To solve this question we need to first look at the relationship between hydroxide and hydrogen ion concentrations:

where is concentration of hydrogen ions and is the concentration of hydroxide ions. Solve for :

We can solve for the concentration of hydrogen ions in this solution by writing the dissociation of hydrochloric acid in solution:

The question states that it is a 1M solution. Since and are in 1:1 ratio, the concentration of hydrogen ions will equal the concentration ; therefore, . Note that this won’t be the case if the acid was weak. For a weak acid, the concentration of will be less than the concentration of acid, even if it was 1:1 ratio. This occurs because a weak acid does not dissociate completely.

We can now use the to solve for .

Therefore, concentration of hydroxide ions in this solution is

To solve this question we need to first look at the relationship between hydroxide and hydrogen ion concentrations:

where is concentration of hydrogen ions and is the concentration of hydroxide ions. Solve for :

We can solve for the concentration of hydrogen ions in this solution by writing the dissociation of hydrochloric acid in solution:

The question states that it is a 1M solution. Since and are in 1:1 ratio, the concentration of hydrogen ions will equal the concentration ; therefore, . Note that this won’t be the case if the acid was weak. For a weak acid, the concentration of will be less than the concentration of acid, even if it was 1:1 ratio. This occurs because a weak acid does not dissociate completely.

We can now use the to solve for .

Therefore, concentration of hydroxide ions in this solution is

To solve this question we need to first look at the relationship between hydroxide and hydrogen ion concentrations:

where is concentration of hydrogen ions and is the concentration of hydroxide ions. Solve for :

We can solve for the concentration of hydrogen ions in this solution by writing the dissociation of hydrochloric acid in solution:

The question states that it is a 1M solution. Since and are in 1:1 ratio, the concentration of hydrogen ions will equal the concentration ; therefore, . Note that this won’t be the case if the acid was weak. For a weak acid, the concentration of will be less than the concentration of acid, even if it was 1:1 ratio. This occurs because a weak acid does not dissociate completely.

We can now use the to solve for .

Therefore, concentration of hydroxide ions in this solution is

NADH is the first electron carrier to be oxidized by the electron transport chain, a process that occurs at complex I. FADH2 is oxidized further down the chain in complex II, causing it to produce less ATP on average than NADH.

The question states that the cell undergoes aerobic respiration. This means that the products from anaerobic respiration (glycolysis) will go through Krebs cycle and electron transport chain (aerobic respiration) to generate ATP. Lactate dehydrogenase is an enzyme important for converting the pyruvate molecules (from glycolysis) to lactate and oxidizing NADH. This reaction occurs in anaerobic fermentation when there is tissue hypoxia (decrease in oxygen).

If this inhibitor was placed in a cell that is deprived of oxygen, then there would be a buildup of pyruvate and NADH; however, since the inhibitor is added to cells undergoing aerobic respiration there will be no buildup. The pyruvate and NADH will undergo aerobic respiration and generate ATP. Note that red blood cells (RBCs) are unique in that they only use anaerobic respiration for ATP; therefore, adding lactate dehydrogenase inhibitor to RBCs will lead to a buildup of pyruvate and NADH.

NADH is the first electron carrier to be oxidized by the electron transport chain, a process that occurs at complex I. FADH2 is oxidized further down the chain in complex II, causing it to produce less ATP on average than NADH.

NADH is the first electron carrier to be oxidized by the electron transport chain, a process that occurs at complex I. FADH2 is oxidized further down the chain in complex II, causing it to produce less ATP on average than NADH.

The question states that the cell undergoes aerobic respiration. This means that the products from anaerobic respiration (glycolysis) will go through Krebs cycle and electron transport chain (aerobic respiration) to generate ATP. Lactate dehydrogenase is an enzyme important for converting the pyruvate molecules (from glycolysis) to lactate and oxidizing NADH. This reaction occurs in anaerobic fermentation when there is tissue hypoxia (decrease in oxygen).

If this inhibitor was placed in a cell that is deprived of oxygen, then there would be a buildup of pyruvate and NADH; however, since the inhibitor is added to cells undergoing aerobic respiration there will be no buildup. The pyruvate and NADH will undergo aerobic respiration and generate ATP. Note that red blood cells (RBCs) are unique in that they only use anaerobic respiration for ATP; therefore, adding lactate dehydrogenase inhibitor to RBCs will lead to a buildup of pyruvate and NADH.

NADH is the first electron carrier to be oxidized by the electron transport chain, a process that occurs at complex I. FADH2 is oxidized further down the chain in complex II, causing it to produce less ATP on average than NADH.

The question states that the cell undergoes aerobic respiration. This means that the products from anaerobic respiration (glycolysis) will go through Krebs cycle and electron transport chain (aerobic respiration) to generate ATP. Lactate dehydrogenase is an enzyme important for converting the pyruvate molecules (from glycolysis) to lactate and oxidizing NADH. This reaction occurs in anaerobic fermentation when there is tissue hypoxia (decrease in oxygen).

If this inhibitor was placed in a cell that is deprived of oxygen, then there would be a buildup of pyruvate and NADH; however, since the inhibitor is added to cells undergoing aerobic respiration there will be no buildup. The pyruvate and NADH will undergo aerobic respiration and generate ATP. Note that red blood cells (RBCs) are unique in that they only use anaerobic respiration for ATP; therefore, adding lactate dehydrogenase inhibitor to RBCs will lead to a buildup of pyruvate and NADH.