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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 oxygen-binding capacity, biochemists use a measure called the P50 value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a P50 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 \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \longrightarrow\) \(\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right] .\)

Short Answer

Expert verified
The equilibrium constant \(K_c\) is about \(1.41\) times larger for the aqueous oxygen-binding reaction in fetal hemoglobin compared to adult hemoglobin.

Step by step solution

01

Identify the relationship between the equilibrium constant \(K_c\), P50, and the partial pressure of oxygen \(({\rm P_{O_2}})\), which is as follows: \(50\% = \frac{K_c \cdot ({\rm P_{O_2}})}{1 + K_c \cdot ({\rm P_{O_2}})}\) We will use this formula to calculate the equilibrium constants for both fetal and adult hemoglobin. #Step 2: Calculate the equilibrium constant for fetal hemoglobin#

Substitute the P50 value of fetal hemoglobin (19 torr) into the formula: \(0.5 = \frac{K_{c,f} \cdot 19}{1 + K_{c,f} \cdot 19}\) Solve for \(K_{c,f}\): \(K_{c,f} = \frac{0.5}{19 - 0.5 \cdot 19} = \frac{0.5}{9.5}\) \(K_{c,f} = 0.0526\) #Step 3: Calculate the equilibrium constant for adult hemoglobin#
02

Substitute the P50 value of adult hemoglobin (\(26.8\) torr) into the formula: \(0.5 = \frac{K_{c,a} \cdot 26.8}{1 + K_{c,a} \cdot 26.8}\) Solve for \(K_{c,a}\): \(K_{c,a} = \frac{0.5}{26.8 - 0.5 \cdot 26.8} = \frac{0.5}{13.4}\) \(K_{c,a} = 0.0373\) #Step 4: Calculate the ratio of equilibrium constants#

Find the ratio between the equilibrium constants of fetal and adult hemoglobin: \(\frac{K_{c,f}}{K_{c,a}} = \frac{0.0526}{0.0373}\) \(\frac{K_{c,f}}{K_{c,a}} \approx 1.41\) So, the equilibrium constant \(K_c\) is about \(1.41\) times larger for the aqueous oxygen-binding reaction in fetal hemoglobin compared to adult hemoglobin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant, symbolized as \(K_c\), is crucial in understanding how easily a reaction occurs at a constant temperature. For hemoglobin, the reaction of interest is its ability to bind oxygen.
In simple terms, \(K_c\) tells us how favorably hemoglobin can bind with oxygen to form oxyhemoglobin. A higher \(K_c\) value signals a greater affinity, meaning hemoglobin binds more readily to oxygen. This is particularly useful when comparing different types of hemoglobin, like fetal and adult hemoglobin.
Here's how the equilibrium constant ties into hemoglobin:
  • Hemoglobin's \(K_c\) for oxygen binding demonstrates how effectively it can capture oxygen from the environment.
  • When calculating \(K_c\), we use the P50 value, which represents the partial pressure of oxygen at which hemoglobin is 50% saturated.
  • Fetal hemoglobin has a higher \(K_c\) compared to adult hemoglobin, illustrating its stronger ability to bind oxygen under the same conditions.
Understanding \(K_c\) provides a deeper insight into hemoglobin's performance in oxygen transport.
Oxygen-Binding Capacity
Oxygen-binding capacity refers to the amount of oxygen hemoglobin can hold. This is a vital feature, especially since hemoglobin's main role is to transport oxygen in the blood.
The ability of hemoglobin to bind oxygen is influenced by several factors, including the structure of the hemoglobin and the surrounding environment. Here’s a closer look:
  • Each hemoglobin molecule has four sites where oxygen molecules can attach.
  • These binding sites work cooperatively. This means that once one oxygen molecule binds, it becomes easier for the next one to attach.
  • Fetal hemoglobin is specifically designed to have a higher oxygen-binding capacity. This ensures that the fetus can effectively capture oxygen from the mother’s blood.
This competitive binding ability makes hemoglobin remarkably efficient at transporting oxygen to where it's needed the most in the body.
P50 Value
The P50 value is a measure used to describe the efficiency of hemoglobin’s ability to bind oxygen. Specifically, it indicates the partial pressure of oxygen at which half of the hemoglobin's capacity is saturated with oxygen.
This metric is incredibly valuable as it provides insight into hemoglobin’s affinity for oxygen under varying conditions.
Let's break it down:
  • A lower P50 value means higher affinity, as hemoglobin reaches half-saturation with less oxygen.
  • Fetal hemoglobin has a lower P50 value (19 torr) compared to adult hemoglobin (26.8 torr), reflecting its superior ability to capture oxygen from maternal blood.
  • The difference in P50 values explains why fetal hemoglobin is more effective in oxygen uptake under the same conditions.
By examining the P50 value, researchers and students gain a better understanding of how hemoglobin behaves in biological and medical contexts.

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Most popular questions from this chapter

Consider the reaction $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \underset{4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \Delta H=-904.4 \mathrm{~kJ}}{\rightleftharpoons} $$ Does each of the following increase, decrease, or leave unchanged the yield of NO at equilibrium? (a) increase \(\left[\mathrm{NH}_{3}\right] ;\) (b) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\) c \()\) decrease \(\left[\mathrm{O}_{2}\right]\); (d) decrease the volume of the container in which the reaction occurs: (e) add a catalyst: (f) increase temperature.

Suppose that the gas-phase reactions \(A \longrightarrow B\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{~s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) ? (b) Which is greater at equilibrium, the partial pressure of \(A\) or the partial pressure of \(B\) ?

Ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with halogens \(\left(\mathrm{X}_{2}\right)\) by the following reaction: $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g) $$ The following figures represent the concentrations at equilibrium at the same temperature when \(\mathrm{X}_{2}\) is \(\mathrm{Cl}_{2}\) (green), \(\mathrm{Br}_{2}\) (brown), and \(\mathrm{I}_{2}\) (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3]

Consider the following equilibrium: $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7} \text { at } 700{ }^{\circ} \mathrm{C} $$ (a) Calculate \(K_{p-}\) (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S}\) ? (c) Calculate the value of \(K_{c}\) if you rewrote the equation \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\).

Consider the hypothetical reaction \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{C}(g)\), for which \(K_{c}=0.25\) at a certain temperature. A \(1.00-\mathrm{L}\) reaction vessel is loaded with \(1.00 \mathrm{~mol}\) of compound \(\mathrm{C}\), which is allowed to reach equilibrium. Let the variable \(x\) represent the number of \(\mathrm{mol} / \mathrm{L}\) of compound \(\mathrm{A}\) present at equilibrium. (a) In terms of \(x\), what are the equilibrium concentrations of compounds \(B\) and \(C\) ? (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibriumconstant expression, derive an equation that can be solved for \(x\). (d) The equation from part (c) is a cubic equation (one that has the form \(a x^{3}+b x^{2}+c x+d=0\) ). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\)-axis is the solution. (e) From the plot in part (d), estimate the equilibrium concentrations of \(A, B\), and \(C\). (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

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