspring a + x + b a + a b a + a + a b + b + b + + a b + b + + + + 105 20 32 Calculate distance between genes a and b. Co-transduction Two virus strains Infected bacteria + + + a b c + + + a c b Bacteria Selected Plaques + + + - 4100 a c b a b c None 3990 + c b + b c a 740 a + + b 670 a c + a + c b 160 + + b + b + c 140 a + b a b + c 90 a c + + + c a 110 10,000 Map these three viral genes. Co-transformation + + + Donor bacteria (a b c ) transforms recipient bacteria (a b c ): a+b-c- 180 a-b+c- 150 a+b+c- 210 a-b-c+ 179 a+b-c+ 2 a-b+c+ 1 a+b+c+ 3 Draw the general map of these three genes Hfr Interrupted Mating Hfr Order of Transfer 1 e r i u m b 2 u m b a c t 3 b m u i r e 4 c t e r i u 5 r e t c a b Starting with “B” draw a map of the bacterial genes. On your map draw where each strain inserted and which direction it inserted. Regulation of Gene Expression: Prokaryotes Chapter Fourteen Gene Expression • When a gene is transcribed and translated into functional protein it is expressed • Gene that is not being transcribed, or if it’s mRNA is not being translated, is silent • Not all genes are expressed at all times • In fact there are multiple layers of control of gene expression – tightly regulated • Responsible for development and cell differentiation into different tissues Regulation of Expression • Needs to be tightly controlled for proper function and interaction • Regulation can be mutated, producing: – Too much protein – Too little protein – Protein at wrong time or in wrong location • Phenotypes: – Disease, cancer, cell death External Conditions • Microorganisms change gene expression in response to their environment • Genes can be: – Inducible: turned on in presence of substrate – Repressible: turned off by substrate – Constitutive: always expressed, regardless • Gene expression can be controlled: – Positively: expressed only when stimulated – Negatively: expressed unless shut off Lac Operon • Adjacent regulatory genes encode multiple enzymes used in lactose metabolism • Cell doesn’t need to express these genes unless lactose is present to digest • Presence of lactose is inducer – Gene expression is inducible Lactose Breakdown Lactose is a disaccharide: Breaks down into glucose (preferred food) and galactose (which is converted into glucose) Lac Operon Three enzymes: 1. β-Galactosidase: digests lactose into monosaccharides galatose and glucose 2. Permease: allows lactose to enter cell 3. Transacetylase: removes toxic by- products of lactose digestion • All three genes are contiguous and transcribed as polycistronic mRNA Repressors • Studying mutants in order to determine how the inducer leads to gene expression, discovered only constitutive mutants – Constitutive = always expressed • Therefore, system must normally be repressed and mutants lost repression • Identified two repressor genes: 1. lacI = the repressor 2. lacO = the operator Lac Operon (wt) Regulatory Mutants Other Regulatory Mutants S - C I , I and O are just some of the possible regulatory mutations Regulatory Mutants • Repressor is a trans-acting regulator • Operator is a cis-acting regulator • Repressor is a constitutively expressed gene in bacteria • Inducer (lactose in environment) is actually an inhibitor of an inhibitor • What is advantage to this type of control Summary of Regulatory Mutants O = mutated operator cannot be bound by repressor I = mutated repressor cannot bind operator I = “super-repressor”, mutant repressor cannot be bound by inducer (lactose) • Be able to determine whether genes (Z, Y, A) will be expressed Catabolite Repression • Catabolite is the product of metabolism • If there is enough/too much product, then there’s no reason to make more • Glucose is the main product of digesting lactose – if glucose is present, no reason to digest lactose  turn Lac operon off • Glucose inhibits cAMP • cAMP naturally increases affinity for RNA Pol to Lac’s promoter Catabolite Repression Glucose cAMP cAMP  RNA Pol RNA Pol  transcription • What is the effect on Lactose’s promoter • How does Glucose repress expression of Lac Operon Catabolite Repression Trp Operon • Tryptophan is an important amino acid • Yet, if there is enough Tryptophan in organism’s diet then no reason to make it • Tryptophan represses Trp operon similar to how glucose repressed Lac operon Difference is: • Glucose inhibited a promoting factor • Tryptophan activates a repressor Trp Operon Trp Regulatory Mutations Think about: • Wild type Trp Operon with/without Trp What sort of regulatory mutations could occur and what would be their effect: • What happens when the repressor’s binding site for the Operator is mutated • Repressor’s binding site for Trp mutated • Mutated Operator Attenuation • Upstream of the enzymes that synthesize Tryptophan is a leader sequence • Within the leader is another regulatory sequence known as the attenuator • Attenuator’s encode multiple codons specifying the amino acid being regulated – In this case: Tryptophan – UGG codon Coordinated Expression • Important to know that transcription and translation are occurring simultaneously in prokaryotes • While the mRNA is being made by RNA polymerase, another part of the mRNA is being translated by ribosome • When ribosome comes upon a string of UGG codons, it will be looking for Trp- charged tRNAs Attenuation • When adequate Tryptophan is present in diet/media - charged tRNA’s that match the UGG codon are present • Translation continue through the attenuator sequence • Producing a terminator hairpin which blocks transcription of Trp operon  Tryptophan terminates transcription of the Trp Operon Terminator Hairpin Terminator hairpin and string of U’s Slows RNA polymerase then dissociates RNA Pol from DNA Attenuation • With no, or very small amounts of, Tryptophan in diet/media charged tRNA’s that match the UGG codon are lacking • Translation will not be able to continue through the attenuator sequence • Terminator hairpin is not formed • Transcription continues through operon  Lack of Tryptophan allows transcription of the Trp operon Anti-termination Hairpin RNA polymerase doesn’t dissociate from DNA, Operon expressed Trp Operon Attenuation Lamda Phage • λ is a virus that infects bacteria cells – Bacteriophage • λ chooses between lysogeny and lysis – Temperate phage • Genetic regulation controls expression of pro-lysogenic genes vs. pro-lytic genes • Once again, gene expression is regulated by repressors, influenced by environment Should I stay or should I go The Clash: • “If I go there will be trouble…” • “And if I stay it will be double!” cro v. cI gene expression Regulation of Gene expression Questions Discussion Regulation of Gene Expression: Eukaryotes Chapter Fifteen Gene Expression • When a gene is transcribed and translated into functional protein it is expressed • Gene that is not being transcribed, or if it’s mRNA is not being translated, is silent • Not all genes are expressed at all times • In fact there are multiple layers of control of gene expression – tightly regulated • Responsible for development and cell differentiation into different tissues Regulation of Expression • Needs to be tightly controlled for proper function and interaction • Regulation can be mutated, producing: – Too much protein – Too little protein – Protein at wrong time or in wrong location • Phenotypes: – Disease, cancer, cell death Eukaryotes • How must gene expression be different in eukaryotic organisms – Transcription and translation are separated – mRNA is processed before translation – Multicelluar organisms have different cell types – expression leads to differentiation – Cells interact and signal diverse expression changes continually through time • Overall control is much more complex Eukaryote = Nucleus • Eukaryotic DNA is located within nucleus • Separated from ribosomes that perform translation and produce all proteins • Therefore the proteins directly regulating transcription must be translocated into the nucleus before they can express genes • Same for protein factors that enhance or silence transcription Eukaryotes • Differential gene expression in different cell and tissue types • Also respond to cellular environment ex – White blood cells: express immunoglobins (IGG) to produce variety of antibodies – Pancreatic cells: don’t express IGG, express insulin, but only in response to blood sugar • Loss of regulation leads to developmental defects, diseases and cancer Multiple Layers of Control 1. Chromatin remodeling 2. Transcription 3. Splicing and mRNA processing 4. Transport to cytoplasm 5. Stability of mRNA 6. Translation 7. Post-translational Chromatin Remodeling • Eukaryotes have much more DNA than prokaryotes and DNA is wound around histones into chromatin • When chromatin is tightly packed RNA Pol and other proteins cannot bind DNA – Therefore genes are silenced • Chromatin must be “remodeled” to loosen histones so that other proteins can bind – Genes can then be expressed Chromatin Remodeling Proteins All of these loosen DNA =O Acetylation CH 3- • Addition of an acetyl group to histones lessen histone’s grip on DNA • Histone acetyltransferases (HAT) – Enzymes that add or remove acetyls – Silence or express gene • Histone deacetylases (HDAC) – Enzymes that add or remove acetyls – Silence or express gene CH -3 Methylation • DNA itself can be directly modified by the addition of a Methyl group • Methylation silences gene expression – DNA Methyltransferases Through two different mechanisms: 1. Inhibit the binding of transcription factors 2. Recruit chromatin remodeling proteins that tighten the chromatin of this region Epigenetics • Entire regions of chromosomes can be silenced through acetylation/methylation changes – Affecting multiple genes at once • Induces chromatin remodeling • Heterochromatin = epigenetically silenced because it is so tightly wound – Includes: telomeres, centromeres, and others cis-acting vs. trans Regulation • cis-acting are specific DNA sequences located in and around the genes they are regulating – Promoters – immediately upstream of gene – Enhancers or silencers – further away • trans-acting factors are proteins that encoded by separate genes and then bind to the DNA sequences (on any chromos) – Usually called Transcription Factors Promoters • DNA sequences that serve as: – Recognition site for RNA Pol binding – Necessary for basal levels of transcription • Located immediately upstream (5’) of gene • Contain the transcription start site • And a number of consensus sequences: – TATA box – almost always – CCAAT box, GC repeats, others – often Enhancers Differs from promoters, although these sequences also promote gene expression • Enhancers (and silencers) can be located on either side of gene, away from gene or even in an intron inside gene • Increase transcription beyond basal levels • Time and tissue specific gene expression Silencers • Short cis-acting DNA sequences • Can be found near or far from gene – Any location around or within gene • Act in a tissue or temporal way to regulate gene expression • Only difference from an enhancer sequence is that silencers repress level of transcription off a specific promoter Transcription Factors • cis-acting regulatory sites influence transcription by acting as a binding site for transcription machinery • RNA Pol binds to promoter sequences • Transcription factors are proteins that bind to enhancers and silencers, and either recruit RNA Pol or inhibit it’s binding – Therefore regulating amount of expression Transcription Factors • Transcription factors themselves are regulated so they are expressed based on: – Development time points – In specific tissue types – In response to environmental stimulation • Diverse and complex effects on transcrip: – Enhance or repress transcription – Compete for binding – based on activation, concentration or timing Transcription Factor Domains Transcription factors have two functional domains: 1. DNA binding domain • Area of protein that binds directly to DNA • Must be able to bind to the cis-acting elements upstream of transcription start 2. Activation or repression domain • Area that interacts with RNA Pol, or other proteins, to induce or repress transcription DNA Binding Domains DNA binding domains have certain amino acids and specific shapes to bind DNA: • Helix-turn-helix – Two alpha-helixes separated by specific aa’s • Zinc finger – Contains Cysteines and Histidines that bind Zn+ (cofactor) and can then bind negative DNA • Basic leucine zipper – Leucine’s can “zip” together into a dimer that is basic – interacts with acids, such as DNA Pre-initiation Complex • Some transcription factors assemble at the promoter forming a pre-initiation complex • Platform for RNA Pol to bind DNA of the promoter sequence – Usually at the TATA box Repressors • Specific name for Transcription Factors that inhibit transcription • Perform same function as in prokaryotes • But through different mechanisms: – Blocking an enhancer site – Block TFs from binding – Block the promoter and/or RNA Polymerase • Pre-initiation complex/platform doesn’t form Cell Differentiation Activators and repressors for one specific gene differentially expressed along a developmental axis Differentiation mRNA Processing • In prokaryotes the mRNA is translated immediately and without changes • In eukaryotes mRNA is highly processed before translation can occur • Processing can include: – Caps and tails – Splicing out of intronic sequences – Transporting mRNA out of nucleus to ribosomes in cytoplasm mRNA Stability • All mRNA’s are degraded eventually – Limiting the amount of translation • More stable mRNA = more translation • Therefore, stability of mRNA increases gene expression, leading to more protein • Some methods for controlling stability: – Length of poly-A tail – Inhibiting translation until correct time point – Decreasing RNase’s Alternative Splicing • Using or skipping different combinations of exons will build different proteins from same mRNA sequence – One gene can produce more than one protein RNAi • Small RNA molecules are produced by cell • Bind to single stranded mRNA – Through complementary base pairing • Lead to mRNA being degraded • Therefore, these small interfering RNAs decrease gene expression by inhibiting translation • Known as RNAi – RNA interference Induced RNAi • This process has been hijacked by scientists to remove a single protein – Or decrease it’s expression level – Although RNAi is so effective that most of the time no protein is made • Allows testing of protein’s function – By analyzing mutant phenotypes • Can also be developed as a drug: – Remove abnormal proteins – Lower expression of any overexpressed protein Controlling Translation • RNAi is one example of regulating amount of translation • As with the Lac and Trp operon, the amount of product can also control the level of expression • High concentrations of a protein shut down translation through a feedback loop • Regulatory proteins can bind to mRNA – Rather than DNA – like TFs Post-translational Control • Gene expression can even be controlled after mRNA has been translated into protein • Done by regulating: – Protein modifications  activate or deactivate protein – Protein’s location within the cell – Negative feedback loop – Binding to other proteins  only act as dimer or only once released Protein Folding • Proteins are not able to perform any function immediately after translation • Because they are linear polymers of aa’s • Need to fold into the correct conformation – Sequence  Structure  Function • Proteins that are inhibited from folding, or from folding correctly, are inactive – Therefore, gene is not truly expressed Protein Degradation • Just as with mRNA, proteins can have different levels of stability • Increase stability = longer protein functions • If protein is degraded, function is decreased or lost • Some regulatory proteins add a Ubiquitin tag to proteins so they will be degraded by the Proteasome – Effectively, decreasing gene expression Questions Discussion

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Chapter 5.2, Problem 37E is Solved
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Textbook: Discrete Mathematics and Its Applications
Edition: 7
Author: Kenneth Rosen
ISBN: 9780073383095

Discrete Mathematics and Its Applications was written by and is associated to the ISBN: 9780073383095. The full step-by-step solution to problem: 37E from chapter: 5.2 was answered by , our top Math solution expert on 06/21/17, 07:45AM. The answer to “Let a be an integer and d be a positive integer. Show that the integers q and r with a = dq + r and 0 ? r” is broken down into a number of easy to follow steps, and 28 words. This textbook survival guide was created for the textbook: Discrete Mathematics and Its Applications, edition: 7. This full solution covers the following key subjects: Integer, let, integers, Positive, show. This expansive textbook survival guide covers 101 chapters, and 4221 solutions. Since the solution to 37E from 5.2 chapter was answered, more than 289 students have viewed the full step-by-step answer.

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Let a be an integer and d be a positive integer. Show that