nanoHUB-U Principles of Nanobiosensors/Tutorial 4.1:: Selectivity - Homework Solutions
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[Slide 1] Welcome back. Today we'll be talking about the homework solution you know that we have covered a lot of material in this segment that we have discussed for example various types of selectivity issues in particular three types the selectivity right? incorrect binding for example and then there was this random sequential absorption where the molecule-- the parasitic molecules were sort of landing in random spaces and not really binding to the target receptor molecules but sort of parasitically getting attached to the sensor surface and that created a parasitic signals false positives that we didn't want and of course there was this noise issue that we discussed. So in this homework we'll be talking about a few problems on the selectivity and then there were two or three lectures on genome sequencing. I will run a few things by you so once you have done this homework if you have done it carefully you should have a very deep understanding about various processes involved in selectivity--
[Slide 2]selectivity issues. So the first problem problem 5.1 had to do it had with the
[inaudible] of binding the bottom line-- you understand the problem, right? The bottom line was that if two DNA were perfectly conjugate well they should bind very strongly and that is what you want but it turns out many cases even if there is a parasitic molecule which should not have bound ideally still binds and gives a false parasitic signal so this is what we're trying to understand that how strong are those-- can we increase the temperature of the solution a little bit so that all the parasitic I molecules that got bound can be taken away-- sort of washed away so that's what we're trying to understand. So let's see what the difference difference in binding energy between matched and mismatched DNA.
[Slide 3] So we just gave a specific sequence you're supposed to use a particular tool; which calculates the corresponding binding energy and then essentially you were asked to calculate the corresponding energy now this tool actually implements the same equations, the small analytical equations that I discussed in the class and so therefore although this tool looks like a black box. Actually the equations were already discussed in terms of the melting temperature that you may have-- may remember from the lectures. So therefore you shouldn't really be afraid to ask questions about these-- the results of these 00:03:13,470 --> 00:03:17,260 from using this simulation software so let me
[Slide 4] let me show you some-- show you some examples. For example I gave you two sequences the first sequences this is the receptor molecule, you see starting from five prime, then you have a 22 base pair in here ending with three prime. What does it mean? First of all this is easy right, ATCG so I have a particular sequence this is a-- let's say a potentiometric sensor surface I have a molecule sort of attached and standing straight like that on the sensors surface, so this is the 22. Now what is this fight prime, well if you look at a DNA molecule and I just 00:04:00,170 --> 00:04:03,430 copied it from the web that you will see one end of it-- inside everything is nicely matched but there are five carbon molecules, one, two, three, four, 1,2,3,4 and 5 and this five molecules are the five carbon molecules on one end and then correspondingly in the fifth location there's next phosphate group and similarly on the other end, on the third 75 00:04:30,600 --> 00:04:33,600 position the next phosphate group
[inaudible]. So bottom line is this five prime and three prime simply say is the orientation of the molecule thats this five prime and three prime. and what we asked were their binding and energy associated with this with the another one and this is the bottom one-- this is the target molecule. Now you may be thinking that this doesn't look very complimentary after all there is this T prime, T cannot possibly conjugate with a T but you see the way you have to think about it this five prime on that top always goes three prime so you should be looking from this side onward. the five prime side should be going from this side onward and now you can see it's complimentary because A goes with T, G goes with C T goes with A, and so therefore if you just sort of flip it you will see that secondary sequence of the target sequence. is exactly complimentary this is how it is also five prime, three prime on this side and five prime goes with the three prime and then correspondingly this two chain sort of lineup. So we'll be thinking about how strong is binding is and this is the mismatched pair. So for example in the this is the same sequence as the first one because remember this top one is the receptor right and then this bottom one is the mismatched DNA that is this is parasitic DNA. Once again you can check very quickly that if you start from three prime and matched it with five prime then you will see that there is a T going with this T which will not mind and therefore you could immediately see as you walk down this sort of set of letters that it's not going to bind very strongly with the primary sequence right. So we want to know what is the difference between the target which is this sequence versus the parasitic sequence which is the false signal-- false positive signal and we want to see how strong are the-- what is the difference in the binding energy. Can we selectively bind with each other? So that's the question. So let's see, so once you have sort of put the sequences in the simulator
[Slide 5] and run it, you'll get some numbers and the numbers oh by the way I just want to emphasize that the goal is that if the binding is perfect then there'll be very strong bonds between the receptor and the target and with the imperfect binding there'll be many missing bonds and then therefore this would be a weaker bond and so in a droplet if we increase the temperature by a transistor let's say these ones will debound and will be washed away only the proper signal with remain that's how we enhance selectivity, right? So let's see how the two
[Slide 6] sequences that we talked about, how strong are the binding. So there are-- these you have to look at a little bit carefully, so for example once you simulate the structure you'll get a whole set of numbers and what I-- want to want to-- you to walk you-- I want you-- I want to walk you through the results a little bit. Let's first read this line, it say's delta G is 35.09 kCal per mole. You know mole is Avogadro's number of particles so it is saying that this is the total binding energy of that many molecules. So first of all you know that it one kCal is 4.2 KJ and a hundred kilo jewels per mole is approximately 1 eV per bonded by molecule and one eV is 40 kT so therefore you could immediately convert this number by just using these three sets into the binding energy in terms of KT you can do that now take a look remember that five prime goes with the three prime and you can see that all 22 bounds are now completely bond together and this solid line says that these are strong bonds and the bonds are-- consecutive bonds are all essentially has been aligned they are properly bonded so it's a very strong 35.09 kCal for perfect matching. but the interesting thing is that even when things are perfectly matched there may be many cases where they will not line up properly they will but align very weakly. For example take a look at two, remember this whole page is for perfect binding but perfect binding is not-- although the molecules themselves can in principle bind properly. which is this case there will be many cases where they don't bind properly even in-- even though in principle they could. look at these two in this case delta G is this minuscule and 1.95 kCal per mole and only two base pairs have on property they could in principle bind 22 places but statistically when the molecules are coming in, in contact with each other they can incorrectly bind while these 22 are not bound for example take a look what has happened, that the bottom molecule the target molecule has sort of come and has gotten shifted so there is two places C going with G A going with T the neighbors are bount there are a few other places given by his double darts four other places where they are bound but their neighbors are not. So they can be very weak bonds. and in that case you can see there's only two base pair that have bound strongly neighboring base pairs and the binding energy is very very small and you would say there are whole source of combinations. So therefore two, two in here two, for various types of combinations going through the whole stack in each case you have about one kCal per mole. Very weak bonds. So although in the case in this case we have perfect binding you see but in principle you will see you have a lots of cases where statistically the binding is not so strong so you can count 1,2,3, 4,5,6, 12, 14 15,16 said there is about 18 in here one in 18 cases the binding would be very good this is the challenge, that even when the sequences are complimentary it is difficult to make sure that align properly and they are bound strongly there are a few cases-- in fact, two cases where you have three binding instead of
[inaudible] binding so we have 22 binding, three binding, two cases, and two binding the remaining cases. So only the one of them is what we are really after. Okay. Now this is for the case of perfect match, 00:12:15,110 --> 00:12:20,149
[Slide 7] what about imperfect match? Well for the case of imperfect match again you can go-- remember there's the second sequence with the parasitic molecule this time things are a little bit better because you can see that in the imperfect match although it's imperfect but there are at least five places like the neighbours still find within the long sequence there's good binding there a set of about 6 where the binding is relatively weak and on average you have for the five you have about 6.71 kCal per mole approximately one or slightly more than one kCall per mole and then there are a whole set of spaces where there binding is possible two bindings are possible you can see these two vertical lines-- these two vertical lines. and all these are two bindings and sometimes you may have-- may have-- so this large number of two bindings, one case of three binding and one case of five binding. So tell me this, what will be the difference in the selectivity? In the previous case the best was 22, right. In here also the best is about 5. So I have to make sure that I raise the temperature of the solution after the bindings are done enough so that at least these five bindings, cases are detached. Of course, if I can make sure five binding cases are detached, that means in the previous slide All these ones will also get detached, all the free ones will also get detached. Only thing that will remain is that 22. So therefore these types of calculations are very helpful you see. So that you can decide ahead of time that what type of temperature do you need in order for-- to have high selectivity.
[Slide 8] Okay, Now the next problem we'll be talking about is-- has to do it this space-based selectivity problem you remember that the molecules can random the-- random places not covered by the receptor and you can have a selectivity problem. So let me-- so this is already specified-- the problem is specified, the solution is given,
[Slide 9] but let me show you what we are really after here. So, this is a-- this is a sensor surface we're looking from the top and what you see here in this blue, is initially these were all sort of the receptors the empty blue circles were receptors sitting there on their own. How many receptors? You can count, 1,2,3,4,5,6,,7,8,9,10. So there were ten receptors within this sensor surface, waiting for the target biomolecules to come; and you incubate it for a certain period of time. the target molecules have indeed come and it has gone through-- it has occupied 1,2,3,4,5,6,7,8. Eight out of the ten is now occupied, and what would happen that rest would be occupied by in this place is this red, are the parasitic molecules. The parasitic molecules of course cannot-- would not fit with the-- with the receptor itself so they are sitting in empty places. So let's first say how many solid blue are the target molecule on have we captured. That will be given by Eta divided by pi r squared. Eta is the fraction-- fraction per centimeter squared. The fraction of the area covered by the-- the target molecule and divided by pi r naught squared. R naught is the radius of this target. Now the remaining place is now available for the biomolecules to sit. and in that case it'll be eta times infinity. What is eta times infinity? eta times infinity is when the entire surface is covered then the fraction of area that will be covered by this biomolecule and in this case you know that answer is .54 eta times infinity is about .54. and but of that .54 blue has already taken-- eta has already taken a certain amount of space the remaining space could be filled by the parasitic molecule pi r squared. that gives me how many what is the potential number of parasitic molecules that can sit here. All right, so it's very easy then, so what you need to do is-- for example in eta here, for eta you substitute this quantity in here solve for it and it gives you number of molecules that-- blue molecules that were captured. Divided by number of parasitic molecules that have been captured; and of course this is related to signal to noise ratio and signal to noise ratio of course-- this will-- this is the charge of each target molecule, let's say. each DNA carries 10 charge so sigma mu would be ten for example, and sigma TP is the charge carried by the parasitic molecule. Often it may carry a small amount of charge let's say one. Sigma P would be one, KT is the rate at which things bind and KP is the rate at which the red molecules bind they come and sit on the sensors surface and then of course there is this initial density of the target molecules. The higher the density, the more easier it is for the molecules to sort of, settle on the sensors surface and a naught divided mp, that's a naught over mp. So if you do this substitute this quantity then you can immediately solve for this value n naught and then n naught-- you see everything inside here is null so for example we want to signal to noise ratio of one hundred so s amount of 100 will go here and the charges will go here the KT and KP these things are known and rho T is already given, target density of target molecule. Rho P is the density of the parasitic molecules these are all given, radius unknown, and so you can calculate N naught and the N naught value that you calculate in this case is 5.6 ten to the power ten centimeter squared. So N naught is essentiality the number of places the receptors can essentially come and sit.
[Slide 10] Okay and if you do the corresponding simulation, you do the corresponding simulation, you will see that you would get, you'd get about the about the same number, you can see five times this was done by this analytical calculation 5.6 ten power ten and in this case you see for a signal to noise ratio of 100 you get about the same number and the important thing is that the higher the number better is our ability to selectively bind to this surface-- signal to noise ratio would be much better now one thing I want to emphasize that this N naught essentially involves all these red plus these blue together, right. So in some places only NP is coming in, and in some places only the target molecules are coming in. so the theta is the fraction that is taken up by target molecule and the reds ones are the places that were taken up by taken up by the parasitic molecule.
[Slide 11] Okay, now it is always better that if you had molecules which are our parasitic molecules bigger size because if it has a bigger size then it will not be able to sort of bind as easily and so that is having these-- this parasitic molecule which is a biggest size enhances the signal to noise ratio and that is given by the next problem that you may have solved for the selectivity problem. This is of course using the biosensor lab, Biosensor lab two.