Respiration in Plants Test

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Respiration in Plants Test - 77

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Weekly Quiz Competition

Question 1
1 / -0

Pacemaker enzyme of EMP pathway is

A
Hexokinase

B
Pyruvate kinase

C
Enolase

D
Phosphofructokinase

Solution

EMP pathway is also called as glycolysis. It is the first step of aerobic respiration. In this, 2 hexokinase enzymes are involved, namely phosphofructokinase and pyruvate kinase. The reactions catalyzed by this enzyme are irreversible. Hence, it is also called as key enzyme or pacemaker enzyme. It catalyzes the phosphorylation of glucose, phosphorylation of fructose-6-phosphate and transfer of phosphate from phosphoenolpyruvate to ADP.
Thus, the correct answer is 'Hexokinase.'
Question 2
1 / -0

The number of ATP produced when one molecule of pyruvic acid is used as respiratory substrate in Kreb's cycle?

A
$$15$$

B
$$12$$

C
$$3$$

D
$$36$$

Solution

When one molecule of pyruvic acid is used as respiratory substrate in Kreb's cycle, 15 ATP molecules are produced. In Kreb's cycle, 1 molecule of FADH2 and 4 molecules of NADH are produced. One molecule of FADH2 produces 2 ATP molecules in ETS. Four molecules of NADH produces 4 x 3 = 12 molecules of ATP. Also, 1 molecule of ATP is produced by substrate level phosphorylation. Hence, total number of ATP molecules produced by 1 molecule of pyruvic acid during Krebs cycle = 12+2+1 = 15.
Thus, the correct answer is '15.'
Question 3
1 / -0

What could be the reason for maida flour to increase in size?

A
Fermentation

B
Preservation

C
Fumigation

D
Both A and B

Solution

Fermentation is an anaerobic process by which glucose is broken down into alcohol, carbon dioxide, hydrogen gas. Yeast is used during baking process. This process is responsible for the increasing the size of maida dough because yeast converts sugar into carbon dioxide and ethanol. Carbon dioxide gas is trapped by gluten proteins which causes dough to rise.
Thus, the correct answer is 'Fermentation.'
Question 4
1 / -0

In how many steps, glycolysis can be regulated?

A
One

B
Two

C
Three

D
Four

Solution

Glycolysis is the process of breakdown of glucose into pyruvate to extract energy for various cellular metabolic processes. Enzymes that catalyze the three steps of glycolysis have very large negative G value, hexokinase (or glucokinase) for step 1, phosphofructokinase for step 3, and pyruvate kinase for step 10, are the regulatory steps for allosteric enzyme regulation. Generally, enzymes that catalyzing irreversible steps in the metabolic pathway have the potential for regulatory control because there is some blockage elsewhere in a pathway that leads to the accumulation of intermediates.
So, the correct answer is 'Three'.
Question 5
1 / -0

At how many step there occurs substrate level phosphorylation in glycolysis?

A
One

B
Two

C
Three

D
Four

Solution

The formation of ATP, by the transfer of phosphate from an organic compound to ADP, is called substrate level phosphorylation. In glycolysis, at two steps substrate level, phosphorylations occur.

So, the correct answer is ‘at two steps substrate level phosphorylation occurs in glycolysis’ this should contain the option which is correct.
Question 6
1 / -0

Which one is not correct with given marks?

A
B-carboxylation- fixation of $$CO_2$$ the aldose sugar

B
C- reduction - reverse of glycolysis

C
A- regeneration - reformation of pentose sugar

D
C- reduction - carbohydrate formation

Solution

The Calvin cycle (also known as the Benson-Calvin cycle) is the set of chemical reactions that take place in chloroplast during photosynthesis. It involves 3 steps, they are carbon fixation, reduction and regeneration. RuBP is a ketone containing sugar with 5 carbon atoms. In the first step, The CO2 molecule condenses with ribulose 1,5-bisphosphate to form an unstable six-carbon compound, which is rapidly hydrolyzed to two molecules of 3-phosphoglycerate. Rubisco(ribulose 1,5-bisphosphate carboxylase/oxygenase) shows its oxygenase activity by catalyzing this reaction.
So, the correct answer is 'option A'.
Question 7
1 / -0

Which of the following statements regarding pyruvate during aerobic respiration is incorrect?

A
It is formed by the glycolytic catabolism of carbohydrates in the mitochondrial matrix.

B
It enters mitochondrial matrix and undergoes oxidative decarboxylation.

C
The conversation of pyruvic acid to acetyl CoA is catalyzed by pyruvic dehydrogenase.

D
Two molecules of NADH are produced from the metabolism of one glucose into two molecules of pyruvic acid.
Solution
- Glycolysis is the process in which one glucose molecule is broken down to form two molecules of pyruvate.
- Glycolysis takes place in the cytoplasm of a cell as the first step in cellular respiration.
- If oxygen is available, aerobic respiration will go forward.
- In eukaryotic cells, the pyruvate molecules produced at the end of glycolysis are transported into mitochondria.
- Pyruvate decarboxylation or pyruvate oxidation, also known as the link reaction (or Oxidative decarboxylation of Pyruvate), is the conversion of pyruvate into acetyl-CoA by the enzyme complex pyruvate dehydrogenase complex. So, both statements 2 and 3 are correct.
- Overall, the process of glycolysis produces a net gain of two pyruvate molecules, two ATP molecules, and two NADH molecules for the cell to use for energy.
Hence, statement 1 is incorrect.
Question 8
1 / -0

In anaerobic respiration of yeast

A
$${ O }_{ 2 }$$ is given out

B
$${ CO }_{ 2 }$$ is given out

C
$${ CO }_{ 2 }$$ is taken in

D
$${ O }_{ 2 }$$ is taken in

Solution

Yeast respires anaerobically (in the absence of oxygen) . The process is known as fermentation. Glucose is the respiratory substrate.
The glucose is converted to pyruvate by the process of glycolysis. This process is independent of oxygen. If oxygen is absent, the pyruvate is broken by yeast into ethanol releasing carbon dioxide along with the production of 2 ATP.
So, the correct answer is 'CO$$_2$$ is given out'.
Question 9
1 / -0

How many total dehydrogenetaion occur in citric acid cycle :-

A
One

B
Two

C
Three

D
Four

Solution

Krebs cycle is a stepwise cyclic oxidation process in which four dehydrogenation steps and two decarboxylations steps of active acetate group takes place to produce reduced co-enzymes and Carbon Dioxide.

Step 3: Dehydrogenation
Now isocitrate undergoes dehydrogenation in the presence of an enzyme isocitrate dehydrogenase. Mn2+ ion is required for the functioning of enzyme. Hydrogen given out by isocitrate is picked up by NAD+ (Nicotinamide adenine dinucleotide) to form NADH2. After losing hydrogen, isocitrate is changed into oxalosuccinate (6C).

Step 5: Oxidative Decarboxylation
In this step 5-carbon compound, α – Ketoglutarate undergoes simultaneous dehydrogenation and decarboxylation in the presence of enzyme α – ketoglutarate dehydrogenase complex. This enzyme complex contain TPP, Lipoic Acid , Mg2+ and trans – succinylase. NAD+ and CoA are required. The products formed are 4 – carbon compound succinyl CoA, NADH2 and CO2.

Step 7: Dehydrogenation (Oxidation)
In step 7 of Krebs Cycle 4 – Carbon compound Succinate is oxidized to another 4-carbon compound fumarate with the help of enzyme succinate dehydrogenase and hydrogen acceptor FAD (Flavin Adenine Dinucleotide). The enzyme is attached to inner mitochondrial membrane. It contains or non haem iron (Fe–S) protein. This enables the enzyme to get directly linked to electron transport chain.

Step 9: Dehydrogenation (Oxidation)
With the help of enzyme malate dehydrogenase, Malate formed in step 8 is oxidized to oxaloacetate. NAD+ reduced to NADH2.

So the correct option is 'Four'.
Question 10
1 / -0

Which one among the following is oxidative step of glycolysis?

A
Phosphoglycerate $$\rightarrow$$ phosphoenol pyruvate

B
Glyceraldehyde $$3-$$ phosphate $$\rightarrow$$ $$1, 3-$$bis Phosphoglycerate

C
Phospho enol pyruvate $$\rightarrow$$ pyruvate

D
Fructose $$1,6-$$ bis phosphate $$\rightarrow$$ Triose phosphate

Solution

A. Phosphoglycerate by hydrolysis and enolase enzyme is converted to phosphoenol pyruvate.
B. Glyceraldehyde-3-phosphate is oxidized and with inorganic phosphate converted to 1,3-bisphosphoglycerate.
C. Phosphoenol pyruvate by pyruvate kinase gets converted into pyruvic acid and ATP is released.
D. Fructose 1,6-bisphosphate by the action of aldolase converts into triose phosphate.
So, the correct answer is 'Glyceraldehyde 3− phosphate → 1,3−bis Phosphoglycerate'.

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Respiration in Plants Test - 77

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Respiration in Plants Test - 77

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SHARING IS CARING

Question 1

Hexokinase

Pyruvate kinase

Enolase

Phosphofructokinase

Solution

Question 2

$$15$$

$$12$$

$$3$$

$$36$$

Solution

Question 3

Fermentation

Preservation

Fumigation

Both A and B

Solution

Question 4

One

Two

Three

Four

Solution

Question 5

One

Two

Three

Four

Solution

Question 6

B-carboxylation- fixation of $$CO_2$$ the aldose sugar

C- reduction - reverse of glycolysis

A- regeneration - reformation of pentose sugar

C- reduction - carbohydrate formation

Solution

Question 7

It is formed by the glycolytic catabolism of carbohydrates in the mitochondrial matrix.

It enters mitochondrial matrix and undergoes oxidative decarboxylation.

The conversation of pyruvic acid to acetyl CoA is catalyzed by pyruvic dehydrogenase.

Two molecules of NADH are produced from the metabolism of one glucose into two molecules of pyruvic acid.

Solution

Question 8

$${ O }_{ 2 }$$ is given out

$${ CO }_{ 2 }$$ is given out

$${ CO }_{ 2 }$$ is taken in

$${ O }_{ 2 }$$ is taken in

Solution

Question 9

One

Two

Three

Four

Solution

Question 10

Phosphoglycerate $$\rightarrow$$ phosphoenol pyruvate

Glyceraldehyde $$3-$$ phosphate $$\rightarrow$$ $$1, 3-$$bis Phosphoglycerate

Phospho enol pyruvate $$\rightarrow$$ pyruvate

Fructose $$1,6-$$ bis phosphate $$\rightarrow$$ Triose phosphate

Solution

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