# 2 statistical bioresearch problems

#### Mandokir

##### New Member
Hello guys,

I require your help in my research project.

Problem 1:
We are filtering a suspension of 1ml containing 1250 cells, through a micro-filter containing 1500 wells with pores. This is 80% filter saturation.

If filtration were random, on avarage how manny wells would contain no cells? how manny would contain 1 cell? how manny would contain 2 cells? and 3? and more than 3?

How would this be distributed with 50% filter saturation (750 cells per ml)?

Problem 2:
If we have a 1ml of a suspension containing 1250 cells and a random distribution. How big is the chance that somewhere in that solution an boxed area exists where 2 cells are within a vollume of 1.4*10^6 square micrometers (1.4 nanoliters).

How big would this chance be for a suspension of 750 cells/ml

As i dont have much statistical background i cannot solve these problems, and i have no way of knowing the difficulty of these problems. Maby someone can solve them?

Good luck.

#### chiro

##### New Member
Hey Mandokir.

In a random process, no previous value or previous history will affect the next value. So if you want to look at the probability of any one thing being blocked or passing through, it will be exactly the same as all the others under the random assumption and if you have a probability for one "thing" going through (a batch of stuff or one indivisible "cell" or something like that) then the probability of k going through given that n were attempted to go through has a binomial distribution with P(X = k).

#### Mandokir

##### New Member

Im not sure wether i formulated my first problem correctly, what i want to know is what on avarage the random distribution is of lets say 80 apples being thrown at an area containing 100 containers, each apple must land in a container. So?
How manny containers will on avarage contain 0 apples?
How manny contain 1?
2?
etc?

How can i calculate this?

#### BGM

##### TS Contributor
I have been thinking your problem for a while.

First, if you assuming cell filtration are independent to each other, and each well is equally likely to contain a cell, then the number of cells in each wells
$$(x_1, x_2, ..., x_k)$$
jointly follows a multinomial distribution, see:

http://en.wikipedia.org/wiki/Multinomial_distribution

where n representing the number of cells and k representing the number of wells.

By equally likely assumption, the product can be simplified as

$$p_1^{x_1} \ldots p_k^{x_k} = \frac {1} {k^n}$$

Now you are interested in the expectation of the number of x_i equal to 0, 1, 2, ...

Theoretically it is possible to enumerate all outcomes and do the calculations, but it is quite tedious when n is large (may have some better counting methods, not sure)

I can do a very simple example: consider distributing 2 cells into 3 wells.

You have 3 scenarios (1,1,0), (1,0,1) and (0,1,1) ending up with 1 0s, 2 1s and 0 2s.
Each with probability 2/9

You have 3 scenarios (0,0,2), (0,2,0) and (2,0,0) ending up with 2 0s, 0 1s and 1 2s.
Each with probability 1/9

Therefore, the expectation of the number of wells with
0 cells = 1*2/3 + 2*1/3 = 4/3
1 cells = 2*2/3 + 0*1/3 = 4/3
2 cells = 0*2/3 + 1*1/3 = 1/3

See if it is what you want.

#### Mandokir

##### New Member

I think that solves problem 1, i will take some time to look at it carefully later.
Essentialy what we need to know is wether after a filtration the distribution is random, or not.

I will take some time to rephrase problem two.

Essentialy though that comes down to:

when given a number of random moving objects, in a given space.
how big is the chance that, at anny one time, two or more objects are within a given proximity of one another?

greetings

#### BGM

##### TS Contributor
A very simple assumption is that each objects movement is independent of each other and they are uniformly distributed in the entire space. Then you are just computing the probability that the distance between a particular pair of objects is less than a certain threshold.

Of course a somewhat more sophisticated model could be using Brownian motion to model the paths, and maybe they are allowed to be dependent.

#### Mandokir

##### New Member
do you know of any programm that could do this for me?

#### Mandokir

##### New Member
Hello BGM,

Based on the article you linked i made the following calculation for 5 cells in 10 wells.

1111100000 -> (10!/(5!*5!))*((5!/1!)) = 30240
2111000000 -> (10!/(6!*3!*1!))*((5!/2!)) = 50400
2210000000 -> (10!/(7!*2!))*((5!/2!*2!)) = 10800
3110000000 -> (10!/(7!*2!))*((5!/3!)) = 7200
3200000000 -> (10!/(8!))*((5!/3!*2!)) = 900
4100000000 -> (10!/(8!))*((5!/4!)) = 450
5000000000 -> (10!/(9!))*((5!/5!)) = 10
Total: 100.000 different distributions

0 cells: 30240*5/10 + 50400*6/10 + 10800*7/10 + 7200*7/10 + 900*8/10 + 450*8/10 + 10*9/10 = 59049/100000 = 59.049%
1cell: 30240*5/10 + 50400*3/10 + 10800*1/10 + 7200*2/10 + 450*1/10 = 32805/100000 = 32.805%
2cells: 50400*1/10 + 10800*2/10 + 900*1/10 = 7290/100000 = 7.290%
3cells: 7200*1/10 + 900*1/10 = 810/100000 = 0.81%
4cells: 450*1/10 = 45/100000 = 0.045%
5cells: 10*1/10 = 1/100000 = 0.001%
On average 59.049% of wells will contain 0 cells, 32.805% of all wells will contain 1 cell, etc.
This means that: 32.805/(32.805+7.29+0.81+0.045+0.001) = 80.108% of all wells that contain cells, contain 1 cell.

--

Assuming this is correct, how do i rewrite the outcome that 32.805% of wells contain 1 cell, to: % of cells singly in a well compared to the total amount of cells filtered?

Greetings.

#### BGM

##### TS Contributor
Not sure if I understand the terminologies correctly.

total amount of cells filtered - is it just the total number of cells? (5 in your example)

Suppose you have calculated that on average 32.805% of the wells (out of 10) contain a single cell only (I haven't check any calculations). Since for each singly-filtered wells contain a single cell only, the number of singly-filtered cells is just equal to the number of the singly-filtered wells, and thus on average the number is 32.805% * 10, and the proportion of cells which is singly-filtered is 32.805% * 10 / 6.

#### Mandokir

##### New Member
Not sure if I understand the terminologies correctly.

total amount of cells filtered - is it just the total number of cells? (5 in your example)

Suppose you have calculated that on average 32.805% of the wells (out of 10) contain a single cell only (I haven't check any calculations). Since for each singly-filtered wells contain a single cell only, the number of singly-filtered cells is just equal to the number of the singly-filtered wells, and thus on average the number is 32.805% * 10, and the proportion of cells which is singly-filtered is 32.805% * 10 / 6.
The total number of cells filterd is indeed 5 in this example. I want to know how i can calculate how many of those 5 cells have entered a single well.

32.805% is the amount of wells containing a single cell.
However becouse we filtered 5 cells 32.805%*5 would mean that 1.64025 cells have been singly separated, however this is incorrect. As the calculations state that most of the cells have been singly separated.

Greetings

#### BGM

##### TS Contributor
The figure "32.805%" is out of 10 wells - it should mean on average there are 3.2805 wells out of 10 wells contains a single cell only.

This should equally means that on average 3.2805 cells are singly separated, and therefore 3.2805/5 of filtrated cells are singly separated. (Should be divided by 5, divided by 6 is a typo in the previous post)

#### Mandokir

##### New Member
The figure "32.805%" is out of 10 wells - it should mean on average there are 3.2805 wells out of 10 wells contains a single cell only.

This should equally means that on average 3.2805 cells are singly separated, and therefore 3.2805/5 of filtrated cells are singly separated. (Should be divided by 5, divided by 6 is a typo in the previous post)
I agree, thank you.

I was in doubt becouse i reasoned that the remaining cells that are not separated are divided among wells in diffrent numbers and therefor skew the result.