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Reika [66]
3 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]3 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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In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so
alex41 [277]

Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: V_{1} = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

M_{1} = 0.321 M,       V_{2} = 21.8 mL = 0.0218 L,      M_{2} = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}

Substitute the values into above formula as follows.

[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M

Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.

4 0
3 years ago
Viết công thức hóa học tạo bởi Mg và O
jasenka [17]
Phương trình hóa học: 2Mg+O2->2Mgo
3 0
3 years ago
What type of bond holds the complementary bases together?.
olga55 [171]

Answer:

hydrogen bonding

Explanation:

The two strands of DNA are held together by hydrogen bonds between complimentary nucleotides.

3 0
2 years ago
What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso
Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247

The pH of the buffer can be known as

pH = pK_{a} + log[\frac{[A-]}{[HA]}}]

The concentration of [A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\

Similarly, the concentration of [HA] = \frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236

So, the pH of the buffer is 6.1236.

5 0
3 years ago
How many moles of CaCl2 are contained in 0.448 L of a 0.85 M CaCl2 solution?
kati45 [8]

Answer:

0.3808

Explanation:

number of moles,n=Conc.XVol.

hence 0.85X0.448

3 0
3 years ago
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