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BICH 409 Spring 2024 Assignment 4 200 pts Key Following questions are worth 3 pts each. Use the following saturation curves to answer Questions 1-11. 1.) Which of the curves depicts the protein with the lowest p50? Curve 1 2.) Which of the curves depicts the protein with the greatest oxygen affinity? Curve 1 3.) If Curve 3 corresponds to isolated hemoglobin in a solution containing physiological concentrations of Cl - , CO 2 and BPG at a pH of 7, which curve corresponds to a decrease in the concentration of CO 2 ? Curve 2 4.) If Curve 3 corresponds to isolated hemoglobin in a solution containing physiological concentrations of Cl - , CO 2 and BPG at a pH of 7, which curve corresponds to an increase in the concentration of Cl  ? Curve 4 5.) Which of the above curves corresponds to myoglobin in a solution containing physiological concentrations of CO 2 and BPG at a pH of 7.0? Curve 1 6.) Which of the above curves corresponds to myoglobin corresponds to an increase in the pH from 7.0 to 7.6? Curve 1 7.) If Curve 3 corresponds to isolated hemoglobin in a solution containing physiological concentrations of Cl - , CO 2 and BPG at a pH of 7, which curve corresponds to the dissociation of hemoglobin into its component subunits? Curve 1 8.) If Curve 3 corresponds to isolated hemoglobin in a solution containing physiological concentrations of Cl - , CO 2 and BPG at a pH of 7.0 , which curve corresponds to pH 7.4? Curve 2 9.) If Curve 3 corresponds to isolated hemoglobin in a solution containing physiological concentrations of CO 2 and BPG at a pH of 7, which curve corresponds to a decrease in the concentration of BPG? Curve 2 10.) Hb Kansas, 102(G4)  Asn  Thr, disrupts a hydrogen bond that stabilizes the R-state. If Curve 3 corresponds to wild type hemoglobin in a solution containing physiological concentrations of Cl - , CO 2 and BPG at a pH of 7, which curve corresponds to Hb Kansas? Curve 4 11.) Hb Yakima, 99(G1)  Asp  His, disrupts a hydrogen bond that stabilizes the T-state. If Curve 3 corresponds to wild type hemoglobin in a solution containing physiological concentrations of Cl - , CO 2 and BPG at a pH of 7, which curve corresponds to Hb Yakima? Curve 2

BICH 409 Spring 2024 Assignment 4 200 pts Key 12.) (3 pts) How would the p50 of hemoglobin be affected when the partial pressure of CO 2 increases from 20 to 30 torr. The p50 of hemoglobin would increase. 13.) (3 pts) How is the majority of CO 2 generated by respiring tissue transported to back to the lungs? As Bicarbonate 14.) (3 pts) How does hemoglobin carry CO 2 from the tissues back to the lungs? Carbamoylation of the N-termini. As Carbaminohemoglobin 15.) (3 pts) Hb Rainier, 145(HC2)  Tyr  Cys, The mutant cysteine forms a disulfide bond with another cysteine of the  subunit that stabilizes the R-state. How would this mutation affect the oxygen affinity and hill coefficient of Hb Rainier? Hb Rainier would have an abnormally high oxygen affinity. Normally quaternary interections between subunits are noncovalent and reversible, But since disulfide bonds are covalent and irreversible this mutation would significantly decrease the hill coefficient towards a Hill coefficient of 1. 16.) (3 pts) Where does the proton that produces the Bohr effect bind to hemoglobin? The Histidine 146 of the  subunits 17.) (3 pts) Hb Cowtown, 146(HC3)  His  Leu. What is the role of His-146 of the  –subunits? How would this mutation affect the oxygen affinity and allosteric regulation of Hb Cowtown? Histide-146 binds the proton that produces the Bohr effect. This mutant would diminish the allosteric regulation of hemoglobin by hydrogen ions and would increase the oxygen affinity of this mutant. 17.) (3 pts) A monomeric oxygen-binding protein has a fractional saturation of 0.35 when pO 2 = 9.3 torr. What is the value of p50 for this protein and what is the value of the hill coefficient n? Show your work for partial credit. n =1 p50 = 17.27 torr 18.) (3 pts) A monomeric oxygen binding protein has a p50 of 13.5 torr. What partial pressure of oxygen would generate a fractional saturation of 0.75 and what is the value of the Hill coefficient? Show your work for partial credit. n =1 pO 2 = 40.50 torr 19.) (3 pts) A monomeric oxygen binding protein has a p50 of 31.92 torr. If the partial pressure of oxygen is 12.33 torr, what is the fractional saturation and the value of the Hill coefficient? Show your work for partial credit. n =1 Y = 0.28

BICH 409 Spring 2024 Assignment 4 200 pts Key 24.) (a) (2 pts) Describe the structural differences between fetal hemoglobin and adult hemoglobin. Adult Hemoglobin has an     quaternary structure Fetal Hemoglobin has a     quaternary structure The only difference between the beta subunits and the gamma subunits is that His 143 has been replaced with a serine. (b) ( 2 pts) Explain how the structural differences effect oxygen and 2,3-bisphosphoglycerate binding. This His-143 of the  subunits bonds to BPG by electrostatic interactions. Since in fetal Hb this histidine is replaced with a serine, 2 electrostatic interactions involved in binding BPG are lost resulting in weaker BPG binding. (c) ( 2 pts) What impact do these differences have on the delivery of oxygen to the fetal red blood cells. Since BPG is a negative allosteric effector of Hb and fetal Hb has weaker BPG binding, Fetal Hb has a higher oxygen affinity than adult Hb. 25.) (3 pts) Cigarette smokers expose themselves to carbon monoxide. What is the effect of carbon monoxide on oxygen transport? How do Cigarette smokers’ bodies adjust to accommodate the increased carbon monoxide? Carbon monoxide binds to the irons of hemoglobin diminishing the oxygen carrying capacity of the blood. The body adapts to the diminished oxygen carrying capacity by elevating the BPG concentration in the red blood cells. 26.) ( 14 pts) Some racers in swimming sprints hyperventilate right before the buzzer initiates the race. (a) How would hyperventilation affect the concentration of O 2 , HCO - 3 , dissolved CO 2 , pH and oxygen affinity of hemoglobin? O 2 increases, HCO - 3 , decreases, dissolved CO 2 Decreases , The pH Increases and oxygen affinity of hemoglobin increases. (b) What advantage or disadvantage could the sprinter gain by hyperventilating? By hyperventilating the swimmer saturates his or hemoglobin with oxygen. The disadvantage is raising of the blood pH which in turn increases the oxygen affinity of the hemoglobin delivering less oxygen to the respiring tissues. (c) As the swimmer sprints, the swimmers muscle tissue is actively respiring, how does this affect the concentration of O 2 , HCO 3 , dissolved CO 2 , pH and oxygen affinity of hemoglobin? As the swimmer sprints, oxygen is consumed as the glycogen stores are converted into dissolved CO 2 . Thus O 2 decreases, HCO - 3 increases, dissolved CO 2 increases, the pH decreases and oxygen affinity of hemoglobin decreases. (d) Some racers in swimming sprints do not breath during the sprint, they hold their breath. How would holding your breath affect the concentration of O 2 , HCO 3 , dissolved CO 2 , pH and oxygen affinity of hemoglobin? O 2 decreases, HCO - 3 increases, dissolved CO 2 increases, the pH decreases and oxygen affinity of hemoglobin decreases. (e) What happens when a swimmer hyperventilates before the start of the race and holds his or her breath during the swimming sprint. Does the swimmer gain an advantage or disadvantage by this approach? By hyperventilating the swimmer would increase the plasma pH and saturate hemoglobin with oxygen. As he holds his breath and swims, the oxygen concentration and pH decrease. The two effects cancel themselves out for a short period of time.

BICH 409 Spring 2024 Assignment 4 200 pts Key Work the following old multiple choice exam questions worth 2 points each. The reaction coordinate diagram shown to the left is for the mitochondrial enzyme ATP synthase. Use this diagram to answers Questions 1 – 4 1.) From this diagram you can conclude: (A) ATP synthesis is endergonic. (B) The reaction is at equilibrium. (C) ATP synthesis is exergonic. (D) ATP synthesis is endothermic. (E) ATP synthesis is exothermic. 2.) How many intermediates are generated in the enzyme catalyzed pathway? (A) 4 (B) 3 (C) 2 (D) 1 (E) 0 3.) The rate determining step for ATP synthesis: (A) cannot be determined from a reaction coordinate diagram. (B) is the binding of ATP. (C) is the release of ATP from the enzyme. (D) is the generation of ATP. (E) is the binding of ADP and Pi. 4.) The rate determining step for the reverse reaction converting ATP into ADP and Pi is: (A) cannot be determined from a reaction coordinate diagram. (B) the release of ADP and Pi. (C) the binding of ATP. (D) the release of ATP from the enzyme. (E) the generation of ADP and Pi.

BICH 409 Spring 2024 Assignment 4 200 pts Key 11.) (A) (3 pts) Consider the following reaction occurring at 298K:  G o ’ =  16.7 kJ/mol, calculate K’ eq   52 . 845 298 / 3145 . 8 / 700 , 16 '                          K K mol J mol J RT G e e K o (B.) ( 3 pts) k 1 = 6 X 10 -14 M -1 s -1 , calculate k 2 . 1 1 17 1 1 14 1 2 2 1 10 1 . 7 52 . 845 10 0 . 6 52 . 845              s M s M K k k k k K (C) ( 3 pts) Hexokinase is an enzyme that catalyzes this reaction. When the hexokinase is present the rate of k 1 is 1.3 X 10 -3 M -1 s -1 . What is the equilibrium constant for the enzyme catalyzed reaction? K’ eq = 845.52 Enzymes do not affect thermodynamic state functions such as equilibrium constants and standard free energy changes. (D) ( 3 pts) Calculate the catalytic power of hexokinase. Catalytic Power = k cat /k uncat = 1.3 X 10 -3 M -1 s -1 /6.0 X 10 -14 M -1 s -1 = 2.17 X 10 10 (E) ( 3 pts) Calculate k 2 for the hexokinase catalyzed reaction. 1 1 6 1 1 3 1 2 2 1 10 54 . 1 52 . 845 10 3 . 1 52 . 845              s M s M K k k k k K +ADP k 1 k 2 O OH OH OH OH CH 2 OH +ATP O OH OH OH OH CH 2 O P O - O O - Glucose Glucose6-phosphate

BICH 409 Spring 2024 Assignment 4 200 pts Key Given the following reaction: k 1 k 2 A+ B P + Q + R The rate constants k 1 and k 2 were measured with the units of Molar concentrations and seconds for time. The numerical values of k 1 and k 2 are given below. 12.) ( 2 pts) Assign the correct units for k 1 = 851.37___ M -1 s -1 ________ 13.) ( 2 pts) Assign the correct units for k 2 = 36,239.45 __ M -2 s -1 _________ 14.) ( 2 pts) K’ eq = __ 0.02349 __________ 15.) ( 2 pts) When the enzyme is added the rate of k 1 is 4.8 x 10 8 . What is the equilibrium constant for the enzyme catalyzed reaction? ___ 0.02349 ________________ 16.) (2 pts) Calculate the catalytic power of the enzyme. ___ 563,797 __________________ 17.) ( 2 pts) Calculate k 2 for the enzyme catalyzed reaction.( be sure to report the units) k 2 = ___ 2.04 x 10 10 M -2 s -1 Following Questions worth 2 pts each. 18.) For an enzyme that displays Michealis-Menten kinetics, What is the reaction velocity, v (expressed as a percentage of V max ) when the substrate concentration is equal to .65K m ? v = 39.39 %V max. 19.) For an enzyme that displays Michealis-Menten kinetics, What is the reaction velocity, v (expressed as a percentage of V max ) when the substrate concentration is equal to .27K m ? v = 21.25 %V max. 20.) For an enzyme that displays Michealis-Menten kinetics, What is the reaction velocity, v (expressed as a percentage of V max ) when the substrate concentration is equal to 6K m ? v = 85.71 %V max. 21.) For an enzyme that displays Michealis-Menten kinetics, What is the reaction velocity, v (expressed as a percentage of V max ) when the substrate concentration is equal to 3/5K m ? v = 37.50 %V max. 22.) For an enzyme that displays Michealis-Menten kinetics, What is the reaction velocity, v (expressed as a percentage of V max ) when the substrate concentration is equal to 32 K m ? v = 96.96 %V max. 23.) For an enzyme that displays Michealis-Menten kinetics, What is the reaction velocity, v (expressed as a percentage of V max ) when the substrate concentration is equal to 2/7K m ? v = 22.22 %V max. 24.) For an enzyme that displays Michealis-Menten kinetics, What is the reaction velocity, v (expressed as a percentage of V max ) when the substrate concentration is equal to the 0.20K m ? v = 16.67 %V max. 25.) Expressing the substrate concentration as a factor multiplied by the K m , What substrate concentration would produce a velocity of 0.67Vmax? S = _ 2.00 _ K m 26.) Expressing the substrate concentration as a factor multiplied by the K m , What substrate concentration would produce a velocity of 0.35Vmax? S = 0.5385 K m

BICH 409 Spring 2024 Assignment 4 200 pts Key Using the following kinetic mechanism. E + S ES E + P k 1 k 2 k 3 An enzyme catalyzed reaction obeying Michealis-Menten kinetics has the following rate constants: k 1 = 4.8 X 10 6 M -1 s -1 k 2 = 8.5 X 10 2 s -1 k 3 = 7.4 X 10 7 s -1 32.) ( 3 pts) Calculate K s for this enzyme. Be sure to report the units. K s = k 2 /k 1 = 1.77 X 10 -4 M 33.) (3 pts) Calculate K m for this enzyme. Be sure to report the units. K m = (k 2 + k 3 )/k 1 = 15.41 M 34.) (3 pts) Calculate k cat for this enzyme. Be sure to report the units. k cat = k 3 = 7.4 X 10 7 s -1 35.)( 3 pts) Calculate the specificity constant for this enzyme. Be sure to report the units. specificity constant = k cat / K m = 4.802 X 10 6 M -1 s -1 36.)( 3 pts) If the enzyme concentration is 3.8 X 10 -9 M, calculate V max for this enzyme. Be sure to report the units. V max = [E 0 ] X k cat = 0.2812 M/s 37.) (3 pts) If the enzyme concentration is 3.8 X 10 -9 M, What concentration of substrate would generate a velocity equal to 0.75V max ? Be sure to report the units. [S] = vK m /( V max -v) = 0.75 X 0.2812 M/s X 15.41M/( 0.2812 M/s -0.75 X 0.2812 M/s) = 0.75 X 15.41M/( .25)=3X(15.41) = 46.23 M 38.)( 3 pts) If the enzyme concentration was increased to 5.4 X 10 -8 M, what would be the value of the V max ? Be sure to report the units. V max = [E 0 ] X k cat = 3.996 M/s K m = = 15.41 M