Chapter 8

Question 1

a general profile of bacteriophage Lambda

Answer 1

linear dsDNA, 40-60kb, ds-DNA with ss-DNA ends (cos), transcriptional regulation of gene expression (immediate early, early, late genes)

Question 2

what happens when the genome is injected to the cell?

Answer 2

DNA circularizes upon entry via cohesive ends.

Question 3

infection of E.coli by phage lambda can have two possible outcomes:

Answer 3

the lytic growth cycle or prophage state

Question 4

phages can either:

Answer 4

grow lytically or exist as a repressed prophage are called temperate phages

Question 5

phage adsorption and DNA entry depends on

Answer 5

cellular proteins involved in sugar transport

Question 6

at the tip of the tail is a [ ] protein that binds to an E.coli outer membrane protein, [ ]

Answer 6

J and LamB

Question 7

injection of DNA through the tail and into the cell requires

Answer 7

the cytoplasmic membrane components of mannose permease

Question 8

transcription by host RNA polymerase then begins at two promoters, P[L] and P[R], located on

Answer 8

either side of the repressor gene, cI

Question 9

the lambda lytic transcription by

Answer 9

termination and antitermination of RNA synthesis at specific sites on the genome

Question 10

transcripts from P[L] and P[R] expresses

Answer 10

the N and cro genes; these “immediate-early” transcripts end at transcription terminators t[L]1 and t[R]1.

Question 11

N protein is a transcriptional antiterminator that

Answer 11

modifies RNA polymerase so that transcripts initiated at P[L] and P[R] are no longer terminated.

Question 12

For N to function, it binds to

Answer 12

specific mRNA sequences, called NUT sequences.

Question 13

antitermination of t[L]1 extend P[L] through

Answer 13

the recombinase genes gam, bet, and exo, the site-specific recombination genes xis and int across the locus for site-specific recombination, attP, all the way to Ea47

Question 14

Q is

Answer 14

a transcriptional antiterminator

Question 15

P[R]’ transcript constitutively expresses a short transcript that ends at

Answer 15

t[R]’

Question 16

Q antiterminates P[R]’ transcript, leading to

Answer 16

production of a very long “late” gene transcript of the cell lysis, head, tail, and tail fiber assembly genes

Question 17

the CI repressor blocks expression of the lytic program by

Answer 17

regulating three nearby promoters: P[L], P[R], P[RM]

Question 18

when lambda infects active cells in logarithmic growth, the phage grows

Answer 18

lytically in most of the cells

Question 19

when lambda infects metabolically sluggish, stationary phage cells, most of the infected cells are

Answer 19

lysogenized. the CI protein produced to shut down transcription of the lytic genes.

Question 20

three repressor-binding sites, called operators, overlap

Answer 20

the P[L] and P[R] promoters

Question 21

CI binds to the operators as a dimer with the following affinities:

Answer 21

O[R]1>O[R]2=O[R]3

Question 22

a second promoter, called P[RM] for repressor maintenance, is located near

Answer 22

P[R] but initates transcription in the opposite direction; this promoter directs transcription of cI mRNA

Question 23

P[RM] is a weak promoter, but in a prophage, P[RM] is activated to produce

Answer 23

sufficient CI repressor to shut down P[R] and P[L]

Question 24

CI bound to O[R]2 is physically near P[RM] and forms favourable contacts with

Answer 24

RNA polymerase; making P[RM] a better promoter

Question 25

when the CI concentration is too high,

Answer 25

O[R]3 is also occupied and transcription from P[RM] is turned off until the CI concentration returns to a normal level

Question 26

Cleavage of CI repressor in cells with damaged DNA leads to

Answer 26

prophage induction

Question 27

induction occurs in cells that have a DNA damage caused by

Answer 27

UV light

Question 28

when damaged nucleotides are removed by

Answer 28

excision repair system of SOS genes

Question 29

the SOS genes of undamaged cells are repressed by

Answer 29

the cellular LexA repressor

Question 30

excision repair excise ss-DNA oligonucleotides, which in turn bind to and activate

Answer 30

the cell's RecA recombination protein such that it acquires a "coprotease" activity

Question 31

the LexA repressor has a weak autoprotease activity that is enhanced by

Answer 31

binding to the activated RecA protein, with the result that the LexA repressor cleaves itself into two domains and loses repressor activity, allowing induction of SOS gene transcription

Question 32

CI's weak autoprotease activity is stimulated by

Answer 32

activated RecA, leading to linker cleavage that inactivates CI

Question 33

Inactivation of CI leads to

Answer 33

transcripiton from P[R] and P[L]

Question 34

N-antiterminated transcription leads to expression of both

Answer 34

Xis and Int, which in turn sponsor excision of the prophage DNA from the bacterial chromosome, and the lytic transcription pattern ensues.

Question 35

the Cro repressor suppresses

Answer 35

CI synthesis and regulates early gene transcription

Question 36

the dimeric Cro protein is the repressor that

Answer 36

establishes and controls a lytic infection by turning off the synthesis of CI.

Question 37

Cro binds to the same operators as CI repressor, but with different affinities

Answer 37

O[R]3>O[R]2=O[R]1

Question 38

early during lytic growth, either during an infection or induction of a lysogen, Cro binds to

Answer 38

O[R]3 and blocks P[RM]

Question 39

Binding to O[R]3 and blocking P[RM] prevents

Answer 39

CI expression and channels the infected cell to lytic growth.

Question 40

later during lytic growth, sufficient Cro accumulation fills

Answer 40

O[R]2 and O[R]1, repressing P[R] and P[L], reducing synthesis of early proteins

Question 41

Sufficient CI ensures that

Answer 41

P[R] and P[L] are turned off and P[RM] is turned on

Question 42

Insufficient CI results in

Answer 42

the lytic program

Question 43

the amount of CI produced from P[RE] depends in

Answer 43

turn on the level of the P[RE] activator protein, CII

Question 44

the cII gene is located just downstream from

Answer 44

cro

Question 45

the cII gene is transcribed from

Answer 45

P[R] promoter and sponsored by N antitermination

Question 46

P[RE], like P[RM] is a weak promoter that is activated by

Answer 46

CII

Question 47

since the level of CI produced from P[RE] depends on the level of CII, the lysis/lysogeny decision depends on

Answer 47

the intracellular concentration of CII protein

Question 48

CII activates two additonal promoters:

Answer 48

P[int] and P[AQ]

Question 49

P[int] is

Answer 49

a CII-dependent promoter for encoding Int, the site-specific recombinase that inserts the lambda prophage into the bacterial chromosome

Question 50

P[AQ] for

Answer 50

encoding an antisense transcript of Q that lowers Q expression, delaying the late gene expression

Question 51

the level of CII is determined with inactivation by

Answer 51

host cell proteases and lower stability in metabolically active cells than in inactive cells and higher stability by CIII

Question 52

CIII is for

Answer 52

inhibiting the host proteases

Question 53

Int is a phage protein that

Answer 53

catalyzes crossing-over between the phage attP site, near the int gene, and the bacterial attB site, located between the gal and bio operons

Question 54

attP is deisgnated

Answer 54

POP' to symbolize its three parts, where O being core squence that is found in both attP and attB

Question 55

Int binds to two different DNA sequence motifs within

Answer 55

the arms and the core of attP

Question 56

Int binds strongly to

Answer 56

the arm sequence that has binding sites for the E.coli DNA bending protein, IHF

Question 57

IHF is for

Answer 57

bending the arms to allow Int to make bridging-contacts with the core sequences

Question 58

an intasome is

Answer 58

a complex nucleoprotein structure

Question 59

three proteins that are involved in the reverse reaction to generate a circular DNA molecule from the integrated lambda phage are

Answer 59

Int, IHF, Xis

Question 60

the attP-attB exchange during integration shuffles the att sequences, creating

Answer 60

attL (BOP') and attR (POB') sites at the left and right junctions between the prophage DNA and the bacterial chromosome

Question 61

When an integrated prophage is induced, N-antiterminated mRNA from PL produces

Answer 61

Int and Xis

Question 62

lambda DNA replication is directed by

Answer 62

O and P, but carried out by host cell proteins

Question 63

Expression of the early genes O and P leads to

Answer 63

lambda DNA replication

Question 64

The O protein binds to

Answer 64

multiple sites in the replication origin, ori lambda, forming a nuceloprotein structure, the O-some

Question 65

Forming the O-some, the two strands of an adjacent AT-rich segment of the DNA become

Answer 65

unpaired

Question 66

the P protein recruits

Answer 66

the E.coli replication Helicase DnaB to the O-some

Question 67

In hijacking DnaB, P binds so tightly that

Answer 67

E.coli chaperone proteins DnaJ and DnaK are needed to release it from the O-some

Question 68

Growing points for DNA replication are set up using

Answer 68

E.coli replication proteins

Question 69

RNA primers are laid down on the separated DNA strands, and DNA polymerase extends

Answer 69

these primers to begin copying the DNA

Question 70

Bidirectional replication produces

Answer 70

progeny circles at early times, prior to the lysis/lysogeny decision

Question 71

during lytic growth, there is a shift to

Answer 71

rolling circle replication

Question 72

rolling circle replication produces

Answer 72

multimeric ds-DNAs, called concatemers, which are the DNA packing substrate

Question 73

assembly of lambda heads involves

Answer 73

chaperones and scaffolding proteins

Question 74

DNA is inserted into performed proheads by

Answer 74

an ATP-dependent mechanism

Question 75

Concatemeric lambda DNA generated by rolling circle replication is cut

Answer 75

to generate unit-length virion DNA molecules during packing of the DNA into the proheads shell

Question 76

cutting requires the introduction of nicks, staggered

Answer 76

12 bp apart, so that the ss-cos ends are made

Question 77

terminase is

Answer 77

the viral cutting enzyme containing Nu1 and A subunits

Question 78

Nu1 binds to

Answer 78

an anchoring site cosB

Question 79

protein A cleaves

Answer 79

the DNA at an adjacent nicking site, cosN

Question 80

following nicking of a concatemer, terminase separates

Answer 80

the cohesive ends, remaining bound to the end of the DNA at cosB

Question 81

DNA-protein complex docks at

Answer 81

the portal vertex of a proheads, and DNA packing proceeds

Question 82

translocation of the DNA into the proheads is powered by

Answer 82

ATP hydrolysis

Question 83

during translocation of the DNA into the proheads, the E shell subunits undergo

Answer 83

rearrangement, causing the shell to expand and become more angular

Question 84

As the shell is filled, the next cos site of the concatemer approaches

Answer 84

the terminase located at the portal

Question 85

cutting the second cos sequence requires

Answer 85

recognition of cosQ, a site located just upstream of the nicking site cosN, and this completes packing of a chromosome

Question 86

terminase undocks from the filled head, remaining bound to

Answer 86

the end of the concatemer to form a new complex that is ready to bind to another proheads

Question 87

at the end of lytic growth, the host cell is

Answer 87

lysed to release progeny phages

Question 88

two lambda gene products govern peptidoglycan hydrolysis:

Answer 88

protein R and S

Question 89

Protein R attacks

Answer 89

the cell's peptidoglycan layer

Question 90

Protein S provides

Answer 90

a route through the cytoplasmic membrane for R

Question 91

Protein Rz and Rz1 are required for

Answer 91

lysis of cells in the presence of high concentration of divalent cations