Download 1. The graph shows the relative levels of Cdk1 and cyclin B

Chem*3560
1.
Question set 5
Feb 2-6, 2004
The graph shows the relative levels of Cdk1
and cyclin B concentration as well as
Cdk1-Cyclin B kinase activity as a function
of progress through the cell cycle. If the
Cyclin B activates the kinase, why does the
activity curve not match the variation in
concentration of cyclin B?
Although cyclin B is required for activity of
Cdk1, it is not the only factor determining
activity. Initially, the CDK carries a phosphate
on Tyr 15, which inactivates by placing a
negative charge near the ATP binding site
(ATP4– has 3 phosphate groups giving it
negative charge). The Cdk1 must also be
phosphorylated on Thr 160 for full activity.
Thus in the early phase, cylcin B can be present but kinase activity is absent or weak
Thr 160 phosphorylation is initiated by cyclin A-Cdk1 which is active in the preceeding phase
(G2). As soon as a few cyclinB-Cdk1 molecules become active, they can phosphorylate their
still inactive neighbours.
Thus kinase activity appears at the end of G2.
Why does the activity curve rise so rapidly?
A specific phosphatase removes the Tyr 15 phosphate, which greatly increases kinase activity.
The phosphatase is stimulated by being phosphorylated by already active cyclin B-Cdk1. This
results in escalating mutual activation of the phosphatase and kinase enzyme pools.
Rapid activation of cyclin B-Cdk1 is important because the onset of M phase is very rapid and
M phase has a short duration (1 hr for a 24 hr cell cycle).
Why does the cyclin B concentration drop so rapidly at the end of M phase?
Cyclin B-Cdk1 activates a specific ubiquitin ligase that tags the cyclin B with ubiquitin. Proteins
that have been tagged with ubiquitin are subjected to rapid degradation by an ATP dependent
protease system called the proteasome. ATP is consumed to unfold the target polypeptide,
since normally folded protein molecules are resistant to simple proteases.
The disappearance of cyclin B is a necessary event to allow completion of M phase.
Chem*3560
2.
Question set 5
Feb 2-6, 2004
If trypsin is inhibited by secretory pancreatic trypsin inhibitor, how is it possible to
activate trypsinogen and the other pancreatic proenzymes?
The amount of inhibitor is only a small fraction of the total trypsinogen available. The objective
is to mask the activity of the odd molecule of trypsin that might be accidentally activated
prematurely. Otherwise, because trypsin can activate trypsinogen, there is potential for an
escalating series of activations that would be catastrophic for the pancreas.
When the zymogens are secreted into the duodenum, they encounter enteropeptidase, which is
an membrane-anchored enzyme on the extracellular surface of cells lining the duodenum.
Enteropeptidase can activate enough trypsinogen to overwhelm the trypsin inhibitor. Once this
happens, activation becomes a cascade-like process.
3.
The following is a list of mechanisms for regulating or controlling enzyme activity:
a) masking the catalytic site with a peptide loop or chain
b) putting the catalytic amino acids out of ideal alignment
c) presence of a charge that repels substrate
d) presence of a charge that attracts substrate
Which of these mechanisms is used by each of the following?
i) phosphofructokinase 1
A charge that repels substrate is found in the T-state: Glu 161 (negative) repels
fructose-6-phosphate 2-.
A charge that attracts substrate is found in R-state: Arg 162 (positive) attracts
fructose-6-phosphate.
ii) trypsin/trypsinogen
The catalytic amino acids Ser 195 and Gly 193 are badly aligned in trypsinogen. The
oxyanion hole that stabilizes the transition state of the reaction is absent.
iii) cyclin dependent kinase
The Thr 160 loop masks the substrate site in the absence of cyclin
When Tyr 15 is phosphorylated, its negative phosphate repels negative ATP
iv) pepsin/pepsinogen
The catalytic site is masked by binding the N-terminal loop of 44 amino acids
v) protein kinase A
The catalytic site of the C subunit is masked by binding to the autoregulatory site of
the R subunit.
vi) phosphorylase b kinase
The catalytic γ -subunit is masked by the autoinhibitory loop of the β -subunit. In the
presence of Ca2+, the δ -subunit (calmodulin) masks the autoinhibitory loop.
Chem*3560
4.
Question set 5
Feb 2-6, 2004
Why should cells want to commit suicide?
In a multicellular organism, the benefit of the organism as a whole has higher priority than life of
individual cells. The following are reasons why apoptosis may be initiated.
The cell has completed its useful role in development of the embryo or organism, and now its
resources can be shared among its neighbours.
The cell has a viral infection that can be detected by cells of the immune system. Since viruses
require host machinery for propagation, the infection can be forestalled by quickly destroying
infected cells.
The cell has a defect in DNA which has stalled the cell cycle while repairs are undertaken. If
this takes a long time, there's a high probability that damage is too extensive to repair safely.
(Cancer cells can arise from major rearrangements of DNA due to damage).
The cell is a tumour cell which is expressing unusual proteins on its surface that mark it for killing
by lymphocytes.
The cell is a tumour cell that has migrated to a new location in the body and expresses surface
proteins which are incompatible with the organ it now resides in.
5.
What different roles do caspases 3, 8 and 9 play?
Caspase 3 is the major executioner caspase, attacking key elements of the cell involved in DNA
replication as well as cytoskeletal proteins and key regulators of these systems. Attack on
ICAD, Inhibitor of Caspase Activated DNase, destroys the ICAD protein and liberates an
active DNAase that shreds cellular DNA into fragments. Once this step is reached, cell death is
irrevocable.
Caspase 8 has a DED domain on its procaspase, so is activated by binding to adapter proteins
that link it to receptors in the cell membrane. When an external ligand binds, the receptors
cluster, allowing procaspase 8 to self-activate. Caspase 8 goes on to activate the other
caspases.
Caspase 9 has CARD domain on its procaspase, so is activated by internal signals in the cells
that lead to cytochrome c release from mitochondria. Once active, it goes on to activate other
caspases.
6.
What is the advantage of storing glucose in a branched polysaccharide like glycogen
versus a linear polysaccharide like amylose?
Branches give rise to multiple non-reducing chain ends per molecule. Since glucose addition
occurs at the non reducing ends, this multiplies the effective susbtrate concentration of the
glycogen.
Chem*3560
7.
Question set 5
Feb 2-6, 2004
How does caffeine affect the level of cyclic AMP in a cell?
Caffeine inhibits cyclic AMP phosphodiesterase, which is responsible for converting the
messenger molecule cyclic AMP into inactive 5'-AMP. If the cyclic AMP level does not fade
away, weak epinephrine signals can build up and maintain a prolonged state of high activity of
glucose-mobilizing enzymes
8.
What role do each of the following play when they bind to glycogen phosphorylase?
A) ATP
ATP is a negative allosteric effector that helps hold phosphorylase b in the T state, opposing the
effect of positive effectors.
B) AMP
AMP is a positive allosteric effector that can switch phosphorylase b into the active R state.
Ser-PO4 2– seems to take over this role in phosphorylase a. The AMP/ATP balance reflects the
cells energy status, and if [AMP] levels rise, glycogen breakdown is urgently needed to resupply
the glycolysis pathway. This is particularly significant in muscle, where ATP consumption rates
vary widely and can suddenly build up.
C) pyridoxal phosphate
Pyridoxal phosphate is a coenzyme required for the catalytic reaction. The phosphate group is
used in successive catalytic steps as general acid and general base. A phosphate with pKa
close to neutral pH is needed, and pyridoxal phosphate provides this
D) glycogen
Glycogen occupies the substrate site. Phosphorylase binds tightly to glycogen throughout its
catalytic cycle, because the “product” is also the next substrate, so does not have to leave the
catalytic site (until a branch point is reached).
E) glucose
Glucose is a negative effector for phosphorylase a, and can hold phosphorylase a back in the
T-state.
F) Glucose-6-phosphate
Glucose-6-phosphate is a negative effector that competes with AMP binding, thus opposing the
positive effect of AMP. If Glucose-6-phophate levels are high enough for this to happen,
there’s less immediate need for glycogen breakdown.
G) glucose-1-phosphate
Glucose-1-phosphate can bind to the catalytic site; it is the product under physiological
conditions, but can become the substrate if the glucose-1-phosphate concentration is artificially
raised or phosphate concentration lowered.
Chem*3560
9.
Question set 5
Feb 2-6, 2004
When glycolysis starts from glycogen instead of from glucose, only one ATP is needed for
activation while 4 ATP are made in the glycolysis reactions. Are you really getting one
extra ATP for free?
The overall cost of adding one glucose to glycogen is 2 ATP: one to convert glucose to
glucose-6-phosphate, and an ATP equivalaent, UTP to make the UDP-glucose that glycogen
synthase uses as substrate.
Therefore on an overall basis, the cost of using glycogen is actually one ATP more than using
individual glucose molecules for glycolysis. However, this cost can be pre-paid at a time when
the cell has surplus ATP.
When the demand occurs for glycolysis to make ATP, the cost at that time for glycolysis
activation starting from glycogen is only one ATP, for the phosphofructokinase step. So at the
time of energy demand, the cost is 1 ATP and the yield is 4 ATP. This is particularly important
for muscles that use anaerobic glycolysis, since the net yield is 3 ATP instead of 2 ATP.
10.
Why is the net yield of glycolysis 2 ATP for one glucose converted to 2 molecules of
lactate while the net cost of gluconeogenesis is 6 ATP for each glucose made?
Some overall negative free energy change is necessary to drive a reaction pathway such as
glycolysis in a given direction. The free energy change for converting glucose to 2 lactate is
about -200 kJ/mol glucose; the free energy of ATP hydrolysis (at cellular concentrations) is
about -50 kJ/mol ATP (Chem 2580, lectures on glycolysis). So for glycolysis, where 2 ATP
are made, the overall net free energy change is about -100 kJ/mol of glucose used, and this
drives the reactions in the desired direction. In gluconeogenesis, 6 ATP are used to convert 2
lactates to glucose, so the overall free energy change is also -100 kJ/mol of glucose made, and
this drives the reaction sequence in the opposite direction.
If energy required and energy released were more in balance, ∆G = 0, which represents a state
of equilibrium. Since backward reaction exactly balances forward reaction, no net conversion
of reactant to product occurs at equilibrium. Very “efficient” in energy terms, but nothing
actually gets done.