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Micro Bio 315 CSUN > exam 1-2 > Flashcards

exam 1-2 Flashcards

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Biology 101 flashcards
1
Q

Spontaneous Generation

A

Microbes could arise spontaneously

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2
Q

Do microbes cause disease?

A

Germ theory say disease are caused by microbes

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3
Q

Robert Koch

A

Demonstrate anthrax chain of infection Determined the bacterium that causes tuberculosis Koch’s postulates are criteria to establish a link b/w a microbe and disease Microbe must be absent in healthy people Microbe must be isolated from host and purely grown When introduced to healthy host, the same disease is seen Microbe must be isolated from knew host and be the same strain

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4
Q

Kochs postulate

A

Criteria for establishing a causative link between an infectious agent and a disease

    1. The microbe is found in all cases of the disease but is absent from healthy individuals
    1. The microbe is isolated from the diseased host and grown in pure culture.
    1. When the microbe is introduced into a healthy, susceptible host (or animal model), the host shows the same disease.
    1. The same strain of microbe is obtained from the newly diseased host. When cultured, the strain shows the same characteristics as before.
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5
Q

Louis Pasteur

A

Discover that weaken strains can cause immunity of the disease Weakening strains differs, for instance rabies required complex heat treatment and inoculations

  • Showed that microbes do not grow in liquid until introduced from the outside
  • Developed the ”swan neck” flask – allowed airflow, but prevented entry of dust that carried microbes
  • Remained free of growth for many years
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6
Q

Geochemical cycling:

A

interconversion of inorganic and organic forms of N, S, P etc.

Winogradsky and others showed that bacteria perform unique roles in geochemical cycling the global interconversion of inorganic and organic forms of nitrogen, sulfur, phosphorus, etc.

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7
Q

Domains of life evolution

A

Microbes were seen as neither plants nor animals Monera were divided into two groups Fungi was added as another eukaryotic kingdom Endosymbiosis theory Carl Woese Proposed 16s rRNA gene sequencing which is found in all life Found that archaea ribosomes were more similar to euk

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8
Q

•Describe cell membrane structure and what affects membrane fluidity

A

Structure that defines the existence of the cell Proteins embedded anchor membranes to envelop

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9
Q

Cell Wall

A

Aka the sacculus consists of single interlinked molecule Withstands intracellular turgor and maintains shape Composed of Peptidoglycan -Parallel (disaccharides) glycan chain crossed linked with peptides of four aa -Peptides contains two aa in D form

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10
Q

gram-positive

A
  • Gram-positive – thick cell wall (e.g. firmicutes)
  • S-layer • Made of protein (contains large pores)
  • Thick cell wall • 3-20 layers of peptidoglycan • Interpenetrated by teichoic acids for strength
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11
Q

• Gram-negative

A

– thin cell wall (e.g. proteobacteria)

  • Outer membrane (OM)
  • Covers peptidoglycan layer
  • Confers defensive abilities and toxigenic properties on many pathogens
  • Thin peptidoglycan layer • 1-2 sheets
  • Periplasm – area between membranes
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12
Q

Recall the structure of the bacterial nucleoid

A

region that extends throughout the cytoplasm Excludes ribosomes DNA forms 50 domains/loops and each domain is supercoiled

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13
Q

Cell Division

A
  • Replication begins at the origin of replication
  • The DNA double helix unzips and forms two replication forks
  • At each fork DNA is synthesized by DNA polymerase with the help of accessory proteins (replisome)
  • As the termination site is replicated, the two replisomes separate from the DNA
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14
Q

POLAR AGING

A

old poles degrade slightly increasing chances of lysis different poles might differ in resistance to antibiotics

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15
Q

Chemotaxis

A
  • Chemotaxis is the movement of a bacterium in response to a chemical gradient
  • Attractants cause CCW rotation “Run”
  • Repellants cause CW rotation “Tumble”

The alternating runs and tumbles cause a “random walk”

• Receptors detect attractant concentrations

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16
Q

archaeal membrane

A

Uses L-glycerol ether linkages: more stable isoprenoid chains branches at 4th C fatty acid chains linked covalently (tetraether) fatty acids have cyclopentane rings  

Some do have a cell wall – composed of pseudomurein or pseudopeptidoglycan

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17
Q

Heterotrophs:

A

break down organic compounds

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18
Q

autotrophs:

A

fix CO2 into complex molecules

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19
Q

phototrophs

A

: absorb light

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20
Q

chemotrophs:

A

electron donor are oxidized

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21
Q

• List the various kinds of differentiated cells that bacteria can produce

A

Bacteria faced with environmental stress undergo complex molecular reprogramming that includes changes in cell structure

  • Examples include:
  • Endospores of gram-positive bacteria
  • Heterocysts of cyanobacteria
  • Fruiting bodies of Myxococcus xanthus
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22
Q

biofilms:

A

surface attached, collaborative communities

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23
Q

•Explain how temperature affects growth

A

temperature + growth temperature denatures enzymes or decrease fluidity and enzymatic activity

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24
Q

psychrophiles

A

(0-20C), proteins are flexible, mem are fluid, antifreeze proteins

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25
Q

mesophiles:

A

(15-45C) lab strains

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26
Q

thermophiles:

A

(40-80C) enzymes have low [glycine] chaperone refold proteins stabilize DNA rigid membranes

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27
Q

barophiles:

A

high pressure growth

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28
Q

osmolarity

A

aquaporins allow water to traverse faster

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29
Q

viroids

A

infects plants and RNA genome if the infectious particle

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30
Q

Differentiate between sterilization, disinfection, antisepsis, and sanitation

A
  • Sterilization – killing of all living organisms
  • Disinfection – killing or removal of pathogens from inanimate objects
  • Antisepsis – killing or removal of pathogens from the surface of living tissues
  • Sanitation – reducing the microbial population to safe levels
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31
Q

Growth and oxygen

A

Strict aerobes – only grow in oxygen

  • Microaerophiles – grow only at lower oxygen levels
  • Strict anaerobes – die in the least bit of oxygen
  • Facultative anaerobes – can live with or without oxygen
  • Aerotolerant anaerobes – grow in oxygen while retaining anaerobic metabolism
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32
Q

filamentous:

A

helical tube around the genome

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33
Q

ICTV

A

genome composition, capsid symmetry, envelope, size and host range

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34
Q

lytic cycle

A

immediate reproduction of phage after infection early genes- produces phage components late genes - capsid is first assembled lysis/burst

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35
Q

lysogenic cycle

A

both genomes combine same type virions can’t infect the bacteria DNA is replicated with host reactivation of prophage can be trigger by stress transduction can accur

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36
Q

slow-release

A

phage reproduce without killing host slow assembly of phage extrude through envelope without lysing

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37
Q

viral life cycle

A

host recognition and attachment entry either through genome injection, endocytosis, or envelope fuse genome replication –+ssRNA can be translated by host ribosomes —ssRNA create mRNA and progeny genomes Exit through lysis or budding (host membrane surround capsid)

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38
Q

Differentiate gene, operon, and regulon

A

Gene; units of information composed of a sequence of DNA nucleotides

Operon : gens existing in tandem with other genes in a unit

Promoter : DNA control sequence that launch RNA synthesis

Regulon – collection of genes and operons at different positions in the chromosome that have a unified biochemical purpose

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39
Q

Recall the structure of DNA

A

4 nucleotides linked by a phosphodiester backbone Deoxyribose 2’carbon position Nucleobase attatched to carbon 1 Phosphodiester bond links 3’ carbon of one ribose to 5’ carbon of next ribose

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40
Q

Explain how bacterial chromosomes are compacted to fit into the cell •

A
  • Boundaries of each loop are defined by anchoring proteins called histone-like proteins
  • DNA ends must be tethered to form supercoils
  • Supercoils are introduced by • Cleaving both strands at one site in the molecule
  • Enzymes that change DNA supercoiling are called topoisomerases
  • Type I – typically single proteins that cleave only one strand of a double helix • Relieve or unwind supercoils
  • Type II – have multiple subunits that cleave both strands of the DNA molecule • Introduce negative supercoils • DNA gyrase
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41
Q

replication initiation

A

– melting (unwinding) of the helix and loading the DNA polymerase enzyme complex

• In E. coli initiation is activated by the protein DnaA and inhibited by the protein SeqA

DNA methylation controls timing of SeqA binding

• A DNA helicase (DnaB) and a DNA helicase loader (DnaC) bind to the DnaA-bound region

Clamp loading complex loads DNA polymerase to the ssDNA

• Replisome • Two DNA polymerase III, DNA primase (DnaG), and helicase (DnaB )

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42
Q

replication Elongation

A
  1. Elongation – sequential addition of deoxyribonucleotides to a growing DNA chain, followed by proofreading
  • DNA Pol III Alpha subunit – DNA synthesis activity • Epsilon subunit – DnaQ – proofreading activity that corrects mistakes and improves fidelity
  • DNA Pol I 5’-to-3’ exonuclease activity or RNaseH cleaves RNA primers
  • Pol I uses the 3’-OH end of DNA as a primer to fill in the gap
  • DNA ligase with energy from NAD forms the phosphodiester bond
43
Q

replication termination

A

Termination – the DNA duplex is completely duplicated, the negative supercoils are restored, and key sequences of new DNA are methylated

 10 ter seq in e. coli

 Terminus utilization substance ( TUS) binds to TER and stops DNAB Helicase

 Replicated cromosomes are linked ( catenane) • XerC and Xer D cut and rejoin

44
Q

• Describe two ways plasmids can replicate

A

o Bidirectional Replication starts at a single orgin and moves in two directions simultaneously o

Rolling circle –Unidirectional  RepA – replication initiator (encoded by plasmd genes) binds to orgin and nicks onestrand  makes a new + strand

45
Q

• Explain how restriction enzymes are used in biotechnology

A

o Palindromic sequences o Cleave phosphodiester back bone of opposite strands

46
Q

Describe the structure and functions of the RNA polymerase holoenzyme components

A

o RNA polymerase holoenzyme ( RnaP ) = Core polymerase + sigma factor

o Core polymenrase = 2 alpha subunits + 1 beta subunit + 1 beta prime subunit + I omega subunit ( not required )

47
Q

• Explain how sigma factors help control gene expression

A

o sigma factors guide RNA polymerase to the beginings of genes – to the promoter o in E. Coli the housekeeping sigma factor is RPOD sigma 70

o Promoter-start transcription noted as +1

 Consensus sequence – similarities among different prometers DNA sequences

48
Q

Transcription

A

• Three stages

  1. Initiation – RNA polymerase (RNAP) binds, melts open DNA helix, catalyzes placement of first RNA molecule
  2. Elongation – sequential addition of ribonucleotides to the 3’- OH end of a growing RNA chain
  3. Termination – sequences at the end of the gene trigger release of the polymerase and the completed RNA molecule
49
Q

• Describe the structure and function of tRNAs

A

o Anti codon loop caries the anti codon which is complementary to the codon

o 3’ end acceptor end attatches amino acid o aminoacyl-tRNA synthase –attatch amino acid

50
Q

o Initiation translation

A
  • IF3 binds to the 30s separateing 2 subunits
  • IF1 and mRNA bind-IF1 blocks the A site
  • IF2 complexed with GTP- brings in initiatior fMET-TRNA
  • GTP is hydrolyzed –IF1 and 2 are released • 50s docks to 30s
51
Q

o Elongation tanslation

A

 Elongation factors are complexed to GTP

 Aminoacyl-tRNA binds to A site • Bound to EF-Tu-GTP

 Peptide bond forms in the P position • 23s peptidyltransferase

 ttranslocation –movement to next codon • EF-G-GTP

52
Q

o Termination translartion

A

 Stop codon moves to A site

 Release factors enter A site

 Peptidyltransferase activity releases peptide from trna in the P site

 Rf3 enters and triggers release for RF1 or 2

 Ribosomes recycling factors RRF enters A site with EF-G-GTP

 GTP hydrolysis undocks ribosomal subunit

 IF3 binds 30s preventing 50s from redocking

53
Q

o Streptomycin

A

 Binds to the 30s interferes with codon anti codon recognition translation

54
Q

o Tetracycline

A

 Binds to the 30 s subunit and blosck aminoacyl trna binding to the A site translation

55
Q

o Chloramphenicol

A

 Inhibits peptidyl transferase activity of the 23s rRna trANSLATION

56
Q

o Erythromycin

A

 binds to the 23s rrna and interferes with translocaton

57
Q

• Summarize the mechanisms that deliver proteins to the inner membrane

A

 Tagged with hydrophobic n terminal • Bound by the signal recognition particle (srp)

58
Q

• Define horizontal gene transfer and compare the three mechanisms that give rise to it •

A

o Movement of genes between species

 Transformation • Importing free DNA into bacterial cells

 Conjugation • Cell cell contact initiated by pilus

 Phage transduction • Generalized transduction o Uses rolling circle amplification Cuts at pac site • Specialized transduction o Phage lamda

59
Q

Compare and contrast generalized and site-specific recombination •

A

o generalized  requires that the two recombining molecules to have considerable stretch of homologous genes

o site specific  requires very lttle sequence homology recognized my recognition enzyme  dedicated enzymes recognize sequence and catalyze event (att sites)

60
Q

o Point muttion –

A

change in a single nucleotide

61
Q

o Insertion

A

addition of one or more nucleotide

62
Q

o Tautomeric shifts

A

 Change in bonding amino and keto group

 C can bind to A

63
Q

o Deamination reaction

A

 Changes C to U  Damage by reactive O2

64
Q

o Error proof repair

A

 Methyl mismatch repair

65
Q

 Photo reactivation

A

• Photolyase binds pyrimidine dimer and cleaves the ring

66
Q

 Nucleotide excision repair

A

• Excises patch of 12-14 • Gap is repaired by dna POL 1

67
Q

 Base excision repair

A

• Cuts out base instead of nucleotide

68
Q

 Recombinational repair

A

• When both strands are damandged • DNA pol III skips over leaves gap • Rec A binds the gap and initiates genetic change • Other repair mechanisms can come in and repair

69
Q

sos repair

A

• When extensive damage • Maintains circular chromosome but sacrafises accuracy • Sloppy repair enzymes no proofreading

70
Q

 Nonhomologous end joining

A

• Doule stranded breaks o Ku And LigD recombine double break o Error prone

71
Q

• Differentiate between nonreplicative and replicative transposition

A

o Nonreplictave  transposable elements jumps one site to another

o replicative  transposable element is copied one copy remains in original site

72
Q

Gene expression – Levels of contro

A
  • Alteration of DNA sequence – random or programmed changes in DNA sequence to activate or disable a particular gene
  • Control of transcription – regulated by protein repressors, activators, and sigma factors – and RNAs!
  • Control of mRNA stability – levels of specific mRNA molecules are regulated by RNase activity (degradation) • Translational control – translation by ribosomes can be regulated by sequestering RBS
  • Posttranslational control – once proteins are made, their activity can be controlled by modifying protein structure
73
Q

• Use the tryptophan operon to explain the mechanism of transcriptional attenuation

A

 DNA region between operator and Trp E • Short peptide • Back to back trp codons a stop codon and 4 complementary nucleotide stretches

 1:2 and 3:4 loops • 3:4= instrisic terminator hairpin • high trp

 2:3 = anti terminator • prevents terminator from forming • low trp ribosome stalls at trp codon at region 1 so 2:3 form

74
Q

o Recall major forms of catabolism

A

 sugars, fats/lipids, protein/peptides,/ aromatic/lignin

 Fermentation: electrons from substrate are put onto organic products

 Respiration: electrons are transfer to inorganic electron acceptor

 Photoheterotrophy: use light energy and organic carbon for catabolism

75
Q

o Glycolysis (EMP)

A

 Stage 1 investment of 2 atp to phosphorylate glucose

 Stage 2 energy yielding phase generates pyruvate (x2)

 NET: 2 ATP 2NADH

76
Q

o ED pathway

A

 Different route to catabolize

 Allows enteric bacteria to use intestinal mucus

 Shortcut for the EMP

 Net 1ATP 1NADH 1NADPH

77
Q

o PPP

A

 Key intermediate : ribulose 5-phosphate

 Starts off with ED pathway bur second oxidation of NADP+ =loss of carbon as CO2

78
Q

• Recall the roles that the TCA cycle bypass play in cell metabolism

A

o Bypass Prevents the loss of carbons

79
Q

• Differentiate among the types of metabolism that use ETS

A

o ORGANOTROPHY  Organic molecule donate elecrton ( glucose)

o Litrotrophy  inorganic substrate ( fe2+ h2s h2)

o Phototrophy  Light absorbtion excites elecroons

o Respiration  If electron donated to ets ultimately reduce a TEA

80
Q

• Identify which redox couples are better electron donnors or acceptors

A

o More – values represent stronger electron donors o More+ values represent stronger electron acceptor

81
Q

o Cytochromes ets

A

 Colored proteins that absorb visible light which shifts when there is a change in the redox state

 Pass electron sequentially from each protein complex to the next stronger electron acceptor ending in o2

82
Q

o Cofactors

A

 Energy transitions are mediated by cofactors small molecules that associate with the protein

 Small energy transitions typically involve

  • Metal ions such as iron r copper coordinated with amino acid residues
  • Conjugated double bond and heterometric rings
83
Q

• MOBILE ELECTRON CARRIER

A

o Quinone can recive 2e- from subrtate along with 2H+

84
Q

• Define anerobic respiration and dissimilatory denitrification

A

o Takes place where oxygen is scarce

o E. coli possesss several different terminal oxidoreductases to reduce alternative TEA

 Inorganic nitrate nitrite

 Organic – fumarate o Use the strongest donors and electron acceptors available

85
Q

• Identify molecules that can serve as terminal electron acceptors on anerobic respiration

A

o Fumarate, thiosulfate, nitrite, nitrate, oxygen* and metals

86
Q

• Identify possible initial electron donors to an ETS in lithotrophy

A

o Iron , nitrogen, sulfur, hydrogen* CO2( methanogenesis)

87
Q

• Describe the structure and function of the cyanobacterial antenna complex and reaction center

A

o Antenna complex

 Many molecules of chlorophyll arranged like a satellite dish

o Reaction center

 Protein complex in which the chlorophyll photoexcitation connects to the ETS

88
Q

o PSI

A

 Receives electrons associated with hydrogens from H2S HS H2 or Fe2+

 Peak absorption 840 400-550nm

 2 e- reduce NADP+

89
Q

o PSII

A

 Returns an electron from the ETS to bacteriochlorophyll

 800-1100m lower energy

 Not enough energy to reduce nadp+

90
Q

o Oxygenic Z pathway

A

 2 pairs of electrons from water molecules generate O2..

 Homologs of PSI (700) and PSII ((680)

 Shorter wave lengths so more energy to spit water

91
Q

• Outline the key steps of the Calvin cycle

A

Stage 1 carboxylation and spliting 6C–> 2(3C) o Stage 2 PGA to G3p o Stage3 regenerating ribulose 1,5 bisphosphate

92
Q

 Rubisco

A

 Rubisco mediates the condensation of ribulose 1,5 bisphosphate with CO2 and H20 • Most abundant protein on earth

93
Q

• Describe the structure and functions of carboxysomes

A

o Many organisms contain rubisco complex in polyhedral structures called carboxysomes

  • Found in Co2 fixing lithotrophs cyanobacteria, and chloroplast
  • Takes up HCO3 which is converted to Co2 by carbonic anhydrase
  • Co2 then fixed by rubisco
94
Q

• Name alternative CO2 fixation pathways and their key products

A

o Reductive / Reverse TCA cycle

-Anaplerotic reactions -Small co2 fixations o Reductive Acetyl

–CoA pathway

  • Two CO2 molecules arecondesed through converging pathways to for the acetyl group of acetyl CoA
  • Used by anerobic soil bacteria methanogens

o The 3- hyproxypropionate cycle -Co2 is fixed by acetyl-CoA into 3-hydroxypropionate

95
Q

• Recall the mechanism of action of nitrogenase

A

o Electrons in Fe protein are passed to FeMo protein o FeMo protein bind H2+

o N2 binds active site displacing H2

o N2 is then reduced

96
Q

• Decribe adaptaions that protect nitrogenase from oxygen

A

o Protective proteins -Stabilize nitrogenase and prevent attack by oxygen .

  • Proteins stabilize nitrogenase from O2
  • Cyanobacteria fix N2 at night -Specialized cells
97
Q

• Describe how ammonium is assimilated from amino acid biosynthesis

A

o ATP helps 2-oxoglutarate condense ammonium into glutamine

o NADPH helps 2-oxoglutarate condense ammonium into glutamate

98
Q

Compare and contrast facilitated diffusion and active transport

A

Passive diffusion

  • Some molecules pass freely through membranes (O2, CO2)
  • Follows concentration gradient

Facilitated diffusion

  • Transporters pass material into/out of cell
  • Follows concentration gradient
99
Q

• Contrast virions, viroids, and prions

A

• Virion (virus particle) – consists of a nucleic acid genome (DNA or RNA, single or double stranded) • Has a protein coat (capsid)

Viriods – RNA genome is itself the infectious particle • Infect plants

• Prions – protein only (no nucleic acid)• Abnormal protein structure • Cause of “mad cow” disease

100
Q

Lac Operon – no lactose

A
  • Absence of lactose – lac operon is transcribed at extremely low levels (<10 molecules LacZ per cell)
  • Repressed by LacI • Tetramer of LacI repressor proteins bind to two lacO of the promoter • Causes the DNA to loop and prevents RNAP access
101
Q

Lac Operon – high lactose

A
  • Presence of lactose – 100-fold higher expression
  • Basal level of lacZYA expression allows LacY to be expressed and let in lactose
  • Low concentration of LacZ does not completely cleave the glycosidic bond of lactose; rearranges it to form allolactose
  • Allolactose binds LacI repressor and unlocks the prote
102
Q

Lac Operon – cAMP activation

A
  • Cyclic AMP (cAMP) accumulates when a cell is starved for carbon
  • cAMP binds to dimeric CRP (cAMP receptor protein)
  • cAMP-CRP complex can bind to specific DNA sequences located near many bacterial genes
  • cAMP-CRP binds to DNA • Usually acts as an activator
103
Q

Lac Operon – glucose repression

A
  • Glucose represses the lac operon
  • Glucose is the favored carbon source
  • Enzymes for glycolysis are always being produced
  • If there is glucose, no need to use lactose
  • Glucose indirectly prevents induction of lacZYA by keeping lactose out of the cell
104
Q
A

Micro Bio 315 CSUN (2 decks)

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