Chapter 14 - AllosteryTrue/False and Multiple Choice (7) 1) Which of the following can result in an allosteric modulation of activity: a) covalent modification such as phosphorylation or acetylation b) oligomerization c) binding of a ligand d) stabilizing an alternative conformation e) all of the above 2) The transcription of a gene is controlled by a transcription factor binding either Glucose or Lactose. When there is 0.2mM of either Glucose or Lactose in the cell, the gene is transcribed at about 10% of the maximum. When there more than 2mM Glucose in the cell, the gene is fully induced. However, when there is 2mM Lactose in the cell, there amount of transcription is approximately half of the maximum. At 40mM of either Glucose or Lactose in the cell, the gene is fully induced. The transcriptional regulation is likely: a) ultrasensitive with respect to both Glucose and Lactose b) ultrasensitive with respect to Lactose but not Glucose c) graded with respect to Glucose d) ultrasensitive with respect to Glucose but not Lactose e) hyperbolic with respect to Glucose but not Lactose 3) In allosteric proteins, ligand binding can only result in positive cooperativity True/False 4) Which of the following is not a known allosteric effector of hemoglobin: a) oxygen b) bisphosphaoglycerate c) low pH d) myoglobin e) CO2 5) Bacterial chemotaxis involves only random, brownian movement True/False the response to an input is _______ than expected from the graded.the maximum value for a dimeric protein is 2 a) only i b) i and iii c) ii and iii d) all of the statements are true Fill in the Blank (5) 1) When a system is ultrasensitive. hyperbolic response. the allosteric hemoglobin proteins of all organisms are tetrameric assemblies True/False Which of the following statements about the Hill coefficient are true: i. it is the steepness of the log-log binding isotherm at the half saturation point ii. (sharper) 2) Oxygen biases the equilibrium of hemoglobin towards the _____ state (R or relaxed) 3) The extent of flagellar clockwise rotation is governed by the ___________________ of the protein CheY (phosphorylation) 4) The __________ histidine residue of hemoglobin sense the effective size of the heme iron atom (proximal) 5) An effect of co-localization is to greatly increase the local __________ of two proteins (concentration) Long Answer/Essay (12) .6) 7) Because of the importance of the F-helix in oxygen binding. allosteric systems have a Hill coefficient of exactly 1 iii. 3-1) 2 = 58 nM . Each subunit contains a binding site for a Xenon gas molecule.8 b) nh > 1.023)) = 0.1) A dimeric glucokinase has a binding site for glucose in each subunit. The K D for the first binding event is 1mM and the KD for the second event is 10uM.01/1)) = 1.3. The KD for the first binding event is measured at 23nM. therefore it is negatively cooperative 3) A dimeric enzyme with two identical binding sites has a Hill coefficent of 1.5uM. therefore it is positively cooperative 2) A dimeric hemoglobin is isolated from a fish. a) What is the Hill coefficient? b) Is this protein positive or negatively cooperative with respect to glucose binding? Answer: a) nh = 2 / (1+√(KD2/KD1)) = 2 / (1+√(0. If the KD of the first binding site is 200nM.1)2 = 200x(2/1.5/0. a) What is the Hill coefficient? b) Is this protein positive or negatively cooperative with respect to Xenon binding? Answer: a) nh = 2 / (1+√(KD2/KD1)) = 2 / (1+√(3. The KD for the second binding event is measured at 3. what is the KD of the second binding site? Answer: KD2 = KD1x(2/nh .15 b) nh < 1. What is the fraction of the protein is bound at 10nM concentration of ligand? Answer: a) KD2 = KD1x(2/nh .12 f = 2. KD2 = 2 nM) at 15nM concentration of ligand? Answer: f/(1-f) = ([L]/KD1 + ([L]/KD1)* ([L]/KD2))/(1+[L]/KD1) = (15/20 + (15/20)* (15/2))/(1+15/20) = 3. A scientist measures the K D1 for the first binding site at 25nM and the Hill coefficient as 1.4/1.4) What is the ratio of bound to unbound receptor (f/(1-f)) for a dimeric protein with two binding sites (KD1 = 20nM.56))/(1+10/25) = 2.9 .4 f = 0.6. a) Sketch the binding isotherms of log(f/(1-f)) vs log([L]) b) What is the Hill coefficient? .12/3.29 6) The first and second binding sites of a positively cooperative allosteric dimeric protein have KDs of 100mM and 10uM respectively.4 = 0.1)2 = 25x(2/1.56 nM f/(1-f) = ([L]/KD1 + ([L]/KD1)* ([L]/KD2))/(1+[L]/KD1) = (10/25 + (10/25)* (10/1.2.68 b) f/(1-f) = [L]/KD = 10/25 = 0.1)2 = 1.12 f = 2.12f = 2.12 f = .12 . a) What fraction of the protein is bound at 10nM concentration of ligand? b) A single point mutation abolishes all cooperativity in the protein such that the protein binds its ligand with an apparent K D equal to 25nM.64 5) A dimeric allosteric protein is isolated.12f 3. 18 .Answer: a) b) nh = 2 / (1+√(KD2/KD1)) = 2 / (1+√(10-5/10-1)) = 1. a) Sketch the binding isotherms log(f/(1-f)) vs log([L]) b) What is the Hill coefficient? Answer: a) b) nh = 2 / (1+√(KD2/KD1)) = 2 / (1+√(10-2/10-4)) = 0.98 7) The first and second binding sites of a negatively cooperative allosteric dimeric protein have KDs of 100uM and 10mM respectively. which reduces affinity for oxygen. At 37C.25 Since homolog B. The K D of the first binding site of Homolog B is measured at 4mM.41 9) A cyclist is interested in cheating in a race by delivering more oxygen to his muscles. Removal of BPG will result in a sharper switch between saturated and empty hemoglobin.5 = ([L]/KD1 + ([L]/KD1)* ([L]/KD2))/(1+[L]/KD1) 0.25* (1/KD2))/(1. the dimeric acetyl transferase enzyme AAC displays positive cooperativity at low temperatures and negative cooperativity at high temperatures. What is the Hill coefficient of Homolog B? Answer: For homolog A: f/(1-f) = [L]/KD = 1/4 = 0. WIthout this effect. f/(1-f) = 0. At 10C. that reducing the BPG concentration in his blood cells should be good for his performance. 10) Due to a fascinating coupling between folding and ligand binding.67mM nh = 2 / (1+√(0. a) What is the Hill coefficient at each temperature? b) What is the change in fraction bound at 50uM ligand between 10 and 37C? . The cyclist reasons that since Bisphosphoglycerate (BPG) stabilizes the “T” state of hemoglobin. Homolog A is a monomer that binds glucose with a KD of 4mM. the binding isotherm will not match the difference in oxygen concentration between the arterial and venous blood and likely lead to suboptimal delivery of oxygen to his muscles. binds twice as much glucose.25) KD2 = 0.25 + 0. KD2 = 230uM.5 f/(1-f) = 0. How might removing BPG might have a detrimental effect on delivery of oxygen to his muscles? Answer: BPG acts as an allosteric effector for hemoglobin. AAC has a K D1 = 330uM and KD2 = 130uM.5 = (1/4 + (1/4)* (1/KD2))/(1+1/4) = (0. At 1mM concentration of glucose.67/4)) =1. AAC has a KD1 = 200uM. Homolog B is a positively cooperative dimer. it is found that Homolog B binds twice as much glucose as homolog A.8) Two homologs of a protein are isolated. NSMB.(Source: Frieburger et al. more of the protein is bound to ligand. 2011.18 f = 0. This alters the interfacial interactions between the alpha and beta .24 f = 0.97 b) For 10C f/(1-f) = ([L]/KD1 + ([L]/KD1)* ([L]/KD2))/(1+[L]/KD1) = (50/330 + (50/330)* (50/130))/(1+50/330) = 0. Affinities have been altered).15 For 37C f/(1-f) = ([L]/KD1 + ([L]/KD1)* ([L]/KD2))/(1+[L]/KD1) = (50/200 + (50/200)* (50/230))/(1+50/200) = 0.18/1.18 = 0. 4% more enzymes are bound at the higher temperature.24 = 0. 11) How does CO2 directly and indirectly stabilize the “T” state of hemoglobin in venous blood? Answer: CO2 directly stabilizes the “T” state by reacting with the N-terminal amino groups of the protein.19 Thus. Answer: a) For 10C: nh = 2 / (1+√(KD2/KD1)) = 2 / (1+√(130/330)) =1.24/1. This is driven by the favourable change in KD1 at higher temperatures.23 For 37C: nh = 2 / (1+√(KD2/KD1)) nh = 2 / (1+√(230/200)) = 0. Even though AAC is negatively cooperative at higher temperatures. which targets them to the plasma membrane in a cell at 25C. stabilizing the “T” state relative to the “R” state. Without these interactions. the “T” state is stabilized. CO 2 indirectly stabilizes the “T” state by reducing the pH within red blood cells. 13) A scientist finds two pathways. dependent on distinct kinase activities. One pathway responds to elevated levels of salt and the other responds to elevated levels of caffeine. 12) Two proteins are modified by myristoylation. The reduction in pH leads to the Bohr effect where beta subunit His 146 becomes protonated. This disrupts interactions centred around the ion pair with beta Asp 94 and an interaction to alpha Lys 40. Explain which pathway likely contains a single kinase and which pathway likely contains a kinase cascade similar to the MAP-kinase pathway [NaCl] mM Hsp90 transcription 1 0 2 1 3 1 4 40 .subunits. Assuming that any favourable mutation would increase their interaction energy by 4 kJ/mol .5 * ln (0. This changes their effective local concentration from 10nM to 1uM. which activate a stress response in yeast.001) = -23 kJ/mol If each mutation added 4 kJ/mol. the at least 6 mutations would be required to result in as high an effective affinity. How many favourable mutations would have to occur in the absence of co-localization to result in an equivalent effective affinity as observed when the two proteins are co-localized? Answer: ∆∆G = RT ln(10-8x 10-8)/(10-6x 10-6) = 2. Below is a table listing the relative concentrations of salt or caffeine and the measured transcription of Hsp90. Both pathways result in transcription of the chaperone Hsp90. which likely has a kinase cascade. This situation is observed only for the salt response.[NaCl] mM Hsp90 transcription 1 0 5 100 10 100 100 100 1000 100 [caffeine] mM Hsp90 transcription 1 0 2 0 3 1 4 1. and not for the caffeine response. Ultrasensitivity resulting from the coupling of multiple kinases converts a graded input signal into a sharp switch.1 5 2 10 10 100 50 1000 100 Answer: Only the salt response is ultrasensitive (in that it undergoes a complete activation in a ~2-fold rather than ~100-fold concentration change). which is more hyperbolic. .