Exam 2: Study Guide Flashcards

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1
Q

Cell cycle definition

A

for cells that are undergoing continuous cell division, the cell cycle maps out each step that occurs within a cell from one division to the next.

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2
Q

Steps of the cell cycle

A

G1
s
G2
M
G0

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3
Q

G1

A

gap 1 stage between M and S

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4
Q

S

A

synthesis
DNA is replicated/duplicated in preparation for cell division

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5
Q

G2

A

gap 2 stage between S and M

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6
Q

M

A

mitosis, process of cell division, chromosomes are equally divided, 1 cell is split into 2 daughter cells

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7
Q

G0

A

gap 0 stage; when cells get out of the active cell cycle and decide not to divide at that moment in time

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8
Q

Quiescence

A

temporary no cell division

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9
Q

senescence

A

permanent no cell division

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10
Q

Checkpoint definition

A

control mechanisms at different point in the cell cycle that ensure the next step in the process does not proceed until necessary prerequisites are fulfilled.

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11
Q

Significance of checkpoints

A

ensures cell division does not proceed if the cell is not in the right condition to divide

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12
Q

R checkpoint occurrence

A

near the end of G1

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13
Q

R checkpoint importance

A

the most important for cancer prevention

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14
Q

Prior to R checkpoint

A
  • cells are able to exit the cell cycle with relative ease
  • cells are able to exit the cell cycle in response to extracellular signals
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15
Q

After R checkpoint

A

Once a cell moves past the R checkpoint it has made a decision to replicate its NA and it will undergo cell division, unless something major occurs

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16
Q

Cyclin Dependent Kinases (CDK) definition

A

protein complexes that are responsible for progression through a specific stage in the cell cycle

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17
Q

What activates CDK

A

only active when bound to a specific cyclin protein

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18
Q

Cyclin presence in cell cycle

A

only present at specific parts of the cell cycle

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19
Q

CDK presence in cell cycle

A

present throughout the cell cycle

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20
Q

Cyclin D

A

unique because its expression is dependent on extracellular events

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21
Q

Cyclin D complexes with…

A

CDK 4/6

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22
Q

CDK4/6 is required for…

A

progression through G1 and past the R checkpoint

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23
Q

Cyclin D expression depends on…

A

extracellular events

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24
Q

Examples of extracellular events that cyclin D expression depends on

A
  • Growth Factor RTK signaling -> RAS -> MAPK -> transcription of cyclin D1
    • cell division is activated in response to growth factors
    • normal cells are dependent on growth factors to undergo cell division

-Attachment to the ECM -> transcription of cyclin D1
- cell division can only occur when a cell is properly attached to the ECM
- normal cells are only capable of anchorage dependent growth

and more

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25
Q

significance of cyclin D

A

cyclin d expression is heavily regulated because as soon as cyclinD-CDK4/6 is active, cell division will occur

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26
Q

Significance of retinoblastoma (RB)

A

is important for progression past the R checkpoint

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27
Q

Function of RB

A

the protein that inhibits cell cycle progression, it must be inactivated for a cell to undergo cell division

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28
Q

When RB is active

A

RB binds to E2F transcription factors and inhibits the transcription of target genes

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29
Q

When RB is inactive

A

RB no longer binds to E2F transcription factors and target genes are transcribed

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30
Q

E2F target genes generally promote…

A

S phase entry including DNA polymerase and the next cyclin, CyclinE

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31
Q

What controls RB activity?

A

phosphorylation

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32
Q

active RB

A
  • de-phosphorylated
  • mono-phosphorylated
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33
Q

dephosphorylation of RB

A

dephosphorylated by protein phosphatases, this occurs at the beginning of G1

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34
Q

mono-phosphorylation of RB

A

RB is phosphorylated by CyclinD-CDK4/6 (G1 CDK), this occurs throughout the progression of G1

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35
Q

Inactive RB

A
  • hyperphosphorylated
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36
Q

Hyperphosphorylation of RB

A

phosphorylated by CDK2/CyclinE (Late G1, early S CDK)
(rest of G1, s phase, G2, and mitosis)

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37
Q

What causes progression past the R checkpoint and S-phase entry

A

growth factor signaling -> cyclin D1 -> RB

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38
Q

CDK inhibitors definition

A

proteins that inhibit cyclin-CDK4/6 complexes by inhibiting their function or inhibiting their formation

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39
Q

Examples of CDK inhibitors

A

1) p16 inhibits CyclinD-CDK4/6
2) P21CIP is induced in response to DNA damage and inhibits CyclinE-CDK2 + CyclinA-CDK2 (this will stop the progression through the end of G1 and S phase so cells will stop where they are at in the cell cycle and wont go through with DNA replication until DNA is repaired

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40
Q

what can inhibit CDK inhibitors

A

growth promoting signals to promote the progression through the cell cycle
(ex: mitogens -> Pi3k -> AKT -> CDK inhibitor

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41
Q

What is TP53?

A

tumor suppressor gene

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42
Q

How is TP53 different from a normal tumor suppressor gene?

A

behaves in a dominant negative fashion
- only needs 1 loss of function allele to see an effect on transformation

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43
Q

What does TP53 function as? what affect does this have?

A

a tetramer
- this is why 1 mutant allele can have a dominant negative effect on the normal allele, because all you need is 1 mutant peptide in a complex with 3 normal peptides for there to be no TP53 function.

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44
Q

What would happen if TP53 is completely depleted?

A

it would have no effect on transformation, because only the normal TP53 peptides are made and present in the cell, therefore all tetramers are made up of 4 normal TP53 peptides

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45
Q

TP53 is usually made and degraded at ___ rates, this means there are usually ____ levels of
TP53

A
  • high
  • low
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46
Q

What does low levels of TP53 allow for?

A

allows for the cell to be able to rapidly increase the cellular concentration of TP53 in response to physiological signals

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47
Q

TP53 function

A

TP53 tetramer functions as a transcription factor

48
Q

TP53 negative regulation

A

TP53 activates the transcription of MDM2
- MDM2 binds to TP53 and causes ubiquitination of TP53, and subsequent degradation of TP53 via the proteasome

49
Q

Physiological signals that activate TP53

A
  • agents that induce DNA damage (x-rays, UV, certain chemo drugs)
  • stress (low O2, increase in reactive oxygen species)
50
Q

What does TP53 do in response to stress or DNA damage?

A

can induce growth arrest, senescence, or apoptosis

51
Q

How DNA damage induces the accumulation of TP53

A
  • DNA damage activates ATM/ATR kinases
  • these kinases phosphorylate TP53, this causes stabilization of TP53 protein levels because it is no longer able to be degraded by the proteosome
  • ATM/ATR kinases also phosphorylate and inactive MDM2, this increases TP53 protein levels because MDM2 is not longer able to ubiquitinate it.
52
Q

How does TP53 induce cell cycle arrest?

A
  • TP53 activates the transcription of a cyclin-dependent kinase inhibitor, P21.
  • P21 inhibits the cyclin-dependent kinases involved in progression through S-phase, g2, and M
  • at the same time TP53 activated transcription of DNA repair enzymes.
  • result, if there is DNA damage in cells during cell division, TP53 protein levels increase, the cell cycle is paused, and the cell can repair its DNA before cell division can proceed.
  • high levels of cyclin-dependent kinase inhibitors such as P21 can also cause a cell to undergo permanent cell cycle arrest (senescence) this typically occurs if a cell has irreparable levels of DNA damage
53
Q

apoptosis

A

programmed cell death

54
Q

function of apoptosis

A

protecting an organism from rogue cells that are at risk of transformation

55
Q

molecular triggers of apoptosis

A
  • first triggered by cytochrome C being released from the inter-mitochondrial membrane and now accumulating in the cytoplasm
  • proteins that make a pore in the mitochondrial membrane (BAX/BAK) are normally inactivated by BCL2 and other anti-apoptotic proteins
  • under apoptotic conditions, pro-apoptotic proteins (BIM/BAD) bind to BCL2 and other anti-apoptotic proteins; allows BAX/BAK to make a pore and cytochrome C to release into the cytoplasm
  • TP53 induces apoptosis by upregulating the transcription of pro-apoptotic proteins that interact with BCL2
  • UV induced DNA damage and loss of anchorage to the ECM can also induce pro-apoptotic proteins that interact with BCL2
56
Q

What happens after cytochrome C is released?

A

cytochrome c forms a complex with APAF1 proteins = apoptosome
- the apoptosome activates an initiator caspase
- eventually an executioner caspase is activated
- eventually the cell is phagocytosed and further degraded by another cell

56
Q

What are caspases?

A

proteases that break down proteins and peptides
- further cleave and activate more caspases

57
Q

executioner caspase

A

cleave death substrates - critical components of a cell
Ex:
- nuclear lamins: breakdown of the nucleus
- cytoskeleton: breakdown of cellular support structures which results in blebs that protrude from the cell membrane

58
Q

Ways cancer cells have found to inactivate apoptosis

A
  • inactivation of TP53 pathway
  • alterations that inhibit mitochondrial pore formation
59
Q

What is a telomere

A

end of a chromosome

60
Q

What do telomeres consist of in humans?

A

consist of repeats (TTAGGG) repeated over and over again
- single stranded 3’ overhang which invades the upstream chromosome and base-pairs with it; this forms a t-loop structure

61
Q

What does t-loop structure do?

A

protects the ends of chromosomes from appearing like a double-stranded break (DSB); and being joined together via DSB repair proteins

62
Q

With every cell division, telomeres get ____ and _____

A

shorter and shorter

63
Q

Why do telomeres shorten with age?

A

because DNA polymerase is not able to fully replicate the 5’ end of a chromosome during S phase
- offers a mechanism to count how many times a cell has undergone a cell division

64
Q

Normal cells are _____

A

mortal - have a finite number of cell divisions; when they reach this finite number they undergo replicative senescence.

65
Q

replicative senescence

A

the specific senescence that occurs when a cell reaches their maximum number of cell divisions, due to short telomeres

66
Q

markers of senescence

A
  • flat, long, large cell morphology (“fried egg”)
  • Have B-galactosidase enzyme activity, when give the proper substrate it will convert it into a blue product and cells appear blue
67
Q

What molecule induces replicative senescence?

A

TP53

68
Q

What happens if TP53 is not functional?

A

cells will continue to divide and telomeres will get very short
- when telomeres are short, they are no longer able to form a loop structure and they look the same as a double-stranded break

69
Q

What happens if double stranded break occurs?

A
  • then cells will fuse the ends of chromosomes together- end to end fusions
70
Q

What happens if end-to-end fusions occur between sister chromatids during cell divison?

A

they are physically broken apart from each other at random locations during anaphase. This causes one daughter cell to have extra copies of some genes and the other daughter cell to have less copies of some genes.

71
Q

What does repeated end-to-end fusions result in?

A

widespread genomic chaos

72
Q

crisis

A

a cell state caused by genomic damage due to short telomeres. Eventually, the cell will undergo apoptosis.

73
Q

What must cancer cells gain in order to be capable of continuous cell divisons?

A

immortality

74
Q

What does telomerase do?

A

adds bases to the 3’ end of telomeres and makes them longer

75
Q

What barrier do cells have to bypass in order to be capable of immortal growth? How do they bypass it?

A

stress associated senescence barrier
- they have to find a way to inhibit the retinoblastoma pathway

76
Q

In order for cells to become fully immortal they must:

A
  1. inactivate the retinoblastoma pathway to avoid stress-associated senescence
  2. reactivate telomerase to avoid the effects of short telomeres (replicative senescence or crisis induced apoptosis)
77
Q

How to calculate time for cancer formation?

A

you cannot calculate the exact time it takes for cancer to form because it depends on the accumulation of random events/mutations in 1 cell

78
Q

Estimations of how long cancer formation takes?

A
  • for lung cancer about 35 years after cigarette consumption rates go up so do lung cancers rates (this suggests it takes 35 years for lung cancer to form as a result of smoking cigarettes)
  • estimates also differ between types of cancer
79
Q

Evidence of cancer multi-step progress

A

it takes so long to form; idea is that multiple unlikely events have to occur for tumor formation
- a spectrum of tissue morphologies exist between normal and tumor

80
Q

Hyperplasia

A

individual cells look normal, but there is increased cell division and the tissue looks a bit thicker than normal

81
Q

dysplasia

A

loss of well-ordered single layers of cells; tissue morphology also starts to look subtly different than normal

82
Q

in-situ cancer

A

growth looks more severely abnormal; at this point it is still confined within the tissue of origin and considered benign because it doesn’t have the ability to move to a different place in the body. Typically, these are not diagnosed as cancer

83
Q

metastatic

A

severe abnormal growth in the tissue; evidence of malignancy - cells have migrated out of the tissue of origin and into distant sites in the body.

84
Q

What accumulates as cancer progresses?

A

mutations

85
Q

Cancer progression involves _______

A

repeated clonal expansions

86
Q

clonal expansions

A

when one cell happens to get a mutation that increases its fitness relative to other cells around it

87
Q

each mutation must…

A

be complementary, act on a separate molecular pathway, and confer increased fitness relative to the prior mutation

88
Q

peto’s paradox

A

large animals are composed of more cells than smaller animals; therefore they undergo more cell divisions
- animals that live longer undergo more cell divisions than smaller animals
- since cell divisions increase your risk for random mutations due to errors in DNA replication; larger animals that live for a long time should have more cancer than small animals that have short lifespans
- but this isn’t the case

89
Q

Why is petos paradox a paradox?

A

because large animals with long lifespans have evolved more anti-cancer mechanisms
- small animals only need mutations in 2 genes for cancer to occur
- large animals need mutations in a minimum of 5 genes for cancer to occur
- Ex: during the course of human evolution the regulation of telomerase has evolved to be
transcriptionally repressed in most somatic cells. This is an anti-cancer mechanism that has
evolved in humans but is not present in mice as mice express their telomerase gene pretty
readily

90
Q

mutations

A

changes to DNA sequence

91
Q

Mutations have the potential to ____ oncogenes and _____ tumor suppressor genes

A
  • activate
  • inactivate
92
Q

What leads to cancer formation?

A

the random chance of accumulating a specific set of cancer-causing mutations in the same cell that leads to cancer formation

93
Q

mutagens

A

agents that cause mutations in our DNA

94
Q

external mutagen

A

UV light, carcinogens in cigarette smoke or our food

95
Q

Internal mutagen

A

ROS made from mitochondria during cell respiration; errors in DNA replication

96
Q

What does tissue design do?

A

protects stem cells from potential cancer-causing agents

97
Q

stem cells

A

replenish our cells for the rest of our lives
- very important to protect DNA in stem cells from being mutated

98
Q

mechanism to protect stem cells

A
  • they undergo minimal cell divisions and are therefore exposed to less potential errors from DNA replication
    • the transit amplifying cell state is what undergoes lots of cell divisions and these will eventually differentiate
  • they are physically protected within their tissue environment
    • Ex: stem cells in the colon crypts are in a cave with secreted mucous; this physically protects the stem cells from exposure to the harsh conditions of the lower intestine
99
Q

DNA polymerase proofreading mechanism

A

has a proofreading mechanism to go back and fix any errors in DNA replication (if it placed a nucleotide with an incorrect base in the newly synthesized strand; it can move backward, degrade the newly synthesized strand, place the correct nucleotide base, and continue making the new strand)

100
Q

mismatch repair enzymes

A

can fix mistakes in DNA synthesis that DNA polymerase may have missed
- generally works on regions with repeated sequences
- DNA polymerase will tend to slip and get out of proper alignment which can lead to longer or shorter versions of the repeats in the newly synthesized strand. (DNA polymerase proofreading cant detect these errors)

101
Q

mismatch repair proteins fix these mismatch errors by:

A
  1. mutSalpha
  2. mutLalpha
  3. repair DNA synthesis will follow to complete fixing the strand
102
Q

MutSalpha

A

scan and locate the error

103
Q

MutLalpha

A

identify the newly synthesized strand and will degrade it at the site of the error

104
Q

endogenous mutagens produce…

A

lesions in the DNA with minimal effect on the DNA helix structure

105
Q

Base excision repair fixes endogenous lesions by:

A
  1. DNA glycosylate
  2. AP endonuclease
    3a. short path repair
    3b. long patch repair
106
Q

DNA glycosylate

A

cleaves the glycosyl bond linking the nitrogenous base with the rest of the nucleotide

107
Q

AP endonuclease

A

cuts out the rest of the nucleotide

108
Q

short patch repair

A

sometimes DNA polymerase beta and ligase fill in the gap and fix the missing covalent bond between the newly synthesized nucleotide and the next nucleotide

109
Q

long patch repair

A

sometimes a strand displacing polymerase delta will follow DNA polymerase beta and extend the strand with the originals error by several more nucleotides.
- this will create a flap of extra nucleotides which will be cut away by an endonuclease
- ligase will fix the missing covalent bond between the newly synthesized nucleotide and the next nucleotide

110
Q

exogenous mutagens produce…

A

lesions in the DNA with a large effect on the DNA helix structure

111
Q

nucleotide excision repair fixes exogenous lesions by:

A
  1. cleaving the strand with the error about 24 nucleotides on the 5’ side and about 5 nucleotides on the 3’ side
  2. DNA polymerase delta will fill in the gap
  3. DNA ligase will make the missing covalent bond between the newly synthesized nucleotide and the next nucleotide
112
Q

mutations in DNA repair enzymes contribute to…

A

increase cancer risk

113
Q

individuals with Xeroderma pigmentosa have…

A

mutations in their nucleotide excision repair proteins
- when they are exposed to UV light, lesions in their DNA are not repaired and their cells accumulate multiple mutations
- as a result, individuals with xeroderma pigmentosa have 200-fold increased risk of getting skin cancer before the age of 20

114
Q

Syndromes associated with increased cancer risk are caused by…

A

mutations in DNA repair enzymes
- BRCA1/2 are involved in repair of the double stranded breaks via homologous recombination. Individuals with mutated BRCA1/1 genes are at increased risk of getting breast cancer

115
Q

DNA repair genes are classic ______ and an individual needs a _______ in both copies of the allele for a cell to show the phenotype of loss of DNA repair.

A
  • tumor suppressor genes
  • loss of function mutation