Chapter 11 - Replication Is Connected to the Cell Cycle

Question 1

Frequency of Replication =

Answer 1

Rate of Cell Growth

Question 2

Number of replication cycles is

Answer 2

connected with Cell Division.

Question 3

Total time of Replication

Answer 3

60 minutes = 40 minutes for DNA Replication + 20 minutes lapse for chromosome segregation.

Question 4

If cell division < 60 minutes

Answer 4

A new round of replication must be initiated before
the previous replication round is completed.
Daughter cell is produced with replication already occurring with the presence of multiforked
chromosomes.

Question 5

Cell-septum components

Answer 5

Same components as cell envelop: inner membrane, outer

membrane (Gram-negative), and peptidoglycan disaccharides.

Question 6

MreB

Answer 6

Forms cylindrical filament structures along the inner membrane and associates with the peptidoglycan synthesis
machinery, RodA & PBP.

Question 7

FtsZ (C)

Answer 7

In cytosol; Initiates peptidoglycan synthesis with the recruitment of the pre-septal complex.
Recruits protein complex for the septum synthesis.
FtsZ forms a ring at midcell (defines septum position); this formation is the Z-ring or septal ring.
The GTPase activity of FtsZ is required for oligomerization of the monomeric FtsZ.

Question 8

Env family

Answer 8

Hydrolyze the peptidoglycan to separate the daughter strands.

Question 9

ftsZ mutations

Answer 9

Multinucleated, long filaments form when septum formation is inhibited but replication is unaffected.

Question 10

Minicells

Answer 10

Septum formation is too rapid or not mid-cell.
Minicells different from anucleated mutants - have the same size as wild-type because septum formation is intact.
Observed with ftsZ over-expression.

Question 11

Z-ring

Answer 11

Z-ring formation is the rate-limiting step in septum formation.
The Z-ring is a dynamic structure with continuous exchange of components. Critical components include cytoskeletal
proteins that link the Z-ring to the membrane.

Question 12

ZipA

Answer 12

Integral membrane protein in the inner membrane.

Question 13

FtsA

Answer 13

Membrane-recruited cytosolic protein.

Question 14

FtsW

Answer 14

Integral membrane protein in the inner membrane.
Recruits FtsI to enhance peptidoglycan production, which in turn pushes inner membrane out and constricts the outer membrane.

Question 15

FtsI/PBP3

Answer 15

Membrane-bound protein with transpeptidase

activity to promote peptidoglycan synthesis.

Question 16

MinD

Answer 16

Activates MinC, which in turn prevents FtsZ polymerization.

Question 17

Number and location of septa is

Answer 17

Determined by the ratio of MinE/MinC,D.
Dynamic movement of the Min proteins in the cell sets up a pattern in which inhibition of Z-ring assembly is highest at the poles and lowest at midcell.

Question 18

MinC/D

Answer 18

is a division inhibitor whose action is confined to the poles
by MinE.

Question 19

Noc/Slm

Answer 19

Ensure septum location with respect to the chromosome in the cell. Prevent septation (and consistently Z-ring
formation) from occurring in the space occupied by the bacterial chromosome (nucleoid occlusion).

Question 20

SlmA

Answer 20

Binds to specific DNA sequences (SlmA binding sites).

Inhibits FtsZ polymerization.

Question 21

Intermolecular recombination

Answer 21

Merges monomers into dimers.

Question 22

Intramolecular recombination

Answer 22

Releases individual units from oligomers.

Question 23

Xer site-specific recombination system

Answer 23

Targets a particular sequence to revert the dimeric

circle.

Question 24

How does the cell know that a dimer has formed due to

recombination?

Answer 24

No recombination allows segregation of the chromosome
and eventually daughter cells.
Recombination confines the movement of the integrated
chromosomes.

Question 25

Xer site-specific recombination system

Answer 25

Includes: 2 Recombinases: XerC and XerD; the dif gene sequence near the replication termination region.

Question 26

Recombinases

Answer 26

Bind to the dif site and form Holliday junction.

Question 27

FtsK

Answer 27

Resolves the junction, is located at the septum.

Question 28

MukB

Answer 28

Essential component of the partitioning apparatus that recondenses the daughter nucleoids.

Question 29

Proper Partitioning requires:

Answer 29

1. Disentangling of the daughter chromosomes.

2. Segregation of the daughter chromosomes.

Question 30

Cyclin-dependent kinases (Cdk)

Answer 30

Key mediator of cell cycle progression in eukaryotic cells.
Set of serine/threonine protein kinases that directly contribute control cell cycle progression. Inactive unless bound by Cyclins.

Question 31

Cyclin proteins

Answer 31

Key mediator of cell cycle progression in eukaryotic cells. Required to activate Cdk proteins; Cdk are

Question 32

Retinoblastoma Protein (Rb)

Answer 32

Key mediator of cell cycle progression in eukaryotic cells.
Inhibits gene transcriptional events required for progression from G1 to S-phase; activity is regulated through phosphorylation/de-phosphorylation PTM.

Question 33

Cdk inhibitors (CKI)

Answer 33

Set of inhibitor proteins that negatively regulate the cyclin/CDKs.

Question 34

Cdk-activating kinases (CAK)

Answer 34

A set of activator proteins that positively regulate the cyclin/Cdks.

Question 35

Interplay between Cyclin/Cdk/Rb

Answer 35

1. Cyclin D is expressed.
2. Cyclin D > CKI in the cell.
3. Cyclin D/Cdk4/6 complex forms.
4. Cyclin D/Cdk4/6 à p-Rb.
5. p-Rb released from complex with E2F.
6. E2F initiates transcription of genes that
   promote entry into S-phase and Cyclin E.
7. Cyclin E/Cdk2 complex form to maintain p-
   Rb and continue progress to S-phase.
8. Cyclin E levels decline and Cyclin A levels
   increase.
9. Cylin A/Cdk2 complex forms to carry cells
   into G2 and inhibit E2F-mediated
   transcription.
10. Cyclin A levels decline and Cyclin B levels
    increase.
11. Cyclin B/Cdk1 complex is required for
    mitosis.

Question 36

Two factors dictate progression of the cells through the cell cycle:

Answer 36

1. Cell growth/size

2. No damaged DNA.

Question 37

p53

Answer 37

Activated by DNA damage with activation of the ATR/ATM
kinases.
Activated p53 prevents entry into the S phase of the cell cycle; stimulates DNA repair (GADD45 gene expression); can initiate cell apoptosis (inhibiting Rb).

Question 38

Function of growth factor

Answer 38

Cause dimerization of its receptor and subsequent phosphorylation of the cytoplasmic domain of the receptor.

Question 39

Function of growth factor receptor

Answer 39

Recruits the exchange factor SOS to the membrane to activate RAS.

Question 40

Function of activated RAS

Answer 40

Recruits RAF to the membrane to become activated.

Question 41

CNK

Answer 41

Serves as a molecular scaffold platform for KSR-mediated

RAF activation on the membrane after being recruited by active RAS-GTP.

Question 42

The function of RAF

Answer 42

Initiates a phosphorylation cascade leading to the phosphorylation of a set of transcription factors that
can enter the nucleus and begin S phase.

Question 43

Most of eukaryotic cells are

Answer 43

Are not growing or in the G0 stage. Exception: stem cells, embryonic cells, and cancer cells.

Question 44

Replication and cell division is controlled via

Answer 44

Signal transductions pathways

Question 45

Signal transduction pathway

Answer 45

The process by which a stimulus or cellular state is sensed by and transmitted to pathways within the cell. Example: The EGFR Signal Transduction Pathway.

Question 46

Oncogene

Answer 46

A gene that when mutated may cause cancer.