Bio/statistics Handout 5 How matrix products arise

· Genetics: Suppose that a given stretch of DNA coding for cellular product is very mutable, so that there are some number, N, of possible sequences that can appear in any given individual (this is called ‘polymorphism’) in the population. To elaborate, a strand of DNA is a molecule that appears as a string of small, standard molecules that are bound end to end. Each of these standard building blocks can be one of four, labeled C, G, A and T. The order in which they appear along the strand determines any resulting cellular product that the given part of the DNA molecule might produce. For example, AAGCTA may code for a different product than GCTTAA.

As it turns out, there are stretches of DNA where the code can be changed without damage to the individual. What with inheriting genes from both parents, random mutations over the generations can then result in a population where the codes on the given stretch of DNA vary from individual to individual. In this situation, the gene is called ‘polymorphic’. Suppose that there are N different possible codes for a given stretch. For example, if one is looking at just one particular site along a particular DNA strand, there could be at most N = 4 possibilities at that site, namely C, G, A or T. Looking at two sites gives N = 4´4 = 16 possibilities.

Lets label the possible sequences by integers starting from 1 and going to N. At any given generation t, let p_j(t) denote the frequencey of the j’th sequence appearing in the population at that generation. Given that the j’th sequence appears in an individual, there is some probability, P(i|j), that the i’th sequence appears in their heir. Here, let us assume that the population is constant and that these P(i|j) are independent of the generation number. (This is all very simplistic except in very special circumstances.) In this situation, we can say the following: The probability of any given sequence, say i, appearing in generation t+1 can be written as a sum:

p_i(t+1) = P(i|1) p₁(t) + P(i|2) p₂(t) + · · · + P(i|N) p_N(t)

(5.1)

This is to say that the probability of sequence i appearing in an heir is equal to the probability that sequence 1 appears in the parent times the probability of a mutation that changes sequence 1 to sequence i, plus the probability that sequence 2 appears in the parent times the probability of a mutation that changes sequence 2 to sequence i, etc.

Now,we can write this using our matrix formalism by thinking of the numbers {p_j(t)}_1≤j≤N as defining a column vector, (t), in R^N, and likewise the numbers {p_i(t+1)}_1≤i≤N as defining a second column vector, (t+1) in R^N. If I introduce the N´N matrix A whose entry in the j’th column and i’th row is P(i|j), then (5.1) says in very cryptic shorthand:

(t+1) = A(t) .

(5.2)

Now, we can sample the population at time t = T = now, and thus determine (T), or at least the proxy that takes p_i(T) to be the percent of people in the population today that have sequence i. One very interesting question is to determine (T´) at some point far in the past, thus T´ << T. For example, if we find (T´) such that all p_i(T´) are zero but a very few, this then indicates that the population at time T´ was extremely homogeneous, and thus presumably very small.

To determine (T´), we use (5.2) in an iterated form:

(T) = A(T-1) = AA(T-2)) = AAA(T-3) = ··· = A···A(T´) ,

(5.3)

where the final term has T-T´ copies of A multiplying one after the other.

On a similar vein, we can use (5.2) to predict the distribution of the sequences in the population at any time T´ > T by iterating it to read

(T´) = A···A(T)

(5.4)

Here, the multiplication is by T´-T successive copies of the matrix A.

By the way, here is a bit of notation: The sequence {(t)}_t=0,1,… of vectors of probabilities is an example of a Markov chain. In general, a Markov chain is a sequence of probabilities, {P(0), P(1), P(2), …, } where the N’th probability P(N) depends only on the probabilities with numbers that are less than N.

· How bacteria find food: If you put certain sorts of bacteria on one side of a petri dish and put some sugar some distance away on the other, the bacteria will migrate towards the sugar. Apparently, the sugar diffuses in the petri dish, so that there is slight concentrations everywhere, with most of the concentration in the original spot. The bacteria are sensitive to the different levels at their front and rear ends and tend to move in the direction of the greater level. At the expense of borrowing from Math 21a, the bacteria sense the direction of the gradient of the sugar concentration. Even so, their movement from low to high concentration has a certain randomness to it; at any given step there is some probability that they will move in the wrong direction.

What follows is a simplistic model for this: Imagine bacteria moving in steps along the x-axis along the segment that stretches from x = 1 to x = N. Here, N is some large integer. Suppose the sugar placed initially where x = N, and the bacteria is placed initially at x = 1. Our model also supposes that the bacteria moves one unit per second, with probability q Î (0, 1) of moving to the right at any given step, and probability 1-q of moving to the left unless it is at the end of the interval. If it is at the x = 1 end, it moves to the right with probability q and stays put with probability 1-q. If it is at the x = N end, it stays put with probability q and moves to the left with probability 1-q. In our model, q is independent of position, x Î {1, . . . , N}. Such would be roughly the case in the real petri dish where the sugar concentration gradient is relatively constant. We should take q > in the case that the bacteria is attracted to the sugar.

For each time step t, and j Î {1, . . . , N}, let p_j(t) denote the probability that the bacteria is at position j at time t. For example, p₁(0) = 1 and p_j(0) = 0 for j > 1. Our model then says that the probability of finding the bacteria at position j ≠ 1 of N at time step t > 0 is equal to the probability that it is at position j-1 at time t-1 times the probability that it moves one step to the right, plus the probability that it is at position j+1 at time t-1 and moves one step to the left. Thus,

p_j(t+1) = q p_j-1(t) + (1-q) p_j+1(t) when 2 ≤ j ≤ N-1

(5.5)

By the same token,

p₁(t+1) = (1-q) p₁(t) + (1-q) p₂(t) and p_N(t+1) = q p_N-1(t) + q p_N(t) .

(5.6)

Let us introduce the vector (t) in R^N whose j’th component is p_j(t). Then (5.5) and (5.6) assert that (t+1) is obtained from (t) by the action of a linear transformation: (t+1) = A(t), where A is the N´N matrix whose only non-zero entries are:

A₁₁ = 1-q, A_jj-1 = q for 2 ≤ j ≤ N, A_jj+1 = 1-q for 1 ≤ j ≤ N-1, and A_NN = q

(5.7)

For example, in the case that N = 4, the matrix A is

(5.8)

In any case, we can iterate the equation (t+1) = A(t) to find that

(t) = A···A(0)

(5.9)

where A···A signifies t copies of A multiplied one after the other. By the way, a common shorthand for some n copies of any given matrix, M, successively multiplying one after the other is Mⁿ.

· Growth of nerves in a developing embryo: Here is a very topical question in the study of development: Nerve cells connect muscles and organs to the brain. When an organism develops, how do its nerves know where to connect? A given nerve cell stretches out extremely long and thin ‘appendages’ called axons. Any given axon may end abutting a muscle cell, or another nerve cell in a chain of nerves that ends in the brain. How do the axons ‘know’ where they are supposed to attach? A simple model proposes that the tip of the growing axon in an embryo is guided by chemical gradients in much the same fashion as the bacteria in the previous discussion is guided to the sugar.

· Enzyme dynamics: An enzyme is a protein molecule that facilitates a chemical reaction that would take a long time to occur otherwise. Most biological reactions are facilitated by enzymes. One typical mode of operation is for the enzyme to facilitate a reaction between molecules of type a and molecules of type b to produce a new molecule of type g. It can do this if it simulaneously attracts the a and b molecules. In doing so, the enzyme holds the two sorts near each other for enough time that they can bind together to form the more complicated g molecule. If this g molecule does noot have a strong attraction to the enzyme, it will break away and so free the enzyme to attract another a and another b to make more g.

Now, it is often the case that the enzyme only works efficently if it is folded in the right conformation. Folded in the wrong way, the parts of the molecule that attract either a or b molecules might find themselves covered by parts that are indifferent to a or b. Due to random collisions and other effects, any given molecule, and thus our enzyme, will change its fold configuration. Suppose that there is some probability, q, that the enzyme is folded correctly to attract a and b in any given unit of time, and thus probability (1-q) that it is not. Suppose in addition, that when folded correctly, the enzyme makes 2 units of g per unit of time. Meanwhile, suppose that the freely diffusing g falls apart or is degraded by other enzymes at a rate of 1 unit per unit time.

For any integer j, let p_j(t) denote the probability of finding level j of g after some t units of time. Then the scenario just outlined finds

p_j(t+1) = q p_j-1(t) + (1-q) p_j+1(t) when j ≥ 1 and p₀(t+1) = (1-q)(p₁(t) + p₀(t))

(5.10)

as long as a and b are well supplied.

Exercises:

Exercises 1-3 concern the example above where the position of the bacteria at any given time t is labeled by an integer, x(t), in the set {1, . . . , N}.

1. Suppose T is a positive integer.

a) What is the sample space for the collection of positions, {x(0), x(1), . . . , x(T)}, of the bacteria at the times t = 0, 1, …, T.

b) Fix some j Î {1, …, N} and t Î {0, 1, …, T-1} and let A denote the event that x(t+1) = j. For each i Î {1, …, N}, let A_j denote the event that x(t) = i. Explain why the equation in (5.5) has the form P(A) = P(A|A₁)P(A₁) + ··· + P(A|A_n)P(A_N).

2. Recall that events A and B are said to be independent when P(AÇB) = P(A)P(B).

a) Explain why this definition is equivalent to the assertion that the conditional probability of A happening given B is equal to the probability of A happening.

b) Suppose that k Î {2, . . . , N-1} and that t ≥ 2. Derive a relation between the respective probabilities that x(t) = k-1 and x(t) = k+1 so as to insure that the event that x(t+1) = k is independent of the event that x(t) = k-1.

3. Suppose now that q₊ and q_- are non-negative numbers whose sum is less than 1. Change the bacteria model so that if the bacteria is at position k Î {2, . . . , N-1} at time t, then it moves to position k+1 with probability q₊, to position k-1 with probability q_- and stays put with probability 1-q₊-q_-. If the bacteria is at position 1 at time t, then it stays put with probability 1-q₊ and moves to position 2 with probability q₊. If the bacteria is at position N at time t, then in stays put with probability 1-q_- and moves to position N-1 with probability q_-

a) Write down the analog of (5.5) for this model.

b) Write down the analog of the matrix in (5.8) for the N = 4 case.

c) Write down the probability that the bacteria is at position N at t = N-1 given that the bacteria starts at position 1 at time 0.

4. Write down a version of (5.10) that would hold in the case that the enzyme when folded correctly produces some L = 1, or L > 2 units of g per unit time.