Practice Midterm 1 Solutions - p S aLUTIbNS YOUR NAME HERE...

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Unformatted text preview: . p S aLUTIbNS YOUR NAME HERE October 4, 2006 Chem 120B Midterm $1 I Definitions and Useful Formulas: Boltzmann distribution: Partition function: Average energy: (E) = — (alanflQ)NV 1 * 6<E> ~ <(6E)2> “(TUW‘ kw Noninteracting, indistinguishable molecules: Heat capacity: Equipartition principle: n (E) = EkBT, where n is the number of quadratic contributions to the energy. Boltzmann’s definition of entropy: S 2 kB In W For an open cylinder with radius 'r and length L: volume = 7rr2L, surface area = 27rrL Some properties of logarithms and exponentials: lnm+lny=lnzcy, alnmzlnxa eaeb = ea+b, This exam consists of 2 questions (both with several parts) on 8 pages. All concern a double—stranded nucleic acid. A small portion of its chemical structure is shown below. ' 1. The double helix structure is stabilized by binding of bases along one molecule to complementary bases along the other molecule. The picture below highlights this , base—pair binding. \—__\,_—_/ \— _J bound base pairs unbound base pairs (zero energy, multiplicity=1) (energy = 6 per base pair, multiplicity = wo per base pair) In this problem we will assume that each pair of aligned bases is statistically inde— pendent. When a pair is unbound, its energy is higher by an amount 6, but it can access wo times more microscopic states than when it is bound. 10 pts (i) Calculate the partition function for a single pair of bases. w" ”a! :x ”m -___,. M ”M fifidil‘flfl (,deczwmilj , (3 *‘ ’, 10 pts (ii) Calculate the partition function for a sequence of N bases. Note that different ' base pairs can be distinguished by their position in the sequence. Q t: q "l (FAWMWMWMWMW ‘ “:0 . {a M ’ ‘1 Cl ‘l’ l3ch 3 NW4) w 1. » w~ —« w « ; g " “T "J 6.“ i l/c {Ll/«5‘5 ‘5 “M71? “‘" "’ M" g: 5’) 3 it ”1% timing-J sen“ <1: P; W‘Cf N/z (iii) From the partition function you calculated in part (ii), calculate the average 10 pts. base—pairing energy for a sequence of N bases as a function of temperature. - _ amt ' <E> i: <56 >«,M ,A c ii Mild l M056“) W M, v 9 crrnrwmw ,r,.,,r-fl,,.,...¢w. r G if 1 («we g a . T1) < E > "S: flip—rm n, J m (l 2, \ I 1+ f: ”,3, 10 pts 2. Under most conditions phosphate groups along a nucleic acid backbone are nega— tively charged. In this problem you will consider the double helix to be a rigid rod of length L, with evenly spaced negative charges: The counterions that balance this negative charge are present in the surrounding so- lution. Each positive ion is attracted to the nucleic acid through a potential energy (which is the sum of attractions to all phosphate groups). As shown above, 1‘ is the distance from the ion to the rod; 0 and T0 are positive constants. (i) Consider a single ion at a fixed distance 7' from the rod. With this distance fixed, the ion is confined to the surface of a cylinder with radius 'r' and length L. What is the number of microscopic states, W(r), available for a particle on the surface of this cylinder? (Only the dependence on r is important. You can neglect all other V factors.) 10 pts (ii) Calculate the probability P(r) of finding an ion at a distance 7' from the rod. Do not worry about interactions between ions, and do not attempt to normalize the distribution. Write your answer in terms of 7', temperature T (or inverse tempera- ture ,6), and the constants a and r0. ,EgL/kflf) ‘PW may “(5 affix (VA/{1}) a; are/3 10 pts (iii) Your result from part (ii) should depend sensitively on the quantity a/kBT. Sketch the distribution P(7‘) below for a/kBT > 1 (for example, ,60 = 3) and for a/kBT < 1 (for example, flcr = 0). I ‘ ”,7 f , ~ , X I a "2 V, ' , PO”) at: 7r" Mime» x l 700,)0L If ‘WIMJU Q<X<( (l A BG>1 BO'<1 5'30"“) P(r) P(r) 10 pts (iv) When ions are highly concentrated around the nucleic acid, it is reasonable to replace the potential u('r) with an effective quadratic energy 1 1 ueg(r) = 51973 = 5k (9:2 +312), Where k: is another positive constant. Use the equipartition theorem to estimate the heat capacity due to motion of a single ion in this situation. Note that motion in ' the z—direction is unrestricted. Explain your reasoning. (Hint: When counting an ion’s degrees of freedom, use Cartesian coordinates.) Wile. am, 1W“; Ifeflbwt, 51.5] H 2 K5 4‘ PE’ if: (P; + Pf + P23 + é: Ia. (wail Vie. Cara/x S; W LL {WW onL. (5‘ S77 ’ (W175 Ai‘Ejmfl’L’Sfi FA ”LL l/wtcil Hquk ) WC 3” <5? : 5111‘ (v) When ions are only weakly attracted to the nucleic acid, the attractive potential u(r) is negligible, so that Uefi‘ (r) = 0. According to the equipartition theorem, What is the heat capacity due to motion of a single ion in this situation? Explain your reasoning. M. W) '20) H4 iCW m” Mi) Mw__wumm_._m__ : ’ V ” {‘3 n { 0. KM) L2 giver) Jim. Scam. $64956. M W'Cj rte n :2 » (flax-3:2} \i _ ) 7‘ a, \( , AZ kl; 10 pts (Vi) In the graph below sketch the heat capacity due to motion of a single ion as a function of temperature. Your plot should reflect your results from parts (iv) and (v), as well as your understanding of heat capacity at low temperature. Mark the temperatures T = O and T = a/kB on the horizontal axis. 10 pts (Vii) Consider N cations Whose attraction to a nucleic acid effectively confines them within a cylinder of radius R and length L. If the nucleic acid binds to a positively charged surface (such as a phospholipid membrane), these ions are “released” into the surrounding solution, whose total volume is V: * * Calculate the change in entropy when this occurs, using the following approxima— tions: Initially, the system consists of N noninteracting particles in a cylindrical “box”, with an effective potential that is zero inside the cylinder and infinite outside the cylinder. Similarly, the final state corresponds to a system of N noninteracting particles in a box with volume V. .1” W» N»! ...
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