Biochemistry II

30. If 0.1 M aqueous solutions of the following pairs of substances are mixed together, which pair could result in the formation of a precipitate? (a) Na 2 S + FeCl 3 (c) KCl + Al(NO 3 ) 2 (b) NaOH + KNO 3 (d) Ni(NO 3 ) 2 + Mg(NO 3 ) 2 31. What is the pH of a buffer prepared by dissolving 0.35 mol of CH 3 NH 3 Cl (methylamine hydrochloride) in 1.00 L of 1.1M CH 3 NH 2 (methylamine). Kb = 4.4×10 −4 . (a) 1.66 (b) 2.86 (c) 10.15 (d) 11.14 32. Consider a buffer prepared by dissolving 0.35 mol of CH 3 NH 3 Cl (methylamine hydrochloride) in 1.00 L of 1.1 M CH 3 NH 2 (methylamine) If 10 mL of 0.01 M HCl is added to this buffer solution, the pH of the solution will ____ slightly because the HCl reacts with the ____. (a) increase, OH− (c) decrease, CH 3 NH 2 (b) increase, CH 3 NH 2 (d) decrease, CH 3 NH 3 + 33. Which of the following substances, when added to a solution of hydrofluoric acid, could be used to prepare a buffer solution? (a) HCl (b) NaNO 3 (c) NaF (d) NaCl 34. Which of the following pairs can make a buffer? (a) NH 3 + NaAc (b) NaNO 3 + NH 4 Cl (c) HAc + NaAc (d) NaOH + NaCl 35. Determine the pH of a solution prepared by mixing 45 mL of 0.183 M KOH with 65 mL of 0.145 M HCl. (a) 1.31 (b) 2.92 (c) 0.74 (d) 1.97 36. Determine the pH of a solution prepared by mixing 50 mL of 0.125 M KOH with 50 mL of 0.125 M HCl. (a) 6.3 (b) 7.0 (c) 8.1 (d) 5.8 37. Consider an experiment where 35.0 mL of 0.175 HAc is titrated with 0.25 M NaOH. What is the pH at the equivalence point of this titration? The Ka for HAc is 1.8×10 −5 . (a) 5.12 (b) 2.87 (c) 11.13 (d) 8.88 38. 30.0 mL of 0.20 M NH 3 is titrated with 30.0 mL of 0.10 M HCl. What is the pH at this point? (Kb = 1.8×10 −5 ) (a) 9.26 (b) 4.74 (c) 8.95 (d) 9.56 39. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M KCOOH (pKa = 3.77)? 40. Calculate the pH of the 0.300 M NH 3 /0.360 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution (Ka = 5.56 × 10 −10 )? 41. 100 mL of 0.100 M HNO2 is titrated with 0.100 M NaOH. Find the pH after 25 mL, 100 mL, and 140 mL of NaOH has been added (Ka = 4.5 × 10 −4 ). 42. What is the solubility of silver chloride in g/L (Ksp = 1.6 × 10 −10 )? ANSWERS: 1b, 2a, 3c, 4d, 5b, 6c, 7c, 8d, 9d, 10c, 11d, 12d, 13b, 14a, 15b, 16b, 17a, 18d, 19c, 20c, 21b, 22b, 23a, 24d, 25c, 26a, 27d, 28d, 29b, 30a, 31d, 32c, 33c, 34c, 35d, 36b, 37d, 38a,39) 4.01 40) 9.17, 9.20 41) 2.9, 8.0, 12.2 42) 1.8 × 10−3 g/L.

BIOCHEM | 2501 | Biochemistry II | exam | 30. If 0.1 M aqueous solutions

Mod_2_Learning_Obj..docx.docx-Module 2 Learning Objecve CHAPTER 8

Module 2 Review Acvity Names__________________________________ You are given an unknown culture of bacteria and want to ﬁnd out more about it. You test the ability of it to grow on glucose, lactose, and sucrose by inoculang sugar broth tubes and incubang them at diﬀerent temperatures both with oxygen and without oxygen. See the results below. 4°C aerobic 4°C anaerobic 30°C aerobic 30°C anaerobic 50°C aerobic 50°C anaerobic Glucose No growth No growth Growth No growth No growth No growth Lactose No growth No growth No growth No growth No growth No growth Sucrose No growth No growth Growth No growth No growth No growth 1. In terms of nutrion, which carbon, energy, temperature, oxygen, and pH requirements does this organism have? Chemoheterotroph, mesophile, aerobe, and neutrophile 2. Below map out how the bacterium will process the sugars to make ATP. Sucrose will be cleaved into glucose and fructose and will ﬂow into glycolysis along with glucose. Lile bit of ATP will be produced during glycolysis in addion to 2 pyruvate per glucose molecule and NADH. Pyruvate is further oxidized to acetyl- CoA which will go through the Kreb’s cycle to produce CO2, NADH, FADH2, and a lile bit of ATP. The NADH and FADH2 that are carrying the electrons will dump them oﬀ at the electron transport chain where oxygen will accept the electrons to be reduced to water. This generates an electrochemical gradient which produces a lot of ATP through oxidave phosphorylaon. 3. You also observed that the organism grew pink at 30C but not at 37C. Describe in terms of gene expression why this might be the case with two diﬀerent condions? Genes are expressed into proteins depending on environmental condions. In this example, at a cooler temperature the genes needed to make pink proteins are expressed. At a warmer temperature, those genes are NOT expressed. 4. Looking at the pink operon below. Label the promoter, operator, and genes on the operon and describe what the role of each are during transcripon. What is the funcon of an operon? Promoter- the ﬁrst sequence that indicates the start of transcripon and where the RNA polymerase will bind Operator- the next sequence in which the repressor will bind to turn transcripon “oﬀ”. This is the sequence that regulates transcripon of this operon. Genes – are p, i, n, and k. These will be transcribed into RNA. Remember all the genes in an operon are transcribed onto a single RNA transcript. k n i p

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