Myoglobin and Hemoglobin

1 ANSWERS Supplementary Problem Set: Myoglobin and Hemoglobin Reminder: Work assigned problems in Chapter 7 1. Where is myoglobin found and what is its biological function? Answer the same question for hemoglobin. Myoglobin in found in skeletal and heart muscle and is a short-term storage protein for oxygen as the oxygen diffuses from the blood stream into the mitochondria and into parts of the cell where oxygen is required for metabolic processes. Myoglobin molecules bind and release oxygen rapidly aiding their transport of oxygen within the cell. Hemoglobin is found in red blood cells and transports oxygen from the lungs to all tissues in the organism. 2. Derive the equation for the proportion of myoglobin that is bound to oxygen (1 ) using the notation MbO 2 and Mb. Substitute for Mb in terms of MbO 2 and show (derive) the alternative form of the equation with the notation [O 2] and P 50. (See lecture notes) 3. Repeat #2 for hemoglobin. 4. Choose one equation from #2 or #3 and show the derivation for the Hill equation. Indicate what form this equation has (liner, hyperbolic, sigmoidal) and sketch the Hill plot for the equation you choose labeling the axes properly. (See lecture notes) 5. What is the numerical value for n in the equation for 1 for myoglobin and how is this value determined from the Hill plot? What does this value of n represent for myoglobin? (See lecture notes) The numerical value for n in the 1 equation for myoglobin is 1.0. This value is determined from the slope of the hill plot (log 1 versus log pO2). The fact that the Hill plot is linear with slope 1.0 is interpreted to mean that only one O 2 binds to myoglobin (n = 1.0) and that there is one value for P 50 (Kd). For myoglobin, the number of sites is only one because there is only one heme in myoglobin! 6. At low pO 2 and at high pO 2, the value of n in the Hill equation is 1.0, but at intermediate values of pO 2, the value is ~ 3.0. What property of hemoglobin accounts for the change in n going from low pO 2 to intermediate pO 2 to high pO 2? Hill equation for Hb: log (bound/free) = n log(pO2) - log (Kd (=1/P50)) The value of n changes from 1.0 to 3.0 in going from the low pO2 to intermediate pO2 because after the first O 2 is bound, the affinity of Hb for additional O 2 increases (P 50 increases, K d (=1/P50) decreases) so that the y-intercept has to change as a reflection of the increase in affinity of the other heme sites. Additional perspective: Hemoglobin is a tetramer having four binding sites for oxygen. The affinity of each heme for O 2 changes (increases) upon the binding of (or with the presence of) O 2 at other heme sites. This means that the value of P 50 increases (K d decreases) as successive O 2 are bound to the hemes. At low pO2, it is likely that only one site per hemoglobin tetramer (on the average) is bound to O 2 and each of these sites has an identical low affinity for O 2, in other words, the same low P 50. This situation gives rise to the portion of the Hill plot where the value of n = 1.0 and a y-intercept showing a lower P 50 value. At higher values of pO2, three sites per hemoglobin (on the average) will be bound to O 2 so that when the fourth O 2 binds, it binds to a high affinity site on each tetramer where the molecules of Hb have the same affinity for O 2, or same high P 50 value. This situation also gives a Hill plot with slope of 1.0 but with a higher y-intercept giving a higher P 50 value. In the intermediate pO2 range, there are mixtures of hemoglobin tetramers with 1, 2, 3, and 4 O 2 molecules bound so that the measured value of the slope (n) represents a transitional slope between the low affinity sites (with low P 50) and the high affinity sites (with low P 50). Since the Hill plot is derived from an approximate (simplified) equation for binding of O 2 to hemoglobin, the interpretation cannot be exact. Note that the value of n for the Hill plot for hemoglobin never reaches 4 as implied by the simplified chemical reaction (and its accompanying equations for 1 and the Hill equation) so that the simplified chemical reaction (and its accompanying Hill equation and plot) are not an exact description of the binding to O 2 to hemoglobin! See #7. 2 ANSWERS Supplementary Problem Set: Myoglobin and Hemoglobin 7. How does the two state model for binding of O 2 to hemoglobin explain the sigmoidal dependence of the binding of O 2 to this protein? The two-state model for the binding of O 2 to hemoglobin explains the sigmoidal dependence of the binding plot because it accounts for the low affinity of Hb for O 2 at low pO2 where the T-state, or deoxy state is the predominant form as well as the transition to the high affinity for O 2 at high pO2 where the R-state, or oxy state, is the predominant form. Additional perspective: If it were the only form present, the T-state would give a "low-lying" hyperbolic plot; similarly if it were the only form present, the R-state would give a "high-lying" hyperbolic plot (see plot at right). Because it includes the transition between the low affinity state (T-state) and the high affinity state (R-state), the two-state model explains the sigmoidal appearance of the binding plot! A sigmoidal plot can actually be derived from the proper mathematical weighting of the concentration of these two forms as the pO2 changes, as well as a proper assignment of two different values of P 50 that would be assigned to them. Needless to say, your text book does not present this mathematical analysis! 8. Describe the changes that happen to the following groups or interactions in hemoglobin when oxygen binds to deoxyhemoglobin. a. Salt bridge between groups on the H and F-helices: H-asp-CO2-.......+HN