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63 notecards = 16 pages (4 cards per page)

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Microbiology Test 1

front 1

external signals for sporulation

back 1

desiccation, starvation, cell density

-->sigmaF is released -->activates sigmaE -->signal from mother cell triggers sigmaG in endospore and sigmaK in mother

front 2

endospore formation is regulated by

back 2

sigma factors

front 3

Transcription

3. Termination

back 3

Sequences at the end of the gene trigger release of the polymerase and the completed RNA molecule

front 4

Transcription

2. Elongation

back 4

The sequential addition of ribonucleotide from nucleoside triphosphate to the 3’OH end of the growing chain

front 5

Transcription

1. Initiation

back 5

Bind of the RNAP to the beginning of the gene, followed by opening the DNA helix and start of RNA synthesis

front 6

Transcription

back 6

Transcription (DNA to RNA) is carried out by RNA polymerase.

• RNA polymerase uses DNA as template.

• RNA precursors are ATP, GTP, CTP, and UTP.

• chain growth 5′ to 3′, just as in DNA replication

  • only one strand transcribed
  • no priming needed

front 7

Which statement is TRUE regarding protein synthesis?

back 7

The 23S rRNA catalyzes peptide bond formation in the growing polypeptide chain.

front 8

Transport to Periplasm via TAT system

back 8

used for folded proteins

  • Signal sequence RRXFXK • TAT System
  • Powered by Proton Motive Force (PMF)

front 9

Transport of synthesized protein to periplasm

back 9

Sec dependent

protein wraps around SecB--> goes over to SecA and SecYEG(embedded in membrane) -->protein is pushed through SecYEG with help from SecA and ATP and goes into periplasm -->LepB cleaves the signal sequence -->protein is folded in periplasm

front 10

SRP

back 10

Signal Recognition Particle

Transport to membrane: Protein is synthesized & localized

Protein is unfolded

front 11

polysome

back 11

clusters of ribosome bound to an mRNA

front 12

Translation

3. Termination

back 12

  • Stop codon (UAA, UGA, UAG)
  • release factor (either RF 1 or RF 2) enters A site, no tRNA for stop codons
  • Peptidyltransferase is activated and releases the completed protein from tRNA in the P site
  • Ribosome recycling factor (RRF) & EF-G undocks 50S from 30S
  • IF3 removes uncharged tRNA & mRNA

front 13

Translation

2. Elongation

back 13

  • EF-Tu-GTP binds tRNA, binds A site, GTP hydrolysis
  • Peptidyltranferase makes peptide bond
  • EF-G-GTP, hydrolysis of GTP, tRNA moves to P site

front 14

Translation

1. Initiation

back 14

Specific factors:

-E.coli- IF1, IF2 & IF3
-M. jannaschii- 6 IF proteins

front 15

An operon is a useful genetic element, because it

back 15

allows coordinated expression of multiple related genes in prokaryotes.

front 16

Transfer RNA (tRNA) molecules

back 16

function to transfer the correct amino acids to the ribosome during translation.

front 17

Shine–Dalgarno sequence

back 17

(5'-AGGAGGU-3')

Initiation is at any Shine– Dalgarno sequence; the mRNA can therefore be a polycistronic mRNA that codes for several polypeptides.

front 18

What is an operon and why is it beneficial to link expression of certain genes?

back 18

An operon is a cluster of genes that are under the control of a single promoter and are therefore cotranscribed as a single, polycistronic mRNA molecule. This arrangement allows related genes to be expressed in a coordinated fashion

front 19

structure of an ORF

back 19

Ribosomal binding site-Start codon-coding sequence-stop codon

front 20

Global regulatory systems include all of the following EXCEPT

transformation

anaerobic respiration

amino acid synthesis

heat shock response

back 20

amino acid synthesis

front 21

Attenuation is a type of regulation that can control

back 21

transcriptional activity exclusively

front 22

The function of a kinase is

back 22

phosphorylation

front 23

Which of the following do NOT bind to the promoter sequence during regulation?

activators

inducers

None of these bind directly to the promoter sequence

repressors

back 23

none of these

front 24

The lac operon is an example of ________ control in which the presence of an ________ is required for transcription to occu

back 24

negative / inducer

front 25

recombination-associated chi sites accumulate in the _____ strand

back 25

leading

front 26

Gene strand bias results in more genes in the _____ strand

back 26

leading

front 27

The Par system is necessary for

back 27

distribution of genetic material in replicating Caulobacter.

front 28

GcrA

back 28

Promotes growth of stalk, cell elongation and division

  • GcrA levels rise and promotes elongation phase of chromosome replication, cell division & growth of the the stalk on immobile daughter cell
  • GcrA levels fall, and high levels of CtrA reappear in the daughter cell

front 29

CtrA

back 29

Transcriptional regulator for flagellar synthesis; represses GcrA expression & DNA replication

  • CtrA is activated by phosphorylation • CtrA~P activates genes need for flagella & represses synthesis of GcrA & • CtrA~P inhibits the initiation of DNA replication by binding the origin.
  • CtrA gets degraded by a specific protease resulting in increasing DnaA levels
  • Absence of CtrA allows for DNA replication to initiate • DnaA is then also degraded by a protease

front 30

Caulobacter crescentus

back 30

  • Gram-negative
  • α-proteobacteria
  • Aquatic bacterium with differential life styles – Swarmer – Stalked/attached
  • Good to study cell cycle in

front 31

The curved-rod shape typical of Caulobacter is due to the proteins ____ and ______

back 31

crescentin and MreB

front 32

  • Can phenotype change without a change in genotype?

back 32

  1. Phenotypic changes often occur without a genotypic change when an organism is placed in a different environment. This could be due either to changes in gene expression or abiotic (physical or chemical) factors that alter a gene product.

front 33

  • Does the phenotype of an organisms change when the genotype changes?

back 33

  1. Phenotypic changes do not always accompany genotypic changes. Genotypic changes that do not alter the phenotype could be due to point mutations that do not result in changes to the primary structure (amino acid sequence) of a protein (i.e., silent mutations)

front 34

Phenotype

back 34

  • The observable characteristics of an organism based on its genetic composition.

front 35

  • Genotype

back 35

  1. The complete genetic makeup of an organism. Its precise determination would require genomic sequencing

front 36

Replication errors --> correction and repair

back 36

-Proofreading (make sure A & T together and C & G together)

-Thiamine dimer (detects damaged bases -->cuts nucleotides-->replaces nucleotides that were cut-->nucleotide linkage)

front 37

septation is directed by ____

back 37

FtsZ

front 38

OriC

back 38

origin of replication

front 39

After DNA helices has moved approx. _____ bases, another ___ primer is synthesized on each lagging strand

back 39

1,000

RNA

front 40

G & C have ___ hydrogen bonds

back 40

3

front 41

A & T have ___ hydrogen bonds

back 41

2

front 42

proofreading begins at time of _____ insertion

back 42

nucleotide

front 43

Tus protein

back 43

binds terminus and blocks progress of the replication fork

front 44

DNA ligase

back 44

ligA, ligB

seals nicks in DNA

front 45

DNA polymerase I

back 45

polA

excises RNA primer and fills in gaps

front 46

dnaQ

back 46

proofreading subunit

front 47

dnaE

back 47

polymerase subunit

strand elongation

front 48

Dimerization subunit (Tau)

back 48

dnaX

holds together the 2 core enzymes for the leading and lagging strands

front 49

holA-E

back 49

clamp loader

loads Poly III onto sliding clamp

front 50

processivity

back 50

an enzymes ability to catalyze consecutive reactions w/o releasing its substrate

front 51

dnaN

back 51

sliding clamp (beta clamp)

holds Poly III on DNA

involved in processivity; processivity factor for DNA polymerase

binds DNA polymerase III and prevents dislocation

front 52

exonuclease activity

back 52

3' to 5'

front 53

DNA polymerase III

back 53

main polymerizing enzyme; the main replication polymerase

subunit for proofreading (dnaQ)

front 54

dnaG

back 54

Primase

priming; primes new strands of DNA

RNA primer synthesis

rifampin resistant RNA polymerase

front 55

ssb

back 55

single-strand binding protein

cover and protect the unreplicated single strand

front 56

dnaB

back 56

helicase

unwinds double helix at replication fork

prepriming

DNA-dependent rNTPase

front 57

dnaC

back 57

helicase loader; loads helicase at origin

open the DNA helicase protein ring and place the ring around the single-stranded DNA at the origin. Loading of the DNA helicase leads to the release of the helicase loader

front 58

dnaA

back 58

origin binding protein; binds oriC

binds origin of replication to open double helix

initiation

dnaB loading

binding of dnaA leads to the strand separation at the 13-mer repeats

front 59

DNA gyrase

back 59

gyrAB gene

replaces supercoils ahead of replisome

front 60

one helical turn has ____ base pairs

back 60

10

front 61

Pyrimidines

back 61

1 ring

thymine and cytosine

front 62

Purines

back 62

2 rings

Adenine and Guanine

front 63

MinD

back 63

required for correct placement of division site