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Reika [66]
2 years ago
6

Select all statements that correctly describe hemoglobin and myoglobin structure. a. Molecular oxygen binds irreversibly to the

Fe ( II ) Fe(II) atom in heme. b. Each hemoglobin or myoglobin molecule can bind four oxygen molecules. c. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron. d. Each iron atom can form six coordination bonds. One of these bonds is formed between iron and oxygen. e. Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron ( Fe ) (Fe) atom. f. Hemoglobin is a heterotetramer, whereas myoglobin is a monomer. The heme prosthetic group is entirely buried within myoglobin.
Chemistry
1 answer:
DIA [1.3K]2 years ago
7 0

Answer:

c. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron.

e. Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron ( Fe ) (Fe) atom.

f. Hemoglobin is a heterotetramer, whereas myoglobin is a monomer. The heme prosthetic group is entirely buried within myoglobin.

Explanation:

The differences between hemoglobin and myoglobin are most important at the level of quaternary structure. Hemoglobin is a tetramer composed of two each of two types of closely related subunits, alpha and beta. Myoglobin is a monomer (so it doesn't have a quaternary structure at all). Myoglobin binds oxygen more tightly than does hemoglobin. This difference in binding energy reflects the movement of oxygen from the bloodstream to the cells, from hemoglobin to myoglobin.

Myoglobin binds oxygen

The binding of O 2 to myoglobin is a simple equilibrium reaction:

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Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

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The equation of the reaction is given as :

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 [Pb²⁺] =   s

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K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

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