Photosynthesis in Higher Plants Test

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Photosynthesis in Higher Plants Test - 84

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following statement is/are correct?
(A) All green plants can prepare their own food.
(B) Most animals are autotrophs.
(C) $CO_2$ is not required for photosynthesis.
(D) Oxygen is liberated during photosynthesis.
Choose the correct answer from the options given below.

A
(A) and (D)

B
(B) only

C
(B) and (C)

D
(A) and (B)

Solution

All green plants can prepare their own food by the process called photosynthesis. Photosynthesis takes in the carbon dioxide produced by all breathing organisms and reintroduces oxygen into the atmosphere. Photosynthesis is the process used by plants, algae, and certain bacteria to harness energy from sunlight and turn it into chemical energy used by animals (heterotrophs).
So, the correct answer is option A.
Question 2
1 / -0

For the synthesis of one molecule of sucrose, how many ATP molecules are required in reduction step of the Calvin cycle?

A
24 ATP

B
18 ATP

C
2 ATP

D
12 ATP

Solution

The ATP molecule required in reduction step of Calvin cycle is 24.
Correct option is 24.
Here, 6 ATP is required for 3 carbon sugar. In sucrose 12 carbon are present .
So, 6 x 4=24 ATP.
Hence, the correct answer is 24.
Question 3
1 / -0

During chemiosmotic synthesis of ATP, protons diffuse through $CF_{0}$ channels that activates ATPase enzyme. As a result, one molecule of ATP is formed when ________ passes through ATPase.

A
$4H^{+}$

B
$H^{+}$

C
$2H^{+}$

D
$6H^{+}$

Solution

During chemiosmotic synthesis of ATP, protons diffuse through CFo channels that activates ATPase enzyme. As a result, one molecule of Atp is formed when two hydrogen ions pass through ATPase.
So, the correct option is ' $2H^+$ .
Question 4
1 / -0

Read the given statement and select the correct option.
Statement 1: Carboxylation is the most crucial step of Calvin where $CO_{2}$ is utilised for the carboxylation of RuBP.
Statement 2: Carboxylation is catalysed by the enzyme RuBisCO which results in the formation of two molecules of 3PGA.

A
Both statement 1 and 2 are correct

B
Statement 1 is correct but statement 2 is incorrect

C
Statement 1 is incorrect but statement 2 is correct

D
Both statement 1 and 2 are incorrect

Solution

Carboxylation is the fixation of $CO_{2}$ into a stable organic intermediate. It is the most crucial step of the Calvin cycle where $CO_{2}$ is utilized for the carboxylation of RuBP through the use of ATP and NADPH generated by the light reaction. This reaction is catalysed by enzyme RuBisCO which results in the formation of two molecules of 3-phosphoglyceric acid or PGA, which is the first stable product of photosynthesis.
Question 5
1 / -0

How many ATP and $NADPH_{2}$ are respectively produced in the process of photorespiration?

A
2 and 4

B
1 and 2

C
4 and 6

D
0 and 0

Solution

Photorespiratory pathway (or $C_{2}$ pathway) results in the release of $CO_{2}$ with the utilization of ATP. In the photorespiratory pathway, three is no synthesis of ATP or NADPH.
So the correct option is D.
Question 6
1 / -0

In calvin cycle, the fixation of CO2 into a stable organic intermediate is called...

A
Carboxylation

B
Decarboxylation

C
Kreb's Cycle

D
Electron Transport

Solution

The first step of the Calvin cycle is Carboxylation, where CO $_{2}$ is fixed into a stable organic intermediate. During the process, CO $_{2}$ is used for the carboxylation of RuBP, via RuBP carboxylase as the catalyst and resulting in the formation of two molecules of 3-PGA.
So the correct option is 'Carboxylation'
Question 7
1 / -0

The use of [ 14C] radioactive carbon by M. Calvin in algal photosynthesis studies provided

A
That CO2 will combine with H2O to produce sugar.

B
That ATP and NADPH were used in the biosynthetic phase.

C
That first CO2 fixation product was 3-carbon organic acid.

D
All of the above

Solution

The use of radioactive carbon by M.Calvin in algal photosynthesis studies :
Correct option is (D).
In C3 pathway carbon dioxide is converted into sugar and further ATP and NADPH were used . The first stable compound in the C3 pathway is 3-Carbon organic compound that is 3 phosphoglyceric acid (3PGA).
Hence, The correct option is D.
Question 8
1 / -0

Light energy is not used directly or indirectly in which of the following option?

A
Splitting of ${ H }_{ 2 }O$

B
To develop proton gradient during photophosphorylation

C
${ CO }_{ 2 }$ reduction in dark reaction

D
To develop proton gradient during oxidative photophosphorylation

Solution

NADH and FADH $_{2}$ transfer their electrons to molecules near the beginning of the transport chain. As electrons are passed down the chain, they move from a higher to a lower energy level, releasing energy. Some of the energy is used to pump H $^{+}$ ions, moving them out of the matrix and into the intermembrane space. The energy released in these reactions is captured as a proton gradient, which is then used to make ATP in a process called chemiosmosis. Together, the electron transport chain and chemiosmosis make up oxidative phosphorylation.
So, the correct option is 'To develop proton gradient during oxidative photophosphorylation'.
Question 9
1 / -0

What is the product of oxygenase acitivity of Rubisco?

A
Phosphoglycerate

B
Phosphoglycolate

C
Both A and B

D
Phosphoglyceraldehyde

Solution

Addition of molecular oxygen to ribulose-1,5-bisphosphate produces 3-phosphoglycerate (PGA) and phosphoglycolate (PG). PGA is the normal product of carboxylation and productively enters the Calvin cycle. Phosphoglycolate, however, inhibits certain enzymes involved in photosynthetic carbon fixation.
So, the correct option is 'Phosphoglycolate'.
Question 10
1 / -0

Which one option have correct mark for A, B, C, D & E:
A B C D E
1 LHC II LHC I FRS Pheophytin NADPH
2 LHC II LHC I Chl a without mg FRS NADPH
3 LHC I LHC II Pheophytin FRS ATP
4 LHC II LHC I FRS Phephytin ATP

A
1

B
2

C
3

D
4

Solution

It is the process of non-cyclic photophosphorylation. An electron released during photolysis of water is picked up by Photosystem II which contains light-harvesting complex (LHC II). While passing over cytochrome complex, the electron loses sufficient energy for the synthesis of ATP. The electron is then picked up by PS I having LHC I. Plastocyanin extrudes the electron which then passes through ferredoxin. Phaeophytin on accepting electrons becomes strong reducing agent which reduces NADP+ to NADPH.

So, the correct answer is '1'.

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	A	B	C	D	E
1	LHC II	LHC I	FRS	Pheophytin	NADPH
2	LHC II	LHC I	Chl a without mg	FRS	NADPH
3	LHC I	LHC II	Pheophytin	FRS	ATP
4	LHC II	LHC I	FRS	Phephytin	ATP

Photosynthesis in Higher Plants Test - 84

Result

Photosynthesis in Higher Plants Test - 84

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Question 1

(A) and (D)

(B) only

(B) and (C)

(A) and (B)

Solution

Question 2

24 ATP

18 ATP

2 ATP

12 ATP

Solution

Question 3

4H+4H^{+} 4H+

H+H^{+} H+

2H+2H^{+} 2H+

6H+6H^{+} 6H+

Solution

Question 4

Both statement 1 and 2 are correct

Statement 1 is correct but statement 2 is incorrect

Statement 1 is incorrect but statement 2 is correct

Both statement 1 and 2 are incorrect

Solution

Question 5

2 and 4

1 and 2

4 and 6

0 and 0

Solution

Question 6

Carboxylation

Decarboxylation

Kreb's Cycle

Electron Transport

Solution

Question 7

That CO2 will combine with H2O to produce sugar.

That ATP and NADPH were used in the biosynthetic phase.

That first CO2 fixation product was 3-carbon organic acid.

All of the above

Solution

Question 8

Splitting of H2O{ H }_{ 2 }OH2​O

To develop proton gradient during photophosphorylation

CO2{ CO }_{ 2 }CO2​ reduction in dark reaction

To develop proton gradient during oxidative photophosphorylation

Solution

Question 9

Phosphoglycerate

Phosphoglycolate

Both A and B

Phosphoglyceraldehyde

Solution

Question 10

1

2

3

4

Solution

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$4H^{+}$

$H^{+}$

$2H^{+}$

$6H^{+}$

Splitting of ${ H }_{ 2 }O$

${ CO }_{ 2 }$ reduction in dark reaction