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mcdb 101a > Midterm 2 > Flashcards

Midterm 2 Flashcards

(95 cards)

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1
Q

Prophage/lysogen

A

a phage that has inserted itself into the bacterial chromosome

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2
Q

lysogenic cycle

A

process of insertion and excision of a bacteriophage into its host genome

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3
Q

virulent phage

A

phage that has only a lytic cycle (ex T4)

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4
Q

temperate phage

A

phage that has both lytic and lysogenic stages (ex lambda)

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5
Q

immunity

A

cell containing a lysogenic phage can’t be “superinfected” with additional phage

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6
Q

how does lambda genome present?

A

linear or circular

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7
Q

what does the promoter PL make (early transcription)?

A

N
-if N reaches a certain concentration it prevents the termination of the 2 mRNAs

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8
Q

what does the promoter PR make (early transcription)?

A

Cro

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9
Q

PR+PL

A

lytic growth (strong promoter)

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10
Q

Prm

A

lysogenic growth (weak promoter)

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11
Q

what happens when PR expresses Cro?

A

-Cro has the highest affinity to OR3->blocks Prm
-RNA binds to PR as a result->Cro keeps getting produced and stays in lytic cycle

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12
Q

what happens when Prm expresses CI?

A

-Cl binds to O1 at the highest affinity->blocks PR
-RNA pol binds to Prm as a result->makes CI and stays in lysogenic cycle

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13
Q

Cro

A

-transcriptional inhibitor at Prm
-blocks production of CI
-transcriptional inhibitor of PL
-blocks production of N, CIII, integrase

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14
Q

Cl

A

-transcriptional inhibitor of PR
-blocks production of Cro and lytic cycle
-auto-regulates own transcription

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15
Q

lysogenic/lytic decision in rich conditions

A

-lytic cycle
-high [HfIA and HfIB]
-low [CIII] bc HfIA and HfIB depletes CIII, therefore depleting CII
-no integrase
-Cro is present

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16
Q

lysogenic/lytic decision in poor conditions

A

-lysogenic cycle
-low[HfIA and HfIB]
-high [CIII], CII is preserved, activating Pre and Pi
-integrase and CI is present
-no Cro

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17
Q

what is the function of int?

A

required for the integration of lambda into the host

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18
Q

how is lambda integrated into the bacteria?

A

-via site-specific recombination
-recombination occurs btwn attP on phage and attB btwn gal and bio operons on bacteria
-requires IHF (integration host factor)

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19
Q

what do the attB site and attP site have in common

A

only 15 bases of common sequence

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20
Q

integration steps

A

1)CI inhibits production of Cro by binding to Or3 (folding the genome)
2)endonuclease (lambda Int) cuts at attP and attB sites
-they are complementary
3)strands recombine and form a single piece of DNA

integration and excision are the reverse of each other

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21
Q

what are the required proteins for excision/integration?

A

-excision requires lambda Int, IHF, and lambda Xis
-integration requires lambda Int and IHF

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22
Q

how is the lysogenic state maintained?

A

-the genes need to be silent bc it would kill the host too early
-the binding of CI to OR3
-optamers bend DNA to make PL and PR overlap
-inhibits production of Cro

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23
Q

induction of lambda by the SOS pathway

A

-host DNA damage whose repair results in the production of ssDNA triggers SOS system
-ssDNA activates RecA, allowing RecA to cleave CI in half, preventing Cl from binding to the operator site on lambda
-reverses the repression of gene expression of CI and triggers off excision of the prophage from the host genome and entry into lytic cycle

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24
Q

excision

A

-once CI is production is blocked, transcripiton from PL can occur (only happens once lambda has integrated and can make excisionase and integrase)
-excisionase and integrase are required for excision of the prophage, along with IHF
-excision involves recombination btwn attL and attR sites which form the junction btwn prophage and host DNA

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25
regulation of integrase and excisionase
-initial PL transcript containing int and xis forms a stem-loop structure "sib" that targets the int-xis mRNA for degradation -lambda is integrated into host DNA, stable stem-loop is formed that isn't a target for RNAse-dependent mRNA degradation due to separation of int from sib -no xis transcript is produced from Pi during integration bc Pi is located within xis
26
PR
Cro (early), CII (late)
27
PL
N(early), CIII (late)
28
PR'
lysis proteins, structural proteins
29
Pre
promoter for repressor establishment: CI
30
Prm
promoter for repressor maintence: Cl
31
Pi
promoter for integrase
32
Paq
promoter for anti-Q -prevents early lysis of host -overlaps Q and is ~200 bp in length (doesn't produce functional protein) -delays production of Q
33
lytic cycle
-PR transcript includes mRNA that encodes antiterminator Q -Q is required for transcription from PR', allowing phage structural proteins and lysis proteins to be made -many copies are packaged into phage heads -once new phage is assembled, the host cell is lysed and releases the phage to infect other cells -theta and rolling circle rep are used during lytic cycle
34
lacZ
encodes beta-gal
35
lacY
encodes permease
36
lacI
inhibits lacZ,Y,A expression when lactose is absent -trans
37
lacO
similar to mutations that inhibit the ability of LacI to regulate expression of Z,Y, and A -cis
38
lacP
required for lacY,Z, A expression
39
what do polysistroic mRNAs exhibit?
-they exhibit polarity -a mutation in a gene encoded by a polycistronic mRNA will often block translation of downstream genes in the same RNA molecule (nonsense mutation) -polarity is generally seen with mRNA molecule that utilize rho-dependent transcription termination -prevents synthesis of proteins that would have no function
40
IPTG
inducer of beta-gal -not a substrate
41
X-gal
-artificial substrate that makes blue when cleaved by beta-gal -not and inducer of beta-gal
42
what happens when there is no lactose
no allolactose to bind to lac repressor proteins, so repressor proteins (lacI) remain bound to the operator (lacO), preventing the transcription of genes
43
what happens when there is lactose
allolactose can bind to lac repressor proteins, removing it from the operator and allowing transcription of genes to occur
44
Is mutation
-alters the conformation of the lac repressor so it can't bind to the inducer
45
I-d mutation
-forms abnormal tetramer that can't bind to lacO -single I-d mutant combines with 3 I+ and forms a non-functional repressor
46
how does a cell switch from growth in the absence of lactose to growth in its presense?
bacterial cells are leaky and small amounts of lactose can enter the cell without permease. this induces the production of permease, allowing more lactose to ender the cells
47
can lacY- cells produce beta-gal?
-small amount of lactose that leaks into Y- cells is insufficient to produce enough beta-gal to generate blue colonies on X-gal plates
48
catabolite repression
decrease in glucose-> increase in cAMP -cAMP binds to CAP site, facilitating the binding of RNA pol to
49
what is the purpose of cap anchors?
bind to the alpha C-terminus domain of RNA polymerase -sigma factor and alpha C-terminus makes a more stable interaction with DNA-> higher change that RNA pol will start transcribing -involved the bending of DNA, exposing -35 and -10 regions
50
positive control (transcriptional regulation)
an activator is required for transcription to occur ex:CAP
51
negative control (transcriptional regulation)
repressor is normally present to block transcription unless activated by the presence of a metabolite ex: LacI
52
alpha complementation and blue-white cloning
lacZ alpha peptide + beta-gal with mutation in N-terminal= beta-gal activity=blue colonies gene insert + beta-gal with mutation in N-terminal= no beta-gal activity=white colonies
53
Luria and Delbruck's fluctuation test
bulk culture of T1 sensitive E. coli -count the colonies of bacterial colonies on each plate
54
adaptation theory of mutation
specific environmental factor that triggers mutation -addition of T1 phage triggers mutation
55
mutation theory of mutation
mutation occurs spontaneously -addition of T1 phage doesn't trigger mutation
56
characteristics of poisson distribution
-can be used to determine the probability of infrequent whole number events -events must be independent -poisson distributions are skewed towards zero -events that follow a poisson distribution are considered to be random
57
mutation rate
probability (based on poisson distribution) that a cell will experience a mutation in one generation
58
mutation frequency
-measured number of mutations per cell per generation -less accurate than calculating the mutation rate due to influence of jackpots -determined by divided the number of mutations by number of cells in the population
59
mutation
alteration of DNA sequence that encodes a gene or RNA molecule
60
point mutation
single base-pair (bp) change
61
substitution
replacement of single bp
62
insertion/deletion
addition/removal of 1 or more bp of DNA. if this occurs within a gene it will usually result in a frameshift mutation
63
transition
change of purine to another purine, or pyrimidine to pyrimidine
64
transversion
change of purine to pyrimidine, or pyrimidine to purine
65
silent
no change in amino acid -AAG to AAA still codes for lysine
66
neutral
same type of amino acid -GAG to GAT changes Glu to Asp
67
missense
different type of amino acid -TGG to CGG changes to Trp to Arg
68
nonsense
changes amino acid to stop codon -AGA to TGA
69
reverse mutations
second mutation that returns phenotype of mutant organisms to wild type
70
true reversion
2nd mutation that returns a mutant genotype back to its original sequence -occurs at same nucleotide where original mutation occurred
71
pseudoreversion
2nd mutation that compensates for the original mutation by partially or totally restoring the phenotype to wild type -aka supressor mutation -doesn't always affect same nucleotide
72
intragenic suppression
1)reversion by downstream deletion within the gene 2)2nd site missense: 2nd mutation returns the folding of the protein to normal
73
nonsense suppressor (extragenic suppression)
nonsense mutation can be corrected by a mutation in a tRNA gene that results in a change in its specificity such that an amino acid at a stop codon
74
missense suppressor (extragenic suppression)
mutant tRNA that inserts the correct amino acid at a mutant codon
75
frameshift suppressor (extragenic suppression)
mutant tRNA that reads 4 bases as a codon -may or may not insert correct amino acid
76
spontaneous mutations
mutations that occur under normal environmental conditions -don't result from exposure to toxic chemicals or high energy electromagnetic radiation
77
amino bases
cytosine and adenine -tautomerize into imino bases
78
keto bases
thymine and guanine -tautomerize into enol bases
79
depurination/depyrimidation
spontaneous covalent bond breakage can result in loss of a base, creating a variety of point mutations
80
deamination
CG to TA -caused by nitrous acid
81
oxidative damage
-oxygen molecules may be added to nucleotide bases as a result of interactions with reactive oxygen species (ROS) -superoxide radical (O2-) -hydroxyl radical (*OH) -hydrogen peroxide (H2O2) GC to TA
82
UV light
makes intrastrand crosslinks btwn adjacent pyrimidines forming cyclobutane dimers - GC to AT, GC to TA, AT to GC, GC to CG
83
ionizing radiation
-generates oxygen radicals -breaks covalent bonds, causing loss of DNA bases, resulting in mispaired bases -can produce interstrand covalent cross links -can cause deletions if breaks occur in both DNA strands
84
5-BU
-common keto form of 5-BU binds to adenine -ionized form binds with guanine
85
nitrous acid
-G to xanthine, which binds with cytosine (only 2 H-bonds, weaker) -C to Uracil, binds to adenine (CG->TA) A to hypoxanthine, binds to cytosine (AT->GC)
86
hydroxylamine
C to hydroxylaminocytosine, binds to adenine -CG->TA
87
MMS
G to O-methylguanine, binds to thymine -GC->AT
88
proflavin
results in single bp insertions and deletions -very dangerous
89
aflatoxin
GC to TA -binds to G and deletes it -transversion
90
mutagen
chemical that causes mutation
91
carcinogen
chemical that causes cancer
92
Ames test
-used to determine if chemical compound is mutagenic -based on reversion of 2 point mutations in the histadine biosynthesis pathway of Salmonella typhimurium
93
modified ames test
1)mix rat liver extract with his minus strain of Salmonella 2)spread mixture on histidine deficient media 3)place filter disk on plate and add suspected mutagen to disk 4)set up control plate identical to test plate except that no test compound is added to disk 5)test is considered positive if more colonies result from bacteria that were exposed to the suspected mutagen than on control plates
94
mutation avoidance mechanisms
1)enzymes that detoxify oxygen radicals -superoxide dismutates: converts O2- to H2O2 -catalase: converts peroxide to H2O and O2 2)MutT protein converts 8-oxydGTP to 8-oxydGMP
95
direct reversal of DNA damage
1)photoreactivation -photolyase binds to cyclobutane photodimers in the presense of visible light and converts the thymidine dimer back to 2 monomers 2)alkyltransferases -remove alkyl group from bases

mcdb 101a (3 decks)

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