In $$C_4$$ plants, Calvin cycle operates in

Grana of bundle sheath chloroplasts

Grana of mesophyll chloroplasts

Stroma of mesophyll chloroplasts.

With high RuBP carboxylase efficiency

With PEP carboxylase in bundle sheath cells

MCQ Questions for Class 11 Medical Biology Photosynthesis In Higher Plants Quiz 8

The movement of ten electrons from mitochondrial NADH

$_2$ molecules produces ____ water molecules.

0%
20
0%
15
0%
10
0%
5

Explanation

The electron transport chain is a series of electron transporters embedded in the inner mitochondrial membrane that shuttles electrons from $NADH_2$ and $FADH_2$ to molecular oxygen.
In the process, protons are pumped from the mitochondrial matrix to the intermembrane space, and oxygen is reduced to form water.
Here, one water molecule is formed by two electrons. So, we can say that ten electrons from mitochondrial NADH molecules produce 5 water molecules.

So, the correct option is D.

Chemiosmotic hypothesis given by Peter Mitchell proposes the mechanism of

0%
Synthesis of ATP.
0%
Synthesis of FADH $_2$ .
0%
Synthesis of NADH.
0%
Synthesis of NADPH.

Carbon dioxide joins the photosynthetic pathway in

0%
PS I
0%
PS II
0%
Light reaction
0%
Dark reaction

Explanation

Photosynthesis is divided into two reactions:-

Light reaction:-It is depended on light and occurs in grana fraction of chloroplast and in this reaction includes all those activities, which are dependent on light. Assimilatory powers like ATP and NADPH $_2$ are mainly produced in this reaction.
Dark reaction:-The pathway by which all photosynthetic eukaryotic organisms ultimately incorporate CO $_2$ into carbohydrate is known as carbon fixation, photosynthetic carbon reduction or dark reaction.

So, the correct option is 'Dark reaction'.

The

$C_4$ -plants are different from the

$C_3$ -plants with reference to the

0%
The substance that accepts $CO_2$ in carbon assimilation.
0%
Types of end product of photosynthesis.
0%
The number of NADPH that are consumed in preparing sugar.
0%
Types of pigments involved photosynthesis.

Explanation

The CO $_2$ acceptors in the C $_3$ cycle is RuBP (Ribulose-1,5-bisphosphate), RuBP is a 5 carbon compound. The first CO $_2$ acceptor in the C $_4$ cycle is PEP (Phosphoenolpyruvate) which is a 3 carbon compound.
C $_3$ plants do not have bundle sheath in leaves while C $_4$ plants have.
First, stable product in C $_3$ plants is 3 carbon compound while in C $_4$ plant is 4 carbon compound.

So, the correct option is 'the substance that accepts CO2 in carbon assimilation.'

Formation of ATP in photosynthesis and respiration is an oxidation process which utilises the energy from

0%
Cytochromes
0%
Ferredoxin
0%
Electrons
0%
Carbon dioxide.

Pigment acting as a reaction centre during photosynthesis is

0%
Carotene
0%
Phytochrome
0%
Chlorophyll a of $P_{700}$
0%
Cytochome.

Explanation

The single chlorophyll a forms reaction centre and is different for both photosystems. In PS I the reaction centre has an absorption peak at 700 nm, hence is called

$P_{700}$ and for PS II it is 680 nm and is called as

$P_{680}$ .

So the correct option is 'Chlorophyll a of

$P_{700}$ '.

Which one of the following is the common product of both respiration and photosynthesis?

0%
ATP
0%
Quinone
0%
Cytochrome
0%
Chlorophyll

Maize, Sugarcane and some other tropical plants have high efficiency of

$CO_2$ fixation because they operate

0%
Calvin cycle
0%
Hatch-Slack cycle
0%
TCA cycle
0%
PP pathway

Explanation

Most of the plants that are adapted to dry tropical regions have the

$C_{4}$ pathway. These plants are called

$C_{4}$ plants. Sugarcane, Maize, Sorghum, Corn, etc., are the examples of these plants. These plants have high efficiency

$CO_{2}$ fixation because they operate Hatch-Slack pathway. It is also known as

$C_{4}$ pathway. It is a cyclic process. It occurs in chloroplasts. It was discovered by two Australian scientists Hatch and Slack. Carbon dioxide fixation is done in this pathway. The primary

$CO_{2}$ acceptor is a three-carbon molecule phosphoenolpyruvate (PEP) and it is present in mesophyll cells. The enzyme that catalyzes this

$CO_{2}$ fixation is PEP carboxylase or PEPcase. The mesophyll cells of

$C_{4}$ plants lack the enzyme RuBisCO. The 4-carbon oxaloacetic acid (OAA) is formed in the mesophyll cells which is further converted into the malic acid or aspartic acid and then transported into bundle sheath cells. In the bundle sheath cells, these

$C_{4}$ acids are broken down to release

$CO_{2}$ and a three-carbon molecule.

Which one of the following does not perform

$C_4$ photosynthesis?

0%
Saccharum
0%
Zea mays
0%
Triticum aestivum (=T. vulgare)/Crotalaria)
0%
Euphorbia milli (=E. splendens).

Explanation

Characteristics of C

$_3$ and C

$_4$ plants:-

C $_3$ cycle use Calvin cycle while C $_4$ plants use Hatch-slack pathway.
In C $_3$ plants first stable compound is PGA while C $_4$ use oxaloacetic acid.
C $_3$ plants are found in cold and wet areas and C $_4$ plants are found in dry and hot areas.
These are abundant in temperate areas while these are mainly found in tropical areas.
Examples of C $_3$ plants are wheat, oats, rice while of C $_4$ plants are maize, sugarcane etc.

So the correct option is 'Triticum aestivum (=T. vulgare)/Crotalaria)'.

Organelles having enzymes involved in photorespiration are

0%
Mitochondria, chloroplasts and ribosomes
0%
Mitochondria, peroxisomes and chloroplasts
0%
Mitochondria, nucleus and ribosomes
0%
Mitochondria, peroxisomes and glyoxysomes.

Explanation

This process occurs when the concentration of oxygen increase and carbon dioxide decrease and its substrate is glycolate. RubisCO, catalase (CAT) and GOX, and GDC are the main enzymes in chloroplasts, leaf peroxisomes, and mitochondria from mature leaves, respectively, and support the major role of the photorespiratory

$C_2$ cycle in leaf metabolism.

So the correct option is 'Mitochondria, peroxisomes and chloroplasts'.

Photorespiration occurs in

0%
Ribosomes
0%
Mitochondria
0%
Peroxisomes
0%
Both B and C

Explanation

Photorespiration is a process which involves loss of fixed carbon as

$CO_{2}$ in plants in the presence of light. It is also known as

$C_{2}$ cycle. It is initiated in chloroplasts, but it occurs in peroxisomes. This process does not produce ATP or NADPH and is a wasteful process. Photorespiration occurs usually when there is the high concentration of oxygen. Under such circumstances, RuBisCO, the enzyme that catalyzes the carboxylation of RuBP during the first step of Calvin cycle, functions as an oxygenase. Some

$O_{2}$ does not bind to RuBisCO and hence

$CO_{2}$ fixation is decreased. The RuBP binds with

$O_{2}$ to form one molecule of PGA (3C compound) and phosphoglycolate (2C compound) in the pathway of photorespiration. Rather, it results in the release of

$CO_{2}$ with the utilization of ATP. It leads to a 25 percent loss of the fixed

$CO_{2}$ .

$O_{2}$ is first utilized in chloroplasts and then in peroxisomes.

Synthesis of

$ADP + Pi \longrightarrow ATP$ in grana/photosynthesis is

0%
Phosphorylation
0%
Photophosphorylation
0%
Oxidative phosphorylation
0%
Photolysis

Explanation

Photophosphorylation is the process in which light energy is converted into chemical energy through the production of ATP. It is the process of formation of ATP from ADP and inorganic phosphate (IP) utilizing light energy. It occurs in grana of chloroplast during photosynthesis. The process of reduction of

$NADP$ into

$NADP + H^{+}$ may be denoted as electron transport system (ETS). The flow of electrons through ETS is linked to photophosphorylation. Electron transport chain is a series of electron carriers over which electrons pass in a downhill journey releasing energy at every step that is used in generating an electrochemical proton gradient which helps in synthesizing ATP. Based on the path of electrons, associated photophosphorylation can be identified as non-cyclic and cyclic phosphorylation.

$C_4$ plants, Calvin cycle operates in

0%
Stroma of bundle sheath chloroplasts
0%
Grana of bundle sheath chloroplasts
0%
Grana of mesophyll chloroplasts
0%
Stroma of mesophyll chloroplasts.

Explanation

In mesophyll cells, carbon dioxide is accepted by PEP and form oxaloacetic acid and further form malic acid which enters the chloroplasts of bundle sheath cells and undergoes decarboxylation yielding pyruvic acid and CO

$_2$ . It released in the stroma of bundle sheath cell to bind with RuBisCO and undergoes Calvin cycle to form glucose.

So the correct option is 'stroma of bundle sheath chloroplasts'.

For synthesis of a molecule of glucose, the requirement of ATP and NADPH is respectively

0%
15 and 10
0%
33 and 22
0%
12 and 8
0%
18 and 12.

Explanation

For every CO

$_2$ molecule entering the Calvin cycle, 3 molecules of ATP and 2 NADPH are required. To make one molecule of glucose six CO

$_2$ molecules required and the cycle turns six times.

So the number of total ATP=3X6=18.

The number of total NADPH=3X4=12.

So the correct option is '18 and 12'.

$C_4$ plants belong to

0%
Gramineae
0%
Monocots
0%
Dicots
0%
Both monocots and dicots

Explanation

$^4$ C plants are those plants that contain bundle sheath and mesophyll cells in leaves and their first stable compound in photosynthesis is four carbon compound oxaloacetic acid and mainly found in the dry and hot climate and constitute both monocots and dicot plants.

So the correct option is 'Both monocots and dicots'.

Sugarcane shows high efficiency of

$CO_2$ fixation because it performs

0%
Calvin cycle
0%
EMP pathway
0%
Hatch and Slack pathway
0%
TCA cycle

Explanation

Most of the plants that are adapted to dry tropical regions have the

$C_{4}$ pathway. These plants are called

$C_{4}$ plants. Sugarcane, Maize, Sorghum, Corn, etc., are the examples of these plants. Sugarcane is also a

$C_{4}$ plant, so it has high efficiency of

$CO_{2}$ fixation because it performs Hatch-Slack pathway. It is also known as

$C_{4}$ pathway. Carbon dioxide fixation is done in this pathway. The primary

$CO_{2}$ acceptor is a three-carbon molecule phosphoenol pyruvate (PEP) and it is present in mesophyll cells. The enzyme that catalyzes this

$CO_{2}$ fixation is PEP carboxylase or PEPcase. The mesophyll cells of

$C_{4}$ acids are broken down to release

$CO_{2}$ and a three-carbon molecule.

Path of dark reaction of photosynthesis was traced through the use of

0%
$^{32}$ P
0%
$^{14}$ CO $_2$
0%
$^{18}$ O $_2$
0%
X-rays

Explanation

Calvin and his co-workers by using the isotope of C that is

$^{14}C$ to trace the path of the carbon fixation in light reaction. Due to his discovery, the cycle is known as the Calvin cycle which is the common path for all plants in the photosynthesis.

So the correct option is '

$^{14}C$ '.

First product of photorespiration is

0%
Phosphoglycolate
0%
Glycolate
0%
Glycine
0%
None of the above

Explanation

Photorespiration is the uptake of oxygen and release of carbon dioxide in light and results from the biosynthesis of glycolate in chloroplasts and subsequent metabolism of glycolate acid in the leaf cell which provide the first product as two molecules of phosphoglycolate which undergoes various step reaction to form three phosphoglyceric acid in C

$_3$ plants by entering Calvin cycle.

So the correct option is 'Phosphoglycolate'

$C_4$ cycle was discovered in

0%
Groundnut
0%
Sugarcane
0%
Chrysanthemum
0%
Apple

Non-cyclic photophosphorylation is performed by

0%
Dark reaction
0%
PS I
0%
PS II
0%
Both B and C

$C_4$ cycle is

0%
Adjunct to Calvin cycle
0%
Independent cycle
0%
With high RuBP carboxylase efficiency
0%
With PEP carboxylase in bundle sheath cells

Explanation

$C_4$ cycle the Calvin cycle is carried out in bundle sheath cell and is the essential and compulsory part of photosynthesis in every green plant but Hatch and Slack cycle is added to it in which carbon dioxide is trapped in mesophyll cells and lead to the formation of oxaloacetic acid by PEP in presence of PEP carboxylase which increases affinity of carbon dioxide and reduces photorespiration. Thus it is an extra and essential part for Calvin cycle.

So the correct option is 'Adjunct to Calvin cycle'.

$C_4$ cycle is connected with

0%
Respiration
0%
Photosynthesis
0%
Transpiration
0%
Growth and development

Explanation

In some plants that grow in dry and hot tropical environment develop a way of photosynthesis name as Hatch and Slack or C

$_4$ cycle in which the first stable molecule of carbon is oxaloacetic acid which is a four-carbon compound and has different anatomy of leaves known as Kranz anatomy.

So the correct option is 'Photosynthesis'.

Balance between

$CO_2$ and

$O_2$ is maintained by

0%
Transpiration
0%
Photosynthesis
0%
Photorespiration
0%
$C_4$ pathway.

Which one is a

$C_4$ plant

0%
Maize
0%
Sugarcane
0%
Sorghum
0%
All of the above.

Explanation

The first CO $_2$ acceptor in the C $_4$ cycle is the PEP

(Phosphoenolpyruvate) which is a 3 carbon compound.

C $_4$ plants have bundle sheath cells.
The stable compound is 4 carbon compound.
These are found in dry and hot tropical regions.
The examples are maize, sugarcane, sorghum, corn, pineapple etc.

So the correct option is 'All of the above.'.

$C_4$ plants differ from

$C_3$ plants with respect to

0%
First product
0%
Substrate which accepts carbon dioxide
0%
Number of ATP molecules consumed
0%
All of the above

Explanation

In C $_3$ plants first product is a 3-C compound 3-phosphoglyceric acid while in C $_4$ plants, a 4-carbon compound (oxaloacetic acid) is the first stable compound.
The substrate which accepts CO $_2$ in C $_3$ plants is a 5 carbon compound RUBP (Ribulose-1,5-bisphosphate) and in C $_4$ plants it is PEP (Phosphoenol pyruvate), a 3-C compound.
C $_3$ cycle requires 18 ATP and 12 NADPH molecules to synthesize one molecule of glucose and C $_4$ cycle requires 30 ATP molecules to synthesize one molecule of glucose

So the correct option is 'All of the above'.

$C_4$ plants have higher net photosynthetic rate as they have

0%
No photorespiration
0%
PEP as $CO_2$ acceptor
0%
Kranz anatomy
0%
Photosynthesis even at low light intensity.

Explanation

$C_4$ plants photosynthetic rates is higher than the C

$_3$ plants as they have PEP as CO

$_2$ acceptor and transfer it to RuBisCO so it is not able to bind with oxygen and in continuation carbon dioxide fixation takes place. So they do not have photorespiration.

So the correct option is ' No photorespiration'.

Photorespiration is affected by

0%
Temperature
0%
Light intensity
0%
$CO_2$ and $O_2$
0%
All of the above

$C_4$ pathway of

$CO_2$ fixation was discovered by

0%
Blackman
0%
Hatch and Slack
0%
Emerson and Arnold
0%
Arnon

Primary carboxylation occurs in

$C_3$ and

$C_4$ plants with the help of

0%
PEP carboxylase and pyruvate carboxylase
0%
PEP carboxylase and RuBP carboxylase
0%
RuBP carboxylase and PEP carboxylase
0%
RuBP carboxylase and pyruvate carboxylase.

Explanation

Primary carboxylation in $C_3$ plants is done by RuBP in the presence of enzyme RuBP carboxylase.
In $C_4$ plants, it is done by the PEP molecule with the help of PEP carboxylase.

So the correct option is 'RuBP carboxylase and PEP carboxylase'.

$C_4$ plants differ from

$C_3$ plants in

0%
Number of ATPs consumed per hexose molecule
0%
Type of end product
0%
Type of pigments
0%
Substrate for $CO_2$ in carbon assimilation.

Explanation

Carbon dioxide fixation of

$C_4$ plants require more ATP than the that by

$C_3$ plants. This additional ATP is needed for conversion of pyruvic acid into phosphoenol pyruvic acid and its transport.

So the correct option is 'Number of ATPs consumed per hexose molecule'.

Which one of the following contain agranal chloroplasts?

0%
$C_3$ plants
0%
Succulents
0%
$C_4$ plants
0%
Hydrophytes

Explanation

Bundle sheath cell of C

$_4$ plants contain larger chloroplasts in size and arranged centripetally and also contain starch granule but lack grana. This is the reason these chloroplasts are agranal as they lack grana.

So the correct option is 'C

$_4$ plants'.

RuBisCO content of chloroplast is atleast

0%
20%
0%
5%
0%
11%
0%
16%

Explanation

Rubisco enzyme is located in the chloroplasts' stroma, consisting of at least 15% and up to 50% of the chloroplast protein. The high concentration of RuBisCO indicates a significant role for the plant's metabolism as it the major acceptor of carbon dioxide in

$C_3$ plants and require in

$C_4$ plants for carried out Calvin cycle. In plants, it is a large protein with a molecular weight of 560,000 Dalton.

So the correct option is '16%'.

ATP formation in chloroplast and mitochondrion is explained by

0%
Cholodny-Went model
0%
Chemi-osmotic theory of Mitchell
0%
Munch's mass flow theory
0%
Relay pump theory of Godlewski

Special feature of

$C_4$ plants is

0%
Thin cuticle
0%
Multilayered epidermis
0%
Kranz anatomy
0%
Both A and B

Explanation

The particularly large cells around the vascular bundles of the C

$_4$ plants pathway are called as bundle sheath cells and leaves which have such anatomy are said to have bundle sheath cells and is called Kranz anatomy which means wreath and is a reflection of the arrangement of cells.

So the correct option is 'Kranz anatomy'.

Condition favouring cyclic photophosphorylation is

0%
Anaerobic environment
0%
Aerobic and low light intensity
0%
Aerobic and optimum light
0%
Anaerobic and low light intensity

Explanation

Cyclic Photophosphorylation: It is a process of photophosphorylation in which an electron expelled by the excited photo center( PS I) is returned to it after passing through a series of electron carriers. It occurs under conditions of low intensity, wavelength longer than 680 nm and when CO

$_2$ fixation is inhibited. Absence of CO

$_2$ fixation results in non- requirement of electrons as NADPH

$_2$ is not being oxidized to NADP

$^+$ .

Aerobic: Presence of O

$_2$ and absence of CO

$_2$ .

Anaerobic: Absence of O

$_2$ and presence of CO

$_2$ .

So, the correct answer is 'Aerobic and low light intensity'.

$C_3$ plants, first carbon dioxide acceptor enzyme is

0%
PEPCO
0%
RuBP carboxylase/Rubisco
0%
RuBP oxygenase
0%
Oxidase.

First stable product of

$C_4$ /HSK photosynthesis is

0%
Phosphoglyceric acid
0%
Ribulose 1, 5-biphosphate
0%
Malic acid
0%
Phosphoglyceraldehyde.

Scientist awarded Nobel Prize in 1960 for tracing the path of carbon in photosynthesis was

0%
Hatch
0%
Huber
0%
Calvin
0%
Ruben

ATP molecules required to synthesise one molecule of glucose by

$C_4$ pathway are

0%
12
0%
18
0%
24
0%
30

Carbon dioxide is fixed in

0%
Light reaction
0%
Dark reaction
0%
Aerobic respiration
0%
Anaerobic respiration

Anatomy of C

$_4$ plant leaf shows

0%
Presence of peroxisomes
0%
Presence of bundle sheath cells
0%
Absence of mitochondria
0%
Absence of bundle sheath cells

In Maize, mesophyll cells perform photosynthetic cycle

0%
$C_4$
0%
$C_3$
0%
$C_2$
0%
$C_1$ .

Which of the following is a

$C_4$ plant?

0%
Sugarcane
0%
Tomato
0%
Mango
0%
Apple

Ferredoxin is

0%
Phenol
0%
Protein
0%
Fat
0%
Carbohydrate

Non-cyclic photophosphorylation involves

0%
PS I only
0%
PS II only
0%
Stroma matrix
0%
Both A and B

Which one is a

$C_4$ plant?

0%
Potato
0%
Mustard
0%
Onion
0%
Wheat

Explanation

In C $_3$ plants first product is 3 carbon compound called phosphoglyceric acid (PGA) while in C $_4$ plants, the first product is 4 carbon compound called the oxaloacetic acid (OAA).

A. Potato- C $_3$ plant.

B. Mustard- C $_3$ plant.

C. Onion- C $_4$ plant.

D. Wheat- C $_3$ plant.

Hence, the correct answer is 'Onion'.

$C_4$ and

$C_3$ pathways of CAM plants are separated by

0%
Bundle sheath
0%
Mesophyll and bundle sheath cells
0%
Mesophyll and bundle sheath chloroplasts
0%
Time

Explanation

Certain plants, especially succulents which grow under extremely xeric conditions, fix carbon dioxide in dark. These plants are termed as CAM (Crassulacean Acid Metabolism) plants. In these plants, stomata remain open at night but closed during the day (in light).

1. In this CO

$_2$ combine with PEP to form OAA in presence of PEP carboxylase like C4 plants.

2. OAA converted to malic acid in the presence of NADH. Malic acid produced in dark as a result of acidification.

3. Malic acid converted to pyruvic acid and evolve CO

$_2$ during the day when stomata are closed. This process is termed as deacidification.

The CO

$_2$ released by deacidification accepted by RuBP and fixed to carbohydrate by C

$_3$ cycle.

Hence, the correct answer is 'Time'.

In photosynthetic electron transport system, Mn ions are associated with

0%
$CF_0 - CF_1$ complex
0%
Cyt $b_6$ - Cyt f complex
0%
Oxygen evolving complex
0%
Plastoquinone.

The sequential order in electron transport from PS II to PS I of photosynthesis is

0%
FeS, PQ, PC and Cytochrome
0%
FeS, PQ. Cytochromes and PC
0%
PQ, Cytochromes, PC and FeS
0%
PC, Cytochromes, FeS, PQ

Explanation

In electron transport process the excited electron from P $_{680}$ (PS II) is accepted by primary acceptor quinone, which sends them to an electron transport system consisting of plastoquinone, cytochrome complex and plastocyanin.

The excited electron absorbed by P $_{700}$ (PS I) reaction centre, then transfer to the ferredoxin (Fe), then the reduced Ferredoxin donates the electron to NADP $^+$ .

The ${NADP}^+$ take an electron from ferredoxin, protons from the medium and get reduced to NADPH in presence of enzyme ferredoxin- NADP- reductase.

Hence, the correct answer is 'PQ, cytochromes, PC and FeS'.

PEP, the first

$CO_2$ acceptor in

$C_4$ cycle is

0%
3-C compound
0%
4-C compound
0%
5-C compound
0%
6-C compound.

Explanation

$C_4$ plants, phosphoenol pyruvate-a 3 carbon compound accepts co2 and forms oxaloacetic acid which is a 4 carbon compound.

So, the correct option is 'Option A'.

CBSE Questions for Class 11 Medical Biology Photosynthesis In Higher Plants Quiz 8 - MCQExams.com

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Practice Class 11 Medical Biology Quiz Questions and Answers

CBSE Questions for Class 11 Medical Biology Photosynthesis In Higher Plants Quiz 8 - MCQExams.com

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Practice Class 11 Medical Biology Quiz Questions and Answers

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