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Transcript
Hemoglobin: the “molecular lung” by Thomas A. Huber Department of Biology Mercer University The red blood cells (erythrocytes) in our first story in lecture are about 33% by weight a single protein: the tetrameric protein hemoglobin. Hemoglobin (Hb) is part of a sophisticated O2 delivery system that transports O2 to the tissues of our bodies via the vascular system and also transports CO2 away from those same tissues. The presence of Hb increases the amount of O2 that can be transported per liter of blood about 100-fold. Similarly to how we breathe into our lungs O2-rich air and breathe out CO2-enriched air, the Hb protein alternates between an O2-enriched form and a CO2-enriched form. Therefore, it has been called the “molecular lung”. Structurally, Hb is a heterotetramer; it has four (tetra-) subunits (-mer) and they are not all identical (hetero-). More specifically, Hb is a “homodimer of heterodimeric polypeptides” – its subunit designation is best written as (αβ)2, but is often given as α2β2. The α and β subunits are structurally and evolutionarily related to each other and to the monomeric protein myoglobin (Mb). Your goal in this recitation is to improve your understanding of how the structure of Hb is related to its functions of O2 and CO2 binding and transport. Before coming to recitation, you should read pages 941-943 in your text and go to the following two web site pages; this should all take about 20-25 minutes: https://www.youtube.com/watch?v=LWtXthfG9_M http://biochem.web.utah.edu/iwasa/projects/hemoglobin.html You learned in BIO 211 that one way to analyze the interaction between an oxygenbinding protein – like hemoglobin or myoglobin – and O2 is by looking at oxygen dissociation curves. Since myoglobin is a monomeric protein with a single heme group it can only bind one O2 molecule at a time. The association reaction for this is given at the top of the next page: Mb O2 Mb O2 Ka However, Hb is a tetrameric protein and can bind up to four O2 molecules at a time. The association reaction (assuming that all four O2 bind or associate simultaneously and with the same affinity) would be as follows: Hb 4 O2 Hb O2 4 Ka Given the above reactions, how would we write the equations that express the value of each association constant? Write your answers in the space below. For myoglobin: For hemoglobin: Ka Ka Answer: Answer: The shapes of the oxygen dissociation curves of Mb and Hb result from the fact that the ability of each protein to bind O2 depends on (1) the structure of the protein, (2) the structure of O2, and (3) the partial pressure of O2. The oxygen dissociation curves for Mb and Hb are shown in Figure 1. Figure 1. Oxygen dissociation curves for the proteins myoglobin and hemoglobin. The curve for Mb is a rectangular hyperbola; the curve for Hb is sigmoidal. For Mb, P50 equals Kd (= 1/Ka); for Hb, P50 would be equal to the product of the four individual Kd values, one associated with each subunit. The figure is taken from http://www.colorado.edu/intphys/Class /IPHY3430-200/image/figure13g.jpg . From the two oxygen dissociation curves in Figure 1, it is clear that at any one partial pressure of O2, Mb is more or less (circle one) saturated than Hb. This means that myoglobin’s affinity for O2 is higher or lower (circle one) than hemoglobin’s affinity for O2. We might conclude from this that when comparing two oxygen dissociation curves, the one with the greater value of P50 indicates a protein or circumstance of higher or lower (circle one) affinity for O2. Not only is there a difference in the affinity for O2 between Mb and Hb, but the shapes of their oxygen dissociation curves are quite different (see Figure 1). Myoglobin’s shape is a rectangular hyperbola; hemoglobin’s shape is sigmoidal. This difference in shape is a result of hemoglobin’s subunit positive cooperativity. That is, when one subunit of Hb binds an O2 molecule, it makes it easier for the next subunit to bind an O2 molecule. Similarly, when one subunit releases an O2 molecule, it makes it easier for the next subunit to release an O2 molecule. Another way of stating this is to say that the affinity of Hb for O2 increases as it binds O2 molecules or decreases as it releases O2 molecules. Positive cooperativity is indicated by (1) increasing values of Ka (association equilibrium constant) for the binding of subsequent O2 molecules, (2) increasingly exergonic O2-binding reactions, and (3) sigmoidal dissociation curves. Allosteric regulators of hemoglobin: DPG Figure 2 shows the effect of increasing concentrations of 2,3-diphosphoglycerate (DPG or BPG) on hemoglobin’s oxygen dissociation curve. This allosteric regulator was discussed on a web site page you reviewed before completing this recitation. Figure 2. The effect that an increasing concentration of 2,3DPG has on the oxygen dissociation curve of hemoglobin. One molecule of 2,3-DPG can bind to one Hb tetramer, but only tightly if Hb is in the T state. Therefore, 2,3-DPG stabilizes the T state of Hb and makes it easier to unload oxygen gas. The figure is from http://www.coheadquarters.com/P ennLibr/MyPhysiology/lect5p/lec t5.04.htm . At lower concentrations of DPG, the oxygen dissociation curve is more hyperbolic; at higher concentrations of DPG, the oxygen dissociation curve is more sigmoidal. From the data summarized in Figure 2, we could conclude that increasing concentrations of DPG increases or decreases (circle one) the affinity of Hb for O2, and this makes it easier or harder (circle one) to unload the O2 in the tissues. The concentration of DPG stays relatively constant at 4.7 mM (about the same concentration as hemoglobin’s) in the blood plasma unless one goes to a place of higher elevation where the air is “thinner” and the partial pressure of oxygen gas is lower. At first you might experience shortness of breath at the higher elevation, but after a day or two you start feeling more “normal.” This is because your body acclimates to the higher elevation (in part) by increasing the concentration of DPG in the plasma so that it’s easier to deliver the appropriate amount of oxygen gas to the tissues. Having the genetic information that codes for the components responsible for regulating DPG concentration at varying altitudes is an adaptation that we have that influences our fitness if our environment changes. There are other responses that allow us to acclimate to higher elevations (such as producing more erythrocytes) that take longer – usually several weeks – to complete. For this reason, if athletes train at high elevation for several weeks before competing at a lower elevation, they might have an advantage before these physiological changes reverse. Allosteric regulators of hemoglobin: CO2, H+, Cl- and temperature Although the concentration of DPG only changes slowly (in days), several other allosteric regulators of Hb change in concentration or value on much shorter time scales and are consequently more important to understanding how hemoglobin functions as a “molecular lung”. Chief among these are the concentrations of CO2, H+ and Cl-. The effects of two of these regulators (and the effect of temperature) are shown in Figure 3 on the next page and are sometimes called the Bohr effect, after Christian Bohr, the father of the physicist Niels Bohr, who first reported them in 1904. From Figure 3 we can conclude that increasing the concentration of CO2 or H+ – or increasing the temperature – will increase or decrease (circle one) the affinity of hemoglobin for O2, making it easier or harder (circle one) to unload O2 in the tissues. The effects of increasing Cl- concentrations would be similar to those of increasing CO2 or H+ concentrations. That is, the curve would shift to the right. Figure 3. The effect that increasing or decreasing the concentration of H+ or CO2 – or the temperature – on the oxygen dissociation curve of Hb. The figure is from http://www.octc.kctcs.edu/GCaplan/anat2/no tes/FG28_12.jpg . Hemoglobin’s two stable conformations As you saw in the web page animation, Hb has two possible stable conformations. These are referred to as the T state (T for “tense”) and the R state (R for “relaxed”). In Figure 4 below, the T state is on the left and the R state is on the right. We will not concern ourselves with all of the differences between the two states, but notice (It will help if you “zoom” this picture up to 200% before looking …) that amino acid 97 in β1 in the T state “fits” between amino acids 44 and 41 of α2, but in the R state it has “ratcheted” down (in the picture) to between amino acids 41 and 38. Likewise, β2 has “ratcheted” up (in the picture) on α1 when moving from the T state to the R state. Figure 4. T state (left) and R state (right) conformers of hemoglobin. The T state conformation is stabilized by increasing the plasma concentrations of CO2, H+ and Cl- or by decreasing the concentration of O2. The R state conformation is stabilized by increasing the plasma concentration of O2 or by decreasing the concentrations of CO2, H+ and Cl-. The T state is more common in the tissue capillary beds and the R state is more common in the capillary beds surround the alveoli of the lungs. These pictures are from the chapter on hemoglobin in the text by Voet, Donald and Judith Voet. 2004. Biochemistry, 3rd edition. John Wiley & Sons, Inc. Conformations “in between” these two stable conformations are not sterically possible. T state hemoglobin has a low affinity for O2 and is often called deoxyHb; R state hemoglobin has a high affinity for O2 and is often called oxyHb. But what usually causes the shift in conformation if hemoglobin is in the T state? In the lungs, O2 molecules are plentiful and one binds to the Fe2+ ion of an α subunit’s heme group of a deoxygenated T state hemoglobin tetramer (the heme groups of the β subunits are sterically blocked by a valine residue in T Figure 5. This figure shows the effect of O2 state hemoglobin). When this happens, the 2+ binding to the heme group. The T state is electronic properties of the Fe change, and shown in blue; the R state is shown in red. the iron ion becomes smaller in volume and When O2 binds to Fe2+, the heme group can now be accommodated into the center of becomes less “domed” and the Fe2+ moves the heme group (see Figure 5). But the into the center of the heme. This pulls His F8 proximal histidine (His F8 in Figure 5) is towards the heme. In turn, His F8 pulls Helix F, which triggers a general T state to R state bonded to the iron, and when the iron ion conformational change. This picture is from moves so does the helix (Helix F in Figure 5) the chapter on hemoglobin in the text by Voet, of which it is a part. This change in the Donald and Judith Voet. 2004. Biochemistry, position and orientation of Helix F eventually 3rd edition. John Wiley & Sons, Inc. results in this subunit shifting from the t state (a lower case “t” is used when referring to a single subunit instead of the whole hemoglobin molecule) to the r state. But because this α subunit interfaces with other subunits (see Figure 4), it tends to convert them from their t states to their r states, too. The R state hemes of β subunits are not sterically blocked by valine because it has moved aside. This is the essence of cooperativity! What happens in one subunit affects the other subunits. In this case, when one α subunit converts to the r state (which has a higher affinity for O2 gas) the entire tetramer tends to convert to the R state. Now each of the other subunits has a higher affinity (higher value of Ka) for O2. Relating structure to function … finally! So, what is “tense” in T state hemoglobin and what is “relaxed” in R state hemoglobin? Oddly enough, this language seems to refer to the C-terminal amino acid residues of the four subunits. When X-ray crystallography is used to obtain electron density maps of T state hemoglobin, the C-terminal amino acid residue of each subunit appears to form definite bonds with other parts of the protein. But in R state hemoglobin, the C-terminal amino acid residues appear as “blurs” and do not seem to form definite bonds. Apparently, the bonds that stabilize T state Hb are broken in order to make the conformational shift to R state Hb. Figure 6 below shows the bonds that occur near the C-terminal amino acid residue – Arg 141 – of an α subunit and the C-terminal amino acid residue – His 146 – of a β subunit when Hb is in the T state. Figure 6. This figure shows the bonds that stabilize the C-terminal amino acid residue (Arg 141) of the α1 subunit of T state hemoglobin (left) and those that stabilize the C-terminal amino acid residue (His 146) of the β2 subunit (right). This picture is from the chapter on hemoglobin in the text by Voet, Donald and Judith Voet. 2004. Biochemistry, 3rd edition. John Wiley & Sons, Inc. Looking at Figure 6 (left), identify the type of bond between the following functional groups or ions: ____________________ N-terminal amino group of α2 and Cl____________________ phenolic hydroxyl group of Tyr 140 of α1 and carbonyl oxygen next to Val 93 of α1 ____________________ carbonyl oxygen of β2 and R-group of Arg 141 of α1 ____________________ C-terminal carboxyl group of α1 and R-group of Lys 127 of α2 If each hydrogen bond represents 30 kJ/mol and each ionic bond (“salt bridge”) represents 20 kJ/mol, how much energy is involved in the bonds shown in 6a? Answer: When these bonds form, energy is released or required for bond formation (circle one). When these bonds form, the free energy of the hemoglobin decreases or increases (circle one) and this decreases or increases the stability of the Hb. In Figure 6b you can see that the R-group of His 146 can form an ionic bond (“salt bridge”) with the R-group of Asp 94 in T state hemoglobin if His 146 is in the His1+ form (that is, if it is protonated). If the pKa value for the R-group of His 146 is 7.10, then what fraction of the His is protonated at pH = 7.40? At pH = 7.20? Show work! The following equation will be useful: Show work! [ His 0 ] pH pKa log [ His1 ] The pH of the blood entering active muscle capillary beds is about 7.40, and as it moves through the capillary beds the pH is lowered to about 7.20. And, as you have calculated above, at a blood pH of 7.20 a greater or lesser (circle one) fraction of the His 146 will be protonated and this means more or fewer bonds can form with the Rgroups of Asp 94 and this will stabilize or destabilize the T state of hemoglobin. The pH of the blood decreases because active muscle produces CO2 gas, and the action of the enzyme carbonic anhydrase (located inside of erythrocytes) allows this CO2 gas to react quickly with water to produce carbonic acid (see Figure 42.30 in your text).The carbonic acid dissociates spontaneously to form H+ and bicarbonate anions. This is shown in the equation below. CO2 H2 O H2 CO3 H HCO3 carbonic anhydrase The protons produced are the reason the pH decreases in the blood. The bicarbonate ions produced in the erythrocytes leave via Band III in exchange for Cl- anions. This is the reason [Cl-] increases in the erythrocytes. Therefore, in the capillary beds of active muscle, the concentrations of CO2, H+ and Cl- all rise inside of erythrocytes. And, as we saw in Figures 3 and 6 above, this means that hemoglobin will tend to covert to the T state and release or bind (circle one) O2. A Brief Look at the Thermodynamics of O2 binding The table to the right provides data from a paper in which researchers measured the changes in free energy, enthalpy and entropy when hemoglobin binds one, two, or four O2 molecules (apparently, in 1987 it was very hard to get these values for when Hb binds three O2 molecules). Note that numbers in the brackets provide the average value for each O2 that Hb binds. Is the binding of O2 by Hb thermodynamically favorable? Yes or No (circle one). How do you know? Answer: How is positive cooperativity expressed by comparing the average values (in the brackets) of ΔGº for each O2 bound? Answer: This table is from Parody-Morreale, Antonio, Charles H. Robert, Gary A. Bishop, and Stanley J. Gill. 1987. Calorimetric Studies of Oxygen and Carbon Monoxide Binding to Human Hemoglobin. The Journal of Biological Chemistry 262: 10994-10999. Is the reaction enthalpically or entropically driven (circle one)? Is cooperativity primarily a result of differences in entropy changes or enthalpy changes (circle one)? For your choice, explain the source of the difference that appears to contribute to cooperativity. (Hint: Remember what happens in the T state to R state change!) Answer: Since 1987, the value of ΔGº for the last O2 bound (on the average value for each O2 bound basis) has been measured and is about -8.52 kcal/mol. Using this value and the average value of -5.8 G o RT ln Ka kcal/ mol for the first O2 bound (from the table on the last G o page), quantify the increased affinity for the last O2 bound or Ka e RT compared to the first O2 bound. (Hint: The affinity in each case is given by the Keq constant, in this case an association constant (Ka). Using the equation to the right, you should be able to calculate each Ka. Since we are working with association constants, next determine the ratio of Ka for the last O2 bound divided by the Ka for the first O2 bound. This will give you the value of the “-fold increase”. Don’t worry about units, since (in this case) they will cancel each other out. Some constants you will need are R = 1.9872 × 10-3 kcal/mol·K and T = 25º C = 298.15 K. Show work! The End of the Story … Back to the Lungs When the deoxygenated blood is pumped to the capillary beds located in the alveoli of the lungs, it is then in an O2-rich, but CO2-poor, environment. So, you start exhaling CO2 gas and the reactions shown in the box at the top of the last page reverse. This means that as you exhale CO2 gas carbonic acid concentrations decrease and bicarbonate reenters the erythrocytes (in exchange for Cl- ions that now exit) and combine with protons to form more carbonic acid (which continues to form more CO2 and H2O, etc.). Therefore, the concentrations of CO2, H+ and Cl- in the erythrocytes are all decreasing at the same time that O2 concentrations are now high. The result of this is that Hb molecules start converting to the R state and quickly bind four oxygen gas molecules each. The formation of the bond between each O2 and its heme iron releases or requires (circle one) energy.