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Quiz 5 for Physics 176: Answers Professor Greenside The following equations may be useful: E = - ln( Z ) ∂β , Z N = Z N 1 N ! , Z 1 = V Z int V Q , V Q = h 2 πmkT 3 . (1) E = X s E ( s ) p s , p s = e - β [ E ( s ) - μ A N A ( s ) - μ B N B ( s )] Z , Z = X s e - β [ E ( s ) - μ A N A ( s ) - μ B N B ( s )] . (2) 1. (10 points) A certain olfactory receptor in your nose has three “slots” such that one slot can bind a single molecular odorant of type A with energy E A < 0 while each of the other two slots can independently bind a single molecular odorant of type B , with binding energy E B < 0. (The energy of the receptor is zero if nothing binds to it.) The receptor is in contact with a reservoir (the air inside your nose) whose temperature is T and whose chemical potentials for molecules A and B are μ A and μ B respectively, which you can assume to be constant. Write down an expression for the probability for an A odorant and a single B odorant to be bound to this receptor. Answer: The probability is given by 2 e - β ( E A + E B - μ A - μ B ) 1 + e - β ( E A - μ A ) + 2 e - β ( E B - μ B ) + 2 e - β ( E A + E B - μ A - μ B ) + e - β (2 E B - 2 μ B ) + e - β ( E A +2 E B - μ A - 2 μ B ) . (3) The first step in calculating any partition function is to identify the states s of the system. Here there are eight: (a) No molecules are bound to the receptor so E = 0 and N A = N B = 0. (b) An A molecule is bound to the receptor so E = E A , N A = 1, and N B = 0. (c) A B molecule is bound to the receptor so E = E B , N A = 0, and N B = 1. The degeneracy for this state is 2 since there are two separate slots that each can bind a B molecule. (d) An A molecule and a B molecule are bound to the receptor so E = E A + E B and N A = N B = 1. This state has a degeneracy of 2 since the B molecule can bind to either of the two B sites. (e) Two B molecules are bound to the receptor so E = 2 E B , N A = 0, and N B = 2. (f) Finally, an A molecule and two B molecules can bind to the receptor so E = E A + 2 E B , N A = 1, and N B = 2. Once the states s have been enumerated and the corresponding energies and particle numbers identified, the answer Eq. (3) is found by direct evaluation of the Gibb’s sum in Eq. (2) by evaluating successive Gibb’s factors in turn, using the known values of E s , N A ( s ), and N B ( s ). A few clever students realized that, because the three binding sites are independent and distinguishable, the partition function for the three binding sites factors like this: Z ABB = Z A Z 2 B (4) = 1 + e - β ( E a - μ A ) · ‡ 1 + e - β ( E B - μ B ) · 2 (5) = 1 + e - β ( E A - μ A ) + 2 e - β ( E B - μ B ) + 2 e - β ( E A + E B - μ A - μ B ) + e - β (2 E B - 2 μ B ) + e - β ( E A - μ A +2 E B - 2 μ B ) , (6) which is the same as the grand partition function Z in the denominator of Eq. (3). 1

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Although Eq. (3) is a formal answer, it doesn’t give much insight about how the probability depends on the air in the nose that the receptor is exposed to. Assuming that molecules of a certain kind (e.g., odorant A) form an ideal gas with partial pressure P so that PV = NkT where V is the volume of the gas (volume of nasal cavity) and N is the number of molecules of a particular type, we can use Eq. (6.93) on page 255 of Schroeder to relate the chemical potential of the molecule to the partial
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