Exam (elaborations) TEST BANK FOR Molecular Driving Forces Statistical Thermodynamics in Biology, Chemistry, Physics, and Nanoscience 2nd Edition By Bromberg

Exam (elaborations) TEST BANK FOR Molecular Driving Forces Statistical Thermodynamics in Biology, Chemistry, Physics, and Nanoscience 2nd Edition By Bromberg, Dill, Stigter (Solution Manual) Chapter 1 Principles of Probability 1. Combining independent probabilities. You have applied to three medical schools: University of California at San Francisco (UCSF), Duluth School of Mines (DSM), and Harvard (H). You guess that the probabilities you’ll be accepted are: p(UCSF) = 0.10, p(DSM) = 0.30, and p(H) = 0.50. Assume that the acceptance events are independent. (a) What is the probability that you get in somewhere (at least one acceptance)? (b) What is the probability that you will be accepted by both Harvard and Duluth? (a) The simplest way to solve this problem is to recall that when probabilities are independent, and you want the probability of events A and B, you can multiply them. When events are mutually exclusive and you want the probability of events A or B, you can add the probabilities. Therefore we try to structure the problem into an and and or problem. We want the probability of getting into H or DSM or UCSF. But this doesn’t help because these events are not mutually exclusive (mutually exclusive means that if one happens, the other cannot happen). So we try again. The probability of acceptance somewhere, P(a), is P(a) = 1 − P(r), where P(r) is the probability that you’re rejected everywhere. (You’re either accepted somewhere or you’re not.) But this probability can be put in the above terms. P(r) = the probability that you’re rejected at H and at DSM and at UCSF. These events are independent, so we have the answer. The probability of rejection at H is p(rH) = 1 − 0.5 = 0.5. Rejection at DSM is p(rDSM) = 1 − 0.3 = 0.7. Rejection at UCSF is p(rUCSF) = 1 − 0.1 = 0.9. Therefore P(r) = (0.5)(0.7)(0.9) = 0.315. Therefore the probability of at least one acceptance = P(a) = 1 − P(r) = 0.685. 1 (b) The simple answer is that this is the intersection of two independent events: p(aH)p(aDSM) = (0.50)(0.30) = 0.15. A more mechanical approach to either part (a) or this part is to write out all the possible circumstances. Rejection and acceptance at H are mutually exclusive. Their probabilities add to one. The same for the other two schools. Therefore all possible circumstances are taken into account by adding the mutually exclusive events together, and multiplying independent events: [p(aH) + p(rH)][p(aDSM) + p(rDSM)][p(aUCSF) + p(rUCSF)] = 1, or equivalently, = p(aH)p(aDSM)p(aUCSF) + p(aH)p(aDSM)p(rUCSF) +p(aH)p(rDSM)p(aUCSF) + · · · where the first term is the probability of acceptance at all 3, the second term represents acceptance at H and DSM but rejection at UCSF, the third term represents acceptance at H and UCSF but rejection at DSM, etc. Each of these events is mutually exclusive with respect to each other; therefore they are all added. Each individual term represents independent events of, for example, aH and aDSM and aUCSF. Therefore it is simple to read off the answer in this problem: we want aH and aDSM, but notice we don’t care about UCSF. This probability is p(aH)p(aDSM) = p(aH)p(aDSM)[p(aUCSF) + p(rUCSF)] = (0.50)(0.30) = 0.15. Note that we could have solved part (a) the same way; it would have required adding up all the appropriate possible mutually exclusive events. You can check that it gives the same answer as above (but notice how much more tedious it is). 2 2. Probabilities of sequences. Assume that the four bases A, C, T, and G occur with equal likelihood in a DNA sequence of nine monomers. (a) What is the probability of finding the sequence AAATCGAGT through random chance? (b) What is the probability of finding the sequence AAAAAAAAA through random chance? (c) What is the probability of finding any sequence that has four A’s, two T’s, two G’s, and one C, such as that in (a)? (a) Each base occurs with probability 1/4. The probability of an A in position 1 is 1/4, of A in position 2 is 1/4, of A in position 3 is 1/4, of T in position 4 is 1/4, and so on. There are 9 bases. The probability of this specific sequence is (1/4)9 = 3.8 × 10−6. (b) Same answer as (a) above. (c) Each specific sequence has the probability given above, but in this case there are many possible sequences which satisfy the requirement that we have 4 A’s, 2 T’s, 2 G’s, and 1 C. How many are there? We start as we have done before, by assuming all nine objects are distinguishable. There are 9! arrangements of nine distinguishable objects in a linear sequence. (The first one can be in any of nine places, the second in any of the remaining eight places, and so on.) But we can’t distinguish the four A’s, so we have overcounted by a factor of 4!, and must divide this out. We can’t distinguish the two T’s, so we have overcounted by 2!, and must also divide this out. And so on. So the probability of having this composition is " 9! 4!2!2!1! #  1 4 9 = 0.014. 3 3. The probability of a sequence (given a composition). A scientist has constructed a secret peptide to carry a message. You know only the composition of the peptide, which is six amino acids long. It contains one serine S, one threonine T, one cysteine C, one arginine R, and two glutamates E. What is the probability that the sequence SECRET will occur by chance? The S could be in any one of the 6 positions with equal likelihood. The probability that it is in position 1 is (1/6). Given that S is in the first position, we have 2 E0s which could occur in any of the remaining 5 positions. The probability that one of them is in position 2 is (2/5). Given those two letters in position, the probability that the 1 C is in the next of the 4 remaining positions is (1/4). The probability for the R is (1/3). For the remaining E is (1/2), and for the last T is (1/1), so the probability is (1/6)(2/5)(1/4)(1/3)(1/2) = 1/360 = " 6! 1!2!1!1! # −1 . 4. Combining independent probabilities. You have a fair six-sided die. You want to roll it enough times to ensure that a 2 occurs at least once. What number of rolls k is required to ensure that the probability is at least 2/3 that at least one 2 will appear? q = 5 6 = probability that a 2 does not appear on that roll. qk = probability that a 2 does not appear on k INDEPENDENT rolls. P(k) = 1 − qk = probability that at least one 2 appears on k rolls. For P(k)  2 3 , 1 − qk  2 3 =) qk  1 3 =) k ln q  ln  1 3  =) k  ln(1/3) ln(5/6) = 6.03 Approximately six or more rolls will ensure with probability P  2/3 that a 2 will appear. 4 5. Predicting compositions of independent events. Suppose you roll a die three times. (a) What is the probability of getting a total of two 5’s from all three rolls of the dice? (b) What is the probability of getting a total of at least two 5’s from all three rolls of the die? The probability of getting x 5’s on n rolls of the dice is  1 6 x  5 6 n−x n! x!(n − x)! Note that this is a “2-outcome” problem (getting a 5 or not getting a 5). It is not a “6-outcome” problem. (a) So the probability of two 5’s on three dice rolls is  1 6 2  5 6 1 3! 2!1! =  1 36   5 6  3 = 15 216 = 6.94 × 10−2 (b) The probability of getting at least two 5’s is the probability of getting two 5’s or three 5’s. Since these two situations are mutually exclusive, we seek p(two 5’s) + p(three 5’s) =  1 6 2  5 6  3! 2!1! +  1 6 3  5 6 0 3! 3!0! = 15 216 + 1 216 = 16 216 = 7.41 × 10−2 5 6. Computing a mean and variance. Consider the probability distribution p(x) = axn, 0  x  1, for a positive integer n. (a) Derive an expression for the constant a, to normalize p(x). (b) Compute the average hxi as a function of n. (c) Compute 2 = hx2i − hxi2 as a function of n. (a) Z 1 0 p(x) dx = 1 =) Z 1 0 axn dx = axn+1 n + 1  1 0 = a n + 1 = 1 =) a = n + 1. (b) hxi = Z 1 0 xp(x) dx = Z 1 0 (n + 1)xn+1 dx = (n + 1)xn+2 n + 2 ! 1 0 = n + 1 n + 2 . (c) hx2i = Z 1 0 x2p(x) dx = (n + 1) Z 1 0 xn+2 dx = (n + 1) xn+3 n + 3 ! 1 0 = n + 1 n + 3 . So 2 = hx2i − hxi2 =  n + 1 n + 3  −  n + 1 n + 2 2 . 7. Computing the average of a probability distribution. 6 Compute the average hii for the probability distribution function shown in the figure below. 0 1 2 3 4 i P(i) 0.0 0.2 0.4 0.3 0.1 A simple probability distribution. hii = X4 i=0 ip(i) = 0(0.0) + 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 3 8. Predicting coincidence. Your statistical mechanics class has twenty-five students. What is the probability that at least two classmates have the same birthday? If you first find the probability, q, that no two students have the same birthday, then the quantity you want is p(2 students have same birthday) = 1 − q The probability that a second student does not have the same birthday as the first is (364/365). The probability that the third student has a birthday different than either of the first two is (363/365), and so on. It is like a sequence problem in which each possible 7 birthday is one card drawn out of a barrel. The probability that no two people have the same birthday, out of m people, is: q =  364 365   363 365   362 365  · · · 365 − (m − 1) 365 ! In factorial notation, q = N! (N − m)!Nm where N = 365. (Incidentally, this expression is identical to the expression for excluded volume in the Flory–Huggins model of polymer solutions (see Chapter 31)). Using Stirling’s approximation x!  (x/e)x, we get q = (N/e)N  N−m e N−m Nm Collecting together terms in e and dividing the numerator and denominator by NN gives q = e−m  1 − m N N−m Substituting m = 25 students and N = 365 gives q = 0.4163, so p = 1 − q = 0.5837 There is a better than 50% chance two students will have the same birthday! 8 9. The distribution of scores on dice. Suppose that you have n dice, each a different color, all unbiased and six-sided. (a) If you roll them all at once, how many distinguishable outcomes are there? (b) Given two distinguishable dice, what is the most probable sum of their face values on a given throw of the pair? (That is, which sum between two and twelve has the greatest number of different ways of occurring?) (c) What is the probability of the most probable sum? (a) 6 on one die 6 × 6 on two dice ... 6n on n dice. (b) Number of ways a sum can occur most probable sum 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 10 11 12 (1,1) (1,2) £ 2 (1,6) £ 2 (3,4) £ 2 (2,3) £ 2 (2,5) £ 2 (1,3) £ 2 (1,4) £ 2 (2,2) sum of 2 dice When dice show different numbers, there is a degeneracy of two. When each of the dice has the same number, the degeneracy equals one. (c) probability of 7 = p(7) = number of ways of getting 7 total number of ways of all outcomes p(7) = 6 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 1 6 9 10. The probabilities of identical sequences of amino acids. You are comparing protein amino acid sequences for homology. You have a twenty-letter alphabet (twenty different amino acids). Each sequence is a string n letters in length. You have one test sequence and s different data base sequences. You may find any one of the twenty different amino acids at any position in the sequence, independent of what you find at any other position. Let p represent the probability that there will be a ‘match’ at a given position in the two sequences. (a) In terms of s, p, and n, how many of the s sequences will be perfect matches (identical residues at every position)? (b) How many of the s comparisons (of the test sequence against each database sequence) will have exactly one mismatch at any position in the sequences? (a) For comparing one sequence, each position assumed independent, the probability of a perfect match of all n residues is pn = (number of matched seqs/number of total seqs) =) number of matches in s sequences = spn. (b) n − 1 positions match, so the probability is pn−1; one position doesn’t match which has the probability (1 − p); and there are n different positions at which the mismatch could occur, therefore = spn−1(1 − p)n Note in general, for k matches: (1) P(k) = spk(1 − p)n−k n! k!(n − k)! . 10 11. The combinatorics of disulde bond formation. A protein may contain several cysteines, which may pair together to form disulfide bonds as shown in the figure below. If there is an even number n of cysteines, n/2 disulfide bonds can form. How many different disulfide pairing arrangements are possible? 1 3 4 2 5 6 This disulfide bonding configuration with pairs 1-6, 2-5, and 3-4 is one of the many possible pairings. Count all the possible pairing arrangements. Number the individual sulfhydryl groups along the chain. The first sulfhydryl along the sequence can bond to any of the other n − 1. This removes two sulfhydryls from consideration. The third sulfhydryl can then bond to any of the remaining n − 3. Four sulfhydryls are now removed from consideration. The fifth can now bond to any of the remaining n − 5 sulfhydryls, etc., until all n/2 bonds are formed. Thus the total possible number of arrangements of disulfide bonds is a product of n/2 terms: D(n) = (n − 1)(n − 3)(n − 5) · · ·1 Another approach gives an expression that is easier to calculate. Consider placing the sulfhydryls in