The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each \(\mathrm{Hb}\) can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) -binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygenbinding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of 19 torr, and adult hemoglobin has a P50 value of 26.8 torr. Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \longrightarrow\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) .

We are given the following information: - Fetal hemoglobin \(P50\) value: 19 torr - Adult hemoglobin \(P50\) value: 26.8 torr

The reaction equation is given as \[4 O_2(g) + Hb(aq) \rightleftharpoons [Hb(O_2)_4(aq)]\] The expression for the equilibrium constant \(K_c\) for this reaction is: \[K_c = \frac{[\mathrm{Hb}(O_2)_4]}{[\mathrm{Hb}](\mathrm{P}_{O_2})^4}\]

At 50% saturation, half of the hemoglobin molecules are bound with oxygen, and half are unbound. Therefore, we have: For fetal hemoglobin, \(\frac{1}{2}[\mathrm{Hb}] = [\mathrm{Hb}(O_2)_{4}] \Rightarrow [\mathrm{Hb}] = 2[\mathrm{Hb}(O_2)_{4}]\) \(P_{O_2} = 19~\text{torr}\) For adult hemoglobin, \(\frac{1}{2}[\mathrm{Hb}] = [\mathrm{Hb}(O_2)_{4}] \Rightarrow [\mathrm{Hb}] = 2[\mathrm{Hb}(O_2)_{4}]\) \(P_{O_2} = 26.8~\text{torr}\)

We will now express \(K_c\) in terms of the given quantities and then find the ratio of \(K_c\) for fetal and adult hemoglobin. For fetal hemoglobin, \[K_{c_{\text{fetal}}} = \frac{[\mathrm{Hb}(O_2)_4]}{[\mathrm{Hb}](\mathrm{P}_{O_2})^4} = \frac{1}{2} \cdot \frac{1}{(19 \ \text{torr})^4}\] For adult hemoglobin, \[K_{c_{\text{adult}}} = \frac{[\mathrm{Hb}(O_2)_4]}{[\mathrm{Hb}](\mathrm{P}_{O_2})^4} = \frac{1}{2} \cdot \frac{1}{(26.8 \ \text{torr})^4}\] Now, we find the ratio between \(K_{c_{\text{fetal}}}\) and \(K_{c_{\text{adult}}}\): \[\frac{K_{c_{\text{fetal}}}}{K_{c_{\text{adult}}}} = \frac{1/(2 \cdot (19 \ \text{torr})^4)}{1/(2 \cdot (26.8\ \text{torr})^4)} = \left(\frac{26.8}{19}\right)^4\] Using a calculator, we get \[\frac{K_{c_{\text{fetal}}}}{K_{c_{\text{adult}}}} \approx 3.95\] So, the equilibrium constant \(K_c\) for the aqueous reaction is approximately 3.95 times larger for fetal hemoglobin than for adult hemoglobin.