Question

Problem #1

1. A) A closed box has 5 molecules and is then partitioned instantaneously into 4 equally sized compartments. What is the probability of there being 4 particles in the top left compartment?

2. B) Now imagine the same experiment as part A, but with 3 million molecules. What is the range of molecules that are in each partition 95% of the time?Hint: ±2 standard deviations is 95% of the normal distribution.

Problem #2

The game of UC poker is similar to poker where players are dealt 5 cards at random, except now from a 56 card deck instead of a 52 card deck. The extra kind of card is the Chancellor (C) card, which is between the King (K) and Ace (A). So the ordering of the face cards are 2-10, J, Q, K, C, A (with four possible suits for each card). We are interested in calculating the probability of various winning combinations. To do this, we must calculate the number of ways we can be dealt a particular hand divided by the total number of possible ways we can be dealt 5 cards from the deck of 56.

1. A) What is the denominator? Hint: This could be thought of as calculating the multiplicity (W) of a protein with 56 binding sites and 5 molecules binding to it.

2. B) What is the probability of a four of a kind (meaning having a hand that has all four suits of a particular card)?

3. C) What is the probability of a full house (three of a kind and two of a kind)?

Problem #3

A protein has 12 phosphorylation sites that are equally likely to be phosphorylated. There are 4 specific sites that if are all phosphorylated, lead to protein activity. If we have a population of the protein with 6 randomly phosphorylated sites, what percentage of these proteins that are activated?

Problem #4

Imagine a protein exists in 4 conformations where the energies are each separated by 1RT (you can assume here ∆U = ∆H). When crystallized the protein moves completely to the lowest energy state, calculate the ∆G of crystallization from one mole of hypothetical protein at equilibrium. Hint: First calculate change in ∆H and ∆S separately.

Answer 1

Ans 1/ part - A

Since each particle can go to any one of the 4 compartments, there are 4 ways in which a particle can go to any one of the 4 compartments. Hence there are $4^{5}$ ways in which 5 molecules can be placed in the 4 compartments.

Number of ways in which 4 molecules out of 5 can go to the top left compartment is $_{4}^{5}\textrm{C}$

Now the remaining 1 particle is to be placed in remaining 3 compartments and this can be done in $3^{1}$ ways.

Hence the total number of favourable cases = $_{4}^{5}\textrm{C}\times 3^{1}$

Required probability $=\frac{_{4}^{5}\textrm{C}\times 3^{1}}{4^{5}}$

Ans 2A

Since one hand consists of 5 cards the exhaustive number of cases is $_{5}^{56}\textrm{C}$

Ans 2B

The number of ways in which 4 cards of a particular number can come $=_{4}^{4}\textrm{C}$

The number of ways in which balance 5-4 = 1 card can come in one hand out of balance of 56-4=52 cards is $_{1}^{52}\textrm{C} = 52$

Since there are 14 cards of different numbers and 4 cards of a particular number can come by the principle of counting the total number of favourable cases of getting 4 cards of a particular number

$= _{4}^{4}\textrm{C}\times 52\times 14 =1\times 52\times 14$

Required probability

$= \frac{1\times 52\times 14}{_{5}^{56}\textrm{C}}$

Ans 2C

To get 3 cards of same kind by the principle of counting is

$= _{3}^{4}\textrm{C}\times 13 =4\times 13 =52$

To get 2 cards of another number (apart from the above number as there is only 4 cards of a particular number) is

$= _{2}^{4}\textrm{C}\times 12 =6\times 12 =72$

Required probability

$= \frac{52\times72}{_{5}^{56}\textrm{C}}$

Ans 3C

The required percentage is

$= \frac{_{4}^{4}\textrm{C}\times_{2}^{8}\textrm{C}}{_{6}^{12}\textrm{C}}\times 100$

Ans 4

OUT OF SUBJECT