MCB 102 Yildiz Problem Set 1 solutions

MCB 102 Spring 2024 Yildiz Problem Set 1 Solutions 1. If the increase in entropy indicates the direction of spontaneous change, how can a system ever undergo a process that results in a decrease in the entropy of the system? The only way for a system to decrease its entropy spontaneously is to couple to a surrounding that is increasing its entropy. The surrounding must provide an external entropy increase that ensures that the total entropy of the system plus the surrounding increases. 2. Two sets of molecules are mixed into a system at time zero. After 10 seconds, equilibrium has been reached. At what time was the entropy of the system maximal? 10 seconds (equilibrium) 3. In a hypothetical protein, a histidine sidechain is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine such that the value of ∆G 0 for deprotonation is 15 kJ•mol –1 at pH 7.0 and 293 K (calculated using the biochemical standard state). What is the pKa of this histidine at equilibrium? Hint: The standard state for [࠵? ! ] is taken as 10 -7 M. ࠵?࠵?࠵? ! ⇄ ࠵?࠵?࠵? + ࠵? ! [His]/[His+] = exp(–∆G 0 /RT) = exp(–15,000/(293 × 8.31)) = 0.002118 pKa = pH – log([His]/[His+]) = 7 – log 10 (0.002118) = 9.67 4. Baker’s yeast, Saccharomyces cerevisiae , is responsible for beer, bread, and many of the most important discoveries in molecular genetics and cell biology. a. This nifty organism is spherical and 4.0 μm in diameter. What is the volume of a single S. cerevisiae cell, in units of liters? r = D/2 = 2.0 x 10 -6 m ࠵? = " # ࠵?࠵? 3 = 33 x 10 -18 m 3 Since 1 L = 10 -3 m 3 V = 33 x 10 -15 L b. When S. cerevisiae is growing on glucose, hexokinase isoenzyme 2 (hxk2) plays a key role in metabolism by phosphorylating glucose. There are an estimated 114,000

copies of hxk2 per yeast cell, making it one of the most abundant proteins in yeast. What is the molar concentration of hexokinase? [࠵?࠵?࠵?2] = ࠵? ࠵? ࠵? $ = 1.14 ࠵? 10 % ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? 33 x 10 &’% L 6.02 ࠵? 10 (# ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?/࠵?࠵?࠵?࠵? = 5.6 ࠵?࠵? where N is the number of molecules in a cell, V is the volume of a cell, and N A is the Avogadro number. In practice, the concentration in the cytosol is somewhat higher than this because of the substantial volume taken up by the vacuole and nucleus, which needs to be subtracted from the total cell volume to get the volume of cytosol. 5. The intracellular pH of a yeast cell is maintained in part by phosphate ion buffering. Phosphoric acid has three protons and therefore three pKa values, 2.1, 7.2, and 12.7. At neutral pH = 7.0, how much of each species (H 3 PO 4 , H 2 PO 4- , HPO 4 2- , and PO 4 3- ) is present? Ignore trace amounts of less than 1 percent. The pKa values of 2.1 and 12.7 are so far from 7.0, so amounts of H 3 PO 4 and PO 4 3- will be far less than 0.01. Using the Henderson-Hasselbalch equation ࠵?࠵? = ࠵?࠵?࠵? + ࠵?࠵?࠵? [HPO " (& ] [H ( PO " & ] 7.0 = 7.2 + ࠵?࠵?࠵? [HPO " (& ] [H ( PO " & ] [HPO " (& ] [H ( PO " & ] = 10 &).( = 0.63 On a percent basis, [HPO 4 2- ] = 100 x (.63/1.63) = 39%, so [H 2 PO 4 - ] = 61%. We are just below the pKa, so it makes sense that there is a little more of protonated form. 6. A little bit north of here, in Napa Valley, S. cerevisiae is busy fermenting some of the world’s best wine. The pH of the medium drops as fermentation proceeds. The pH of the cell can also eventually drop. a) If the pH drops to 6.5, now how much of each phosphate species is present? The pKa values of 2.1 and 12.7 are still so far from 6.5, so amounts of H 3 PO 4 and PO 4 3- will be far less than 0.01. Using the Henderson-Hasselbach equation ࠵?࠵? = ࠵?࠵?࠵? + ࠵?࠵?࠵? [HPO " (& ] [H ( PO " & ]

Acetate ion is the conjugate base of acetic acid. Therefore, acetate ions react in equilibrium reaction with water and form hydroxide ions and acetic acid. Most acetate ions stay in the acetate form. CH 3 COO − + H 2 O ⇌ CH 3 COOH + OH - K eq = [OH - ][CH 3 COOH] [CH 3 COO - ] = ࠵? - If we multiply both sides of the equation with [H + ], ࠵? - = [OH - ][H + ][CH 3 COOH] [CH 3 COO - ] [H + ] [OH - ][H + ] = K w = 10 &’" [CH 3 COOH] [CH 3 COO - ] [H + ] = 1 ࠵? . ࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? Therefore, ࠵? - = K w 0 " = ’) #$% ’.+" 2 ’) #& = 0.57 ࠵? 10 &3 This is a very low equilibrium constant. Therefore, acetate ion concentrations remains nearly constant: [CH 3 COO - ] ≈ 0. 1 M The reaction produces the same concentration of [ OH - ] and [CH 3 COOH]: [ OH - ] = [CH 3 COOH] K b = [ OH - ] ! 0.1 = 0.57 x 10 -9 X[࠵?࠵?−] ( = √0.57 ࠵? 10 &’) [ OH - ] = 0.75 ࠵? 10 &% [OH - ][H + ] = 10 &’" [H + ] = 10 &’" 0.75 ࠵? 10 &% = 1.3 ࠵? 10 &3 ࠵?࠵? = −࠵?࠵?࠵? [H + ] = 8.88