nanoHUB-U Principles of Nanobiosensors/Tutorial 5.1: Integrated Sensor Systems - Homework Solutions
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[Slide 1] Ok, the third thing was about genome sequences. I want to give two examples then you'd be done for the day. The example this involved that how long does it take to sequence the genome because this is sort of an important number if it took each one of them took one year, then of course we'd be in big trouble but-- so we want to quickly, take a look and we are saying here that assume that the loading efficiency about 20 percent so 20 percent of the-- the genome sequence to remember the beads sitting in various places 20 percent of them has actually had DNA in it so that's what the loading efficiency means
[Slide 2] So lets see so you remembered the problem we're trying to calculate is two things one is that once you have a DNA you cut them up and multiple pieces and then-- you amplified them so the question we are asking how big it has to be I said it has to be about 100 to 1,000 we'll see what the exact number it is and second is that once all these-- reactions takes place if each reaction has a certain amount of time lets say 4 seconds then the entire thing would take about 16 seconds for each base pair and if you have a certain number to read 123 456 and so on and so forth then you'd have to my multiply with that many numbers so let's see how long it takes for genome sequencing
[Slide 3] So I said that if it is possible to read it by 20 percent Let's say we have hundred-million sequences at the same time so what would be that total so let's say we have 100 million transistors right 100 million transistors and of them only 20 percent sort of can read it, so how many we have 20 million active sequencers and if the DNA is 4 billion-- million base pair long, you have to divide by 20 million so each one of them that read length in each well be about 200 now if each read takes about 10 seconds let's say and we have 4 base pairs that we have to sort of try before we decide find out exactly which one it is then it will take 40 seconds and 40 seconds multiplied by-- there are two hundred of them so total of about 8,000 seconds that means just a couple of hours so you can just do it in a couple of hours and remember then you have to do it a couple of times so that you can align them and find out exactly the right order Let's say you want to do them five-- reads so you can do it in a half a day the entire sequencing that previously had taken 10 years you can do the whole thing in any given machine You can do one or two every every hour-- every day in terms of genome sequencing. So very very efficient as you can see Actually the main problem is that even after the read is done in order to do this computer alignment, that takes much longer so these days is the computation that's the bottleneck, not the reading of the genome itself.
[Slide 4] Alright. And then finally I'll show you some examples because about-- the dynamics of genome sequencing using the Ion-torrent
[Slide 5] So you remember that when we are discussing in the class in the lecture that the this is the bead and the DNA were decorated around it and then they were diffusing out the question we ask that how long does it take for these beads-- for the proton to come to the sensor surface and these were sort of the symbols right these-- sort of the spikes were sort of-- the signal reaching the the sensor surface so we did a we calculation I said numerically If you have a certain size and these are experimental data the symbols and the continuous line They will match very very well but of course I didn't really solve the problem exactly its a three-dimensional problem there's this sphere and all sorts of complications so you can do it numerically that's not a problem So-- what I want to show you here is that you can get the essence that why that why there is this particular shape Right? A peak and then things go down by solving a
[Slide 6] much simpler problem. And the much simpler problem has to do it-- Just use-- inject-- the-- DNA or-- the proton at a given point and then you solve for the diffusion and see how the molecule essentially arrived at the sensor surface so the solution was already given and the solution has this particular form now why this
[Slide 7] solution has this two components you have this this component why and these two components very briefly, so you could just take the solution and do it but the way the solution was done that assume that this is your sensor and this is the point where the blue--is where you inject Biomolecules and then they diffuse but if you say and that this doesn't absorb anything, this biomolecule-- could just go and hit the sensor surface and turn back then the net flux through this red line will be 0 because nothing is going in and this is a difficult problem to solve except that once you realize that you can in fact remove the sensor completely. Put a image source on the other side and then solve the whole thing because in the infinite media the solution is known this is the solution solution is known for the simply add these two solutions and from-- therefore from 0 to all positive x if you add the two solutions they will always be correct so this is just a mathematical trick to solve the diffusion equation but the bottom line is you can see how it works so this is
[Slide 8] the concentration-- concentration of the biomolecule so you inject something here and we did .1 R squared over D some time constant depend R is from sensor to the injection point then you with the that peak which was the delta function has gone down and there is a small amount of-- a tiny amount of biomolecules are protons which have reached the sensor surface as time goes on, these are time snap shots you see that the peak will gradually go down and some of the molecules which have sort of they'll be more and more molecules reaching on the other side after-- random work and the concentration of particles will go up and eventually what will happen that if you to wait longer than the peak is completely go away the concentration will saturate if you wait even farther long that entire-- this whole concentration will go down as everything will be leaving through the-- through the other side through the open door sort of
[Slide 9] So that is what is being shown for example at a later time .5 the whole curve is going down and as you can see that this magnitude continue to go down as as a function of time
[Slide 10] And so that you can immediately see that why the curve was up like this, that because this is the way in the particles start arriving on the surface and why it goes down because eventually everything will leave because this door is open so even when you put the pulse here in the beginning a fraction of them will go to the other side but after getting reflected the whole thing-- will gradually gradually get out to this explain the shape that we saw
[Slide 11] in the Ion-torrent curve, right? And you can do the other extreme also you can assume that every particle that comes there on the sensor surface is captured by the surface-- molecules, that is just another extreme
[Slide 12] in that case all you have to do is make sure that these two things come with the negative sign so that the concentration in the middle always stay equal to 0 and instead of cosine hyperbolic that for now you have sine hyperbolic
[Slide 13] Again you can do this so whole thing the same except the only difference you see in the sum concentration that this concentration is always 0 because it is sort of removing from the mobile population everything that comes there like a perfect blue it is removing all the biomolecules and so therefore this concentration is always 0 and as the whole peak is gradually going down large fraction of the biomolecule are getting captured and of course the fraction of the amount escaping to the other side and so gradually the peak is going down
[Slide 14] and of course if you run it for a little bit longer time .2 .5, .25, .3, then gradually you can see the pulse fail pulse will go away and again once again you will see the very similar
[Slide 15] type of time response there'll be a peak at R square over 60 and gradually it will decay, so in practice, in the iron to iron case there was a the genome sequencer case it was sort of in-between molecules are getting reflected but at the same time some fraction of it were being bond by the OH bonds, you may remember from the ph sequencers and so therefore it has a response which is somewhere in in between so that's it so in this-- homework therefore we have seen a couple of things the innergetics of the binding is something that you need to go to the table a little carefully things would be become clear space selectivity again something, then solve one or two problems carefully that works out and this genome sequence that problem where did we integrate everything is actually very easy we can take example type problems just like the one-- the two problems i showed and you can essentially understand most of the essential features let it be simply okay that's it thanks