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Use Theorem and the result of exercise 10 to prove that if

Discrete Mathematics with Applications | 4th Edition | ISBN: 9780495391326 | Authors: Susanna S. Epp ISBN: 9780495391326 48

Solution for problem 37E Chapter 5.2

Discrete Mathematics with Applications | 4th Edition

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Discrete Mathematics with Applications | 4th Edition | ISBN: 9780495391326 | Authors: Susanna S. Epp

Discrete Mathematics with Applications | 4th Edition

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Problem 37E

Use Theorem and the result of exercise 10 to prove that if p is any prime number with p ? 5, then the sum of squares of any p consecutive integers is divisible by p.Theorem 1Sum of the First n IntegersFor all integers n ? 1, Proof (by mathematical induction):Let the property P(n) be the equation Show that P(1) is true:To establish P(1), we must show that But the left-hand side of this equation is 1 and the right-hand side is also. Hence P(1) is true.Show that for all integers k? 1, if P(k) is true then P(k + 1) is also true:[Suppose that P(k) is true for a particular but arbitrarily chosen integer k ? 1.That is:] Suppose that k is any integer with k ? 1 such that [We must show that P(k + 1) is true. That is:] We must show that or, equivalently, that [We will show that the left-hand side and the right-hand side of P(k + 1) are equal to the same quantity and thus are equal to each other.]The left-hand side of P(k + 1) is And the right-hand side of P(k + 1) is Thus the two sides of P(k + 1) are equal to the same quantity and so they are equal to each other. Therefore the equation P(k + 1) is true [as was to be shown].[Since we have proved both the basis step and the inductive step, we conclude that thetheorem is true.]

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BIOS 206 –General Genetics Dr. Alan Christensen Chapter 9 Notes “Extranuclear Inheritance” 9.1-9.3 9.1 Organelle heredity involves DNA in chloroplasts and mitochondria • Heteroplasmy –when a few mutants are present, but not enough to be expressed Chloroplasts: variegation in four o’clock plants • Some leaves were white green, and variegated • White leaves lacked chlorophyll to create green color • Inheritance drickly based on ovule source o If from a green leaf, all progeny green • Concluded cytoplasm transmission because pollen has little or no cytoplasm • White leaves from mutation in DNA chloroplasts Chloroplast mutations in chlamydomonas • Has single chloroplast with 75 copies of circular double -strand DNA • First known cytoplasmic mutation, streptomycin resistance (s trR) o Through mt+ and mt-, make equal cytoplasmic contributions o str phenotype transmitted through mt+ parent o Many mutations show the same pattern of single parental inheritance o Fertilization à fusion of 2 cells à meiosis and haploid cells (only show mt+ phenotype) • Mt- parent contributes the mitochrondrial genetic material Mitochondrial mutations: early studies in neurospora and yeast • Mitochondrial mutations transmitted through cytoplasm • Poky –maternally slow-growing mutation in Neurospora crassa o Slow growth = impaired mitochondrial function (missing cytochromproteins) • Cross between poky and wild results in poky, but reciprocal cross makes normal wild colonies • Yeast = saccharomyces cervisiae • Mutation called petite –defieiency in cellular respiration (abnormal electron transport by mitochondria) o Facultative anaerobe –can function with and without oxygen § Without oxygen, it ferments glucose and loss of mitochondrial function o Figure 9-5 • Segragational petites –Mendelian inheritance, result of nuclear mutations for products in mitochondria • Neutral petites –cross with wild = meiotic products (ascospores) only wild or normal colonies o Lack mtDNA completely, need to inherit normal mitochondria that can aerobically respirate o Mitochrondria in yeast inherited from both parental cells • Suppressive petites –cross with wild = diploid zygotes that yield haploid cells with petite phenotype o Dominant-negative mutation –suppresses function of wild mitochondria • Hypotheses for orga nelle DNA 1. Mutant (deleted) DNA in mitochondria (mtDNA) replicates more rapid, thus mutant “takes over”/dominates phenotype 2. Recombination occurs between mutant and wild mtDNA, resulting in errors/disrupting normal mtDNA 9.2 Knowledge of mitochondrial and chloroplast DNA helps explain organelle heredity Organelle DNA and the endosymbiotic theory • Mitochondria DNA looks like bacteria DNA • Much of the DNA originally in mitochondrial genome now in the nucleus where it is transcribed and transported (now less tha n 10%) Molecular organization and gene products of chloroplast DNA • Chloroplasts have DNA and protein -synthesizing apparatus • cpDNA uniform in size, circular, double -stranded, free of euk DNA protein • Compared to nuclear DNA, different density and base compos ition • cpDNA much larger than mtDNA (many genes) o Many noncoding sequences o Duplications also present • rRNA coding genes present (16S and 23S) • tRNA • Independent evolution in chloroplasts • Chloroplast ribosomes have Svedberg coefficient of 70S (bacterial ribosomes) Molecular organization and gene products of mitochondrial DNA • In eukaryotes = double-stranded, closed circle • mtDNA varies among organisms, smaller than cpDNA o No introns o No repetitions o No intergenetic spacer DNA • Replication dependent on enzymes encoded in nuclear DNA • Human mtDNA encodes 2 rRNAs, 22 tRNAs, and 13 polypeptides • Protein synthesizing apparatus and molecular parts for cellular respiration derived from nuclear and mitochondrial genes • mtDNA in vertebrae are different densitie s (H –heavy, L –light) • Mitochondria ribosomes are different from cytoplasm rib.s o Encoded by nuclear genes o Endosymbiont theory confirmed o A polymerase is sensitive to bacterial RNA synthesis antibiotics, not eukaryotic inhibitors. 9.3 Mutations in mitochondrial DNA cause human disorders • mtDNA has gene products with 13 proteins for aerobic cellular respiration • Vulnerable because 1. mtDNA doesn’t have structural protection from mutations from histones 2. DNA repair is limited 3. Mitochondria concentrate mutageneic reactive oxygen species (ROS) through cellular respiration • Toxic and damage proteins, lipids, & mtDNA • 10x more mutations • Variable heritability (heteroplasmy) • Attributable to genetically altered mitochondria: 1. Inheritance must exhibit a maternal rather than Mendelian pattern 2. Disorder must reflect deficiency in bioenergetics function of organelle 3. Must be mutation in one or more mitochondrial genes • Myoclonic epilepsy and ragged -red fiber disease (MERRF) o Maternal transmission o Offspring of affected mothers inheri t, not of affected fathers o Ataxia (lack of muscular coordination), deaf, dementia, epileptic siezures o Blotchy red patches from proliferation of aberrant mitochondria o Brain deficiencies o Mutation in tRNA Lysgene –tRNA that delivers lysine during translation o Heteroplasmy –some normal and abnormal • Leber’s hereditary ooptic neuropathy (LHON) o Maternal inheritance and mtDNA lesions o Sudden bilateral blindness (~27yo) o Mutation disrupting normal oxidative phosphorylation (NADH dehydrogenase) • Kearns-Sayre syndrome (KSS) o Lost vision, hearing loss, heart conditions o Deletions in mtDNA o Progressive symptoms as more mutations accumulate Mitochondria, human health, and aging • Mitochondrial dysfunction associated with anemia, blindness, Type 2 diabetes, autism, atherosclerosis, infertility, neurodegenerative diseases (Parkinson, Alzheimer, Huntington), schizophrenia, and bipolar, and cancers • Carcinogens and UV light cause cancers (don’t know if mtDNA mutations a cause or effect) • Random mutations lead to defectiv e mitochondria, thus aging • Need certain amt of ATP to function, when not enough, aging increases • Nuclear gene mutation to reduce replication of mtDNA results in premature aging and other affects of aging Future prevention of the transmission of mtDNA -based disorders • 1 in 5000 humans have disease of at risk for mtDNA disorder • Can be detected, but no cure • Can prevent transmission from mother to offspring o Insert nuclease of mother into donor egg with normal mtDNA o Mitochondrial swapping o DNA from 3 parents. Ethical Legal BIOS 206 –General Genetics Dr. Alan Christensen Chapter 16 Notes “Regulation of Gene Expression in Prokaryotes” 16.1-4 9.4 Prokaryotes regulate gene expression in response to environmental conditions • Lactose (a galactose and glucose-containing disaccharide) in growth medium of yeast o Organisms synthesize enzymes required for lactose metabolism o When absent, synthesis diminishes to basal level • Bacteria can adapt to enviro producing inducible enzymes depending on chemical substrates present • Continuously produced enzymes called constitutive enzymes • Sometimes gene expression inhibited, usually ending as end products in anabolic biosynthetic pathways o Amino acid, tryptophan, in enviro/culture medium, no reason for organism to make tryptophan and waste enevery o Tryptophan represses transcription of mRNA to make tryptophan enxumes o Repressible, negative control • Under negative or positive control o Negative –genetic expression occurs unless shut of by regulator molecule o Positive –transcription occurs only if regulator molecule directly stimulates RNA production 9.5 Lactose metabolism in E coli is regulated by an inducible system • Jacques Monod (1946) lactose metabolism evidence • In presence of Lactose, concentration of enzymes for metabolism increased from few to thousands per cell (inducible, and induced by lactose) • Genes responsible for related functions are in clusters under control of one regulatory region o Region upstream 5’ of cluster o On same strand as genes –cis-acting site o Cis-acting regions bind to molecules controlling transcription of gene cluster (trans-acting elements) • Decide if transcribed to mRNA • Can be positive (turn on transcription) or negative (turn off) • Lactose or lac, operon -3 structural genes and adjacent regulatory site that decide if lactose needs to be made or not Structural genes • Structural genes –genes coding primary structure of enzyme • 3 in the lac operon o lacZ encodes B-galactosidase –converts disaccharide to monosaccharides o lacY encodes permease –facilitates entry of lactose into bacterial cell o lacA encodes transacetylase –might remove toxic by-products • Joshua Lederberg removed one and saw what happened (lac -) o When cant make b -galactosidase (Z-) or permease (Y-), cant use lactose o Genes closely linked or bordering one another in ZYA order • Transcribed as single -unit (polycistronic mRNA) Discovery of regulatory mutations • Gratuitious incuders –sulfur containing isoproyplthiogalactosides (IPTG) • Natural inducers –doesn’t depend on interaction between inducer and enzyme • Constitutive mutations –enzymes are made no matter what o lacI- close to ZYA –is a repressor gene o Constitutive mutations are identical to laci - in region adjacent to structural genes § lac OC, on operator region • Enzymes made continually, reg ulation lost Operon model: negative control • Operon model –hypothetical mechanism for negative control • Group of genes regulated and expressed as unit • ZYA in lac operon and adjacent sequences (all called operator region) • lacI regulates structural gene transcription making a repressor molecule which is allosteric (reversibly interacts with another molecule , undergoes conformational change in 3D shape, and changes chemical activity) • Repressor normally on DNA sequence of operator inhibiting RNA polymerase • When lactose present, sugar binds to repressor and caused conformational change o Can’t connect to operator o RNA polymerase then transcribes structural genes and enzymes for lactose metabolism • Negative control –transcription occurs when repressor fails to bind to operator • Without lactose, enzymes encoded aren’t needed and expression repressed • With lactose, genes to break down lactose encoded until no lactose left to -ind so Chen repressed again • I and O mutations interfere with molecular interactions and keep transcribing - o I repressor is altered/absent and cant bind to operator, so genes always on o O nucleotide sequence of operator DNA altered and no binding to repressor, genes always transcribed Genetic proof of the operon model • 3 major predictions to predict validity 1. I gene makes diffusible product (trans-acting product) 2. O region involved in regulation but doesn’t make product (cis -acting) 3. O region must be adjacent to structural genes to regulate transcription • F plasmid may have chromosomal genes (F’) - o F cell gets F’, has own chromosome plus 1+ extra genes in plasmid o Host cell called merozygote –cell diploid for certain genes o Can introduce I+ gene to host cell with genotype I -, etc. o Jacob-Monod operon model predicts regulation § Adding I+ to I- restores inducibility (trans) § Adding O+ to OC = no change (cis, needs to be adjacent to structural genes) o DRAW PICTURES (PG 400) • Certain mutations in I gnee should have opposite effect of I o Mutant repressor molecules should be made that ca nt interact with inducer, lactose o Always bind to operator and permanently repressed o I would have no effect on repression S o I discovered. Super repression. o (Different DNA-binding and inducer-binding domains) Isolation of the repressor • E. coli has up to 10 copies of lac repressor • Didn’t know if repressor molecule was a protein or RNA • I mutant used to isolate lac repressor • IPTG binds to repressor (called equilibrium dialysis) o Material that binded to it was isolated and had protein characteristics 9.6 The Catabolite-Activating Protein (CAP) exerts positive control over the lac operon • B-galactosidase cleaves lactose into glucose and galactose • Galactose must be converted to glucose to be used • When lactose and glucose in a medium, CAP acts ascatabolite repression to not process lactose o CAP makes binding more efficient • Transcription is initiated when binding of RNA polymerase and promoter region (upstream 5’ from initial coding sequences) o In lac operon, found between I and O • Without glucose and in inducible conditions, CAP has positive control (binding to CAP site, facilitates RNA -polymerase and transcription) • Max transcription = repressor bound by lactose and CAP bound to binding site • Why does glucose inhibit CAP binding o cAMP and CAP must be bound together to bind at all o Presence of cAMP depends on enzyme adenyl cyclase (ATP to cAMP) o Glucose inhibits adenyl cyclase, reducing cAMP and no binding of CAP • cAMP-CAP complex changes conformation when bound 9.7 Crystal structure analysis of repressor complexes has confirmed the operon model • Repressor –product of I gene, monomer of 360 amino acids • Repressor that binds to inducer is a homotetramer (4 copies of monomer) • Cleaved with protease to make 5 fragments o 4 are N-terminal ends of subunits, bind to operator o 5 derived from COOH-terminal ends, binds to lactose and IPTG o All can bind to 2 symmetrical operator DNA • Operator -21 base pairs, primary operator o Auxiliary operators (401 base pairs downstream from primary, in lacZ) o Another aux. O is 93 BP upstream primary (beyond cap site) • All 3 operators must be bound for max repression • Binding repressor to 2 operator sites distorts DNA conformation o Make repression loop o Promoter inside loop (prevents access by polymerase) o CAP-binding site to interact with RNA polymerase BIOS 206 –General Genetics Dr. Alan Christensen Chapter 17 Notes “Regulation of Gene Expression in Eukaryotes” 9.8 Eukaryotic gene regulation can occur at any of the steps leading from DNA to protein product • Euk cells have more DNA, change structure to modify gene expression • mRNA’s spliced, capped, and polyadenylated to change # of mRNA available for translation • DNA in nucleus, mRNA enters and is regulated by DNA expression • mRNAs have different half -lives (longer than PROK) • Translation rates can be modified 9.9 Eukaryotic gene expression is influenced by chromatin modifications • RNA polymerase II (RNAP II) transcribes Chromosome territories and transcription factories • Interphase nucleus –has chromosome territory and interchromsomal domains • Transcription factory –have active RNA polymerases and are somehow involved in transcription Open and closed chromatin • Groudine and Weintraub isolated nuclei and put them with DNase o Tight bound ones are resistant to DNase I o Concluded transcriptionally active and inactive genes are associated with nucleosomes o Active genes altered conformation Histone modifications and nucleosomal chromatin remodeling • Change composition –promoter either activated or repressed • Histone modification –adding acetyl, methyl, or phosphate groups to amino acid histone tails o Opens conformations o Histone acetylation catalyzed by histone acetyl transferase enzymes (HATs). o Histone deacetylases remove acetate groups from histone tails • Reposition/Remove nucleosomes on DNA o Remodeling complex –SWI/SNF o Loosen DNA and histones o Reorganize nucleosome components o DNA exposed DNA methylation • Often at 5 position of cytosine • Methyl group protrudes into major groove of DNA helix • On cytosine of CG doublets in DNA on both strands • CpG islands –concentrated CpG-rich regions, 5’ end in promoter region • 5% cytosine residues methylated • Based on inverse relationship between degree of methylation and degree of expression o Inert regions heavily methylated • Methylation patterns are tissue specific and heritable o Essential for mammalian development o Can be altered by methylase and demethylase enzymes to activate or silence DNA • 5-azacytidine –nucloside used in place of cytidine during DNA replication o Analog cannot be methylated o Changes pattern of expression and stimulates alleles on inactivated X chrom o Induce expression of genes normally silent • Methylation inhibits binding of transcription factors to DNA 9.10 Eukaryotic transcription initiation requires specific c is-acting sites • Cis-acting sequence –located on same chromosome as gene it regulates • Trans-acting sequence –(DNA-binding proteins/small RNA molecules) influence expression on any chromosome Promoter elements • Promoter –region of DNA that knows transcription machinery and binds to 1+ proteins that regulate transcription o Needed for initiation o Next to genes they regulate o Specify where to start and direction o Promoter elements –short nucleotide sequences that bind t o specific regulatory factors • Core promoter –gets accurate initiation by RNAP II • Proximal promoter elements –modulate efficiency of basal level • Much diversity in promoters (focused or dispersed) • Focused promoters –at single nucleotide ( Trasncription start site) o Major type for lower euk (30% vertebrae) o Associated with genes with highly regulated o Made of 1+ DNA sequence elements with Initatior (Inr), TATA box, TFIIB recognition element (BRE), downstream promoter element (DPE), and motif ten element (MTE) *(only in some) o Role: to bind specific proteins o INR –start site (-2 to +4) § Humans its YYAN / YT (Y=pyrimidine, N=Any nucleotide) o TATA –(-30) TATA / AAR (R=any purine) T o BRE –in core promoters either immediately upstream or downstream TATA box o MTE and DPE –sequence motifs downstream start side (+18 to +27 and +28 to +33) • Dispersed promoters –initiation from # of weak transcription start sites over 50- to 100- region o Transcribed constitutively o Found in CpG islands (influenced by chromatin modifications) o Maybe made out of same stuff as focused but more errors • Proximal promoter elements –upstream TATA and BRE o Act on core-elements to increase transcription o CAAT box –CAAT or CCAAT (70 -80 bp upstream Inr) o Critical for initiation o Murations lower rate • GC box –GGGCGG and one or more at ( -110) Enhancers and silencers • Enhancers –anywhere by or in gene o Cis because function on same chromosome o Enhancers necessary for max transcription o Time and tissue specific expression • Distinguished from promoters by 1. Position not fixed relative to gene 2. Orientation can be inverted with no effect 3. If cloned, transcription enhanced • Enhancer in gene–immunoglobulin heavy-chain gene enhancer (in intron) o Active in cells expressing immunoglobulin genes (tissue specific) • Enhancer downstream is B-globin o In chickens, between B- and E-globin. Embryonic goes to E - and adult goes to B- way • Modular and have diff DNA sequences • SILENCER –repress level of transcription initiation o Cis acting short DNA sequences o Act in tissue- or temporal-specific ways 9.11 Eukaryotic transcription initiation is regulated by transcription factors that bind to cis-acting sites • Transcription factors –transcription regulatory proteins • Activators –increase level of transcription • Repressors –reduce transcription levels • Expressed at certain time or in response to certain stimulus • Different factors can interact and change timing Human metallothionein IIA gene: multiple cis-acting elements and transcription factors • hMTIIA –interaction of many promoter and enhancer elements and Tfactors • hMTIIA product is protein that binds to zinc and cadmium and protects cells from toxic effects o Protect from oxidative stress • Each cis-acting element is short DNA sequence that binds to 1+ Tfactor • hMTIIA has core-promoter in TATA and Inr • Free zinc binds to zinc-finger motifs bind to MREs upstream changing their conformation • HMTIIA can be influenced by external and internal conditions Functional domains of eukaryotic transcription factors • DNA-binding domain –binds to DNA sequences in cis -acting regulatory site o 3D motifs à § Helix-turn-helix (HTH) –pro and euk Tfactors • Genometric conformation • 2 adjacent A-helices separated by “turn” of amino acids • Proteins can bind § Zinc finger • MTF1 in many Tfactors • Clusters of 2 cysteines and 2 histidines repeating • Into loops and interact with DNA sequences § Basic leucine zipper (bZIP) • Zipper allows protein -protein dimerization • When 2 dimerize, leucines “zip” together • Dimer has 2 a-helical regions adjacent to zipper and bind to phosphate and bases • Trans-activating/Trans-repression domain –activate or repress transcription o Interact with other Tfactors or RNA polymerase o 30-100 amino acids (vary) 9.12 Activators and repressors interact with general transcription factors and affect chromatin structure Formation of the RNA polymerase II t ranscription initiation complex • General transcription factors –needed at the promoter to initiate transcription. Together form preinitiation complex (PIC) o Platform for RNAP II to recognize start sites and initiate • RNAP II general Tfactors are TFIID, TFIIB, TFIIA, TFIIE, TFIIF, TIFFH, and Mediator. 1. Forming PIC –all general Tfactors come together around the TATA box and allow transcription • RNAP II leaves promoter and goes down template to elongation complex Mechanisms of transcription activation and repression • DNA loops bing distant enhancer/silencer elements close to promoter • Support from formaldehyde cross linking (proteins crosslinked to DNA, shows enhancer and promoter regions are close when promoter active ) and direct visualization (FISH technizue to show enhancers and promoters close to nucleus) • Recruitment model –enhancer and silencer are donors to increase concentration of regulatory proteins o Increase rate of PIC assembly/stability or increase release of RNAP II from promoter, Tactivato rs increase initiation rate o Coactivators help make enhanceosomes • Chromatin alterations –open or close promoter to interactions with Tfactor • Nuclear relocation –enhancer or repressor looping may relocate target gene 9.13 Gene regulation in a model organism: tran scription of the GAL genes of yeast • GAL gene system -4 structural genes, 3 regulatory • Inducible –regulated by presence/absence of glactose o If gone = not transcribed o If there = transcription • Positive control –activator must be present to turn on gene • HOW GAL1 AND GAL10 ARE UNDER POSITIVE REGULATION • Controlled by UAS G70bp, like enhancers • Constitutively open (DNase hypersensitive) • 4 sites for Gal4 (always occupied) • Gal4p neg regulated by Gal80p • No galactose à Gal80p on Gal4p • Gal3p allows Gal4p to open up 9.14 Posttranscriptional gene regulation occurs at many steps from RNA processing to protein modification Alternative splicing of mRNA • CT/CGRP gene from pre -mRNA o Thyroid, first 4 exons, regulate calcium o Brain, 12356 exons • Proteome –total number of proteins an organism can make • Dscam gene (axon growth) –many different possibilities • Para -6 sites for alt splicing and 11 editing sites o RNA editing –base substitutions after transcription and splicing Alternative splicing and human diseases • Myotonic dystrophy –muscular dystrophy, autosomal dominant o DM1 (way too many copies of CTG repeat in DMPK gene) o DM2 (repeat of CCTG in intron of ZNF9 gene) o Toxic effect of repetitions o Spliceopathies Sex determination in Drosophila: a model for regulation of alternative splicing • Cascade for sex determination, different splicing creates diff sex • Sex lethal (Sxl) –Femalesàactivate gene, Malesànot enough to activate o Cascade begins o Default male splicing • Transformer (tra) gene –in male and females o Femaleàpre-mRNA spliced to make rna to functional TRA protein o Maleàstop codon • Doublesex (dsx) –different transcripts Control of mRNA stability • Steady-state level –rate of transcription and mRNA degradation • Measured in half life • Degraded along 3 pathways o Targeted by enzymes and eat poly-a-tail o Decapping enzymes remove cap o mRNA cleaved § Nonsense mediated decay –terminate at premature stop codon • Stability regulated o Recruit helpful/harmful complexes § Stability –adenosine-uracil rich element (ARE) Translational and posttranslational regulation • P53 –protect normal cells from DNA damage o Normally low levels o DNA damage or stress increase levels • Mdm2 and p53 attached when no stress. Prevents tranctiption o Ubiquitin ligase –breaks down p53 • Presence of p53 triggers negative feedback loop to make more Mdm2, to degrade p53 9.15 RNA silencing controls gene expression in several ways • RNA interference (RNAi) –repress translation and trigger degradation of mRNA • RNA-induced gene silencing –alter chromatin structure and repress transcription • Double strand RNAi was way more powerful in repressing gene Molecular mechanisms of RNA -induced gene silencing • Small interfering RNAs (siRNAs) o Learnear, double strnd in cytoplasm o Virus or transposon o Dicer –double strand RNA cleaver (cleaves to siRNA) • MicroRNAs (miRNAs) o Single strand, in nucleus, Double strand stem -loop o Loop cut off o Move to cytoplasm and cut by dicer to double -strand miRNA • RNAi pathway steps 1. siRNA/miRNA associate with RIS(ilencing)C 2. 2strand RNA denatured and sense strand degraded 3. RNA/RISC complex functional and specific, seek complementary 4. If complementary, cleave mRNA then degraded 5. If not complementary, stay with mRNA and no translation • RIT(ranscription)S(ilencing) o Repress transcription of specific genes o Antisense RNA in RITS targets RITS for specific promoter o Gets enzymes that methylate histones and DNA RNA-induced gene silencing in biotechnology and medicine • Can make defects specifically and cheap • Synthetic siRNA for cultures • Attack bad things with RNAi o Viral infections o HIV o The flu and hep C o Cancers –defects in miRNA expression, treat with siRNA 9.16 Programmed DNA rearrangements regulate expression of a small number of genes Immune system and antibody diversity • Antigen recognition –lock and key binding when see foreign substances • Antigens –molecules, proteins, that cause immune response • Humoral immunity –making immunoglobulins or antibodies that attach to antigens • Imglob made from B lymphocyte or B cell (bone marrow) • Imglob has 2 polypeptides: light chains and heavy chains o Make a Y shape o Has constant recion at C-terminus and variable revion at N- terminus o Recognize 1 specific antigen • Imglob genes not fixed. Change through deletions, translocations, and random mutations to increase amount of antibodies Gene rearrangements in the K light-chain gene • Made in b-cell development • 70-100 Leader and variable regions (50% functional) • LV region after transcription initiation sequence o 6 kb from joining region o close to enhancer and C exon • 1 LV region joins 1 J region • Promoter activated à transcription • Hypermutation –susceptible to high rate of random somatic mutation during B-cell development • There is much diversity in making this molecule, contributed by many things 9.17 ENCODE data are transforming our concepts of eukaryotic gene regulation • Identify all functional DNA sequences • 80% genome has regulatory element or does active transcription o Used to think it was junk Enhancer and promoter elements • Using knowledge, furthering knowledge • 3700 DNA sites bound by Tfactors in all cells • Most promoters in genes • Enhancers influence transcription at any promoter, not just closest one • Unknown things tell enhancer which promoter to go to Transcripts and RNA processing • 75% whole genome can be transcribed • Enhancer RNA –made by active enhancers • Many splicing variants sim ultaneously BIOS 206 –General Genetics Dr. Alan Christensen Chapter 20 Notes “Recombinant DNA Technology” 9.18 Recombinant DNA technology began with 2 key tools: restriction enzymes and DNA cloning vectors Restriction enzymes cut DNA at specific recognition sequences • Degrade DNA • Binds to recognition sequence or restriction site. • Cut both strands by cleaving phosphodiester backbone (“digestion”) • Can cut to certain sizes • # of fragments made by enzyme estimated from # times cuts DNA 4 o 4 base recognition sequence = 4 = every 256 base pairs • Symmetry called palindrome –sequence that reads same on both strands when read at 5’ to 3’ o EcoRI and HindIII make overhanging cohesive ends (“Sticky”) o AluI and BalI cut at same pair, blunt -end fragments • EcoRI –can base pair with complemen tary single-strand of other EcoRI cut fragments o Can anneal –stick together o DNA ligase binds strands together DNA vectors accept and replicate DNA molecules to be cloned • Cloning vectors –accept DNA fragments and replicate them when vectors present o Vector has many restriction sites to insert DNA fragments o Can replicate in host cell to make independent replication of vector DNA and DNA fragment o Selectable marker gene –distinguish cells with vectors and without vectors. Antibiotic resistance gene that encodes protein with visible product (color, fluorescent light) o Specific sequences for sequencing inserted DNA Bacterial plasmid vectors • Plasmids –first vectors o Extrachromosomal o Double-stranded • Plasmids inserted by transformation (treat with calcium with brheat shock for DNA into cells) or electroporation (high-intensity, brief pulse of electricity for DNA into cells) • To clone: o Cut with same enzyme, join, and replicate o Sometimes close back on themselves o Not all cells take up recombinant • How can you tell whic h ones are clones o Selectable marker genes (resistance to antibiotics and genes) blue- white selection § LacZ encodes b-galactosidase § If recombination in LacZ, wont make B- gene § Put on agar plate –ampicillin § Cells without recombination die (no resistance gen e) § Agar plate has X-gal –similar to lactose, cleaved turns blue • If didn’t take up gene, turn blue • Recombinants turn white (clones of each other) • Can only accept small pieces (25kb) Other types of cloning vectors • Bacteriophage λ o Genome mapped and sequenced, added cloning sites • Phage vectors popular –take up 2x than plasmid vectors • BACs and YACs are vectors for cloning large fragments of DNA o BACs –not many copies, accept 100 -300 kb o YACs –telomeres at both ends, origins of replication and centromere; accept 100-1000kb • Expression vectors –ensure mRNA expression Ti vectors for plant cells • Tumor inducing • Insert genes into plant cells and cause tumors (aka crown galls) • Ti plasmid inserted and replicates to make tumors • Restriction site in Ti plasmid to insert foreign DNA o Ti genes removed from vector Host cells for cloning vectors • Yeast used for cloning experiments o Euk but grows like bacteria o Good genetic map and sequenced genome o Can use functional form o Safe to use in vaccines and therapeutic agents 9.19 DNA libraries are collections of cloned sequences Genomic libraries • Overlapping fragments of genome • Includes introns and possibly partial genes • Want to make least amount of clones needed • Used YAC’s to clone DNA genome • Whole-genome shotgun approach –new way of doing it Complementray DNA (cDNA) libraries • Better than genomic • Have DNA copies of noly genes • Easier to identify function of each gene • Isolate mRNA with poly-A- tail o Mix with oligo(dT) primers (short, single -strand sequence of T nucleotides) o Reverse transcriptase –extends primer and synthesize cDNA copy of mRNA o Make mRNA-DNA double-strand hybrid • Inserted into plasmids (add linker sequence) Specific genes can be recovered from a library by screening • Library screening –sort through all genes to find one you wanna look at • Probes –DNA or RNA sequence complementary to sequence in library o Needs to be tagged, radioactive o Sometimes made from other species o Bacteria grown on agar plates, replica made, lyse bacteria cells (denature DNA and bind new DNA) o Heated then cooled quickly to keep single -strands o If sequence is complementary to probe, double-strand will form – hybridization o Incubated and washed to keep what want o Autoradiography –xray film will turn black if radioactive part touches • Genomics –sequenced without making a library 9.20 Polymerase chain reaction is a powerful technique for copying DNA • FAST! • No host cell needed • In vitro • Sensitive and amplifying • • Primers attach to sstrand DNA and copy through 3’ end. This makes another strand. • Fast cuz exponential • Can be made into plasmid vectors 1. Denaturation: dstrand to sstrand DNA at 92 -95°C for 1 min 2. Hybridization/Annealing: temp to 45 -65°C. primer binds to sstrand DNA. Signals where start is for synthesis ] 3. Extension: 65-75°C and DNA polymerase synthesizes new strands (5’ to 3’) • Thermocyclers can be set to perform PCR • Have to use a good protein to withstand temp changes Limitations of PCR • Must know part of sequence • Easily contaminated (don’t wanna replicate bad stuff) • Can’t do super long strands Applications of PCR • Versatile • Diagnose bacteria and viruses • Good in c • Crime scene when sample is small! • RT-PCR –mRNA production , shows levels of expression • qPCR/real-time PCR –measures amount of PCR made. o Uses light to measure o SYBR Green and TaqMan probes 9.21 Molecular techniques for analyzing DNA Restriction mapping • Tells #, order, and distances between restriction enzyme cleavage sites along cloned segment of DNA Nucleic acid blotting • Southern blot –can determine specific DNA fragments • Northern blot –detect specific RNA fragments • Western –detect protein using antibody probe • Eastern –use analyze protein post -translational modification 9.22 DNA sequencing is the ultimate way to characterize DNA structure at molecular level o Dideoxynucleotide chain -termination sequencing or Sanger sequencing § Dstrand DNA converted to sstrands as template § Add a little dideoxynucleotide (ddNTP) • Have 3’ H instead of hydroxyl group • Cant form phosphodiester bond • Sequencing terminates o Each ddNTP labeled with different color in the electrophoresis and autoadiography o Each color goes to machine and converted into sequence o Called computer automated high -throughput DNA sequencing Sequencing technologies have progressed rapidly • Sanger is used for small sequencing • Costly Next-generation and 3 -generation sequencing technologies • It’s a race to see who can do it first • Pyrosequencing –sequence DNA on beads of each well o Sequence put into well o If nucleotide added, releases pyrophosphate which releases light o Lets you know when single nucleotide added to the stra nd • Much faster but hard to store the data • Sequencing by synthesis technique • Cheapter than sanger • TGS -single molecule of sstrand DNA DNA sequencing and genomics 9.23 Creating knockout and transgenic organisms for studying gene function Gene targeting and knockout animal models • KO technology is to determine the function of a gene • Takes a long time to make but once made, can easily be shared with people Making a transgenic animal: the basics • Knock-in animals –have desired transgene Email me if you need any clearifications or want more notes! beccasehnert@gmail.com

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Chapter 5.2, Problem 37E is Solved
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Textbook: Discrete Mathematics with Applications
Edition: 4
Author: Susanna S. Epp
ISBN: 9780495391326

The full step-by-step solution to problem: 37E from chapter: 5.2 was answered by , our top Math solution expert on 07/19/17, 06:34AM. This textbook survival guide was created for the textbook: Discrete Mathematics with Applications , edition: 4. The answer to “Use Theorem and the result of exercise 10 to prove that if p is any prime number with p ? 5, then the sum of squares of any p consecutive integers is divisible by p.Theorem 1Sum of the First n IntegersFor all integers n ? 1, Proof (by mathematical induction):Let the property P(n) be the equation Show that P(1) is true:To establish P(1), we must show that But the left-hand side of this equation is 1 and the right-hand side is also. Hence P(1) is true.Show that for all integers k? 1, if P(k) is true then P(k + 1) is also true:[Suppose that P(k) is true for a particular but arbitrarily chosen integer k ? 1.That is:] Suppose that k is any integer with k ? 1 such that [We must show that P(k + 1) is true. That is:] We must show that or, equivalently, that [We will show that the left-hand side and the right-hand side of P(k + 1) are equal to the same quantity and thus are equal to each other.]The left-hand side of P(k + 1) is And the right-hand side of P(k + 1) is Thus the two sides of P(k + 1) are equal to the same quantity and so they are equal to each other. Therefore the equation P(k + 1) is true [as was to be shown].[Since we have proved both the basis step and the inductive step, we conclude that thetheorem is true.]” is broken down into a number of easy to follow steps, and 244 words. Since the solution to 37E from 5.2 chapter was answered, more than 300 students have viewed the full step-by-step answer. Discrete Mathematics with Applications was written by and is associated to the ISBN: 9780495391326. This full solution covers the following key subjects: true, hand, SIDE, show, integers. This expansive textbook survival guide covers 131 chapters, and 5076 solutions.

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Use Theorem and the result of exercise 10 to prove that if