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Consider two genes controlled by X. Their production rate is g1 = β 1X/(K 1 + X) and g2 = β 2X/(K 2 + X). The questionis consider the same two genes controlle...

Question

Consider two genes controlled by X. Their production rate is g1 = β 1X/(K 1 + X) and g2 = β 2X/(K 2 + X). The questionis consider the same two genes controlled by X with an FM strategy.Here X is either at level 0 or level X high. The fraction of timethat X is high is proportional to the signal s. Show thatproduction of each gene is proportional to signal, but the ratio g1/g 2 is constant as a function of signal.Hint: Consider what would happen at the limits; when X= 0 andX>>>Ki . Give

Consider two genes controlled by X. Their production rate is g 1 = β 1X/(K 1 + X) and g 2 = β 2X/(K 2 + X). The question is consider the same two genes controlled by X with an FM strategy. Here X is either at level 0 or level X high. The fraction of time that X is high is proportional to the signal s. Show that production of each gene is proportional to signal, but the ratio g 1/g 2 is constant as a function of signal. Hint: Consider what would happen at the limits; when X= 0 and X>>>Ki . Give also a mathematical explanation.



Answers

Gene regulation Genes produce molecules called mRNA that go on to produce proteins. High concentrations of protein inhibit the production of mRNA, leading tostable gene regulation. This process has been modeled (see Section 10.3 ) to show that the concentration of mRNA over
time is given by the equation
$m(t)=\frac{1}{2} e^{-t}(\sin t-\cos t)+\frac{1}{2}$
(a) Evaluate lim_{t-0} m ( t ) \text { and interpret your result. }
(b) Use the Squeeze Theorem to evaluate lim $_{t \rightarrow \infty} m(t)$ and $\quad$ interpret your result.

This question asks us about fitness in a situation where the Hamas August dominant BB as a fitness of one headers, I guess, also has a fitness of one. But the home is egas. Recessive has a fitness of zero, so those individuals won't survive so and were asked how, um, the population of Q. How Cuba will change? Because, remember, we do have some of meals here of the little be Leo, so they'll be present in the population for a while to come. So remember that fitness. Yeah, I can be calculated just finding the average for this. Ah, using this formula just an extension of the Hardy Weinberg equation and then queues Quito. And it's the fitness of little bee Elio from another Gina type. Now, remember that this is zero, right? So this whole term is gonna drop off when we plug in the numbers, get 0.9, try square. And of course, this is one That's the value right there. So we know it's one. And then to peak you this 0.95 and 0.5 when they count the 95 and five, Of course from the question not not making those numbers up myself. And then times 1.0 and the course that's plus Euro. And if you do the arithmetic here, um, you got 0.90 to 5 plus 0.0 95 So this is close toe one. So it's 10.99 75 And so that's the average fitness for this population. And then we want to figure out how Q is gonna change over time and their different ways to do this. I'm just gonna do it sort of generative Lee, until I actually think that's probably quicker. Um, where we have this formula, uh, over the average fitness. So this tells us the next generations P values based on this generation's P value times the fitness of B, which, of course, is gonna be one divided by the average fitness. And so that's P crime. You called 0.9 Fine. So that's our current P value times one over the average fitness, and that gives us a P prime value of 0.952 And that's going up, which is what we'd expect, right, because the fitness of those air one, of course, the fitness of the B L. E. A list zero. So we expect more and more big ble Osama population. Remember, we can get people's Q equals one so we can just get Q prime my taking one, um minus 29 5 to So when my next 0.952 and that gives the Cirque you prime value of 0.48 so that numbers dropped from 0.5 and then we can do the same thing. We can call this the P two generation Ah, and just start over here. Well, we might want to re calculate the average fitness. It will change, but it won't have changed much. And so it's not really going. Teoh have, ah market impact on our results. In fact, it won't change that much at all. So we just have 0.9975 again and this gives us RP to value. Uh huh. 0.954 for that number is continuing to rise. And if we take it, just do this attraction like we did above the cue to value is going to be 0.0 for six. The second part of the question changes the, um, terms of the problem a little bit in that it makes the headers. I go, um it gives it a fitness of zero, and you conduce the math and it's exactly the same math. Here. You just calculate the average fitness and then calculate with the P and Q values are. But if this is now zero, that means no individuals who have a little be alil in the population will survive. Right? And so once we've done this after in the next generation, um, P will goto one and Cue will go to zero. So and you could get do this show that show it with the math. And I did actually do this. Um, but you can just sort of reason it out, right? Is this. If any individual who has a little little Pia Leo is lethal, lettuce has a fitness of zero. Then in the next generation, no individuals with the little Delia will exist. And so everybody will have the big Bigby. I'm trying little be illegal. The opilio and the only individuals that will survive have the big be Bigby. And so the B will have ah, frequency of one and Q a high frequency of zero

So we're giving the falling system and part. They were asked to fund all equilibrium. So let's go ahead and set both equations equal to zero. So second equation automatically told us that p and M have to be the same. So Ah, the equation on the left hand side tells us that m equals one over one plus p, but peace The same is m. So I'm going to replace that by m um and I'm gonna cross multiply giving me, um plus I'm squared equals one. So I'm squared plus m minus one equals zero This So am I iss equal to negative 1/2 closer. Minor square root of five over too. And this is just using the quadratic formula, and soapy is also equal to the same thing. So we either have, um we have to go over some points with the plus and the one with the minus. So party. Um, we need to calculate the Jacoby INS matrix. So forget all that. The first equation. Differentiate with respect to em, giving us negative one. Never one to differentiate with respect the p. So we have one plus p to the minus one. So we have negative one over one plus p squared. Now we're gonna go to the second equation, differentiating with respect to em. Give us one. And with respect, the P gives us minus one. So that's our Jacoby in Ah, so part. See, only one of these equilibrium makes sense in part A that we found on. We know that M and P R amounts of protein and Amaar and a in the cells. So these things don't make physical sense to be negative. So only the positive one makes sense. So that's the equal of Rhea. Ah, negative 1/2 plus square root of five over too negative 1/2 plus court if I remember too. So we're gonna, um, look at the stability for this point. So if we take the Jacoby in a valuation at this point, we get negative ones about negative zero points. 382 one negative one. So if we look at the Eiken values for this, if we calculate the determinant of J Moniz Linda, I we get that the argon values our landa is equal to a negative one plus or minus square root of 9 55 I over 50. So we don't really care about this part. We just care about the largest riel part of the eye and values. So that value our is equal to negative one, which is less than zero. So that tells us that these equilibrium is, in fact stable. Still lovely party. We need toe. See what happened around this point. What kind of shape for dealing with. So in order to do that, we need to look at the trace on the determinant of the Matrix. So we have that, uh, the trees of J at this equilibrium point. I'm not gonna read that all out again. Um, as you go to the sum of the diagonals. So negative one minus one gives us negative, too. And the determinant is equal to negative one times negative one, which is one and then minus a negative point 3/8. So, plus two, you're a 20.38 So we have a determinant of 1.38 So now, um, there's that picture in the book with the parabola. Um, with the equation, 14 trace of J squared equals determinant of J. So let's see what side on which side of this problem we fall, so we know our trace we know are determinants. So let's compute 1/4 trace of Jay's. You called a negative two squared. So that's equal to 1/4 of four, which is one which is less than our determinant. So this tells us, why was he in that picture that we're gonna have, um, spirals? So we have stable spirals.

This question asked us to find the rate of change of em or in a concentration is a function of time. What we knows that this means we're essentially finding the derivative m prime of tea because the rate of change is the director. So we start out with 1/2 eat the negative tea, and then we have sign of tea minus co sign of tea. But we could take the derivative of this so we know that the derivative of sine of T is co sign a pity. You know, the derivative of coastline of tea is negative. Sign of T negative negative makes a positive. We're now subtracting 1/2 e to the negative T. We have a sign of T minus coastline of tea, and then plus zero is the derivative. 1/2 is simply zero because it's a constant. This simplifies to negative 1/2 or positive 1/2 each. The negative T Times Co Sign of tea plus sign of tea minus sign of tea. Close co sign of tea giving us eat the negative T co sign of teeth

109 You want us to factor the left side of the equation? Sorry. There's no to there. It's peace. Where'd Waas to be? You plus Hughes. Where'd equals one? So this is the left hand side, and this is the right hand side of the equation. So they want us to factor the left inside. Which is this? I so p the perfect square. Try no meal P plus Hugh wanted. He's weird.


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