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Whitepunk [10]
3 years ago
15

Nitrate acts as a poison by interfering with oxidative enzymes. oxidizing hemoglobin to methemoglobin, which cannot carry oxygen

. reacting with enzymes that break down toxic wastes in the liver. blocking cell protein synthesis.
Biology
1 answer:
xxMikexx [17]3 years ago
7 0

Explanation:

Methemoglobin (metHb) is an oxidized derivative of hemoglobin in which heme iron is in the ferric (Fe3+) or oxidized state rather than the ferrous (Fe2+) or reduced state. Small amounts of methemoglobin normally are formed daily, associated with the release of oxygen from hemoglobin (auto-oxidation).

Methemoglobin cannot bind oxygen, which means it cannot carry oxygen to tissues. It is bluish chocolate-brown in color. In human blood a trace amount of methemoglobin is normally produced spontaneously, but when present in excess the blood becomes abnormally dark bluish brown. The NADH-dependent enzyme methemoglobin reductase (a type of diaphorase) is responsible for converting methemoglobin back to hemoglobin.

Normally one to two percent of a person's hemoglobin is methemoglobin; a higher percentage than this can be genetic or caused by exposure to various chemicals and depending on the level can cause health problems known as methemoglobinemia. A higher level of methemoglobin will tend to cause a pulse oximeter to read closer to 85% regardless of the true level of oxygen saturation. An abnormal increase of methemoglobin will increase the oxygen binding affinity of normal hemoglobin, resulting in a decreased unloading of oxygen to the tissues.

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This very low concentration of the desired product would be unfavorable for glycolysis. In fact the reaction is coupled to ATP h
abruzzese [7]

Answer:

ΔG'° for the coupled reaction = -16.7 kJ/mol

<em>Note: The question is missing some parts. The complete question is as follows:</em>

<em>The first reaction in glycolysis is the phosphorylation of glucose: </em>

<em> Pi+glucose⟶glucose−6−phosphate+H2O </em>

<em>This is a thermodynamically unfavorable process, with ΔG∘′= +13.8kJ/mol. In a liver cell at 37 ∘C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each.</em>

<em>This very low concentration of the desired product would be unfavorable for glycolysis. In fact the reaction is coupled to ATP hydrolysis to give the overall reaction: ATP + glucose → glucose-6-phosphate + ADP + H+</em>

<em>What is the ΔG'° for the coupled reaction?</em>

Explanation:

The coupling of ATP hydrolysis which is a thermodynamically favourable reaction to the phosphorylation of glucose makes it favourable. Since the two reactions constitute a sequential reaction, their standard free energy changes are additive.

For ATP hydrolysis: ATP + H₂O ---> ADP + Pi ; ΔG'°  = -30.5KJ/mol

For phosphorylation of glucose:  Pi + glucose⟶glucose−6−phosphate + H2O  ; ΔG'° = +13.8 kJ/Mol

For the overall reaction: ATP + glucose → glucose-6-phosphate + ADP + H⁺ ; ΔG'° = 13.8 + (-30.5) kJ/mol = -16.7 kJ/mol

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Assuming seed germination.

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