Respiration in Plants Test

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Respiration in Plants Test - 88

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Question 1
1 / -0

Fructose 1 : 6 biphosphate splits into two triose phosphates by enzyme

A
Aldolase

B
Amylase

C
Zymase

D
Lipase

Solution

In the process of glycolysis, the molecule of glucose is phosphorylated twice to form fructose-1,6-biphosphate. It is then acted upon by aldolase and splits into two triose phosphates: 3-phosphoglyceraldehyde and dihydroxyacetone phosphate. Both these compounds are tautomers in dynamic equilibrium and can interchange in presence of phosphotriose isomerase.
So the answer is 'Aldolase'.
Question 2
1 / -0

Number of NADH produced during breakdown of one molecule of glucose to 1 : 3 diphosphoglycerate stage is

A
6

B
4

C
3

D
2
Solution
A.Correct option D
B.Explanation for correct option:
One molecule of glucose forms two molecules of pyruvic acid as the end product of glycolysis.
At the 6$$^{th}$$ step of glycolysis, 2 molecules of 1,3 diphosphoglyceraldehyde convert to 2 molecules of 1,3 diphosphoglyceric acid by converting 2 NAD+ to 2 NADH$$_2$$.
So, 2 molecules of NADH$$_2$$ are produced which yields 6 ATP through ETS during aerobic respiration.
Question 3
1 / -0

Krebs cycle forms an important product

A
Acetyl CoA.

B
ADP.

C
ATP.

D
Water.

Solution

Krebs cycle explains how living organisms generate energy. It describes transfer of adenosine diphosphate (ADP) to adenosine triphosphate (ATP).
So the correct option is 'ATP'.
Question 4
1 / -0

Krebs cycle begins with the reaction

A
OAA + Acetyl CoA

B
Citric acid + Acetyl CoA

C
OAA + Pyruvic acid

D
OAA + Citric acid

Solution

Krebs' cycle is a part of cellular respiration in which the end product acts as a substrate for the next cycle to begin. Oxaloacetic acid, the end product of the cycle again enters the cycle by reacting with acetyl CoA to form citric acid. It releases CoA and the cycle starts with this reaction.
So the answer is 'OAA + Acetyl CoA'.
Question 5
1 / -0

Enzyme pair common to both EMP and $$C_3$$ cycle is

A
Aldolase and enolase

B
Aldolase and triose phosphate isomerase

C
Phosphoglyceromutase and triose phosphate isomerase

D
Cytochrome oxidase and enolase

Solution

In the EMP pathway or glycolysis, a 6-carbon fructose 1,6- biphosphate is formed which is acted upon by aldolase and phosphotriose isomerase to form 3-PGA and DHAP in dynamic equilibrium. This same pair of an enzyme is also employed in the C3 cycle in which a 6-carbon intermediate is decayed by it to form two molecules of 3-PGA.

So the answer is 'Aldolase and triose phosphate isomerase'.
Question 6
1 / -0

Isocitric acid is changed to 2-oxoglutaric acid by

A
Oxidative carboxylation

B
Oxidative decarboxylation

C
Dehydrogenation

D
Hydrogenation and decarboxylation

Solution

Conversion of isocitric acid to 2-oxoglutaric acid in Krebs' cycle occurs in two steps by two different enzymes. Isocitric dehydrogenase converts isocitric acid to oxalosuccinic acid by removing H2 (oxidation). It is then converted to 2-oxoglutaric acid by action of oxalosuccinic decarboxylase by removal of CO2 (decarboxylation).
So, the correct answer is 'Oxidative decarboxylation'.
Question 7
1 / -0

In TCA cycle, conversion of succinyl CoA to succinic acid needs

A
GDP + Pi

B
CoA + GTP + Pi

C
Acetyl CoA + GDP +Pi

D
Acetyl CoA + GTP + Pi

Solution

During the conversion of succinyl CoA to succinic acid needs GDP + Pi which converted to GTP.
So the correct option is 'GDP + Pi.'
Question 8
1 / -0

ETC and TCA enzymes occur in

A
Ribosomes.

B
Endoplasmic reticulum.

C
Mitochondria.

D
Cytoplasm and nucleus.

Solution

Electron Transport Chain and the Tricarboxylic acid cycle is also called as Krebs cycle or Citric acid cycle in which enzymes for this process is found in Mitochondria.
So the correct option is 'Mitochondria'.
Question 9
1 / -0

RQ of sprouting potato is

A
1

B
>1

C
<1

D
Zero

Solution

RQ value = volume of CO$$_2$$ evolved/volume of O$$_2$$ consumed.
Potato contains carbohydrate for which RQ value is 1.
So the correct option is '1.'
Question 10
1 / -0

RQ is less than one for

A
Carbohydrate

B
Organic acid

C
Starch

D
Protein

Solution

R.Q. is the ratio of amount of CO$$_2$$ produced to the amount of O$$_2$$ consumed by the cell for respiration. R.Q. value of protein is 0.8 as the volume of carbon dioxide released is less than the amount of oxygen. Carbohydrates, organic acid and starch have R.Q. of 1 or greater than 1.
So, the correct answer is 'Protein'.

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Respiration in Plants Test - 88

Result

Respiration in Plants Test - 88

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SHARING IS CARING

Question 1

Aldolase

Amylase

Zymase

Lipase

Solution

Question 2

6

4

3

2

Solution

Question 3

Acetyl CoA.

ADP.

ATP.

Water.

Solution

Question 4

OAA + Acetyl CoA

Citric acid + Acetyl CoA

OAA + Pyruvic acid

OAA + Citric acid

Solution

Question 5

Aldolase and enolase

Aldolase and triose phosphate isomerase

Phosphoglyceromutase and triose phosphate isomerase

Cytochrome oxidase and enolase

Solution

Question 6

Oxidative carboxylation

Oxidative decarboxylation

Dehydrogenation

Hydrogenation and decarboxylation

Solution

Question 7

GDP + Pi

CoA + GTP + Pi

Acetyl CoA + GDP +Pi

Acetyl CoA + GTP + Pi

Solution

Question 8

Ribosomes.

Endoplasmic reticulum.

Mitochondria.

Cytoplasm and nucleus.

Solution

Question 9

1

>1

<1

Zero

Solution

Question 10

Carbohydrate

Organic acid

Starch

Protein

Solution

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