Midterm REVIEW cards

Question 1

What is epigenetic regulation?

Answer 1

Transcription in eukaryotes takes place on DNA that is wrapped in chromatin. Chromatin
needs to open for a gene to be activated and transcription to proceed. Chromatin-
mediated regulation is an entirely new mechanism as compared to prokaryotes. It is often
called EPIGENETIC REGULATION of gene expression

Question 2

heterochromatin

Answer 2

regions of chromosomes that are intensely stained
- DNA is more densely packed
- rich in repetitive DNA (Transposons, centromeres
and telomeres)
- not accessible to transcriptional machinery
- Inactive genes are found in heterochromatin

Question 3

Euchromatin

Answer 3

• lightly stained chromosome regions
• Active genes are found in euchromatin
• accessible to transcriptional machinery

Question 4

What kind of RNA does Polymerase I transcribe?

Answer 4

Pre-rRNA (28S, 18S,
5.8S rRNAs)
(These are Ribosome components, used for protein
synthesis)

Question 5

What kind of RNA does RNA pol II transcribe

Answer 5

mRNA - Encodes protein

snRNAs - RNA splicing

siRNAs - Chromatin-mediated repression,
translation control

miRNAs - Translation contro

Question 6

What kind of RNA does RNA pol III transcribe?

Answer 6

tRNAs - Protein synthesis

5S rRNA - Ribosome component, protein
synthesis

snRNA U6 - RNA splicing

7S RNA - Signal recognition particle for
insertion of polypeptides into the
endoplasmic reticulum

Other small stable
RNAs with Various functions, unknown for
many

Question 7

How many polypeptides does RNA pol II consist of?

Answer 7

12

Question 8

The 5 main components of RNA polymerase - and their functions?

Answer 8

1. Clamp domain - opens to accomodate DNA and closes/swings shut during transcription to anchor polymerase to DNA
2. Bridge - closes clamp domain
3. Catalytic centre - where the synthesis of RNA takes place (with the participation of Mg++)
4. Channel - where newly synthezed RNA exits and then is immediately capped by 7MGuanosine
5. Carboxyterminal domain - domain involved in many regulatory interactions and plays a key role in initiation, release and elongation of synthesized mRNAs

Question 9

Carboxyterminal Domain

Answer 9

part of RPB1 subunit

This is a specialized domain not found in any other polymerase, prokaryotic or eukaryotic.
* The CTD is involved in multiple regulatory interactions and plays a key role in the initiation,
release, elongation and processing of the synthesized mRNAs
* The CTD in yeast contains of 26 repeats of Tyr-Ser-Pro-Thr-Ser-Pro-Ser
– in mammals it contains 52 repeats
* The Ser residues in the CTD are phosphorylated upon transition from initiation to
elongation
* The CTD is not structured. It is not shown on the previous two or three slides because it
can not be analysed by crystallography!!

Question 10

Core promoter sequences

Answer 10

TATA box – a tight consensus sequence
o prevalent in highly transcribed genes
- -31 to -26

• Initiator – less conserved element
  o some genes contain Initiator but no TATA
• -2 to +4
• BRE (TFIIB Recognition Element) and/or
  o influence the activity of the promoter
• -38 to -32
• DPE (Downstream Promoter Element)
  o influence the activity of the promoter
  +28 to +32

Question 11

What techniques do we use to analyse these processes?

Answer 11

• DNA sequencing
• PCR
• RT-PCR
• RNA sequencing
  Techniques with antibodies:
• immunofluorescence, immunoprecipitation, ChIP

Question 12

PCR

Answer 12

Polymerase Chain Reaction

PCR: amplifying DNA by repetitive cycles of denaturing and renaturing of
DNA in the presence of thermostable DNA polymerase (Taq polymerase).
* You need two primers that anneal to the ends of the amplified DNA
fragment at 50 - 60°C.
* Taq polymerase synthesizes new strands of DNA starting at the 3’-ends of
the annealed primers (72°C).
* Newly synthesized DNA is denatured at high temperature (95°C).
* The temperature is lowered (50 - 60°C) and more primers anneal to the new
strands.
* The cycle is repeated 30 times. This means that if you start with one
molecule of DNA you will end up with 230 molecules

Question 13

RNA sequencing and qRT-PCR

Answer 13

QUANTITATIVE
techniques for the analysis of gene expression
9
* RNA is isolated from a sample
* RNA is converted to DNA by the use of specific primers directed to
a specific gene and RNA-Dependent-DNA Polymerase
RT-PCR (Reverse Transcription-PCR)
* Use the produced DNA in a PCR reaction with Taq polymerase and
the specific primers directed to a specific gene
* By quantifying the DNA produced by PCR you indirectly quantify
the abundance of the corresponding RNA in the sample

Question 14

RNA sequencing

Answer 14

RNA is isolated from a sample and converted to DNA by the use of random primers and
RNA-Dependent-DNA Polymerase
* Break the produced DNA in small (200 bp) pieces.
* Sequence the DNA by Massive Parallel DNA sequencing
* The produced sequences are analyzed by a software and aligned to the sequence of the
genome (see the animation on the next slides)
* The number of sequences of that align to each locus in the genome are quantified and
then plotted.
* Resolution in the plot is very high, sometimes within a base.
* The plot is giving a quantitative presentation of the levels of transcription at each
position of the genome.

• In lecture 4 (Fig. 9-17) this technique is used for the analysis of uni-directional and bi-
  directional transcription from a eukaryotic promoter

Question 15

What are antibodies?

Answer 15

Antibodies are natural immunoglobulins produced by animals to combat
invading exogenous proteins of any kind
* A specialised class of B-lymphocytes rearrange the Ig genes

• Each B-lymphocyte produces ONE UNIQUE ANTIBODY against an exogenous
  protein (THE ANTIGEN).
• Upon invasion by an antigen the B-lymphocytes recognise this antigen,
  proliferate and produce large amounts of the antibody to destroy it

Question 16

monoclonal antibody

Answer 16

The antibody produced by one B-lymphocyte (one clone)

Question 17

polyclonal antibodies

Answer 17

In the blood of the animal with multiple B-lymphocytes produce multiple
antibodies.

Question 18

Structure of an antibody

Answer 18

ntibodies consist of two heavy and two light chains, joined by disulphide bonds.
Each light and heavy chains consist of a constant domains and a hyper-
variable domain. In the complete antibody the hyper-variable domains of the heavy
and lights chains come together to form the antigen binding site.

Question 19

Epitopes

Answer 19

the part of the antigen that the antibody recognizes and binds to via the antigen binding site/paritope!

epitopes can be:
- Linear, made up of continuous sequence of amino acid
* Conformational made up of animo acids close together in the folded structure of
the protein

Question 20

Production of antibodies against a desired antigen

Answer 20

We can “trick” an animal (mouse, rabbit, sheep, donkey) by injecting it
with a protein (an antigen) of choice. The animal will respond by
producing multiple antibodies against the antigen.
* We can take out some blood from these animals, purify the
immunoglobulins and prepare POLYCLONAL ANTIBODIES AGANST THE
ANTIGEN
* We can isolate single clones of B-lymphocytes, maintain them in culture
to produce MONOCLONAL ANTBODIES
* Now we have highly specific antibodies against a protein of interest.
We can use these antibodies for various techniques of identification and
quantifying

Question 21

Immuno-fluorescence

Answer 21

We can use specific antibodies coupled to
a fluorescent dye and localize the antigen in the cell

1) prepare sample and place on microscope slide
2) incubate with primary antibody
3) incubate with fluorochrome-conjugated secondary antibody: wash away unbound antibody
4) mount specimen and observe in fluorescence microscope

Question 22

Immuno-precipitation

Answer 22

We can hook the antibodies to large
beads and mix them with extract, then wash away the extract.
The antigen (and its associated proteins) remain associated with
the beads via the antibody

1) primary antibody is added to mixed protein solution
2) agarose beads are added which form a complex with ab-antigen
3) centrifuge and wash to separate antigen complex from mix
4) elute antigen (protein of interest) and detect via western blot

Question 23

ChIP

Answer 23

Chromatin Immuno-Precipitation
- multi-step technique that detects the binding of a specific protein to a specific DNA element (or multiple DNA elements) in vivo

Question 24

ChIP steps

Answer 24

1) treat living cells with a membrane-permeating cross-linker such as formaldehyde
2) sonicate to shear cellular chromatin to short fragments and add antibody to pol II
3) immunoprecipitate to isolate pol II cross-linked to dna
4) revers cross-linking, isolate DNA, and subject to massively parallel DNA sequencing

Question 25

Formation of the PIC steps:

Answer 25

1) association of TFIID to the promoter

2) followed by TFIIB and TFIIA

3) RNA polymerase is recruited and The non-phosphorylated CTD
(Carboxy Terminal Domain) of
RNA pol II establishes contact
with several GTFs

4) TFIIE and TFIIH are recruited

5) TFIIH helicase activity facilitates
the opening of the double helix of
DNA: Open PIC

6) RNA Polymerase II initiates
transcription

7) TFIIH kinase activity
phosphorylates the CTD to release RNA polymerase from
the promoter

Question 26

Transition from initiation to elongation steps / pausing steps

Answer 26

After the initial phosphorylation of CTD by TFIIH kinase,

1) two negative
elongation factors, NELF and DSIF, associate with RNA polymerase II

2) they pause it downstream of the initiation site.

3) * At this point another kinase, CDK9/CycT (also called PTEFb (Positive
Transcription Elongation Factor b) phosphorylates the CTD and NELF.

4) * RNA polymerase II with fully phosphorylated CTD is released

5) * Positive elongation factors PAF and Spt16 join the polymerase.

6) * DSIF becomes a positive elongation factor

Question 27

What happens in HIV/AIDs with Tar and Tat

Answer 27

The 5’ end of the short RNA is synthesized before the polymerase pauses
* The short RNA forms a secondary structure called TAR. TAR holds the polymerase and has an
inhibitory effect on Cdk9/CycT.
* HIV1 (human Immuno-deficiency virus) encodes the protein called Tat. Tat binds to TAR.
* Upon cell stress other factors release the polymerase and some Tat is made.
* When bound to TAR, Tat ACTIVATES CDK9/CycT kinase and releases the paused RNA
polymerase II
* Now more Tat is made and viral genes are fully expressed

Question 28

what is the key to HIV latency

Answer 28

HIV is dormant but activates upon stress when the organism is
weak, slowly killing T-cells and abolishing immunity
* The patients die from lost immunity and any opportunistic infections

The Tat protein of HIV holds RNA polymerase II until conditions for the spreading of the virus are favorable

Question 29

TATA-containing promoters vs. CpG islands

Answer 29

At most TATA-containing and TATA-less promoters transcription initiates
and proceeds in one direction
* !!!Your textbook does not have a good illustration on this type of uni-
directional transcription initiation. The figure shown on the next slide
is from the 8th edition of the textbook.

• At other promoters (called CpG islands) transcription initiates in both directions, but the polymerase stalls and falls off DNA in one of the directions and then proceeds in the other direction only

Question 30

WHat are CpG islands

Answer 30

CpG islands occupy the promoters of about 70%
percent of the genes in vertebrates. These genes
are transcribed at a low rate.
* CpG rich DNA contains less nucleosomes and are easier to transcribe.
* At CpG islands transcription initiates at any position (not defined) within the island.
* Transcription from CpG island initiates in both directions, but proceeds in the direction of the
Open Reading Frame (ORF) only.

Question 31

How do we find gene regulatory sequences?

Answer 31

Linker scanning-reporter gene technique for identifying gene regulatory elements on DNA

Question 32

What is linker scanning mutagenesis?

Answer 32

You buy a plasmid that contains a reporter gene, usually a gene that encodes an enzyme (tk in figure 8-17)
The tk ORF (Open Reading Frame) on the plasmid has no promoter.
Upstream of the tk ORF you clone a large piece of DNA that you know is controlling YOUR gene of interest (not tk )
You systematically replace small pieces (10 bp) of the cloned regulatory DNA by a “linker”.
You produce multiple plasmids with linker substitutions (9 in figure 8-17)
You transform each individual plasmid in a cell culture and measure the activity of the reporter (a simple enzyme reaction of cell extract)
Loss of reporter activity means that you have hit a regulatory sequence

Question 33

Reporter plasmids contain:

Answer 33

• an ORF for an enzyme (Luciferase (luc) in this case)
• NO promoter-proximal regulatory sequences
• A multiple cloning site (MCS) for the cloning a regulatory sequence
• Translation initiation elements upstream of the reporter gene
• Poly A+ signal downstream of the reporter gene

Question 34

linker scanning mutation steps

Answer 34

• Eukaryotic promoter region DNA that supports high-level expression of a reporter gene (light purple) is cloned in a plasmid vector.
• Overlapping linker scanning (LS) mutations (insertions of scrambled sequence) are introduced from one end of the region to the other.
• Mutant plasmids are transfected separately into cultured cells.
  The expression/activity of the reporter gene product (thymidine kinase gene) is assayed.

Question 35

Enhancers

Answer 35

distal regulatory elements,

could be upstream or downstream of the initiation site

position independent

orientation independent

Question 36

promoter proximal regions

Answer 36

Function within a short distance (several hundred bps from the transcription initiation site

Immediately upstream or downstream from the initiation site

Position dependent: usually non-functional if moved

Question 37

How do enhancers communicate with promotors

Answer 37

via bending DNA

Question 38

Pioneer transcription factor

Answer 38

Binds to a specific regulatory sequence within the condensed chromatin
Interacts with chromatin-remodeling enzymes and histone acetylases that decondense the chromatin, making it accessible to RNA polymerase II and general transcription factors.

Question 39

Activator proteins

Answer 39

Bind to specific transcription-control elements in both promoter-proximal sites and distant enhancers
Interact with one another and with the multisubunit Mediator complex to assemble general transcription factors and RNA polymerase II (Pol II) on promoters

Question 40

Transcriptional activation

Answer 40

Pol II initiates transcription.
Pol II pauses after transcribing fewer than 100 nucleotides because of action of the elongation inhibitor NELF (negative elongation factor) associated with DSIF (DRB sensitivity-inducing factor).

Question 41

Activators in transcription

Answer 41

Promote association of the Pol II-NELF-DSIF complex with elongation factor P-TEFb (cyclin T-CDK9 [kinase]), which releases NELF
NELF release allows resumption of RNA transcription.

Question 42

How can transcriptional activators and repressors be identified?

Answer 42

by deletion analysis of the protein

Question 43

What are the two main domains of transcriptional activators and repressors?

Answer 43

Transcriptional activators or repressors often have independent DNA-binding domains and effector (activation or repression) domains

Question 44

Details of deletion expereiment for transcriptional activators and repressors

Answer 44

Activators are composed of distinct functional domains.
(a) Reporter-gene construct containing a lacZ reporter gene (encoding β-galactosidase) and a TATA box ligated to UASGAL, a regulatory element that contains several Gal4-binding sites.

The reporter-gene construct and DNAs encoding wild-type or mutant (deleted) Gal4 introduced simultaneously into mutant (gal4) yeast cells.

Activity of β-galactosidase expressed from lacZ was assayed.

High activity if introduced GAL4 DNA encodes a functional protein.
(b) Wild-type and mutant Gal4 — results:
N-terminal end 50 amino acid deletion:
Eliminates UASGAL binding.
Contains DNA-binding site.
C-terminal end deletions:
β-galactosidase expression activation eliminated only when between 126 and 189 or more amino acids deleted from the C-terminal end.
End contains activation domain.
Internal deletions — Gal4 central region is not crucial for its function in this assay.

Question 45

Motif

Answer 45

Motif is a sequence in a protein that, based on similarity with other known functional domains, is likely to produce a distinct 3D structure and a distinct domain.
- WITHIN DOMAINS
ex. zinc-finger, leu zipper

These motifs are easy to recognize by bioinformatic tools

Question 46

What is a domain

Answer 46

DOMAIN is a portion of a polypeptide with a specified function: Transactivation Domain, dimerization domain, dna binding domain, hormone binding domain.
The domains are identified by deletion analysis of actual peptides

Question 47

Zn finger motif

Answer 47

regularly spaced C and H amino acids
have a zn2+ in middle
in DNA binding domains

Question 48

leu zipper motif

Answer 48

is found in protein dimerization domains: regualrily spaced L amino acid residues:

Question 49

HLH motif

Answer 49

helix turn helix motif
found in many dimerization domains

Question 50

Dimers

Answer 50

This means that if two proteins (green and yellow) can combine, they will have three binding sites (gg, yy, gy), three proteins (green, yellow, blue) can have six binding sites (gy, gb, yb, gg, yy, bb)

These six combinations can recognize six DNA binding sites
Adding an inhibitor for one of the three proteins increase the possible outcomes for activation (sort of a ‘fine tuning’)

Question 51

Enhanceosome

Answer 51

Combinatorial binding to DNA-binding proteins from different classes

Eukaryotes have very complex promoters and enhancers.
In these, various transcription activators and repressors can bind

The transcriptional activators or repressors often interact with each other and cooperate to establish a strong activator or repressor site to tightly control the expression of the target genes

Question 52

What are silencers?

Answer 52

There are positions in the genomes where large heterochromatin blocks are established and maintained
We will call these positions “SILENCERS”

Genes that are close to these SILENCERS are strongly repressed
The repression of the genes depends on their position (close to, or within heterochromatin) relative to the silencer and does not depend on their promoters.

Question 53

What are the two mating type loci in yeast?

Answer 53

HML and HMR
they are constantly repressed thanks to Rap1 and Sir proteins

Question 54

How does acetylation of histone tails affect the activity of genes?

Answer 54

Acetylation of the histone tails reduces their positive charge, allows for the opening of chromatin (euchromatin) and for transcription

Deacetylation of histone tails reconstitutes positive charge, strengthens DNA-histone interactions, compacts chromatin(heterochromatin) and suppresses transcription

Question 55

How does Rap1 and Sir proteins regulate the repression of HML and HMR?

Answer 55

Two mating type loci , HML and HMR, are constantly repressed
Genetic studies identified proteins that regulate this repression
Rap1 binds next to the repressed loci and recruits Sir (Silent Information Region) proteins
Sir proteins spread away and cover the chromatin of nearby genes
Sir2 is a HISTONE DEACETYLASE. It deacetylates the histones within HML and HMR
The deacetylated histones bind tighter to DNA and form heterochromatin
If the genes in HML and HMR are moved away from HML and HMR loci, they become active
Conclusion: Histone hypo-acetylation is necessary for positional gene repression

Question 56

Lysine to arginine substitution in histone tails result

Answer 56

PERMANENT GENE REPRESSION

Arginine retains a positive charge only, can not be acetylated.
The DNA-histone interaction is strong, chromatin is compact, gene repression can not be reversed by acetylating the histones
These mutants maintain permanent gene repression at telomeres.

Question 57

Lysine to glutamine substitutions in histone tails affect

Answer 57

CANNOT BE DEACETYLATED - CANNOT BE REPRESSED
Glutamine has a neutral charge, resembles acetylated Lysine, can not be deacetylated to gain a positive charge

Chromatin is decondensed, gene can not be repressed.

These mutations maintain permanent gene activation at telomeres

Question 58

How are genes silenced at the telomeres of S. cerevisiae?

Answer 58

Rap1 binds directly to the telomeric repeats

SIR proteins are recruited by Rap1

Sir proteins spread in the sub-telomeric regions and cover them with condensed hypoacetylated nucleosomes

More detailed:
Many telomeres cluster at the periphery of the nucleus and form large 3D heterochromatin block
RAP1 –
binds HML and HMR silencer sequences that are repeated multiple times at each yeast-chromosome telomere
nucleates formation of heterochromatin
SIR2 – histone deacetylase that removes acetyl groups on histone tail lysines
RAP1 and SIR2, 3, and 4 proteins – bind to each other
SIR3 and SIR4 – bind to histone H3 and H4 N-terminal tails that are maintained in a largely nonacetylated state by SIR2 deacetylase

Question 59

What does that telomere and sir protein light up diagram represent?

Answer 59

(b,c) Yeast cells labeled with:
(b) a telomere-specific hybridization probe and
(c) fluorescent-labeled specific anti-SIR3 antibody
SIR3 – localized in the repressed telomeric heterochromatin. (RAP1, SIR2, and SIR4 also colocalize with the repressed telomeric heterochromatin.)

Question 60

What is rap 1

Answer 60

the REPRESSOR
binds to the silencer region of the dna and recruits the corepressors

Question 61

What are sir 3 and sir 4

Answer 61

co-repressors - they bind to H3 and H4 N-terminal tails (deacetylated)

Question 62

What is sir 2?

Answer 62

a histone deacetylase - removes acetyl groups on histone tail lysines

Question 63

What type of repression is RAP1 and Sir

Answer 63

Position-dependent repression of genes
The repression of the genes depends on their position (close to, or within heterochromatin) relative to the silencer and does not depend on their promoters
these positions are called silencers

If the genes in HML and HMR are moved away from HML and HMR loci, they become active
Conclusion: Histone hypo-acetylation is necessary for positional gene repression

Question 64

Promoter dependents repression and activation

Answer 64

promotor-dependent and position-independent regulation also works via histone acetylation/deacetylation, however different enzymes and different co-repressors and co-activators are employed

EX. Ume6 and Rpd3
The repressors and activators bind to promoters and enhancers and recruit the co-repressors or co-activators

The co-repressors and co-activators do not spread, they stay localized at the regulatory elements

In these situations the position of the genes is not a critical factor; the promotor is.

Question 65

What kind of enzyme complex will Gcn4 recruit? Which specific domain of Gcn4 will interact with this enzyme complex? What would be the consequence of its activity?

Answer 65

1) the SAGA complex - a histone hyperacetylase
2) the activation domain
3) ACTIVATION of the genes with the UAS promoter

Question 66

What kind of enzyme complex will Ume6 recruit? Which specific domain of Ume6 will interact with this enzyme complex? What would be the consequence of its activity?

Answer 66

1) Rpd3L - a histone deacetylase
2) the repression domain (RD)
3) Repression of the genes with URS1 promoter

Question 67

Primary sex traits

Answer 67

changes in the body part of reproduction (menstration)

Question 68

What kind of enzymes do transcriptional activators of SAGA and Gcn5 complex recruit?

Answer 68

Histone acetyl transferases

and a class of co-activators called chromatin remodeling complexes

Question 69

What do chromatin remodeling complexes do?

Answer 69

Chromatin-Remodeling Complexes then use ATP to push nucleosomes along DNA to “open” promoters

ex. SWI and SNF

Question 70

What are SWI and SNF?

Answer 70

Chromatin remodeling factors

SWI/SNF (switching/sucrose non-fermenting) ATPases

Move nucleosomes along the DNA by sliding or transferring them
Shift nucleosomes away from the promoter/enhancer sites and gives transcription factors access to the DNA
Chromatin-Remodeling Factors use ATP to move nucleosomes

Question 71

What is the mediator

Answer 71

an example of a co-activator
(a major “bridging” factor between transcriptional activators, GTFs, histone modifiers and chromatin remodelers)

The Mediator is a huge complex containing variety of proteins including enzymes, proteins that recognize activators, proteins that recognize GTFs and other coactivators
Different mediator complexes contain proteins required for specific genes

The Mediator(s) directly interacts with transcriptional activators
One mediator complex can interact with multiple transcriptional activators
The Mediator(s) directly interacts with GTFs

Question 72

Transmission of memory by epigenetic means

Answer 72

Cell differentiate and select the genes they express.
Then the cells divide multiple times but continue to express the same genes after each cell division.
DNA replication is a major disruptor of chromatin and bound transcription factors.

The “memory” of gene expression is achieved mainly through reconstitution of the same chromatin structure after the passage of the replication fork
We call this epigenetic memory of transcription

Question 73

epigenetic marks

Answer 73

Epigenetic control of transcription is maintained by DNA methylation and methylation and acetylation of histones.

We call these epigenetic marks
Epigenetic marks are faithfully rebuilt during and/or soon after the passage of the replication fork

Question 74

Epigenetic marks on histones

Answer 74

(Histone post-translational modifications (PTMs): Methylation, Acetylation, Phosphorylation

Question 75

which is the only methylated histone site that leads to repression

Answer 75

H3 (K9, K27) Me2/3

Question 76

DNA methylation

Answer 76

Methylation of CpG islands leads to condensation of chromatin
Methylated DNA recruits special class of Me-DNA-biding proteins (MeBPs).
MeBPs recruit factors that deacetylate histones and condense chromatin

IMPORTANT: the methylation of DNA is rebuilt immediately after the passage of the replication forks

Question 77

At what levels do noncoding RNAs regulate gene expression?

Answer 77

At the levels of transcription, RNA processing and translation

Question 78

miRNA

Answer 78

Short interfering RNAs, 22 bp long, that base-pair extensively, but not completely with mRNAs, especially over bases 2 to 7 at the 5’ end of the miRNA (the seed sequence)

this pairing inhibits translation of the target mRNA and targets it for degradation

Question 79

miRNAs

Answer 79

microRNAs, 22 bases long, base pair extensively, but not completely (imperfect match) with mRNA, especially over bases 2 to 7 at the 5’ end of the miRNA

THis pairing inhibits translation of the target mRNA

Question 80

What modifications happen to premRNA to produce a mature mRNA

Answer 80

Pre-mRNA (the primary transcript) is capped, then spliced, then polyadenylated before being exported to the cytoplasm

Question 81

splicing

Answer 81

means linking two ropes without a knot
A large ribonucleoprotein spliceosome complex catalyzes the joining of two exons

Question 82

Alternative splicing

Answer 82

and alternative cleavage at different poly(A) sites yield a significant variety of different mRNAs from the same gene in different cell types or at different developmental stages.

Some of the proteins generated by alternative splicing have drastically different activities

Question 83

Two examples of alternative splicing in in

Answer 83

1) Alternative splicing in neurons
Perception of sound
Neuron connectivity
2) Sex determination in Drosophila

Question 84

Dscam isoform expression in Drosophila neurons

Answer 84

Most extreme example of regulated alternative RNA processing yet discovered
Helps to specify tens of millions of different specific synaptic connections between neurons in the Drosophila brain
(bottom) Dscam neuron cell-surface protein:
10 different immunoglobulin (Ig) domains (ovals)
6 different fibronectin type III domains (rectangles)
One transmembrane domain (yellow)
C-terminal cytoplasmic domain
Fully processed mRNA contains:
One of 12 Ig2 exons
One of 48 Ig3 exons
One of 33 Ig7 exons
One of 2 transmembrane exons
Alternative splicing – can generate 12 x 48 x 33 x 2 = 38,016 possible isoforms

Question 85

Dendrite self avoidance wildtype vs. dscam mutants

Answer 85

Dendrites are branches that extend out of the neural cell bodies
The branches of the same neuron repel each other while overlapping with branches from other neurons
This is controlled by Dscam
Dscam mutants show evidence of reduced dendrite self-avoidance

Question 86

autism vs. normal spicing

Answer 86

In one-third of patients with autism spectrum disorder (ASD), more skipping of cassette exons and microexons are observed than in normals
Abnormalities in the frequency of splicing of alternative 5′ and 3′ splice sites are observed only rarely in ASD.

Question 87

What is sxl

Answer 87

Sxl is an RNA binding protein, acts as a suppressor of splicing
- present in females
it suppresses the splicing of sxl gene btw exons 2 and 3

and it suppresses the splicing of tra protein btw exons 1 and 2

Question 88

Tra

Answer 88

Tra is an RNA binding protein, acts as an activator of splicing

activates splizing of dsx

Question 89

Dsx

Answer 89

is a transcription activator/repressor

activates transcription of genes in females encoding female parts
represses transcription the transcription of those genes in males

Question 90

What other protein regulates splicing of Dsx

Answer 90

Rpb1 (along with tra) form a complex that activates splicing of the Dsx gene
Tra-Rbp1-Tra2 complexes interact with six exonic splicing enhancers in exon 4.
Activates exon 3–4 splicing

Question 91

Lin-4, Let-7

Answer 91

The development of C. elegans is regulated by multiple lineage (LIN) determining and lethality (LET) genes
Interestingly, the genes lin-4 and let-7 genes are NOT encoding for any protein
They produce small RNAs that are complementary to other mRNAs

The imperfect binding of the lin-4 or let-7 micro RNAs (miRNAs) to the 3’-UTR (untranslated region) of the mRNAs dramatically reduce the translation of the mRNAs

Question 92

Difference between miRNAs and siRNAs

Answer 92

miRNA would IMPERFECTLY (there are mismatches) bind to the target mRNA
The binding is usually at the 3’-UTR
The imperfect binding leads to repression of translation

siRNA are produced from cleavage of double stranded RNA
They bind PERFECTLY (no mismatches) to the target RNA and cause its rapid degradation by a mechanism called RNA interference

Question 93

miRNA function in limb development

Answer 93

Dicer gene knockout – eliminates generation of miRNAs in mammals – embryonic lethal
Experiment – Dicer knockout only in limb tissue (13-day mouse embryos): (immunostained for Gd5 protein, a marker of joint formation)
(left) Dicer-expressing control – normal limb development
(right) Dicer-knockout in limbs (Conditional expression of Cre induces dicer gene deletion only in limb cells.)
Result – fundamental pattern maintained but abnormal limb development
Conclusion – Dicer (miRNA) is required for morphogenesis but not patterning of the vertebrate limb

Question 94

miRNA processing in nucleus

Answer 94

miRNA transcription and processing:
RNA polymerase II transcribes primary miRNA transcripts (pri-miRNA) – folds to form double strand region
Nuclear double-strand RNA–specific endoribonuclease Drosha and double-strand RNA–binding protein DGCR8 (Pasha in Drosophila) bind pri-miRNA double strand regions
Drosha cleaves the pri-miRNA – generates a ~70-nucleotide pre-miRNA

Question 95

WHat transports miRNA to the cytoplasm

Answer 95

Exportin 5
– nuclear transporter transports processed pri-miRNA to the cytoplasm

Question 96

RNA interference steps once in cytoplasm

Answer 96

Step 1: dsRNA is cleaved to small (21-28 bp) RNA fragments by a dsRNA-endonuclease called DICER
Step 2: Small dsRNAs bind to RISC (RNA Induced Silencing Complex) and form an inactive RISC complex.
Step 3: Argonaut in RISC is a RNA helicase. It unwinds dsRNA. Active RISC* is formed.
Step 4: RISC*, guided by single-stranded siRNA, cleaves target mRNA (“SLICER” activity by Argonaut)

Question 97

Argonaut

Answer 97

Argonaut (RISC):
An RNA helicase
Removes one of the strands of the dsRNA
Leads the trimmed siRNA to the target mRNA

Question 98

What is PAX6

Answer 98

PAX6 regulates several processes during the development of mammals

PAX6 is a developmentally regulated gene
it encodes a transcriptional activator

It contains three promoters and three transcription initiation sites

It has more than three enhancers

It is expressed at different times and in different organs during embryonal development

It is expressed in certain parts of the brain in adult organisms

The translation of PAX6 mRNAs is regulated by several miRNAs

Question 99

How many promotors does pax6 have?

Answer 99

3 alternative promoters

Question 100

10.5 days after fertilization - the embryo

Answer 100

the gene is inactive in all cells of the body: heterochromatin

Question 101

Close to 10.5 days after fertilization:

Answer 101

pioneer factors open heterochromatin at the PAX6 locus in the pancreas (P) and the lens plate (LP)

In all other organs, PAX6 remains in heterochromatin

Question 102

10.5 days after fertilization:

Answer 102

the pancreatic and lens/cornea enhancers activate PAX6 by bending DNA and come in proximity of the promoter

The pancreatic and the lens/cornea enhancers will be occupied by multiple, but not identical transcriptional activators (enhanceosome)

Transcriptional activators communicate with the GTFs, Mediator, Chromatin Remodeling Complexes and Histone Modification Complexes.

Histones are acetylated
Nucleosomes are displaced from the promoter region

all other organs, PAX6 remains in heterochromatin

Question 103

What happens 10.5 days after fertilization after the displacement of nucleosomes:

Answer 103

Mediator and GTFs communicate
GTFs recruit RNA polymerase II
The CTD of RNA polymerase II is phosphorylated by the TFIIH kinase
The TFIIH helicase activity opens the DNA helix and the RNA polymerase starts transcribing

Question 104

Between 10.5 to 13.5 days after fertilization:

Answer 104

: enhaceosome exchange
The Lens/cornea enhancer stops working
The pancreatic enhancer continues to operate
The retina enhancers start operating. They will bend DNA and come in the proximity of the promoter.

Question 105

After day 13.5:

Answer 105

telencephalon and rhomboencephalon will start working in the brain

Question 106

What is the difference between the transcripts of the different promoters of PAX6

Answer 106

PAX6 can produce three different transcripts from its three promoters
Alternative splicing of these transcripts produces proteins with different function.

Transcript a) is missing exon 1 and the alternative exon α (alpha)

Transcript b) is missing exon 0 and the alternative exon α (alpha)

Transcript c) is missing exons 0 and 1-4 but retains the alternative exon α (alpha)

Exons 1-4 encode a part of the DNA-binding domain of PAX6
Hence, transcripts a) and b) produce similar DNA binding proteins that differ in their DNA-binding domains and bind to different DNA elements
Transcript c) has an entirely different function

Question 107

Enhancers

Answer 107

Range ~50 to 200 bp.
Binding sites for several transcription factors.
Interacting proteins distally bound can produce loops in the DNA between their binding sites

Question 108

β-Interferon enhancer:

Answer 108

Six overlapping binding sites bind two heterodimeric factors:
Jun/ATF-2, p50/RelA (NF-κB), and 2 copies each of the monomeric transcription factors IRF-3 and IRF-7.

Question 109

Hydrophilic hormone pathway

Answer 109

Hydrophilic signaling molecules – (small molecules [adrenaline, acetylcholine], peptides [yeast mating factors, glucagon], and proteins [insulin, growth hormone])
Cannot diffuse across the cell membrane
Step 4: Bind to specific cell-surface receptor proteins – triggers receptor conformational change that activates the receptor
Step 5: Activated receptor activates one or more downstream signal transduction proteins or small-molecule second messengers.
Step 6: Signal transduction proteins or small-molecule second messengers activate one or more effector proteins.
Step 7a: Effector – stimulates modification of specific cytosolic proteins; short-term (sec-min) changes in cellular function, metabolism, or movement
Step 7b: Effector – moves into the nucleus; triggers long-term (hours-permanent) changes in gene expression
Termination or down-modulation of the cellular response –
Step 8: Negative feedback/feedback repression from intracellular signaling molecules
Step 9: Destruction of the extracellular signal

Question 110

Hydrophobic signalling pathway

Answer 110

Hydrophobic signaling molecules – (steroids and related molecules)
Step 1: Diffuse through the plasma membrane
Step 2: Bind to cytosolic receptors
Step 3: Receptor-signal complex moves into the nucleus – binds transcription-control regions in DNA to activate or repress gene expression

Question 111

Types of extracellular signalling

Answer 111

• endocrine
• paracrine
• autocrine
• membrane protein signals

Question 112

Endocrine signaling

Answer 112

endocrine hormones affects target cells some distance from site of synthesis
transported through circulatory system
ex. epinephrine and insulin

Question 113

Paracrine signaling

Answer 113

paracrine hormones affect target cells in close proximity
some may bind to ECM
affect only nearby target cells expressing the receptor

ex. neurotransmitters, growth factors

Question 114

Autocrine signalling

Answer 114

cells respond to signals that they produce
target sites are on the same cell
ex. growth hormones

Question 115

Signalling by plasma-membrane attached proteins

Answer 115

signal neighboring cells by direct contact with surface receptors.

Question 116

What is Dex?

Answer 116

Dexamethasone, activates the Glucocorticoid Receptor

Question 117

What are the three domains of the glucocorticoid receptor?

Answer 117

Activation domain, DNA binding domain and ligand binding domain

Question 118

How does Dex activate the receptor and such…?

Answer 118

Dex = Dexamethasone, activates the Glucocorticoid Receptor
In the absence of Dex the Glucocorticoid Receptor is in the cytoplasm
In the presence of Dex the receptor-hormone complex moves to the nucleus
The control ꞵ-galactosidase is cytoplasmic regardless of the presence of Dex

In more detail:
(–) Hormone – receptor retained in the cytoplasm by interaction between its ligand-binding domain (LBD) and chaperone proteins
(+) hormone –
1) Hormone complex diffuses through the plasma membrane and binds to the receptor ligand-binding domain.

2) Causes a conformational change that releases the chaperone proteins

3) Receptor-hormone complex translocates into the nucleus.

4) DBD binds to response elements
Ligand-binding domain and an additional activation domain (AD) at the N-terminus stimulate transcription of target genes.

Question 119

Membrane receptors

Answer 119

bind only a single type of hydrophilic signaling molecules

All hormone receptors are highly specific for their ligands
High affinity of the hormone-receptor interaction
Each receptor binds only a single type of hormone
The hormone changes the conformation of the receptor sending a signal to the cascade

Question 120

Examples of hydrophilic signaling molecules

Answer 120

small molecules: adrenaline, acetylcholine
peptides: glucagon
proteins: insulin, growth hormones

Question 122

Growth hormone coupled to two receptors findings:

Answer 122

Basically - found that only 8 DISTANTLY SPACED AMINO ACIDS CONTRIBUTE 85% BINDING ENERGY

As determined from the three-dimensional structure of the 1 growth hormone:

2 growth hormone receptor complex, 28 amino acids in the hormone are at the binding interface with the first receptor protein.

To determine which amino acids are important in ligand-receptor binding, researchers mutated each of these amino acids one at a time to alanine and measured the effect on receptor binding.
From this study, it was found that only eight amino acids on growth hormone (green) contribute 85 percent of the energy that is responsible for tight receptor binding; these amino acids are distant from each other in the primary sequence but adjacent in the folded protein.
Similar studies showed that two tryptophan residues (blue) in the receptor contribute most of the energy responsible for tight binding of growth hormone, although other amino acids at the interface with the hormone (yellow) are also important

Question 123

What do receptors do?

Answer 123

Hormone receptors activate Effector Proteins

Often the effector proteins are enzymes that produce small molecules called second messengers

Secondary messengers carry and amplify the signals from activated receptors/hormone complexes to the effector molecules

Question 124

What are secondary messengers?

Answer 124

• small short-lived molecules
• diffuse rapidly
• allow enzymatic amplification

Question 125

What are the water-soluble secondary messengers?

Answer 125

Ca2+ ions, cAMP, and cGMP

Question 126

What are the examples of lipid-soluble hormones?

Answer 126

DAG, IP3

Question 127

cAMP

Answer 127

• type of water-soluble secondary messenger
• Generated from ATP by adenylyl cyclase
• Activates PKA

Question 128

cGMP

Answer 128

• type of water-soluble second messenger
• Generated by guanylyl cyclase
• Activates PKG and specific cation channels

Question 129

IP3 and DAG

Answer 129

• type of lipid-soluble second messengers
• Both made from PIP2 by phospholipase C
• IP3 – opens channels to release Ca2+ from the ER
  DAG – with Ca2+ activates PKC

Question 130

Ca2+ ions

Answer 130

Released from intracellular stores or transported into the cell

Activates calmodulin, specific kinases (PKC), and other regulatory proteins

Question 131

What are the two types of kinases in animal cells?

Answer 131

1. tyrosine (Y) kinases
2. serine (S) / threonine (T) kinases

Question 132

GTPase switch proteins - simple mechanism

Answer 132

Hydrophilic hormones bind to membrane receptors.

These so-called G-protein coupled receptors (GPCR) activate G-proteins (GTPase switch proteins).

G-proteins exchange GDP to GTP leading to a conformational change and activation.

Active G-protein sends a signal
G-protein has intrinsic GTPase activity, which can be accelerated by GTPase-accelerating proteins (GAPs)

Once the signal is sent, the G-protein inactivates itself by hydrolyzing GTP to GDP

Question 133

How is G-protein activity regulated?

Answer 133

G-protein activity is regulated by exchange of GDP to GTP

Association with GTP switches the protein “ON”

After sending the signal, GTP is hydrolysed and the protein is switched “OFF”

Question 134

GTP hydrolysis rate

Answer 134

time G protein remains in the active conformation

Question 135

What is the switching mechanism of monomeric G proteins?

Answer 135

Conformation of the G protein is altered by GTP/GDP binding

Inactive (GDP-bound) G protein is capable of interacting with upstream activators

Active (GTP-bound) G protein is capable of interacting with downstream effector proteins

The downstream effector (often an enzyme) produces SECOND MESSENGER MOLECULES

Switch domain conformation – can bind and activate specific downstream effector proteins

IN THE ACTIVE/ON STATE, THE SWITCHES ARE PULLED IN AND BOUND TO GTP through interactions with the backbone amide groups of the conserved Thr and Gly residues (blue and green switches in diagram)

IN THE INACTIVE/OFF STATE - SWITCH i AND ii ARE RELAXED INTO OFF CONFORMATION - inhibits interaction with downstream effectors

Question 136

What are the 4 elements GPCR signal transduction pathways have in common?

Answer 136

(1) A receptor that contains seven membrane-spanning α helices
(2) A receptor-activated heterotrimeric G protein cycling between GTP-active and GDP-inactive forms
(3) A membrane-bound effector protein
(4) Proteins that desensitize the signaling pathway

Question 137

What are the 7 Human GPCR of pharmaceutical importance?

Answer 137

H2 receptor (histamine is ligand) - affects acid secreting cells of stomach

H1 receptor (histamine is ligand) - causes symptoms of allergy

Serotonin 5Ht 2a and 1a (serotonin is ligand) - causes synaptic transmission btw neurnons in CNS

Angiotensin - that is also ligand - increases blood pressure

B-adrenergic receptor - (epinephrine is ligand) - facilitates respiration

CysLT - (Leukotrienes are ligand) - contracts smooth muscle

Question 138

Synthesis and hydrolysis of cAMP by adenylyl cyclase and phosphodiesterase (PDE)

Answer 138

> 30 different mammalian GPCRs activate Gαs activation of adenylyl cyclase production of cAMP as a second messenger. (Most cell types express one or more such GPCRs.)

Adenylyl cyclase (AC) – catalyzes formation of cyclic cAMP (second messenger) bond from ATP precursor
cAMP phosphodiesterase (PDE) – catalyzes hydrolysis of cyclic bond – AMP (not second messenger)

Similar reactions occur for production and destruction of cGMP second messenger.

Question 139

AC and PDE

Answer 139

adenylyl cyclase - catalyzes FORMATION of cyclic cAMP (a second messenger) from ATP precursor

then, PDE (phosphodiesterase) takes that cAMP and catalyzes HYDROLYSIS of the cyclic bond AMP (not second messenger)

Question 140

What are the steps in the transduction and AMPLIFICATION initiated by epinephrine?
(Amplification of an extracellular signal by a signal transduction pathway involving cAMP and PKA)

Answer 140

Amplification:
Single epinephrine molecule binding to one G protein–coupled receptor – activates up to hundreds of G proteins (amplification)

Each G protein activates 1 AC until G protein hydrolyzes GTP (not amplification).

AC catalyzes the synthesis of a large number of cAMP molecules while activated (amplification).

Two cAMPs activate two PKA catalytic subunits (not amplification).

Activated PKA phosphorylates and activates multiple target proteins (amplification).

Activated target proteins:
Other kinases – kinase cascade (may be multiple steps of amplification; the more steps in such the cascade, the greater the signal amplification)

Enzymes – multiple products (amplification)

Question 141

Cellular responses to hormone-induced rise in cAMP

Answer 141

Epinephrine activation of the fight-or-flight response:

Liver and skeletal muscle – increase in glucose for energy production

Adipose – breakdown of triglycerides for energy production

Cardiac muscle – increased heart rate – O2 and glucose transport to peripheral tissues

Smooth muscle cell relaxation (not shown) – increase blood flow and air intake

Intestine – fluid secretion (use your imagination)

Question 142

What are the 6 steps of the Induction of gene expression by MAP kinase?

Answer 142

Steps 1) MAP kinase in the cytosol phosphorylates and activates the kinase p90RSK,

2) p90RSK then moves into the nucleus and

3) once in the nucleus - phosphorylates the SRF (serum-response factor) transcription factor.

4) MAP kinase translocates into the nucleus and

5) directly phosphorylates TCF transcription factor already bound to the promoter of the c-fos gene.

6) Phosphorylated TCF and SRF together stimulate transcription of c-fos and other genes that contain a promoter SRE
(serum response element) sequence.

MAP kinases phosphorylate over 175 target proteins in the nucleus.

Question 143

General mechanism of the activation of effector proteins by heterotrimeric G proteins:

Answer 143

Step 1: Ligand binding induces receptor activation conformational change.

Step 2: Activated receptor binds to trimeric G protein.

Step 3: Activated receptor GEF activity stimulates Gα subunit release of GDP.

Step 4: GTP binding changes Gα conformation
Dissociates Gβγ
(Gβγ subunit activates other effector enzymes in some pathways (b))
Activates Gα

Step 5: Gα·GTP activates effector enzyme.

Step 6: Gα intrinsic GTPase activity hydrolyzes GTP to GDP – dissociates Gα and turns off effector enzyme. (G protein active for minutes or less)

Question 144

What is the map kinase pathway steps simplified:

Answer 144

1) RTK activation leads to the auto-phosphorylation of the RTK intracellular domain

2) The phosphorylated intracellular domain interacts with adaptor proteins (GRB2, Sos) through the so called SH2 and SH3 domains (standing for Src-Homology Domains)

3) Sos activates Ras (binding of GTP)

4) Ras activates Raf (MAP Kinase Kinase Kinase)

5) Raf activates MEK (MAP Kinase Kinase)

6) MEK activated ERC (MAP Kinase)

Question 145

What are the 3 steps in the activation of Ras?

Answer 145

Receptor tyrosine kinases activate Ras via adapter proteins.

1) Binding of FGF hormone causes receptor dimerization, kinase activation, and phosphorylation of cytosilic receptor tyrosine kinases

A specific phosphotyrosine on an activated, ligand-bound receptor provides a specific site for binding the cytosolic adapter protein GRB2 SH2 domain.

2) GRB2 SH3 domains specifically bind proline domains on SOS, recruiting SOS to the membrane to interact with Ras∙GDP (tethered to the cytosolic surface of the plasma membrane by a hydrophobic farnesyl anchor).

(Binding of GRB2 and Sos couples receptor to inactive Ras)

(3) The SOS guanine nucleotide exchange factor (GEF) activity promotes release of GDP and formation of active Ras∙GTP.

Sos promotes dissociation of GDP from Ras; GTP binds and active Ras dissociates from Sos

Question 146

SH2 domain

Answer 146

Part of Grb2 protein involved in Ras activation

SH2 domain associates with phospho-tyrosine present on ligand-activated receptors of the RTK

Question 147

SH3 domain

Answer 147

part of Grb2 in the Ras activation

domain binds specifically to proline-rich sequence in Sos

Question 148

Grb2

Answer 148

is a protein involved in ras activation
- Has SH2 and SH3 domains
- Joins active receptor (RTK) to cytoplasmic Sos (brining it closer to the plasma membrane)

Question 149

What is Sos

Answer 149

Sos is a GEF involved in Ras activation

Sos gets recruited by SH3 domain of Grb2 to be close to the plasma membrane, where it can promote dissociation of GDP from Ras

Question 150

Ras/MAP kinase pathway steps

Answer 150

Stimulated cell:
Step 1: Active RTK receptor activates formation of active Ras·GTP.
(Ras activated by the exchange of GDP to GTP)

Step 2: Ras·GTP triggers the downstream kinase cascade by interacting with the Raf N-terminal regulatory domain, which results in dephosphorylation of one of the serines, release of 14-3-3, and activation of the Raf kinase activity.
(Active Ras recruits, binds, and activates Raf)

Step 3: Ras GTP hydrolysis to Ras·GDP releases active Raf.

Step 4: Raf phosphorylates and activates MEK.

Step 5: MEK, a dual-specificity protein kinase, phosphorylates MAP kinase (and ERK kinase) on both tyrosine and serine/threonine residues.

Step 6: MAP kinase phosphorylates many different proteins in different cells, including nuclear transcription factors, that mediate cellular responses.

Question 151

What is the only other signal transduction pathway that is activated by RTK that we are studying?

Answer 151

PI-3 kinase! aka Phosphoinositide Signaling Pathway

Question 152

Phosphoinositide Signal Transduction Pathways

Answer 152

RTKs and cytokine receptors can initiate the IP3/DAG signaling pathway by activating phospholipase Cγ.

RTKs and cytokine receptors can initiate another phosphoinositide pathway by activating PI-3 kinase formation of PI 3-phosphate, which bind PH domains in various proteins to form signaling complexes that activate PKB.

Activated PKB promotes survival of many cells by directly inactivating pro-apoptotic proteins and synthesis of pro-apoptotic proteins.

Question 153

What are the names of the produced molecules in the phosphoinositide signalling pathways?

Answer 153

PhosphoInositol 3,4,- biphosphate (IP2)
PhosphoInositol 4,5,- biphosphate (IP2)
PhosphoInositol 3,4,5- triphosphate (IP3

Question 154

How is full activation of PKB achieved?

Answer 154

Full activation is achieved by phosphorylation of the membrane-bound PKB by two other kinases, 3-phosphoinositide dependent protein kinase 1 and 2 (PDK1 and PDK2)

Question 155

What are the steps in the PI3 pathway?

Answer 155

Unstimulated cells: PKB is in the cytosol with its PH domain bound to and inhibiting its catalytic kinase domain.

Hormone stimulated cells:

Inactive PKB is in the the cytosol of unstimulated cells

Step 1: Hormone stimulation leads to activation of PI-3 kinase, which forms PI 3-phosphates
(phosphorylates 3 position of PI to form PI 3,4 bisphosphates)

Step 2: The 3-phosphate group binds the PH domains of PKB (partially active) and PDK2, docking both to the membrane and partially activating PKB

Step 3: PDK1 phosphorylation of the PKB activation loop serine and PDK2 phosphorylation of a PKB C-terminus serine fully activates PKB activity, which induces many cellular responses.

Question 156

General structure of G protein coupled receptors

Answer 156

All G protein–coupled receptors have:

The same orientation in the membrane – N-terminus outside, C-terminus in cytosol
Contain seven transmembrane α-helical regions (H1–H7)
Have four extracellular segments (E1–E4)
Four cytosolic segments (C1–C4)

Question 157

Structure of the β2-adrenergic receptor in the inactive state and in the active state with a bound ligand and with its associated heterotrimeric G protein

Answer 157

Ligand binding causes movement of the receptor TM5 and TM6 helices and changes in the C3 loop to create a surface that forms extensive interactions parts of Gαs
Gαs small conformational change
Lengthening the α5 helix creates a large surface consisting of the N-terminal α-helical segments αN and α5 that bind mainly to receptor TM5 and
TM6.
Triggers release of GDP and Gβγ
(c) Binding of epinephrine to the side chains of multiple amino acids in the interior of the β2-adrenergic receptor

Question 158

What’s the FRET technique?

Answer 158

Forster resonance energy transfer (FRET) technique:
Wavelength of emitted fluorescence changes when two fluorescent proteins interact.
Reveals kinetics of G protein activation

Mechanism of detection of protein interactions IN VIVO

One protein is tagged with Gα-cyan fluorescent protein (CFP)

Another protein is tagged with Gβ-yellow fluorescent protein (YFP, excited by CFP (cyan fluorescent protein) when alpha subunit is bound to beta subunit

On the left: The two proteins are close to each other (INACTIVE RECEPTOR) Then the light emitted by CFP (blue) excites YFP and WE OBSERVE YELLOW LIGHT

On the right: The two proteins are away from each other (ACTIVE RECEPTOR) We observe only the light emitted by CFP (blue) WE ONLY OBSERVE CYAN LIGHT

Question 159

What are inhibitory hormones of cAMP and how do they work?

Answer 159

PGE and adenosine are inhibitory hormones and they bind to the inhibitory receptor which activates G protein complex and binds and causes the inactivation of adenylyl cyclase which causes the down regulation/lack of production of cAMP

(seesaw looking diagram)

Question 160

What are stimulatory hormones of adenylyl cyclase

Answer 160

Epinephrine and Glucagon

these are stimulatory hormones that bind to receptors causes activation of a stimulatory G protein complex that binds and activates adenylyl cyclase and causes production of cAMP

(seesaw looking diagram)

Question 161

Receptor Tyrosine Kinases

Answer 161

Hydrophilic signaling molecules (hormones or peptides) activate Receptor Tyrosine Kinases (RTKs), which phosphorylate downstream targets and initiate a signal transduction cascade

RKTs have a different mechanism of action as compared to GPCRs

Usually the hormone or peptide leads to dimerization of two kinase polypeptides

The dimerization of RTKs leads to cytoplasmic domain autophosphorylation, which recruits proteins that initiate intracellular signaling pathways

Question 162

What are the types of cell-surface receptors?

Answer 162

1) receptor associated kinases
2) cytosolic kinases
3) protein subunit dissociation pathways
4) protein cleavage pathways

Question 163

Receptor-associated kinases (type of cell-surface receptor and signal transduction pathway)

Answer 163

Many receptor cytosolic domains are protein kinases or tightly associate with a cytosolic kinase.

(1) Ligand binding causes receptor dimerization which activates kinase activity:

Kinases directly phosphorylate and activate transcription factors or other signaling proteins.

(2) Many receptors also activate small GTP-binding “switch” proteins such as Ras.

The Ras signal transduction pathway and others activate a kinase cascade, in which one kinase phosphorylates and thus activates (or occasionally inhibits) the activity of multiple copies of another kinase for signal amplification.

G-protein pathway kinases phosphorylate multiple protein targets, including cell type-specific transcription factors.

Question 164

Cytosolic kinases (type of cell-surfacr receptor and signal transduction pathway)

Answer 164

Seven membrane-spanning segments receptors activate GTP-binding Gα proteins.

Gα-GTP activates specific kinases or other signaling proteins.

Question 165

Protein subunit dissociation pathway (type of cell-surface receptor and signal transduction pathway)

Answer 165

Several signaling pathways cause disassembly of a multiprotein complex in the cytosol, which releases a transcription factor that translocates into the nucleus.

Question 166

Protein cleavage pathways (type of cell surface receptor and signaling pathway)

Answer 166

irreversible signaling
Proteolytic cleavage of a receptor releases an active transcription factor.

Question 167

What conditions does the ability of a transcription factor to bind depend on?

Answer 167

The binding sites for the factor and
the gene’s epigenetic state
and on the presence of master transcription factors and other nuclear proteins

Question 168

Signaling molecules that activate RTKs:

Answer 168

are soluble or membrane-bound peptide or protein hormones:
Growth factors:
NGF - nerve GF
PDGF - platelet-derived GF
FGF - fibroblast GF
EGF - epidermal GF

Question 169

Steps of RTK activation:

Answer 169

RTK activation mechanism:
1) (–) ligand: RTKs generally exist as monomers with poorly active kinases.

(2) (+) ligand: Two ligands binding to two RTK extracellular domains homodimerizes the receptors,

bringing together two poorly active cytosolic domain kinases,

which phosphorylate each other on an activation loop tyrosine residue.

The phosphorylated loop moves out of the kinase catalytic site, increasing ATP and/or the protein substrate binding.

(3) The activated kinase phosphorylates several tyrosine residues in the receptor’s cytosolic domain, which provides docking sites for SH2 and other binding domains on downstream signal-transducing proteins.

Question 170

Ligand-induced dimerization of HER1, a human receptor for epidermal growth factor (EGF).

Answer 170

(a) (–) EGF: β-hairpin from domain II that forms the “dimerization arm” binds to domain IV of the same receptor molecule, inhibiting dimerization.

(b) (+) EGF: EGF binds domains I and III, inducing an extracellular domain conformational change that exposes the domain II dimerization arm.

(c) Dimerization arms interactions dimerize two identical ligand-bound receptor monomers.

Question 171

What does the activation of the EGF receptor by EGF result in?

Answer 171

The formation of an asymmetric kinase domain dimer

The EGF receptor kinase domain is activated by phosphorylation of an activation loop tyrosine.

Inactive, monomeric state: activation loop binding in the kinase active site inhibits kinase activation.

Receptor activation mechanism:
The dimer is stabilized by interactions between the juxtamembrane segments of the two receptors.

Dimerization generates an asymmetric kinase dimer in which the donor kinase C-terminal C-lobe binds the acceptor kinase N-terminal N-lobe.

Interactions cause a conformational change that removes the activation loop from the acceptor kinase site, activating its kinase activity.

The active acceptor kinase phosphorylates cytosolic domain tyrosine residues on both receptors

Question 173

What is the cell cycle?

Answer 173

a sequence of ordered events
these events are controlled by CDKs
consists of G1, S, G2 and mitosis phases

Question 174

What regulates the cell cycle?

Answer 174

Cyclin Dependent Kinases (CDK)
The key regulatory subunits of CDKs are called Cyclins
Other regulatory mechanisms involve synthesis and degradation of CDK, inhibitors or activating or inhibitory phosphorylation of the CDK

Question 175

Positive regulators of the cell cycle

Answer 175

Each Cyclin acts at a specific stage of the cell cycle and sets up the transition to the next stage, then gets degraded

Question 176

Negative regulators of the cell cycle

Answer 176

there are surveillance mechanisms that check for the proper progression through the cell cycle.
If anything goes wrong, these mechanisms, called checkpoints, hold the cycle until repair is done

Question 177

Why is the end of G1 so important?

Answer 177

The decision to divide is tightly controlled at the end of G1. Once this decision is made, the cell must divide or die.

Question 178

Interphase

Answer 178

G1+S+G2

Question 179

Interphase checkpoints

Answer 179

Make sure that the stage of the cell cycle is completed

act as brakes on cell cycle progression when events are not completed or when DNA is damaged

Question 180

Budding or Fission Yeasts

Answer 180

models for studying the cell cycle

unicellular organisms, easy to grow and synchronize
rapid cell cycle (2-4 hrs)
cell cycle stage visualized by the morphology of the cell
can be grown as haploid cells (amazingly simple genetics)
conditional loss of function (ts) mutants display distinct morphology

Question 181

Budding yeast cell cycle

Answer 181

G1 – growth to size large enough to form viable bud; irreversible commitment to progression through cell cycle is made at the G1 START point; bud begins to emerge
S – DNA replication (1n to 2n in haploid organism)
G2 – less distinct stage in yeast than in higher eukaryotes
M – intranuclear spindle; nuclear envelope does not break down in yeast

Question 182

Fission yeast cell cycle

Answer 182

cell cycle mutations are easily generated and recognized by differences in cell size

G1 grow longer before cell division.
G1 START point – irreversible commitment to cell division
S − nuclear envelope does not breakdown

Question 183

What have model organisms revealed?

Answer 183

Model organisms have revealed that:

A few master cell cycle regulatory proteins control and coordinate processes such as DNA replication initiation and entry into mitosis.

Master regulators and the proteins that control them are highly conserved across eukaryotic cells

Question 184

Progesterone stimulation of Xenepus oocytes

Answer 184

progesterone stimulates maturation of Xenopus oocytes

Model cell − large enough for biochemical characterization of cell extracts

Progesterone treatment causes G2-arrested Xenopus oocytes to enter meiosis I, a modified form of mitosis.

Homologous chromosome segregation and asymmetric cell division expels one set of chromosomes into the first polar body.

Oocyte arrests in metaphase II to yield an egg.

When fertilized by sperm, egg proceeds through meiosis II anaphase and a highly asymmetric cell division expels one chromatid of each chromosome into a second polar body.

The haploid female and sperm pronuclei fuse to produce a diploid nucleus.

Diploid zygote proceeds through cell cycle to divide by mitosis into two cells.

11 more synchronous divisions produce blastula.

Question 185

Human cells undergoing mitosis visualisation

Answer 185

HeLa Kyoto cells seen under phase contrast microscopy were filmed as they underwent nitosis.

HeLa Kyoto cells − are flat during interphase but round up during mitosis and cell division. Daughter cells flatten again.

Normal primary human cells and other mammalian cells have a finite life span when cultured in vitro.

Normal human cells divide 25–50 times before undergoing replicative senescence.

Immortalized cell lines, some of tumor origin such as the HeLa cells, harbor genetic alterations that enable them to escape replicative senescence and become immortal.

Question 186

What agents are used to synchronyze cells in YEAST

Answer 186

Mating pheromone arrests the cells in late G1

Nocodazole disrupts mitotic spindle – cells arrest in pro-metaphase

Question 187

What agent is used to synchronyze cells in mammals?

Answer 187

Hyproxyurea and nocodazole

Question 188

What is nocodazole

Answer 188

Anti mitotic agent used in yeast and humans that reversibly interferes with polymerization of microtubules

it disrupts the mitotic spindle

Question 189

What is mating pheromone?

Answer 189

a cell synchrony agent used in yeast cells that arrests the cells in late G1

Question 190

What is hydroxyurea

Answer 190

aka thymidine is an antimetabolite medication that inhibits the synthesis of dNTPs -
CAUSES CELL AREST IN EARLY S PHASE
USED IN MAMMALIAN CELL CULTURE

Question 191

What does flow cytometry measure?

Answer 191

the amount of cellular DNA

Question 192

Flow cytometry procedure:

Answer 192

Fix cells by Ethanol (Permeabilizes and preserves cells)

Degrade all RNA (RNAse)

Soak the cells with a DNA staining dye (propidium Iodine)

Wash the cells

Run cell suspension through a capillary tube, one cell at a time

Use lasers to detect the amount of stain (indirect measure of the amount of DNA) in each cell

Plot the results

Question 194

Analysis of DNA content by flow cytometry

Answer 194

Researchers use multiple tools to study the cell cycle.

DNA in yeast cells stained with propidium iodide fluorescent dye can be quantified by flow cytometry.

DNA content analysis shows three populations of cells: cells with unreplicated DNA (1C), cells with replicated DNA (2C), and cells in the process of DNA replication (S phase).

Question 195

Isolation of ts mutants in S. cerevisiae

Answer 195

Isolated temperature sensitive (ts) mutants not growing at higher temperature (arrest with certain morphology)

Treat yeast w mutagen

Grow colonies (23*C)

Check which colonies do not grow at higher temperature (these are called TS - temperature sensitive mutants)

Check morphology of TS mutants
(the morphology tells you at which state of the cell cycle the mutated gene works)

Isolate and analyze the gene

Question 196

ts cdc mutants

Answer 196

temperature sensitive cell division cycle mutants

In ts mutants grown at restrictive temperature, cell morphology (presence and size of bud) indicates at which stage of the cell cycle the mutants gene works. Position in the cell cycle can be determined by morphology

Question 197

cdc mutants in S. pombe

Answer 197

mitosis at size larger than normal
delayed progress through cell cycle
(fission yeast)

Question 198

wee mutants

Answer 198

mitosis at size smaller than normal
advanced progress through cell cycle

Question 199

How do cyclin levels differ at each stage of the cell cycle?

Answer 199

Cyclin levels rise during interphase
peak at mitosis
and drop before anaphase

Question 200

Discovery of cyclin in dividing eggs

Answer 200

samples collected at different times post-fertilization (every 10 min)

identified a protein that was continuously synthesized and periodically destroyed

cyclin destruction occurred just prior to onset of mitotic cleavage (anaphase)

Question 201

Which genes that regulate the cell cycle are homologous?

Answer 201

cdc28, cdc2, Cdk1 and MPF (cdk1)

Question 202

Overview of how CDKs regulate the cell cycle

Answer 202

Early G1 − no cyclin − CDKs are active

Mid-G1 − G1/S phase CDKs activate transcription of genes required for DNA replication

S phase − initiated by SCF ubiquitin-protein ligase ubiquitinylation of S phase CDKs inhibitors, marking them for degradation by proteasomes. S phase CDKs activate DNA replication.

Late G2, mitotic CDKs trigger entry into mitosis.

M Anaphase − anaphase-promoting complex (APC/C) ubiquitinylates:

The anaphase inhibitor protein securin, marking it for degradation by proteasomes, activating securin to degrade cohesin complexes linking the sister chromatids. Separation of sister chromatids initiates anaphase.

Mitotic cyclins, targeting them for degradation by proteasomes. The resulting drop in mitotic CDK activity and protein phosphatase activity results in chromosome decondensation, reassembly of nuclear membranes around the daughter-cell nuclei, and cytokinesis.

Question 203

CDKs are active only in…

Answer 203

CDKs are active only in the stages of the cell cycle they trigger:
G1/S phase CDKs are most active at the G1–S phase transition to trigger entry into the cell cycle.
S phase CDKs are most active during S phase.
Mitotic CDKs are most active during mitosis.

Question 204

APC/C cyclosome

Answer 204

The anaphase promoting complex
ubiquitin protein ligase

• ubiquitinylates proteins, including cyclins, targeting them for degredation by proteosomes
• catalyzes two key cell cycle transitions: anaphase and exit from mitosis

Question 205

What do the graphs showing activity of CDKs and cyclins show?

Answer 205

The activity of the different cyclin-CDK complexes varies in a manner that is very similar, but not exactly identical to the accumulation of the corresponding cyclins:

Conclusion:
There are other regulators of CDK activity

Question 206

What are the 3 main types of regulators of CDK activity?

Answer 206

1) Kinases and Phosphatases
2) inhibitory proteins
3) ubiquitin-protein ligases

Question 207

CDK inhibitors

Answer 207

CDK inhibitors (INKs, CIPs, KIPs) associate with CDK/Cyclin complexes and inhibit them until the previous stage of the cycle is completed.

After the completion, the active CDK/Cyclins phosphorylate the CDK inhibitors. These inhibitors are then degraded. The inhibitor degradation provides a clean molecular trigger of the next stage of the cycle.

Next, the cyclins from the previous stage are degraded via protein ubiquitination.

The protein-ubiquitin ligases are:
SCF - works at the transition from G1 to S phases
APF - works at the onset of mitosis (anaphase)

Question 208

SCF

Answer 208

A protein ubiquitin ligase
works at the transition from G1 to S phases
Degrades phosphorylated S phase inhibitor

Question 209

APC/C proteosome

Answer 209

a protein ubiquitin ligase
used during metaphase/anaphase transition AND at the mitotic exit/end of anaphase
degrades securin
degrades mitotic cyclins

Question 210

What are the regulators of cyclin/CDK activity?

Answer 210

Kinases and Phosphatases
Inhibitory proteins
ubiquitin-protein ligases
and
Phosphoserine/Threonine Binding proteins and modular domains

Question 211

Wee1 kinase

Answer 211

Wee1 kinase phosphorylates Y (tyrosine) and T(threonine) residues on CDK and inhibits the mitotic kinases in S-phase.

The dual-specificity Wee1 kinase inhibits mitotic cyclin-CDK activity by phosphorylation of CDK T14 and Y15 until DNA replication is completed

Question 212

Cdc25 phosphatase

Answer 212

removes the inhibitory phosphorylation of Y and T residues previously phosphorylated by Wee1 and activates Mitotic CDKs

When DNA replication is completed, the dual-specificity Cdc25 phosphatase is activated (by an unknown mechanism) and dephosphorylates CDK T14 and Y15 , reversing the inhibitory effect of Wee1 and activating cyclin-CDK activity.

Question 213

WHat do mitotic cyclins synthesized during S and G2 do?

Answer 213

They activate mitotic CDKs

Question 214

Mitotic cyclin-CDK positive eedback on its own activity:

Answer 214

Phosphorylates Cdc25, activating its CDK-activating phosphatase activity
Phosphorylates Wee1, inhibiting its CDK-inhibiting kinase activity

Question 215

Cdr2

Answer 215

inhibits Wee1 activity when a cell has reached an appropriate size to divide

Question 216

When is the decision to divide made?

Answer 216

The decision whether to divide or not is made in late G1 phase: RESTRICTION (or START) point
This decision is tightly regulated
Cells that never divide (highly differentiated cells) arrest before this START point in G0

Question 217

What proteins and stuff are used to regulate G1-S phase CDK activation?

Answer 217

Whi5
SBF
Rb
E2F

Question 218

How is the G1 to S phase transition regulated in S. cerevisiae/budding yeast?

Answer 218

Whi5 and SBF!

Whi5 is on top of SBF and it is a transcriptional repressor - inhibits SBF activation of the CLN1/2 cyclin gene transcription
Nutrient availability allows translation of Cln3 mRNA, activating Cln3-CDK activity.
Cln3-CDK phosphorylates Whi5, promoting its export from the nucleus.
When 50 percent of Whi5 has been exported, cells transition through START.
Active SBF induces transcription of CLN1 and CLN2 and other genes whose products are needed for DNA replication.
G1/S phase CDKs further phosphorylate Whi5, promoting further CLN1 and CLN2 transcription.
High levels of G1/S phase CDKs initiate DNA replication, bud formation, and spindle pole body duplication

Question 219

Metazoan/vertebrate regulation of G1/S phase transition

Answer 219

G1 − The E2F transcription factor and its regulator Rb control the G1–S Phase transition
Rb binds and inhibits E2F activity.
Growth factors (mitogens) stimulate G1 CDK activity.
Cyclin D–CDK4/6 complexes phosphorylate Rb, releasing E2F.
E2F stimulates transcription of early response genes encoding cyclin E, CDK2, and E2F.
Cyclin E–CDK2 further phosphorylates Rb, initiating a positive feedback loop that leads to a rapid rise in the expression and activity of both E2F and cyclin E–CDK2.
G1/S phase cyclin E/A-CDK2 CDKs support cell passage through the restriction point and initiation of DNA replication and centrosome duplication.

Question 220

Are there cyclins at the end of mitosis?

Answer 220

No! All cyclins have been degraded at the end of mitosis by the APC

Question 221

In yeast, presence of nutrients leads to…

Answer 221

to increased synthesis of cyclin Cln3 in early G1

Question 222

Rb

Answer 222

regulator of E2F transcription factor
binds to E2F and suppreses the transcription of cyclin E or cyclin A in mammals in early G1

Question 223

What phosphorylates Rb?

Answer 223

Cyclin D (CDK4/6)

Question 224

What’s the Sic 1 protein and what process is it involved with?

Answer 224

Sic1 is an S phase inhibitor (it’s attatched to S phase CDK/cyclin) it prevents initiation of DNA replication ununtil the cells have completed all G1 events
1) G1/S phase CDKs assembled in late G1 phosphorylate Sic1 at multiple sites, marking it for degredation
2) SCF ubiquitin-protein ligase ubiquitinylation of SCF and subsequent proteasomal degradation
3) The active S phase CDKs trigger initiation of DNA synthesis by phosphorylating and recruiting MCM helicase activators to DNA replication origins.

Question 225

How many sites do we phosphorylate on Sic and why?

Answer 225

G1/S phase CDK phosphorylation of a single optimal Sic1 site would cause a sluggish G1/S phase transition. G1/S phase CDKs accumulation during G1 would cause slow progressive degradation of Sic1 and initiation of S phase.
 slow Slc1 degradation
 slow S phase CDK activation
 sluggish G1 – S phase transition

(c) Six suboptimal phosphorylation sites ensure that Sic1 is fully phosphorylated and targeted for destruction by SCF only when G1/S phase CDKs have reached high levels. This ensures that Sic1 degradation and DNA synthesis initiation occurs rapidly, when G1/S phase CDKs have accomplished all their other G1 tasks.
Six sub-optimal sites of phosphorylation of Sic1
 fast Sic1 degradation
 fast S phase CDK activation
 optimized G1 – S phase transition

Question 226

What are the molecular mechanisms governing DNA replication

Answer 226

Step 1: Exit from mitosis and early G1 the origin recognition complex (ORC) is recruited and associated with all DNA replication origins

Step 2: Late G1 CDK activity is still low and the Cdc6 and Cdt1 MCM loading factors associate with ORC and load the INACTIVE MCM replicative helicase complex onto each DNA replication origin.
The Pre-replication complex is formed

Step 3: DDK and S phase CDKs are activated. They phosphorylate:
MCM → activation of helicase fires up the origin of DNA replication

Cdc6 and Cdt1 → degradation (after firing there will be no loading of MCM on the newly replicated origins)

CDC45, Sld2 → Recruit GINS to MCM to facilitate elongation
replication.
Step 2: Late G1 - DDK and S phase CDKs are activated.
DDK phosphorylates two subunits of the MCM helicase.
S phase CDKs phosphorylate Sld2 and Sld3.
These phosphorylation events promote recruitment of Cdc45-Sld3 complex and the GINS complex MCM helicase activators to sites of replication initiation, which leads MCM helicases to unwind DNA.
(S phase CDKs prevent reloading of MCM helicases by phosphorylating Cdc6 and Cdt1, promoting their release from the replication origins and degradation by SCF.)
S phase CDKs also phosphorylate MCM helicases, causing the helicases to disengage from the DNA when replication is complete and export from the nucleus.

Step 3: DNA polymerases recruited to origins initiate DNA replication

Question 227

What do Cdc6 and Cdt1 do?

Answer 227

They load MCM onto the origin of replication

Question 228

Why are checkpoints of the cell cycle important?

Answer 228

Checkpoints ensure that the next stage of the cell cycle does not initiate prior to the completion of the preceding one.

Checkpoint pathways arrest the cell cycle progression in response to DNA damage or improper spindle assembly.
so we don’t make bad cells

Question 229

What are the three main parts of checkpoint pathways?

Answer 229

• event sensors
• a signaling pathway
• and effector that halts cell cycle progression

Question 230

What are the 4 steps of DNA damage response system (short answer w the 4 arrows) what’s next…

Answer 230

1) Damaged DNA activates ATM, ATR (protein kinases) and DNA-PK

2) DNA-PK, ATM and ATR phosphorylate and activate the chicks! (Chk1, Ck2) and MK2
(Chk1 and Chk2 are checkpoint kinases)

3) Chk1 and Chk2 inhibit/block cdc25, which induce DNA repair machinery!
(cdc25 normally would activate mitotic CDKs but now that it’s inhibited, it can’t)

4) This inhibits mitotic CDKs and causes cell cycle arrest

Question 231

What do cells require to repress apoptosis for surviva(that binds to surface receptors)

Answer 231

Cells require TROPHIC FACTORS! that bind to surface receptors to repress apoptosis for survival

Question 232

Caspase proteases in apoptosis

Answer 232

they degrade cellular proteins and apoptosis involves the activation of caspase protease

Question 233

BMPs

Answer 233

Bone morphogenic proteins
- expressed by interdigital cells
- induce apoptosis

(Dominant negative type I BMP receptor blocks BMP signal and apoptosis (like in the diagram with the hands and one’s webbed)
(BMPs – not expressed in duck interdigital cells)
Conclusion: BMP signaling mediates interdigital cell death in the embryonic limb.

Question 234

What’s Necrosis?

Answer 234

MESSY CELL DEATH
Different than apoptosis
cell rupture, cellular contents pouring out leading to inflammatory response, auto-immunity (messy cell death)

Question 235

What is difference between necrosis and apoptosis?

Answer 235

Necrosis: cell rupture, cellular contents pouring out leading to inflammatory response, auto-immunity (messy cell death)
Apoptosis: cell membrane remains intact, cell is dismantled and packaged into apoptotic bodies, cellular contents within vesicles, absorbed by macrophages and recycled (clean cell death)

Question 236

What are the main signs of cell death by apoptosis

Answer 236

Mild convolution
Chromatin campaction and margination
Condensation of cytoplasm

Breakup of nuclear envelope
Nuclear fragmentation
Blebbing
Cell fragmentation
INTER_NUCLEOSOMAL DNA FRAGMENTATION IS A FEATURE OF APOPTOTIC CELLS

Apoptotic bodies are produced
Phagocytised

Apoptotic stimuli triggers DNA degrading nucleases that cleave DNA between nucleosomes
Degraded chromatin is packaged into apoptotic bodies
The apoptotic bodies contain less that 2n DNA as compared to normal living cells

Question 237

What are the two assays for the detection of fragmented chromatin that we study?

Answer 237

• Flow cytometry and COMET assay

Question 238

How can you detect apoptotic bodies on a flow cytometry graph?

Answer 238

You detect these apoptotic bodies as a broad peak to the left of the C1 peak since they have much LESS DNA THAN NORMAL CELLS

Question 239

What is the COMET assay process?

Answer 239

Cells are imbedded in agarose
Electric current is passed through the agarose
Similarly to gel electrophoresis, DNA fragments will move because of their charge
Large fragments of DNA (intact chromosomes) do not move
Fragmented chromosomes leave the cell and produce a “comet-like” tail.
The detection of such tails is indicative of apoptosis.

Question 240

What did S. Brenner find?

Answer 240

S. Brenner: introduced Caenorhabditis elegans as a model organism to study development

Small nematode roundworm, ~1.2 mm long, transparent

Consists of 1090 somatic cells produced during development

Of these, 959 cells survive, 131 “doomed” cells die by apoptosis

Question 241

What did John Sulston find?

Answer 241

determined complete cell lineage map

Question 242

CED genes

Answer 242

cell death genes

Question 243

What did R. Horvitz find?

Answer 243

R. Horvitz isolated > 4000 cell lineage (lin) mutants that do not kill some of the 131 cells

He postulated that the mutant genes must be responsible for the killing

Conclusion: Cell death is genetically controlled
- Led to search for cell death (CED) genes

Question 244

Mutations in ced-3 picture meaning

Answer 244

ced-1 mutant: apoptotic cells are not engulfed and digested (arrows)

ced-1/ced-3 double mutant: apoptotic cells do not appear

They do not die!!!
ced-3 is necessary for apoptosis (pro-apoptotic gene)

Question 245

What are the two classes of caspases?

Answer 245

Initiator caspases: process and activate effector caspases
- casp9 (intrinsic)
- casp8 and 10 (extrinsic)

Effector (executioner) caspases: cleave specific cellular proteins leading to apoptosis
- caspase 3, 6, 7
these caspases digest the proteins of the cell and activate nucleases that degrade DNA and chromatin

Question 246

What regulates the release of Cyt C

Answer 246

The release of Cyt C is regulated by pro-apoptotic and anti-apoptotic proteins

Pro-apoptotic proteins: Bak, Bax, Boc

Pro-survival proteins: Bcl-2, Bcl-xl

All these proteins are homologous to Bcl-2 and contain Bcl-2 Homology domains (BH1, BH2, BH3, BH4)

BH3-only proteins: (Bad, Bim, Puma) Regulate pro-survival proteins

Question 247

BH3-only proteins

Answer 247

Bad, Bim and Puma - regulate pro-survival proteins (Bcl-2)

Question 248

Bcl-2 family proteins

Answer 248

Play central anti-apoptotic (pro-survival) or pro-apoptotic roles in apoptosis
Contain one or more functional Bcl-2 homology domains (BH1–4)
All participate in oligomeric interactions.
Three classes – one anti-apoptotic and two pro-apoptotic:
Only some BH3-only proteins contain a transmembrane (TM) domain that anchors other Bcl-2 family proteins in the outer mitochondrial membrane.
Pro-apoptotic proteins Bax and Bak form pores in the outer mitochondrial membrane large enough for cytochrome c escape to the cytoplasm.

Question 249

Difference between intrinsic and extrinsic death pathway

Answer 249

Intrinsic Death Pathway
activated by proteins that reside in the mitochondria (Bak, Bax, Bcl-2, Cytochrome C (CytC)
activated in response to DNA damage, cell stress, lack of adhesion, lack of trophic signals

Extrinsic Death Pathway
activated through cell surface death receptor
activated by direct contact with other cells that direct them to self-destruct by apoptosis

Question 250

How does DNA damage induce apoptosis

Answer 250

DNA damage or ultraviolet irradiation stimulates apoptosis:
- Induces synthesis of Puma BH3-only protein
- Puma binds to Bak, Bax, and Bcl-2.
- Bak and Bax form oligomeric pores that release Cyt c.

Question 251

Cell lack of adhesion to substrate -a

Answer 251

Cell lack of adhesion to substrate – apoptosis in adhesion-dependent cells:
- Disruption of integrin signaling
- Release of the Bim BH3-only protein from the cytoskeleton
- Bim binds to Bak and Bax to promote pore formation and Cyt c release.

Question 252

What’s more abundant, Bcl-2 or Bak proteins?

Answer 252

Pro-survival proteins (Bcl-2, Bcl-x) are abundant. They bind to pro-apoptotic proteins (Bak or Bax, less abundant), prevent oligomerization and close pore

Question 253

What do Bad, bim and puma do?

Answer 253

Bad, Bim, Puma, when activated, displace Bcl-2 from Bak or Bax and allow to open the pore

Question 254

What does Cyt C do when it’s released?

Answer 254

Cyt C is released, binds to Apaf-1 and activates Caspase 9, which activates the executioner caspases

(in intrinsic pathway)

Question 255

General framework for the evolutionary progression of cancer

Answer 255

The progression of cancer from a normal somatic cell to a full-grown tumor can be thought to take place in four stages.
Step 1: Initiation.
Except in rare cases in which cancer is caused by a singular genetic event, such as infection by a tumor virus, we usually cannot pinpoint the initiating event but a somatic mutation is usually suspected.

Step 2: Cancer progression.
Precancerous cells acquire mutations that dysregulate growth-control pathways, causing inappropriate cell proliferation.
Most cancers progress by a process of natural selection, sequentially acquiring multiple such mutations in different pathways.

Step 3: Evasion of cancer cell elimination.
Precancerous cells are normally removed from the body by either apoptosis or immune-surveillance processes.
To progress beyond this stage, tumor cells must acquire additional somatic mutations that allow them to evade these systems.

Step 4: Tumor growth and dispersal.
Continued growth of a tumor requires a blood supply and tumor cells must acquire yet more changes that promote angiogenesis.
To spread throughout the body by metastasis, cells from a solid tumor must acquire the ability to migrate from their original site and adhere at a new location in the body.

Question 256

DNA repair and surveillance molecules

Answer 256

Genome destabilization

P53
Inhibitors of CDKs
Sensors of DNA damage (ATM, ATR)
DNA repair genes

Question 257

Cell cycle regulatory proteins

Answer 257

• G1/S Cyclins or CDKs
• Regulators of G1/S (Rb protein)
• Growth Factor Receptors and Signal transduction proteins

Question 258

Metastatic cancer cells:

Answer 258

Degrade the basement membrane
Migrate on extracellular matrix (ECM) fibers away from the primary tumor to reach the blood vessels

Can be attracted by signals such as epidermal growth factor (EGF), which can be secreted by macrophages

Penetrate the blood vessel endothelial cell layer that forms the vessel walls and enter the bloodstream

Question 259

Carcinoma cells how do they penetrate the ECM and blood vessel wall?

Answer 259

Carcinoma cells penetrate the ECM and blood vessel wall:
Use the actin cytoskeleton to extend invadopodia
Produce matrix metalloproteases and other proteases that degrade basement membranes to open a path for metastasis

Question 260

What is extravasation?

Answer 260

circulating tumor cell adheres to the blood vessel lining in a new location and migrates through it to colonize a new tumor in the underlying tissue.
Fewer than 1 in 10,000 cells that escape the primary tumor survive to form a secondary metastatic tumor

Question 261

Cancer cell genetic characteristics?

Answer 261

Individual chromosome number is altered

Composite chromosomes composed of pieces from different chromosomes

Question 262

Colon mutation on APC tumour - consequence of this mutation

Answer 262

• causes the mutated cell to divide to form a localized polyp of benign tumour cells

Question 263

Events that lead to formation of malignant tumours

Answer 263

• Loss of APC tumor-suppressor gene (chromosome 12)
• activation of K-Ras oncogene (chromosome 12)
• Loss of p53 tumor-suppressor gene (chromosome 17)
• Other changes that in combination can lead to tumor cells invading blood vessels, allowing metastasis to occur

Question 264

TGF-beta receptor. Two points the effect of this active receptor has on the Smad transcription factor

Answer 264

1) The active receptor phosphorylates inactive SMAD in the cytosol, causing it to oligomerize and activate
2) Active Smad translocates into nucleus where it binds to the following genes to induce their transcription

Question 265

What genes are activated by smad?

Answer 265

Smad stimulates expression of:
1) p15 (which inhibids CDK4) - this causes cells to arrest in G1

2) An inhibitor of a protease that degrades extracellular matrix
- cells do not migrate.

Question 266

Secondary sex traits

Answer 266

changes in the body not a part of reproduction (ex. growth of pubic hair, muscle in boys and breasts in girls)

Question 267

menarche

Answer 267

onset of menstruation (usually 12)

Question 268

Cognitive reframing

Answer 268

strategy of dealing with emotions where we learn to look at our experience through a different frame - like turning something negative into a learning experience

Question 269

Delaying gratification

Answer 269

putting off immediate temptations in order to focus on long term goals

Question 270

formal operational thinking

Answer 270

abstract reasoning and logic and higher cognitive abilities (starts to go up after age 12)

Question 271

Lawrence Kohlberg

Answer 271

moral development
learning right from wrong

Question 272

Levels of reasoning:

Answer 272

preconventional
conventional
postconventional

Question 273

Preconventional reasoning

Answer 273

• people reason based on self-interest, avoiding punishment

Question 274

Conventional reasoning

Answer 274

people reason based on social conventions (traditions), can involve authority figures “Because mom said so”

Question 275

Post conventional level

Answer 275

Level of reasoning that considers justice and fairness

Question 276

Sensitive periods

Answer 276

windows of time during which exposure to a specific type of environmental stimulation is needed for normal development of a specific ability
ex. becoming fluent in a language –> should be introduced to that language in really early years

Question 277

assimilation

Answer 277

fitting new information into the belief system you already posses
ex. all girls have long hair - that person has short hair but is a girl! (azlan)

Question 278

accomidation

Answer 278

creative process where people modify their belief structures based on experience ex. that girl has short hair, so girls can have short hair