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For the beam and loading shown, | Ch 9 - 9.133, 6th Edition

Mechanics of Materials | 6th Edition | ISBN: 9780073380285 | Authors: Ferdinand Beer ISBN: 9780073380285 142

Solution for problem 9.133 Chapter 9

Mechanics of Materials | 6th Edition

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Mechanics of Materials | 6th Edition | ISBN: 9780073380285 | Authors: Ferdinand Beer

Mechanics of Materials | 6th Edition

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Problem 9.133

For the beam and loading shown, determine (a) the slope at point A, 615(b) the deflection at point D.

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2/8: Replication Online Lecture See notes from 2/5  Replication allows the organism to grow, must be accurate to create genetically identical daughter cells  Replication begins at the origin of replication (AT rich) o New strand proceeds from 5’  3’ o Template strand replicated from 3’  5’  Bidirectional DNA replication in prokaryotes o 2 original strands are templates for the new daughter strands (semiconservative) o Replication forks will proceed in each direction in bidirectional  2 forks are moving in opposite direction  speeds up replication rate (about 40 min. to replicate)  Multiple replication forks allow for cell to decrease doubling time o DNA Pol III stops when it reaches synthesized DNA o DNA Pol I replaced Pol III, removes RNA primer and adds nucleotides o DNA ligase­ makes last phosphodiester bond sealing nicks (also helps in DNA repair)  DNA pol­ catalyze the addition of nucleotides (dNTPs­ deoxyribonucleotide triphosphate) o DNA Pol III­ primary enzyme replicating chromosomal DNA o Can only add dNTP to existing 3’ –OH (only synthesize DNA from existing template) o Requires an RNA primer to start replication  Errors in replication­ introduced mutations into the system o Low rates of mutations: 10 – 10 errors per base pair due to proofreading by DNA pol III  Replication fork: o Single stranded binding proteins­ stabilize unwound parental DNA o RNA primer­ 11­12 nucleotides complement to DNA  One on leading, multiple on lagging  Primosome­ helicase and primase subcomplex  Chromosome is replicated bidirectionally in prokaryotes 2/10: Microbial Growth Readings: 5.1­5.1, 5.5­5.7 I: Bacterial Cell Division 5.1 Binary Fission  Binary fission­ cell elongates to twice its size and constricts to form two daughter cells o Septum­ inward growth of cytoplasmic membrane and cell wall from opposing directions  Generation time­ time it takes for one cell to divide into two daughter cells  Balanced growth­ each cell receives the same organelles and genetic information 5.2 Fts Proteins and Cell Division  Divisome­ division apparatus in cell, orchestrates synthesis of new cytoplasmic membrane and cell wall (division septum). Proteins involved (Fts­ filamentous temperature­sensitive) o FtsZ­ tubulin, attachment of FtsZ ring around the center of cell, guided by Min proteins to cell midpoint o FtsA­ actin, connects FtsZ ring and recruits other divisome proteins o FtsI­ penicillin binding proteins o ZipA­ anchors FtsZ ring to cytoplasmic membrane o MinD­ localize MinC, oscillates, preventing FtsZ ring from forming o MinE­ sweep MinC/D aside, keeping lowest concentration of these in the middle  septum forms  Cell constricts  FtsZ ring depolymerized  inward growth of cell wall  Short generation times aided by multiple replication forms  each cell has at least 1 (maybe more) copy of genes II: Population Growth 5.5 Quantitative Aspects of Microbial Growth  Exponential growth­ number of cells double in time interval o Log graph­ straight line, cells number is growing, but the rate is constant  N=N 0 n n=3.3(log N – log N0) o N= final number of cells. 0 = initial number of cells. n= number of generations in time t  G­ generation time (g= n ¿ log2 k= ¿  K­ specific growth rate ( g 1  V­ division rate (= ¿g , h1 o Number of generations per unit time 5.6 Growth Cycle  Lag phase­ between inoculation and beginning of growth, transfer from rich culture medium to poorer one  Exponential phase­ healthiest state, most desirable for study  Stationary­ growth is limited due to depletion of essential nutrient or accumulation of waste products, growth rate=0, some growing some dying  Death phase­ slower rate of exponential death than growth, some cells will remain for long periods of time  Events are characteristic of populations of cells, not individual cells 5.7 Continuous Culture  Chemostat­ continuous culture (open system microbial culture of fixed volume) o Growth rate (how fast) and cell density (how many cells per mL obtained) controlled independently. o Dilution rate­ rate at which fresh medium pumped in and spent medium pumped out (affects growth rate)  Too high­ organisms washed out because cannot grow fast enough  Too low­ cells die from starvation because not enough limiting nutrient o Concentration of limiting nutrient­ entering the chemostat, C (affects cell density)  Increasing concentration of a limiting nutrient results in greater biomass but same growth rate  Concentration of incoming nutrient increased at a constant dilution rate, cell density will increase, but growth rate is constant  Varying dilution rate and nutrient level can establish dilute, moderate, or dense cell populations at any growth rate Lecture:  Production of new cell wall material is a major feature of cell division o In cocci, cell walls grow in opposite directions outward from the FtsZ ring o In rod­shaped cells, growth occurs at several points along length of the cell  Preexisting peptidoglycan needs to be severed to allow newly synthesized peptidoglycan to form o Beginning at the FtsZ ring, small openings in the wall are created by autolysins where new cell wall material is added o Wall band: junction between new and old peptidoglycan  Bactoprenol: carrier molecule, insertion of peptidoglycan precursors o Bonds to N­acetylglucosamine/N­acetylmuramic acid/pentapeptide peptidoglycan precursor  Glycolases: enzymes that interact with bactoprenol o Insert cell wall precursors into growing points of cell wall, catalyze glycosidic bond formation  Generation time shorter for prokaryotes than eukaryotes  In a batch culture, growth conditions are constantly changing; it is impossible to independently control both growth parameters  Growth measured by: o Petroff­Hausser counting chamber­ cells counted in small square and multiple by plate size o Serial dilution­plate count x dilution factor = cells per milliliter of original sample 2/12: Factors Affecting Microbial Growth Readings: 27.11­.18 IV: Antimicrobial Drugs 27.11 Synthetic Antimicrobial Drugs  Antimicrobial drugs­ kill or control growth of microorganisms in the host  Selective toxicity­ inhibit or kill pathogens without adversely affecting the host  2 categories of antimicrobial agents: o Synthetic antimicrobial drugs  Growth factor analogs­ interfere with microbial metabolism, prevent the analog from functioning in the cell (sulfa drugs)  Quinolones­ interfere with bacterial DNA gyrase, preventing supercoiling  interfere with DNA packaging (fluoroquinolones­ ciprofloxacin UTI treatment) o Antibiotics (27.12)­ produced by microorganisms, inhibit or kill other microorganisms, 1% are clinically used  Broad spectrum­ effective against gram +/­  Narrow spectrum used when broad does not work 27.13 β­lactam antibiotics: Penicillin’s and Cephalosporin’s  β­lactam antibiotics­ cell wall synthesis inhibitors o Penicillin­ first antibiotic o Cephalosporins­ bind to PBP and prevent cross linking of peptidoglycan, more resistant to β­lactamase which destroys the β­lactam ring 27.14 Antibiotics from Bacteria  Aminoglycosides­ target 30S subunit, inhibits protein synthesis, used when other antibiotics fail  Macrolide­ 20% of total antibiotics  Tetracyclines­ broad spectrum, along with β­lactam drugs, are two most important classes of antibiotics  Novel antibiotics: o Daptomycin­ creates port and induces rapid depolarization of membrane o Platensimycin­ fatty acid biosynthesis, no host toxicity or potential of resistance 27.15 Antiviral Drugs  Azidothymidine­ blocks reverse transcription  Nucleoside reverse transcription inhibitors (NRTIs)­ inhibit elongation of viral nucleic acids, lost potency  Nonnucleoside reverse transcriptase inhibitors (NNRTI)­ inhibits reverse transcription  Protease inhibitors­ prevent viral maturation  Fusion inhibitor­ stops conformational changes necessary for fusion of HIV and T lymphocyte  Virus interference­ infection with one virus interferes with subsequent infection by another virus  Interferons­ prevent viral replication by stimulating production of antiviral proteins in uninfected cells o Highly virulent viruses inhibit host protein synthesis before interferons can be produced o Host­specific 27.16 Antifungal Drugs  Toxic to hos so usually topical application used V: Antimicrobial Drug Resistance 27.17 Resistance Mechanisms and Spread  Antimicrobial drug resistance­ acquired ability of microorganism to resist effects of antimicrobial agent to which it is normally susceptible  R­plasmids contain drug resistant genes, allow horizontal transmission, modify or inactivate drugs and encode enzymes that prevent uptake of drug or actively pump it out  Antibiotic use does not produce resistance, but rather selects for microorganisms with preexisting resistance mechanisms  Overuse of antibiotics accelerates resistance 27.18 New Antimicrobial Drugs  Drug combinations can decrease rate of resistance

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Chapter 9, Problem 9.133 is Solved
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Textbook: Mechanics of Materials
Edition: 6
Author: Ferdinand Beer
ISBN: 9780073380285

Mechanics of Materials was written by and is associated to the ISBN: 9780073380285. Since the solution to 9.133 from 9 chapter was answered, more than 243 students have viewed the full step-by-step answer. This full solution covers the following key subjects: point, Loading, determine, deflection, beam. This expansive textbook survival guide covers 11 chapters, and 1493 solutions. This textbook survival guide was created for the textbook: Mechanics of Materials, edition: 6. The answer to “For the beam and loading shown, determine (a) the slope at point A, 615(b) the deflection at point D.” is broken down into a number of easy to follow steps, and 19 words. The full step-by-step solution to problem: 9.133 from chapter: 9 was answered by , our top Engineering and Tech solution expert on 11/15/17, 02:40PM.

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For the beam and loading shown, | Ch 9 - 9.133, 6th Edition