A viral linear DNA molecule of length, say, 1 is often known...

A viral linear DNA molecule of length, say, 1 is often known to contain a certain "marked position," with the exact location of this mark being unknown. One approach to locating the marked position is to cut the molecule by agents that break it at points chosen according to a Poisson process with rate $\lambda$. It is then possible to determine the fragment that contains the marked position. For instance, letting $m$ denote the location on the line of the marked position, then if $L_{1}$ denotes the last Poisson event time before $m$ (or 0 if there are no Poisson events in $[0, m])$, and $R_{1}$ denotes the first Poisson event time after $m$ (or $I$ if there are no Poisson events in $[m, 1])$, then it would be learned that the marked position lies between $L_{1}$ and $R_{1} .$ Find (a) $P\left\{L_{1}=0\right]$ (b) $P\left\{L_{1}<x\right\}, 0<x<m$ (c) $P\left[R_{1}=1\right\}$ (d) $P\left(R_{1}>x\right\}, m<x<1$ By repeating the preceding process on identical copies of the DNA molecule, we are able to zero in on the location of the marked position. If the cutting procedure is utilized on $n$ identical copies of the molecule, yielding the data $L_{i}, R_{i}, i=$ $1, \ldots, n$, then it follows that the marked position lies between $L$ and $R$, where $$ L=\max _{i} L_{i}, \quad R=\min _{i} R_{i} $$ (e) Find $E[R-L]$, and in doing so, show that $E[R-L] \sim \frac{2}{n \lambda}$.

A viral linear DNA molecule of length, say, 1 is often known to contain a certain "marked position," with the exact location of this mark being unknown. One approach to locating the marked position is to cut the molecule by agents that break it at points chosen according to a Poisson process with rate $\lambda$. It is then possible to determine the fragment that contains the marked position. For instance, letting $m$ denote the location on the line of the marked position, then if $L_{1}$ denotes the last Poisson event time before $m$ (or 0 if there are no Poisson events in $[0, m])$, and $R_{1}$ denotes the first Poisson event time after $m$ (or $I$ if there are no Poisson events in $[m, 1])$, then it would be learned that the marked position lies between $L_{1}$ and $R_{1} .$ Find
(a) $P\left\{L_{1}=0\right]$
(b) $P\left\{L_{1}<x\right\}, 0<x<m$
(c) $P\left[R_{1}=1\right\}$
(d) $P\left(R_{1}>x\right\}, m<x<1$
By repeating the preceding process on identical copies of the DNA molecule, we are able to zero in on the location of the marked position. If the cutting procedure is utilized on $n$ identical copies of the molecule, yielding the data $L_{i}, R_{i}, i=$ $1, \ldots, n$, then it follows that the marked position lies between $L$ and $R$, where
$$
L=\max _{i} L_{i}, \quad R=\min _{i} R_{i}
$$
(e) Find $E[R-L]$, and in doing so, show that $E[R-L] \sim \frac{2}{n \lambda}$.

Key Concepts

Order Statistics

Order statistics involve the analysis of the minimum and maximum values drawn from a sample of random variables. In this problem, combining multiple DNA molecules leads to using the minimum of the right-side cut positions and the maximum of the left-side cut positions to narrow down the region containing the marked site, thereby harnessing properties of order statistics to estimate the uncertainty.

Independent Increments

The property of independent increments means that the numbers of events occurring in disjoint intervals are independent random variables. This is essential in the analysis since it allows one to compute probabilities for one segment of the DNA without considering events in another, simplifying calculations of events before and after the marked position.

Fragment Length Estimation

The method of estimating the length of the DNA fragment containing the marked position involves calculating the expected difference between the minimum and maximum cuts across multiple samples. This estimation benefits from the properties of the Poisson process and order statistics, ultimately showing that the expected fragment length shrinks inversely with the number of samples and the cutting rate.

Poisson Process

A Poisson process is a stochastic process used to model a sequence of events occurring randomly over time or space. In this context, it describes the random positions along the DNA molecule where cuts occur, with the number of cuts in any interval following a Poisson distribution. Its key properties include independent increments and the fact that the number of events in disjoint intervals is independent.

Exponential Distribution

The time or distance between consecutive events in a Poisson process follows an exponential distribution. This is crucial for determining probabilities such as the occurrence of the last cut before a marked position or the first cut after it, reflecting the memoryless property of the exponential distribution.

Transcript

00:02 Okay, this is a problem looking at a restriction map of a segment of dna.

00:08 And i've drawn it out here, and it shows you a segment of dna.

00:13 The numbers underneath represent base pairs.

00:17 So from this mark, from position 1 to position 2, it's 5 ,000 base pairs.

00:24 From position 2 to position 3 ,000, from 3 to 4 ,000 is 4 ,000.

00:30 Of five is 2 ,000 and from 5 to 6 is another 5 ,000 base pairs.

00:37 At positions 1, 3, and 5, we see the word echo, eco, and that is representative of the echor -1 restriction enzyme recognition site.

00:50 So this means that at position 1, at position 3, and in position 5, the enzyme would be able to cut at those places because their recognition sites are present.

01:04 Pst represents the recognition site for a restriction enzyme pst1.

01:10 Those recognition sites are at position 2, position 4, and position 6.

01:16 Finally, the other piece of information we need to know is that our probe, which is represented here in blue, would overlap with this piece of dna.

01:26 So it overlaps starting here and ending over here.

01:31 So now we're going to look at different patients ' dna, and we're going to cut them with certain enzymes, and then try to find the pieces that will anneal or hybridize to our probe.

01:45 So in the first scenario, we have a patient one.

01:50 Patient one has all six restriction enzyme sites.

01:56 And they're saying that you are digesting this dna with the enzyme pst.

02:03 So we're going to digest it at site 2, site 4, and site 6.

02:10 Now we add our probe, and the question is, what size pieces will anneal to our probe? so since we're going to have the dna cut at position 6 and position 4, this will be, this fractal.

02:26 Will be 7 ,000 base pairs.

02:29 Since it overlaps with the probe, 7 ,000 base pairs are going to be one of the pieces of dna we find.

02:39 And we're going to cut from position 4 to position 2, and we're going to create another 7 ,000 base pair piece.

02:49 And it too will anneal to the probe because there is overlaps.

02:53 So for this patient, we will get two pieces that are approximately 7 ,000 base pairs that will anneal to the probe.

03:02 The next question asked about an individual patient two, and patient two has a mutation that eliminates position four.

03:12 So they do not have the recognition sequence for this enzyme pst.

03:16 All right.

03:17 What happens if we digest with pst now? well, we don't have this site any longer.

03:23 So we will cut at position two and we will cut at position six.

03:28 So that means we will create one piece of dna, this length, which will be 3 ,000 plus 4 ,000 plus 2 ,000 plus 5 ,000, which is 14 ,000 base pairs...

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