MOB30306: Control of Cellular Processes and Cell Differentiation (MOB30306)
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Wageningen University (WUR)
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MOB30306: Control of Cellular Processes and Cell Differentiation (MOB30306)
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Exam and self-test questions BIOCHEMISTRY part of Control Cellular Processes and
Cell Differentiation (MOB-30306) 2013-2016
Attention: The exam will consist entirely of 10 “open end” type questions selected from the
list below and a “bonus” question taken from one of the presented case studies.
1. 1
Why is covalent modification advantageous when compared to proteolytic activation?
1
.
Covalent modification is reversible, while the proteolytic activation is not (it is only useful if
the inactivation is supposed to be long-term). Thus, after initiating a specific signaling
pathway, it is necessary to turn the signal of after a specific reaction has taken place. Then,
covalent modification can be performed again (usually it is done by another enzyme) and
brought back to the initial state, while the return to the initial state before the degradation
is not possible, unless a new protein is synthetized (takes more time).
2. 1
Phosphorylation is an extremely effective tool for catalytic control. Explain the reasons.
2 • Fast kinetics – phosphorylation and dephosphorylation can take place in less that a
. second or over hours – timing is adjusted to meet the timing of physiological
processes
• It is suited for amplification of the signal – one kinase can activate hundreds of other
kinases within a second
• It is reversible (switch like responses are possible)
3. 1
How is a kinase cascade activated?
3
. kinase cascade is activated by an external stimuli that activates the first kinase. Kinases are
A
almost always activated by phosphorylation (addiction of phosphate) – phosphorylation
starts a cascade – protein gains capability of phosphorylating another protein, that then gets
the ability to phosphorylate next one and cascade of phosphorylation begins.
4. 3
What are some of the common structural features of the receptors to which signal
4
molecules bind?
.
The molecule needs to have a ligand binding site located on an extracellular membrane, a
hydrophobic domains that cross the plasma membrane and an intracellular domain. Binding
of the receptor must induce conformational changes of the intracellular domain so that the
signal can be transmitted to the cell. The receptor should always have a type of interaction
with a intracellular protein to induce the pathway (e.g. G-proteins coupled with 7TM
receptors).
Notes: almost all receptors are single helixes – they dimerize
1
,5. 3
What is a disadvantage of using common molecules for signaling paths?
5
.
They can be used for many different pathways as well, thus when any mutation occurs (e.g.
in signal switch off function) they can start initiating pathways that should not be taking
place at this specific moment in that cell.
6. 3
What happens when signaling paths are not terminated properly?
6
. signals are not terminated, the cell looses its responsiveness to new signals.
If
What is more a signal transduction that does not have an end and therefore is constantly
active (even if it should not be) may cause highly undesirable consequences – e.g. cancer.
7. 3
How many 7TM membrane receptors exist? What are some of their functions?
7
.
More than 20,000 transmembrane receptors are now known.
They’re functions are transmiting information from diverse signals, such as: protons,
hormones, neurotransmitters, odorants and tastants.
Some of the pathways they regulate:
- Hormone action
- Hormone secretion
- Chemotaxis
- Exocytosis
- Control of blood pressure
- Embryogenesis
- Cell growth and differentiation
- Development
- Smell
- Taste
- Vision
- Viral infection
-
Examples:
• Rhodopsin – senses photons and initiate cascade for visual sensation
• Adrenergic receptor (B2-AR)
2
,8. 3
What is the general mechanism for signal transmission by 7TM receptors?
8
. • Ligand binds into a 7TM receptor.
• Conformational change in the receptors cytoplasmic domain activates a G-protein
(it binds guanyl nucleotides) – instead of binding GDP it starts to bind GTP
• Activated G-protein (GTP) stimulates activity of adenylate cyclase
• Adenylate cyclase catalyzes ATP – cAMP conversion
• cAMP can travel through a cell and start the signaling cascade – activation of protein
kinase A
• activation of protein kinase A stimulates the phosphorylation of many target
proteins – PKA phosphorylates serine and threonine residues on its targets
9. 3
What are some examples of G protein families and functions?
9
. - Gs – stimulates adenylate cyclase
- Gi – inhibits adenylate cyclase
- Gt – stimulates cGMP phosphodiesterase
- Gq – increases IP3 and intracellular calcium levels
- G13 – stimulates Na+ and H+ exchange
• Adrenaline receptor is directly connected to alpha unit of G-protein
• G-alpha Q protein that is activated with association with phospholipase
10. 4
How does binding of epinephrine initiate the cAMP production? Discuss briefly in terms of
0
receptor structure and function.
.
Epinephrine binds to B-adrenergic receptor which results in a change of conformation of its
cytoplasmic part.
This receptor is bound to a G-protein, which subunit alpha (Galpha) binds GDP. Due to
ligand binding with the receptor (hormone-receptor interaction), GDP is exchanged for GTP
binding.
GTP can then bind to adenylate cyclase, activating it. Adenylate cyclase is then capable of
catalyzing a conversion of ATP to cyclic AMP.
A single hormone-receptor complex can stimulate nucleotide exchange in many G-protein
heterotrimers.
3
, 11. 4
How is the hormone-bound activated receptor reset after activation?
1
.
This reset is accomplished by 2 processes:
1) The hormone dissociates, returning receptor to its initial, not-activated state
2) The signaling cascade initiated by hormone-receptor complex activates the kinase
that phosphorylate serine and threonine residues in the carboxyl-terminal tail of the
receptor making it inactive
o Finally after phosphorylation, B-arrestin bind to disactivated receptor and
further diminishes its activity to activate G-proteins
12. 4
How does cAMP affect so many different paths? Is there a single common molecule
2
involved?
.
It is possible due to the fact that most of the effects of cAMP in eukaryotic cells are
mediated by the activation of protein kinase A (PKA).
Thus yes, there is a single common molecule involved.
13. 4
Provide specific cascade examples of the effects of adenylate cyclase activation.
3 - Enhanced degradation of storage fuels
. - Increased secretion of acid by the gastric mucosa
- Dispersion od melanin pigment granules
- Diminishes aggregation of blood platelets
- Induces the opening of chloride channels
In the regulation of glycogen by epinephrine in the muscle cells:
• The ligand binds the B-adrenergic receptor activating the G-protein and the
adenylate cyclase. The cAMP activates PKA, PKA phosphorylates and activates the
phosphorylase kinase which in turn phosphorylates the glycogen phosphorylase
activating the degradation of oxygen
• Odorant is sensed by 7TM receptor, which raises cAMP levels triggering activation
of non-specific ion channel polarizing the membrane of sensory neuron
14. 4
What features do the structures of the various Phospholipase C enzymes have in common?
4
.
Common domains:
- pleckstrin homology domain (PH domain) that recognizes PIP2 and allows the
binding of the PIP2
- EF hands – inactivated by evolution
- C2 domain – binds to phospholipids
- catalytic domain.
4
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