Biology, University Course Notes

Random Walks Class Notes

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Lecture 1

1D Random Walk

1D Random Walk

  • $\delta =$ distance unit
  • $\tau =$ time unit
  • Unbiased random walk: $\mathrm{Prob(left \quad step)} = \mathrm{Prob(right \quad step)} = 0.5$
  • Biased random walk: $\mathrm{Prob(left \quad step)} \neq \mathrm{Prob(right \quad step)}$

Applications of Randomness in Biology

Macromolecule structures

  • Entropy = the number of ways someting can be arranged
  • Equilibrium systems vs. nonequilibrium systems
  • Order of magnitude estimates ⇒ sanity checks
  • Force-extension experiment ⇒ entropy vs. energy

Chemotaxis = movement in response to chemicals

  • E coli example ⇒ flagella rotation
    • Counterclockwise spin ⇒ propel straight
    • Clockwise spin ⇒ “tumble”

Diffusion – biased vs. unbiased

Molecular motors

  • Kinesin on microtubules
  • Polymerase on DNA (during DNA replication)
  • Ribosome on RNA (during RNA synthesis)

Ion channels – amount of time spent open vs. closed

Noisy cellular reaction networks

  • Bursting ⇒ off state vs. on state
  • Time-averaging ⇒ protein translation/degradation
  • Propogation ⇒ gene activation

Reaction-diffusion and biological pattern formation (e.g., B-Z reactions)

Viral evolution – high mutation rate (compared to higher-order organisms)

Extinction

  • Extinction vs. reintroduction of disease in a local population
  • Critical community size = population size needed to sustain infection without exinction
    • Small local populations ⇒ more stochastic

Lecture 2

Review of the average, variance, and standard deviation

Consider the following set of data points: 1, 1, 2, 3, 4, 5, 5, 5, 6, 7, 7, 8

  • Here, the number of data points is $N=12$
  • Average = point that data points are centered about
    • The mean average is defined as follows:

$\boxed{\mu = \left< x \right> = \frac{1}{N} \sum _{i=1}^{N}{ x_i}}$

      • For our $N=12$ data points, $\left< x \right> = 4.5$
    • Alternatively, we can find $\left< x \right>$ using the probabilities associated with each data point, $P(x_i)$
      • For our $N=12$ data points…
        • There are two data points (the first two listed) that have a value of 1
          • ⇒ $P(x_1) = P(x_2) = 2/12$
        • There is one data point (the third listed) with a value of 2
          • ⇒ $P(x_3) = 1/12$
        • There is one data point (the fourth listed) with a value of 3
          • ⇒ $P(x_4) = 1/12$
        • There is one data point (the fifth listed) with a value of 4
          • ⇒ $P(x_5) = 1/12$
        • There are three data points (the sixth, seventh, and eighth listed) with a value of 5
          • ⇒ $P(x_6) = P(x_7) = P(x_8) = 3/12$
        • There is one data point (the ninth listed) with a value of 6
          • ⇒ $P(x_9) = 1/12$
        • There are two data points (the tenth and eleventh listed) with a value of 7
          • ⇒ $P(x_10) = P(x_11) = 2/12$
        • There is one data point (the twelfth listed) with a value of 8
          • ⇒ $P(x_12) = 1/12$
      • We can use this information and to determine $\left< x \right>$ using the following equation:
        • $\boxed{\left< x \right> = \sum _{i=1}^{N}{ x_i \cdot P(x_i)}}$
        • For our specific example, we would write,
          • $\left< x \right> = 1(\frac{2}{12}) + 2(\frac{1}{12}) + 3(\frac{1}{12}) + 4(\frac{1}{12}) + 5(\frac{3}{12}) + 6(\frac{1}{12}) + 7(\frac{2}{12} + 8(\frac{1}{12}) = 4.5$

Variance = measures how far the data points tend to be spread about the mean

  • Mathematical expression:

$\boxed{\sigma ^2 = \left< (x – \mu)^2 \right> = \left< x^2 – 2x\mu + \mu ^2 \right> = \left<x^2 \right> – 2\mu \left< x \right> + \mu ^2}$

  • Recall: $\mu = \left< x \right>$
    • $\sigma ^2 = \left<x^2 \right> – 2 \left< x \right> \left< x \right> + \left< x \right> ^2 = \left<x^2 \right> – 2 \left< x \right> ^2 + \left< x \right> ^2 = $\left< x^2 \right> – (2-1) \left< x \right> ^2 = $\left< x^2 \right> –  \left< x \right> ^2 $
    • Thus, $\boxed{\sigma ^2 = \left< x^2 \right> –  \left< x \right> ^2}$ (or,  $\sigma ^2 = \left< x^2 \right> –  \mu ^2$)
      • IMPORTANTLY, $\left< x^2 \right>$ is DISTINCT from $\left< x \right> ^2$
        • $\left< x^2 \right>$ denotes the average of a set of data points that have been squared from their original value
        • $\left< x \right> ^2$ denotes squaring the average of a set of data points
  • Lets compute the variance for our $N=12$ data set we defined earlier: 1, 1, 2, 3, 4, 5, 5, 5, 6, 7, 7, 8
    • First, we need to find $\left< x^2 \right>$:
      • Let $m = 8$ denote the number of possible values a data point may obtain
      • The mathematical expression for $\left< x^2 \right>$ is…
        • $\left< x^2 \right> = \sum _{j=1}^{m}{ x_{j}^2 \cdot P(x_j)} = \frac{1}{N} \sum _{i=1}^{N}{ x_{i}^2}$
        • Thus, for our example
          • $\left< x^2 \right> = 1^2 (\frac{2}{12}) + 2^2 (\frac{1}{12}) + 3^2 (\frac{1}{12}) + 4^2 (\frac{1}{12}) + 5^2 (\frac{3}{12}) + 6^2 (\frac{1}{12}) + 7(\frac{2}{12}) + 8(\frac{1}{12}) \approx 25.333$
    • Now, we can compute the variance for our data set:
      • $\sigma ^2 = \left< x^2 \right> – \left< x \right> ^2 \approx 25.333 – (4.5)^2 \approx 5.0833$
  • Important notes on the variance formula:
    • The variance formula contains a squared exponent to account for any possible negative data values ⇒ prevents data points from cancelling eachother out
      • Although an operator such as the absolute value initially seems like an appropriate way to describe the variance of a data set, it does’nt always provide “good” answers (i.e., it can cause problems with certain data sets that squaring doesn’t create)
  • Any measure of variance should have squared units with respect to the units used in the data set we are taking the variance of
    • E.g., if our example data set has units in $\mathrm{mm}$, the units of the variance will be $\mathrm{mm^2}$
    • However, interpreting values that has squared units of what we were originally measuring can be counter-intuitive
      • So, we usually calculate something called the standard deviation to get a better intuition when interpreting the variance of a data set
        • Equation: $\boxed{\sigma = \sqrt{\sigma ^2} = \sqrt{\left< x^2 \right> – \left< x \right>^2}}$
        • For our example data set, $sigma = \sqrt{5.08333 \mathrm{mm^2}} \approx 2.2546 \mathrm{mm}$ (units aren’t squared!)
      • Take-home-message: the standard deviation is better for gaining intuition about the spread of data than variance because it gives a value in units rather than units squared

Review of probabilities

Suppose $A$ and $B$ are two independent events (i.e., the chances of event $B$ occuring are independent of whether event $A$ occurs and vice-versa)

Some notations:

  • Let $S$ denote the all possible events that can occur in some sample space
  • $P(A \cap B)=$ the probability of event $A$ AND event $B$ occuring
    • Ven Diagram (shaded region represents the case where both events occur): Venn Diagram AND
    • We can find the probability of multiple independent events occuring by multipliying the probabilities of each individual event together
      • For our events $A$ and $B$, $\boxed{P(A \cap B)= P(A)P(B)}$
    • Example: the probability of flipping two heads in a row if we flip a coin twice
      • Let event $A$ be the event where the first coin toss is heads
        • For an unbiased coin, $P(A) = 0.5$
      • Let event $B$ be the event where the second coin toss is heads
        • For an unbiased coin, $P(B) = 0.5$
      • Since events $A$ and $B$ are independent of one another (our outcome of the second coin toss is independent of the outcome of our first coin toss and vice-versa), we can use our expression for $P(A \cap B)$ to determine the probability of getting two heads:
        • $P(A \cap B) = P(A)P(B) = 0.5 \cdot 0.5 = 0.25$
        • Thus, there is a 25% chance of getting two heads after flipping an unbiased coin twice
  • $P(A \cup B)=$ the probability of event $A$ OR event $B$ occuring
    • Ven Diagram (shaded region represents cases where at least one of our two events of interest occur): Venn Diagram OR
    • We can find the probability of at least one of these two events occuring by summing the individual probabilities and subtracting any double counting
      • For our events $A$ and $B$, $\boxed P(A \cup B) = P(A) + P(B) – P(A \cap B)$
    • Example: if we roll two dice, what is the probability that at least one of die giving a even number?
      • Let event $A$ be the event that the first die gives an even number
        • $P(A) = 0.5$ is the probability of event $A$ occuring
      • Let event $B$ be the event that the sescond die gives an even number
        • $P(B) = 0.5$ is the probability of event $B$ occuring
      • Note: if we just added $P(A)$ and $P(B)$ together to determine $P(A \cup B)$, we would be double-counting the probability associated with both events $A$ and $B$ occuring ⇒ need to subtract $P(A \cap B)$ from our summed probabilities
        • $P(A \cup B) = P(A) + P(B) – P(A \cap B)$
          • First determine $P(A \cap B)$:
            • $P(A) = 0.5$
            • $P(B) = 0.5$
            • $P(A \cup B) = P(A)P(B) = 0.5(0.5) = 0.25$
          • $P(A \cup B) = 0.5 + 0.5 – 0.25 = 0.75$
            • Thus, there is a 75% chance of rolling at least one even number if we roll two die
  • $P(\mathrm{neither} A \mathrm{nor} B) = P(\mathrm{not} A) \cdot P(\mathrm{not} B)$
    • Venn Diagram (shaded region represents all possible events besides events $A$ or $B$): Venn Diagram NOR
    • Note: $P(A \cup B) = 1 – P(\mathrm{neither} A \mathrm{nor} B)$
      • Often times, it is convienent to find the probability of something by considering the probability that the “opposite” of what you are trying to find and subtracting that probability from 1
      • Example: use this strategy to compute the probability of rolling at least one even die if we roll two dice
        • The “opposite” of rolling at least one even number is to roll two odd numbers
          • Let $A$ denote the event where the first die gives an odd number
            • $P(A) = 0.5$
          • Let $B$ denote the event where the second die gives an odd number
            • $P(B) = 0.5$
        • Subtracting the product of the probabilities of our “opposite” events from 1 will give us the total probability of at least one “non-opposite” event occuring
          • $P(\mathrm{at \\ least \\ one \\ even} = 1 – P(A)P(B) = 1 – 0.5(0.5) = 1 – 0.25 = 0.75$
            • 75% chance of rolling at least one even number (agrees with the probability we found for this earlier!)

Sources

  • Dr. Leah Shaw – BIOL 356: Random Walks in Biology, Lecture 1 (College of William and Mary)

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