Respiration in Plants Test

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Respiration in Plants Test - 52

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Question 1
1 / -0

During conversion of pyruvic acid into acetyl CoA, pyruvic acid is

A
Reduced

B
Oxidised

C
Isomerised

D
Condensed

Solution

The pyruvic acid generated in the glycolysis enters into the mitochondrial matrix, undergoes oxidative decarboxylation and forms active acetate in a coordinated series of reactions catalyzed by the multienzyme complex. The reaction requires the participation of an enzyme pyruvate dehydrogenase and several coenzymes including NAD+, coenzyme A, thiamine pyrophosphate and lipoic acid.
Thus, the correct answer is option B.
Question 2
1 / -0

Kreb's cycle produces

A
2CO$$_2$$

B
3CO$$_2$$

C
4CO$$_2$$

D
6CO$$_2$$

Solution

During the process of Krebs cycle, two molecules of carbon dioxide are produced by the process of oxidative decarboxylation.
1. Oxidation of isocitrate to α- ketoglutarate- Isocitrate gives off a pair of hydrogen atoms and a molecule of carbon dioxide and becomes 5-carbon α-ketoglutarate. The pair of hydrogen atoms gives two electrons and one H+ to NAD+ forming NADH + H+. The reaction is catalyzed by enzyme isocitrate dehydrogenase in the presence of manganese ions.
2. Oxidation of α-ketoglutarate to succinyl coenzyme A- In this reaction, coenzyme A reacts with α-ketoglutarate, forming 4-carbon succinyl coenzyme A and releasing carbon dioxide and a pair of hydrogen atoms. The reaction is catalyzed by α-ketoglutarate dehydrogenase complex enzyme. The pair of hydrogen atoms passes two electrons and one H+ to NAD+ forming NADH+ H+.
Thus, the correct answer is option A.
Question 3
1 / -0

Oxidation of pyruvate forms

A
Acetyl CoA

B
NADH

C
$$CO_2$$

D
All the above

Solution

The pyruvic acid generated in the cytoplasm by the process of glycolysis enters into mitochondrial matrix and undergoes oxidative decarboxylation and forms active acetate in a coordinated series of reactions catalysed by multienzyme complex. The reaction requires participation of an enzyme pyruvate dehydrogenase and several coenzymes including NAD+, coenzyme A, thiamine pyrophosphate and lipoic acid. The above equation occurs in 4 steps involving conversion of pyruvic acid to acetyl coenzyme A, one molecule of carbon dioxide is released. The coenzyme NAD+ is reduced to NADH + H+ in the reaction. The acetyl coenzyme A is accepted by Kreb's cycle.
Thus, the correct answer is option d.
Question 4
1 / -0

A complex enzyme system functional in Krebs cycle is

A
Citrate synthetase

B
Isocitrate dehydrogenase.

C
Oxalosuccinate decarboxylase

D
$$\alpha$$-ketoglutarate dehydrogenase

Solution

Alpha-ketoglutarate dehydrogenase is a complex enzyme system functional in Krebs cycle and catalyzes the 5th step of Krebs cycle. In this coenzyme A reacts with α- ketoglutarate forming 4-carbon succinyl coenzyme A and releasing carbon dioxide and a pair of hydrogen atoms. The pair of hydrogen atoms pairs two electrons and one proton to NAD +, forming NADH and H+.
Thus, the correct answer is option D.
Question 5
1 / -0

When succinate is oxidized in Krebs cycle, it's hydrogen is accepted by

A
NAD$$^+$$

B
FAD

C
FMN

D
Fumarate

Solution

The sixth step of Krebs cycle converts succinate into 4-carbon fumarate with the aid of an enzyme succinate dehydrogenase and liberates a pair of hydrogen atoms. The latter pass to FAD forming FADH2.
Thus, the correct answer is option B.
Question 6
1 / -0

Number of ATP molecules synthesised through substrate level phosphorylation during aerobic respiration of one glucose molecule is

A
8

B
6

C
4

D
2

Solution

Substrate level phosphorylation is a process of formation of ATP by the physical addition of phosphate group to ADP. In this process, a phosphate group is taken from an intermediate compound referred to as ADP and given to ADP. In glycolysis, 4 ATP molecules are formed as a result of substrate level phosphorylation.
1. First two ATP are formed when 1,3- diphosphoglycerate changes to 3-phosphoglycerate due to loss of phosphate group.
2. Another two ATP are formed in the last step of glycolysis when 2-phosphoenol pyruvic acid changes to pyruvic acid transferring its high energy phosphate bond to ADP.
Two ATP molecules are formed during Krebs cycle by substrate level phosphorylation when succinyl coenzyme A transfers its high energy to a phosphate group that joins GDP/ADP to form GTP/ATP.
Thus, a total of 6 ATP molecules are formed through substrate level phosphorylation during aerobic respiration of one glucose molecule.
Thus, the correct answer is option B.
Question 7
1 / -0

Hexose monophosphate pathway takes place in

A
Mitochondrial matrix

B
Cristae

C
Cytoplasm

D
E.R

Solution

The reactions of hexose monophosphate shunt occur exclusively in the cytoplasm and is an additional pathway of glucose metabolism which starts with glucose-6-phosphate. About 10-25% of glucose is metabolised by this pathway. In plants, it occurs both in cytosol and plastids. This pathway involves both oxidative and nonoxidative steps. The oxidative steps generate NADPH and ribose-5-phosphate and then non-oxidative steps convert ribose-5-phosphate back into glucose-6-phosphate which can be further utilized for NADPH generations by means of oxidative steps.
Thus, the correct answer is option C.
Question 8
1 / -0

Fat has two components, glycerol and fatty acids. They enter common pathway of respiration as

A
DiHAP and $$\alpha$$-ketoglutarate

B
DiHAP and acetyle CoA

C
Glyceric acid and acetyl CoA

D
Glyceric acid and $$\alpha$$-ketoglutarate

Solution

When fats are the respiratory substrates, they are first converted to fatty acids and glycerol in the presence of enzyme lipase. Glycerol is phosphorylated to glycerol-3-phosphate in the presence of enzyme glycerol kinase.
Glycerol + ATP → Glycerol-3-phosphate + ADP
Glycerol-3-phosphate is then converted to dihydroxyacetone phosphate by enzyme glycerol phosphate dehydrogenase which enters into glycolysis.
Glycerol-3-phosphate + NAD⁺→ Dihydroxyacetone phosphate + NADH + H⁺
^{The fatty acids are converted to acetyl coenzyme A by beta oxidation. The acetyl coenzyme A enters into Krebs cycle and oxidised to carbon dioxide and water.}
^{Thus, the correct answer is option B.}
Question 9
1 / -0

A complex enzyme system of mitochondria functional outside Krebs cycle is

A
Pyruvate kinase

B
Pyruvate dehydrogenase

C
Enolase

D
$$\alpha$$ - Ketoglutarate dehydrogenase

Solution

Pyruvic acid and NADH2 formed anaerobically in the cytoplasm can release their energy only in aerobic conditions and in the presence of enzymes present inside the mitochondria. So they enter mitochondria where oxidative decarboxylation occurs. It occurs in the presence of a hydrogen acceptor (NAD+), pyruvic acid dehydrogenase complex with many coenzymes such as coenzyme A, thiamine pyrophosphate, and lipoic acid. The acetyl coenzyme A produced in the reaction then enters Krebs cycle.
Pyruvate kinase and enolase both are enzymes which catalyze the last two steps of glycolysis and are present in the cytoplasm of a cell. Alpha ketoglutarate dehydrogenase is an enzyme of Krebs cycle and catalyzes the oxidative decarboxylation of alpha ketoglutarate into succinyl coenzyme A.
Thus, the correct answer is option B.
Question 10
1 / -0

Tricarboxylic acids of Kreb's cycle are

A
Succinic acid, Fumaric acid and Citric acid

B
Oxalosuccinic acid, Citric acid and $$\alpha$$-ketoglutaric acid.

C
Citric acid, Isocitric acid and Malic acid

D
Citric acid, Isocitric acid and Oxalosuccinic acid

Solution

Krebs cycle consists of 8 cyclic steps which begin with the reaction of two carbon compound acetyl coenzyme A (produced in glycolysis) with oxaloacetic acid (4C) in the presence of water forming 6C citric acid and freeing coenzyme A for reuse in pyruvate oxidation. The citric acid formed has 3 carboxyl groups and is known as tricarboxylic acid. Citric acid is then reorganized to form the isocitric acid (6C) which retains 3 carboxylic groups. Isocitrate undergoes dehydrogenation to form 6C oxalosuccinate. Thus all the three - citric acid, isocitric acid and oxalosuccinic acid contain 3 carboxylic acids and are called tricarboxylic acids.
Thus, the correct answer is option D.