practice_E1KEY

practice_E1KEY - Professor Lindsay Hinck Bio 110 Fall, 2009...

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Professor Lindsay Hinck Bio 110 Fall, 2009 Midterm I Version A Remember to manage your time wisely. You have 35 problems: 5 True/False (2 points each) and 30 multiple choice (3 points each). This means approximately 3 minutes/ question so you have to move along. Remember to write and bubble in your 1) Name 2) Test Version (A or B) 3) Student ID on the scantron sheet.
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1) Which of the following statements is FALSE? A) Active transport by carrier proteins is saturable under high gradients B) Passive transport by carrier proteins is saturable under high gradients C) Diffusion is not saturable even under high gradients D) Transport mediated by carrier proteins can be coupled to energetically favorable reactions E) Passive transport by channel proteins is energetically unfavorable 2) Where does most new membrane synthesis take place in a eukaryotic cell? A) The outer membrane of the Golgi apparatus B) The inner membrane of the Golgi apparatus C) The plasma membrane D) The inner membrane of the endoplasmic reticulum E) The outer membrane of the endoplasmic reticulum 3-5) (3 parts: 1 point each part) Protein A binds to protein B to form a complex, AB. A cell contains an equilibrium mixture of protein A at a concentration of 1 microM, Protein B at a concentration of 1 microM and the complex AB also at 1 microM. I know you all remember this simple equation from biochemistry, but just to refresh. K (equilibrium constant)=[AB]/[A][B] 3) What is the equilibrium constant, K< for the reaction A+B AB? A) The equilibrium constant, K , equals 10 4 M–1 B) The equilibrium constant, K , equals 10 5 M–1 C) The equilibrium constant, K , equals 10 6 M–1 D) The equilibrium constant, K , equals 10 7 M–1 E) The equilibrium constant, K , equals 10 8 M–1 4) What would the equilibrium constant be if A, B, and AB were each present in equilibrium at a concentration of 1 nanoM? A) The equilibrium constant, K , equals 10 8 M–1 B) The equilibrium constant, K , equals 10 9 M–1 C) The equilibrium constant, K , equals 10 10 M–1 D) The equilibrium constant, K , equals 10 11 M–1 E) The equilibrium constant, K , equals 10 12 M–1 5) At this lower concentration, about how many extra hydrogen bonds would be needed to hold A and B together tightly enough to form the same proportion of the AB complex as formed in part A. Remember that free energy change is related to the equilibrium constant by the equation Δ G o = –2.3 RT log K where R = 1.98 X 10 -3 kcal/kmole and T = 310K. Assume that the formation of 1 H-bond is accompanied by a favorable free-energy change of about 1 kcal/mole. A) 1-3 H bonds
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B) 3-5 H bonds C) 5-7 H bonds D) 7-9 H bonds E) 9-11 H bonds This example illustrates that interacting cellular proteins present at low concentrations need to bind to one another with high affinities if a high proportion of the molecules are to be bound together. A three-order of magnitude decrease in the equilibrium constant corresponds to a free-energy difference of about –4.2 kcal/mole. Thus, effective binding at the lower concentration would require the
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practice_E1KEY - Professor Lindsay Hinck Bio 110 Fall, 2009...

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