The Systems Biology of COVID-19 and the SARS-CoV-2 virus. The class will build a foundation that includes the emergence of complexity, simple biological subsystems, their reductionist and equivalent toy and organ-chip models, and the measurements required to specify model architecture and parameters. Applications to biology, physiology, medicine, chemical and biological defense, pharmacology, drug discovery, and toxicology. UGrad: PHYS 240 01 and BME 290B; Grad: PHYS 326 and BME 395C.
Monday, September 3, 2018
SysBio18 Asgn_2B_Class_4_PPT_01_2018_09_04
Sniffer-buzzer PPT -1: Read, first pass: PPT 01 PPT from J. J. Tyson, K. C. Chen, and B. Novak. Sniffers, buzzers, toggles and blinkers: dynamics of regulatory and signaling pathways in the cell. Curr.Opin.Cell Biol. 15 (2):221-231, 2003. Use to focus your explanations
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Kylie Balotin
ReplyDeleteReading Assignment 2a: Powerpoint Figure Explanations - Figure 1d Perfect Adaptation/The “Sniffer” Model
R – response magnitude – [protein]
S – signal strength – [mRNA]
X – some other species
Rss - stead state solution of dR/dt
The wire diagram shows that S is involved in the synthesis of X and R. X is involved in the removal of R. R will degrade (removal of R).
The middle plot:
Solid line - rate of removal
Dashed line - rate of synthesis
Solid circles - steady state solutions
This is showing that the rate of removal and rate of production are both dependent on S. When you get an increase in the rate of production, you get an increase in the rate of removal (as opposed to Figure 1a where an increase rate of synthesis has no effect on the rate of removal). In the end the steady state value of R does not depend on the rate of production or removal, and therefore, it is not dependent on S.
The far right plot:
S is involved in the synthesis of X and R. There is a slight delay in the synthesis of X when there is an increase in the amount of S, while there is an immediate change in the synthesis of R when the amount of S is added. Therefore, you see a huge immediate jump in the amount of R when the S is added to the system and a graduate increase in the amount of X. As X increases, there is an increase in the removal of R from the system (more degradation of R). Therefore you see the rate of the synthesis of R decrease and then see a gradual increase in the rate that R is removed from the system. With each addition of S to the system, you are also seeing an increase in the amount of X in the system, so there is less R synthesized in the system (a greater amount of S will have less effect with each sequential addition of S to the system). Therefore, the steady state concentration of R will not depend on the concentration of S in the system, but there is a response in the concentration of R where there are sudden changes in the concentration of S in the system.
Figure 2(b): Activator-Inhibitor oscillator
ReplyDeleteR = response
S = signal strength
X = other molecule
Left plot:
Autocatalytic process that promotes the production of inhibitor X, which speeds up R removal. R and X are on both sides of equation 2. So when R builds up, this forces R removal, then X disappears, and R can rise again
Middle figure:
The R-nullcline is where the rate of change = 0.
Red curve is (X,R) pairs where dR/dt = 0
Blue curve is (X,R) pairs where dX/dt = 0
Where the curves intersect is the steady state for the full system.
This system is unstable, so R and X oscillate around the steady state in a stable limit cycle/hysteresis loop. This is called a hysteresis oscillator.
An example of this is cyclic AMP production.
When cAMP binds to the receptor, adenylyl cyclase is produced and more cAMP is produced.
The cAMP binding make the receptor inactive, so the inactive sensor acts as the inhibitor and stops autocatalytic production of cAMP. cAMP is hydrolyzed to 5'-ATP, so the receptor gradually regains its sensitivity and cAMP is synthesized again.
Right plot:
Solid line = stable steady state
Dashed line = unstable steady state
Closed circles = maximum values of R during oscillation
Open circles = minimum values of R during oscillation
S critical 1 and 2 are Hopf bifurcation points
Line in the middle is steady-state
Figure 1c: The “Buzzer”
ReplyDeleteR – response magnitude, [protein]
S – signal strength, [mRNA]
RP – phosphorylated (active) form of response protein
Rss – steady state solution of dR/dt
Left Plot:
The wire diagram shows that S, with ATP, phosphorylates R, forming RP and ADP. RP spontaneously de-phosphorylates with the help of water, releasing the phosphate and returning to state R.
Middle Plot:
Solid line - rate of dephosphorylation of RP
Dashed lines - rate of phosphorylation R
Solid circles - steady state solutions
When rate of phosphorylation increases, rate of dephosphorylation also increases.
Right Plot:
S directs the synthesis of RP. Once S hits a threshold, the response is triggered like a switch, and sustained until the signal, S, is diminished, like a buzzer. When S goes away, the switch goes back to state R spontaneously, making the system reversible.
Figure 1a: Linear Signal-Response Curve
ReplyDeleteS = signal strength (concentration mRNA)
R = response magnitude (concentration protein)
Left Plot:
This molecular wiring diagram shows how the signal element, S, directly stimulates the response element, R. R is thus only controlled by S, and if S is not stimulated, R will not be activated. This directly corresponds to biology, where mRNA is translated at a ribosome into the protein. Thus, mRNA is a required signal to produce the protein.
Center Plot:
Solid Line = Rate of degradation of the protein
Dashed Lines = Rate of Synthesis of protein
#s 1-3 = amount of mRNA, with the line labelled 1 having the smallest concentration of mRNA and 3 having the greatest
This graph shows that both S and R, or mRNA and protein, have directly proportional relationships with the rate of synthesis and degradation, respectively. For example, as the protein concentration is increased, the rate of degradation increases proportionally as well (linear graph). When the mRNA concentration is increased, the rate of protein production increases proportionally as well. The solid circles represent the steady-state solutions, where the rate of production = rate of degradation of the protein. Realistically, there would be a cap off for the increases in synthesis and degradation rates, since the ribosomes/amino-acyl transferases and ubiquitin (respectively) proteins would eventually be saturated, resulting in no increase in production/degradation rates with more substrate.
Right Plot:
This graph shows the linear signal-response curve itself. When the signal strength, or mRNA concentration, is increased, the steady-state concentration of protein increases proportionally. Essentially, the steady-state concentration of protein is solely dependent on the concentration of mRNA. Realistically, homeostasis would cause a feedback mechanism system where more protein concentration would reduce the strength of the signal's effect on the protein production rate.
Figure 2a: Negative-feedback Oscillator
ReplyDeleteS = Primary Signal Component
X = Downstream signal of S
Yp = Active Downnstream signal of X
Rp = Active Response Component, Downstream signal of Yp
Repressor for Activation of X
Left Plot:
This is the network configuration of the negative-feedback oscillator. Where some signal S activates X which activates Y which Activates Response R. R then inhibits the activation of X by S and degrades active X lowering the amount of X able to activate Y then R acting to inhibit the response resulting in a oscillatory behavior. Where increases in R lead to decreases in R and thus increases and decreases in X and Yp. Note Arrows, ->, are activation changing a molecule from one state to another and ---| are inhibition/repression that reduce this transition from one state to another. Two Arrows back and forth are reversible reactions that are effected by connections that shift the rate of the forward and backward reactions.
Middle Plot:
Time-dependant analysis of the system showing how X, YP, and RP change overtime in a oscillatory manner where they each have a minimum value, maximum value and period of oscillation that needs to occur before the same molecule species value can be observed again.
Right Plot:
Phase Plane Analysis of this system to see how Signal S and Response Rp change as function of each other. And since they both are sinusoidal with time then they form a cycle where the over time the current Rp and S changes along the ellipse like curve called a limit cycle as all trajectories that stabilize the system that model can take for different values of S overtime approach the cycle. There can also be unstable values of the signal and depending on the initial amount of S, X, Y and R can tend towards different stable solutions or unstable solutions. Where the dynamical system approaches unstable oscillations that either lead to parts ofthe system decaying to zero or expanding to infinity. Critical values of S tell where the transition occurs from the stable oscillations to unstable oscillations and vice-versa.
Figure 1b: Hyperbolic Response
ReplyDeleteR – response magnitude, [protein]
S – signal strength, [mRNA]
RP – phosphorylated form of response protein (Rt-R)
Rss – steady state solution of dR/dt
Left Plot:
S and ATP phosphorylates R, forming RP and ADP. With H2O, RP de-phosphorylates, releasing the phosphate.
Middle Plot:
Solid line: rate of dephosphorylation of RP
Dashed lines: rate of phosphorylation R
Circles: steady state solutions
Right Plot:
As signal strength (S) increases, the concentration of phosphorylated R (Rp) increases as a hyperbolic response, reaching a max amount of 1 (100% phosphorylated).
Figure 2c: Substrate Depletion Oscillators:
ReplyDeleteS - Signal
X - Intermediate of Signal
R - Response
E - Related Enzyme
EP - Phosphorated Enzyme
Equations:
dX/dT = k1S-k0'X-k0EP(R)X
Where k1 represents the rate constant for the autocatalytic reaction production X. k0' is for the reaction creating R, k0 is for effect of the phosphorylated enzyme on said reaction, and k2 is the for the degradation of R.
dR/dT = k'0X +k0EP(R)X-k2R
Where k2 is the rate constant for the degradation of R.
EP(R) = G(k3R,k4,J3,J4)
Which is the Goldbeter-Koshland function and gives us the amount of phosphorylated enzyme. k4 is a rate constant for the conversion between phosphorylated and un-phosphorylated versions of the enzyme.
This switch-like function results in a build-up of R creating an explosive conversion of X into R.
Middle Plot:
This plot is of the nullclines of R and X (red and blue respectively). That is, the lines where the time derivatives for the concentrations are 0. The nullclines intersect at one steady state (the open dot). This dot is open as the steady state is not stable, and the system falls into a limit cycle (the close loop with arrows). Any system within that limit will begin to oscillate.
Right Plot:
This plot is of the response of the system vs. the signal. At small and large values, the systems responds linearly. Once it hits bifurcation points (Scrit1 and Scrit2), the system begins to oscillate. The filled in circles represent the maximum and minimum values of the oscillations. Open circles represent unstable oscillations (which build up to larger ones).
Figure 1f: Mutual Inhibition by Positive Feedback
ReplyDeleteS: Signal
R: Response
E: Protein
EP: phosphorylated protein
Left Plot: This wiring diagram shows that S activates the Response R. R can then degrade on its own, or be aided in degradation from protein E. However, R also aids in E being phosphorylated into EP. EP is able to be de-phosphorylated back into E, but is not assisted by any other species.
Middle plot: The dotted lines are the signal. They are constant because S is a sole input into the system and is not effected by other species. With a low signals (0.6) there is only one stable steady state. However, when there is a moderate amount of signal (1.2), there are two stable steady states of R. At high signal (1.8), there is only one stable high steady state. This can be explained from the mechanism of the wiring diagram. Once there becomes a sufficient amount of signal S, there is enough response R that turns E into EP, hence leaving less E to degrade R. Further, when there are very high levels of signal, this process is so amplified that there is practically no E to degrade R, so we get more response R.
Right graph: this process follows from the explanation in the middle graph. Once we hit critical values of S, there are jumps between stable steady states of response.