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Hemoglobin and Cooperatively - Exam 3 | Biochemistry 2006 | CHEM 431, Exams of Biochemistry

Material Type: Exam; Professor: Campbell; Class: BIOCHEMISTRY; Subject: Chemistry; University: Jackson State University; Term: Unknown 1989;

Typology: Exams

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Name__________________________________________
CHEM 431/531 - Biochemistry I/Exam III/November 06, 2006
Part I: Hemoglobin and Cooperativity- Answer ONLY 4 out of the 8 questions.
1. Why is it advantageous for hemoglobin to have allosteric properties?
Hemoglobin binds oxygen in a positive cooperative manner. This allows it to become
saturated in the lungs, where oxygen pressure is high. When the hemoglobin moves to
tissues, the lower oxygen pressure induces it to release oxygen and thus deliver oxygen
where it is needed.
2. Explain the structural differences between fetal hemoglobin and adult hemoglobin. Explain
how fetal hemoglobin can bind oxygen released by adult hemoglobin inside the uterus. This
must include a discussion of 2,3 BPG as part of your answer.
Fetal hemoglobin contains two
and two
chains, in contrast to adult hemoglobin with two
and two
chains. The fetal hemoglobin
chain is probably a result of gene duplication
and divergence. The difference in the chains results in a lower binding affinity of 2-3 BPG to
fetal hemoglobin. Thus, the fetal hemoglobin has a higher affinity for oxygen, and the
oxygen is effectively transferred from the mother’s hemoglobin to fetal hemoglobin.
3. Describe the octahedral coordination sphere of the iron ion in hemoglobin and myoglobin.
The Fe +2
ion is coordinated to the four nitrogens in the center of the protoporphyrin of the heme. The fifth
coordination site is occupied by the “proximal histidine” of the globin chain. The oxygen is bound to the
sixth coordination site of the iron.
4. Briefly describe the cause of sickle-cell anemia.
Sickle-cell anemia is a genetic disorder that is the result of a single substitution of β6 Glu with a Val. This
changes a negatively charged side chain to a nonpolar, hydrophobic side chain. This Val binds into a
hydrophobic pocket on the β chain of an adjacent molecule whose β6 Val binds to another molecule, thus
hemoglobin aggregates. These aggregates form long fibers that strain the RBC and force into a sickled
shape. The distorted red blood cells clog capillaries and impair blood flow, resulting in the sickle-cell
crisis. The sickled cells are then destroyed, resulting in the anemia.
5. Describe how carbon dioxide affects the oxygenation of hemoglobin.
Increased levels of carbon dioxide cause hemoglobin to release oxygen. The more active the
tissue, the more fuel is burned and the more CO2 is produced. These active tissue cells have
the greatest need for oxygen to produce more energy. The CO2 combines with the N-
terminal amino groups to form negatively charge carbamate groups. The negatively charge
carbamate groups form salt bridges that stabilize the T-state. Thus, the increase of carbon
dioxide causes the conversion of the R-state to the T-state, releasing the bound oxygen to
the tissues producing the most CO2.
6. Describe the role of 2,3-bisphosphoglycerate in the function of hemoglobin..
2,3-bisphosphoglycerate, 2,3-BPG, is a relatively small, highly anionic molecule found in the RBC. 2,3-
BPG only binds to the center cavity of deoxyhemoglobin (T-state). The size of the center cavity decreases
upon the change to the R-form so that it cannot bind to the R-state. Thus, the presence of 2,3-BPG shifts the
equilibrium toward the T-state. T-state is unstable, and without BPG, the equilibrium shifts so far toward
the R-state that little oxygen would be released under physiological conditions.
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Name__________________________________________ CHEM 431/531 - Biochemistry I/Exam III/November 06, 2006

Part I: Hemoglobin and Cooperativity - Answer ONLY 4 out of the 8 questions.

  1. Why is it advantageous for hemoglobin to have allosteric properties? Hemoglobin binds oxygen in a positive cooperative manner. This allows it to become saturated in the lungs, where oxygen pressure is high. When the hemoglobin moves to tissues, the lower oxygen pressure induces it to release oxygen and thus deliver oxygen where it is needed.
  2. Explain the structural differences between fetal hemoglobin and adult hemoglobin. Explain how fetal hemoglobin can bind oxygen released by adult hemoglobin inside the uterus. This must include a discussion of 2,3 BPG as part of your answer. Fetal hemoglobin contains two  and two  chains, in contrast to adult hemoglobin with two  and two  chains. The fetal hemoglobin  chain is probably a result of gene duplication and divergence. The difference in the chains results in a lower binding affinity of 2-3 BPG to fetal hemoglobin. Thus, the fetal hemoglobin has a higher affinity for oxygen, and the oxygen is effectively transferred from the mother’s hemoglobin to fetal hemoglobin.
  3. Describe the octahedral coordination sphere of the iron ion in hemoglobin and myoglobin. The Fe+2^ ion is coordinated to the four nitrogens in the center of the protoporphyrin of the heme. The fifth coordination site is occupied by the “proximal histidine” of the globin chain. The oxygen is bound to the sixth coordination site of the iron.
  4. Briefly describe the cause of sickle-cell anemia. Sickle-cell anemia is a genetic disorder that is the result of a single substitution of β6 Glu with a Val. This changes a negatively charged side chain to a nonpolar, hydrophobic side chain. This Val binds into a hydrophobic pocket on the β chain of an adjacent molecule whose β6 Val binds to another molecule, thus hemoglobin aggregates. These aggregates form long fibers that strain the RBC and force into a sickled shape. The distorted red blood cells clog capillaries and impair blood flow, resulting in the sickle-cell crisis. The sickled cells are then destroyed, resulting in the anemia.
  5. Describe how carbon dioxide affects the oxygenation of hemoglobin. Increased levels of carbon dioxide cause hemoglobin to release oxygen. The more active the tissue, the more fuel is burned and the more CO 2 is produced. These active tissue cells have the greatest need for oxygen to produce more energy. The CO 2 combines with the N- terminal amino groups to form negatively charge carbamate groups. The negatively charge carbamate groups form salt bridges that stabilize the T-state. Thus, the increase of carbon dioxide causes the conversion of the R-state to the T-state, releasing the bound oxygen to the tissues producing the most CO 2.
  6. Describe the role of 2,3-bisphosphoglycerate in the function of hemoglobin.. 2,3-bisphosphoglycerate, 2,3-BPG, is a relatively small, highly anionic molecule found in the RBC. 2,3- BPG only binds to the center cavity of deoxyhemoglobin (T-state). The size of the center cavity decreases upon the change to the R-form so that it cannot bind to the R-state. Thus, the presence of 2,3-BPG shifts the equilibrium toward the T-state. T-state is unstable, and without BPG, the equilibrium shifts so far toward the R-state that little oxygen would be released under physiological conditions.
  1. What functional role does the “distal histidine” play in the function of myoglobin and hemoglobin? The bonding between the iron and oxygen can be described as a combination of resonance structures, one with Fe2+^ and dioxygen and another with Fe3+^ and superoxide. The “distal histidine” donates a hydrogen bond to this complex stabilizing the complex and inhibits the oxidation of the iron to the ferric state.
  2. Describe the concerted model to explain allosteric cooperative binding. The protein exists in two conformations, a T-state (for tense) that has a lower affinity for the ligand and an R-state (for relaxed) that has a higher affinity for the ligand. In the concerted model, all of the molecules exist either in the T-state or in the R-state. At each ligand concentration, there is an equilibrium between the two states. An increase in the ligand concentration shifts the equilibrium from the T- to the R-state.

Part II: Enzyme Kinetics and Catalytic Strategies - Answer ONLY 6 out of the 10 questions.

9. How do the intermediate steps in multi-substrate enzyme mechanisms (sequential

displacement vs double displacement) differ?

In a sequential displacement reaction both substrates bind, and a ternary complex of all three is

formed. In a double displacement one or more products are released prior to binding of all

substrates. Thus, a substituted enzyme intermediate is formed.

10. How are competitive, uncompetitive, and non-competitive inhibitions kinetically

distinguishable? Explanation must include information about (1) the extent of

inhibitor as it relates to [S], (2) substrate/enzyme/inhibitor complex, and (3) changes

in Km and/or Vmax.

Competitive Inhibition: (1) Competitive inhibition can be overcome by the presence

of large amounts of substrate. (2) The enzyme forms a complex with either the

substrate (ES) or the inhibitor (EI), but not both. (3) The apparent Km is increased

and Vmax is unchanged.

Uncompetitive Inhibition: (1) Uncompetitive inhibition can not be overcome by the

presence of large amounts of substrate. (2) The inhibitor can form a complex only

with the enzyme-substrate (ES) complex. (3) The apparent Vmax decreases and the

apparent KM decreases.

Non-competitive Inhibition: (1) Non-competitive inhibition can not be overcome by

the presence of large amounts of substrate. (2) The enzyme can form a complex with

either the substrate (ES) or the inhibitor (EI) or both substrate and inhibitor at the

same time EIS. (3) The apparent Vmax decreases and KM is unchanged.

11. What is the Gibbs free energy of activation? Explain how enzymes increase reaction rates as

related to Gibbs free energy of activation.

Gibbs free energy of activation is the energy required to transform substrate molecules to the

transition state. Enzymes increase reaction rates by facilitating the formation of the transition

state via a different pathway that has a lower energy of activation without changing the free

energy for the reaction.

  1. You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and in the absence of 10 mM inhibitor. The following data are obtained: app

K M  KM (1 [I]/K )i

[S] mM V 0 (mM/sec) Inhibitor +Inhibitor 0 .1 33 17 0 .2 50 36 0 .5 71 59 1 .0 83 67 2 .0 91 80 5 .0 96 91 10 .0 98 95 a) What is the Vmax in the absence of inhibitor? b) What is the Km in the presence of inhibitor? c) What is the binding constant of this inhibitor? d) What kind of inhibitor is it likely to be? [S] Vo (-inh) Vo (+inh) 1/[S] 1/Vo (-inh) 1/Vo (+inh) 0.1 33 17 10 0.030303 0. 0.2 50 36 5 0.02 0. 0.5 71 59 2 0.0140845 0. 1 83 67 1 0.0120482 0. 2 91 86 0.5 0.010989 0. 5 98 95 0.2 0.0102041 0. 10 98 98 0.1 0.0102041 0.

Problem 18

-0. -0. 0

-6 -4 -2 0 2 4 6 8 10 12 1/[S]

1/V

No Inh Inhibitor Linear (No Inh) Linear (Inhibitor) 1/Vmax = 0. Vmax = 100 KM = (at Vmax/2 which is 50) KM = -1/4.8 = 0.21mM

KM (apparent) = (at Vmax/ which is 50) KM = -1/1.8 = 0.55mM KM(app) = KM(1 +[I]/Ki) 0.55 = 0.21(1+10/Ki) Ki = 6.2 mM