Lack both RuBisCO and PEP carboxylase

MCQ Questions for Class 11 Medical Biology Photosynthesis In Higher Plants Quiz 10

Free radical has electron

0%
Unpaired and extremely reactive
0%
Paired and extremely inactive
0%
Unpaired and extremely inactive
0%
Paired and extremely reactive

Which does not show HSK pathway

0%
Maize
0%
Jowar
0%
Sugarcane
0%
Sunflower

Explanation

HSK pathway or Hatch-Slack pathway is a cyclic process. It occurs in chloroplasts. It was discovered by two Australian scientists Hatch and Slack. It is also known as

$C_{4}$ pathway because it is found in

$C_{4}$ plants. Most of the plants that are adapted to dry tropical region are known as

$C_{4}$ plants. Sugarcane, Maize, Sorghum, Jowar, etc., are the examples of

$C_{4}$ plants. 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.

Sunflower is not a

$C_{4}$ plant, so it does not show HSK pathway.

$C_4$ pathway, PEP carboxylase gains

$CO_2$ from

0%
Pyruvic acid
0%
PEPA
0%
OAA
0%
Malic acid

Explanation

A) Pyruvic acid does not release

${ CO }_{ 2 }$ .

B) Phosphoenol pyruvic acid (PEPA) is converted to OAA in the presence of PEP case.

C) OAA is converted to malate which releases

${ CO }_{ 2 }$ .

D) PEP carboxylase binds to

${ CO }_{ 2 }$ to make oxaloacetic acid (OAA). OAA then makes malate. Malate enters bundle sheath cells and releases the

${ CO }_{ 2 }$

$C_4$ pathway, RuBP accepts

$CO_2$ inside

0%
Xylem parenchyma
0%
Phloem parenchyma
0%
Mesophyll cells
0%
Bundle sheath cells

Explanation

A) Xylem parenchyma is not involved in photosynthesis.

B) Phloem parenchyma is not involved in photosynthesis.

C) Mesophyll cells are rich in PEP which accepts

${ CO }_{ 2 }$ inside the mesophyll cells.

D) Bundle sheath cells are involved in the decarboxylation of

${ CO }_{ 2 }$ . They do not have PEP.

${ CO }_{ 2 }$ is accepted in the mesophyll cells.

So the correct answer is 'Mesophyll cells'.

Presence of bundle sheath is characteristic of

0%
$C_4$ plants
0%
Xerophytic plant
0%
Grass family
0%
$C_3$ plants

Explanation

${ C }_{ 4 }$ plants have bundle sheath cells and this is a characteristic feature.

B) Leaves in xerophytes are reduced or modified into spines or are absent.

C) Bulliform cells are a characteristic feature of the leaves of the grass family.

${ C }_{ 3 }$ plants have only mesophyll cells.
So the correct answer is '

${ C }_{ 4 }$ plants'.

Choose the correct combination of labelling 1-5

0%
1. Mesophyll cell. 2. Bundle sheath cell. 3. Regeneration. 4 Fixation. 5. Decarboxylation
0%
1. Bundle sheath cell 2. Mesophyll cell. 3. Fixation 4. Regeneration 5. Decarboxylation
0%
1. Mesophyll cell 2. Bundle sheath cell 3. Decarboxylation 4. Fixation 5. Regeneration
0%
1. Mesophyll cell 2. Bundle sheath cell 3. Fixation 4. Regeneration 5. Decarboxylation

Explanation

A) 1 corresponds to the mesophyll cell and 2 corresponds to the bundle sheath cell. But 3 corresponds to fixation of carbon dioxide to form a

${ C }_{ 4 }$ acid.
B) 4 corresponds to the regeneration of PEP.
C) 5 corresponds to the decarboxylation of the

${ C }_{ 4 }$ acid in Calvin's cycle.
D) 1 corresponds to the mesophyll cell, 2 corresponds to the bundle sheath cell, 3 corresponds to fixation of carbon dioxide, 4 corresponds to the regeneration of PEP and 5 corresponds to decarboxylation.
So the correct answer is '1. Mesophyll cell 2. Bundle sheath cell 3. Fixation 4. Regeneration 5. Decarboxylation'.
1020414_665965_ans_33150bb72f9149b8bea734a8bf5068de.png

1020414_665965_ans_33150bb72f9149b8bea734a8bf5068de.png

Photooxidation of water results in the formation of

0%
H $^+$ , O $_2$ , ATP
0%
H $^+$ , O $_2$ , e $^-$ , ATP
0%
H $^+$ , O $_2$ , e $^-$
0%
None of the above

Explanation

A) One electron is released per molecule of water that is split. ATP is not released.
B) ATP is not released as a byproduct of the splitting of water.

C) Photooxidation or photolysis of water occurs in the LHC of PS II where water is split by light to liberate

${ H }^{ + },{ O }_{ 2 },{ e }^{ - }$ .

D) Water splits to liberate

${ H }^{ + },{ O }_{ 2 }, { e }^{ - }$
So the correct answer is '

${ H }^{ + },{ O }_{ 2 },{ e }^{ - }$ '.

Identify the correct labeling of the given diagram.

0%
a - PS II, b - PS I, c - $e^-$ acceptor, d - LHC
0%
a - $e^-$ acceptor, b - LHC, c - PS II, d - PS I
0%
a - LHC, b - $e^-$ acceptor, c - PS I, d - PS II
0%
a - PS I, b - PS II, c - $e^-$ acceptor, d - LHC.

Explanation

A) 'a' corresponds to PS II, 'b' corresponds to PS I, 'c' is an

${ e }^{ - }$ acceptor and 'd' is the LHC of PS II.
B) 'a' is PS II and 'b' is PS I.

C) 'c' is an

${ e }^{ - }$ acceptor and 'd' is the LHC of PS II.

D) 'a' corresponds to PS II and 'b' corresponds to PS I.
So the correct answer is 'a - PS II, b - PS I, c -

${ e }^{ - }$ acceptor, d - LHC'.

1020429_665971_ans_ae17ea1f1266470493f1832e4791fa82.jpg

Which is related to light reaction of photosynthesis

0%
In PS II light reaction centre is $P_{700}$
0%
In PS I light reaction centre is $P_{680}$
0%
Photolysis is associated with PS I
0%
PS I and PS II are associated in Z-scheme

Explanation

A) PS II light reaction centre is

${ P }_{ 680 }$ .
B) PS I light reaction centre is

${ P }_{ 700 }$ .
C) Photolysis of water takes place in PS II to liberate oxygen.
D) Both PS I and PS II are associated in Z-scheme (non cyclic photophosphorylation).
So the correct answer is 'PS I and PS II are associated in Z-scheme'.

Identify the correct statement(s) in relation to photosynthesis.
(a) First

$CO_2$ acceptor in

$C_4$ cycle is PGA.
(b) In

$C_3$ plants, the first stable product of photosynthesis in RuBP.
(c) Cyclic photophosphorylation results in the formation of ATP.
(d) Oxygen liberated during photosynthesis comes from water.

0%
a and b alone are correct
0%
a and c alone are correct
0%
c and d alone are correct
0%
b and c alone are correct

Explanation

A) 'a' is incorrect as the first

${ CO }_{ 2 }$ acceptor in

${ C}_{ 4 }$ cycle is PEPcase. 'b' is incorrect as the first stable product of

photosynthesis in

${ C}_{ 3 }$ plants is PGA.

B) 'a' is incorrect as the first

${ CO }_{ 2 }$ acceptor in

${ C}_{ 4 }$ cycle is PEPcase. 'c' is correct as cyclic photophosphorylation results in the formation of the only ATP.

C) 'c' is correct as cyclic photophosphorylation results in the formation of the only ATP. 'd' is correct as oxygen is evolved from the photolysis of water.

D) 'b' is incorrect as the first stable product of photosynthesis in

${ C}_{ 3 }$ plants is PGA. 'c' is correct as cyclic photophosphorylation results in the formation of the only ATP.

So the correct answer is 'c and d alone are correct'.

Which of the following statement is correct?

0%
$C_4$ plants are more efficient than $C_3$ plants
0%
$C_3$ plants are more efficient than $C_4$ plants
0%
Photorespiration is useful process
0%
Photorespiration is absent in $C_3$ plants

Explanation

${ C }_{ 4 }$ plants are more efficient than

${ C }_{ 3 }$ plants as they do not undergo photorespiration. So, the carbon dioxide liberated is recycled back to PEP and is reused in

${ C }_{ 4 }$ cycle.
B)

${ C }_{ 3 }$ plants are less efficient than

${ C }_{ 4 }$ plants as they undergo photorespiratory losses as they lose carbon dioxide which is not recycled, whereas

${ C }_{ 4 }$ plants reuse the carbon dioxide.

C) Photorespiration is a wasteful process because it prevents plants from using their ATP and NADPH to synthesise carbohydrates.

${ C }_{ 3 }$ plants use this energy to regenerate carbon dioxide instead of recycling it.
D) Photorespiration takes place in

${ C }_{ 3 }$ plants whereas it is absent in

${ C }_{ 4 }$ plants.
So the correct answer is '

${ C }_{ 4 }$ plants are more efficient than

${ C }_{ 3 }$ plants'.

Select the incorrect match for

$C_4$ cycle.

0%
Primary $CO_2$ fixation product- PGA
0%
Site for initial carboxylation- mesophyll cells
0%
Primary $CO_2$ acceptor- PEP
0%
$C_4$ plant- maize
0%
Location of RuBisCo- bundle sheath cells

Explanation

A) The primary

${ CO }_{ 2 }$ fixation product is a

${ C }_{ 4 }$ acid.
B) Carboxylation first takes place in the mesophyll cells followed by Calvin's cycle in the bundle sheath cell.

C) The primary

${ CO }_{ 2 }$ acceptor is PEP- Phosphoenol Pyruvate.
D) Maize, pineapple and sugar cane are

${ C }_{ 4 }$ plants.
E) RuBisCo is present in the bundle sheath cells and takes part in Calvin's cycle.
So the correct option is 'Primary

${ CO }_{ 2 }$ fixation product- PGA'.

Which is incorrect regarding C

$_4$ pathway?

0%
The primary $CO_2$ acceptor is PEP
0%
Bundle sheath cells contain PEPcase
0%
Enzyme for $CO_2$ fixation is PEPcase
0%
Mesophyll cells lack RuBisCO

Chemiosmotic hypothesis of ATP synthesis in chloroplasts is based on

0%
Accumulation of $K^+$ ions
0%
Proton gradient
0%
Accumulation of Na ions
0%
Membrane potential
0%
Increase in pH of lumen

The first product of

$C_4$ pathway is

0%
OAA
0%
PGA
0%
PGAL
0%
DHAP.

Explanation

A) OAA - Oxalo Acetic Acid is the first product of the

${ C }_{ 4 }$ pathway.
B) PGA is the first product of the

${ C }_{ 3 }$ pathway.
C) PGA molecules are converted into PGAL by using ATP and NADPH from the light dependent reaction.
D) Dihydroxyacetone phosphate (DHAP) is the anion that is involved in many metabolic pathways, including the Calvin cycle in plants and glycolysis. It is the phosphate ester of dihydroxyacetone.
So the correct answer is 'OAA'.

Bundle sheath cells

0%
Lack RuBisCO
0%
Lack both RuBisCO and PEP carboxylase
0%
Are rich in RuBisCO
0%
Are rich in PEP carboxylase

Identify the correct statement(s) in relation to

$C_4$ photosynthesis.
(a) Kranz anatomy is an essential feature for

$C_4$ plants.
(b)

$C_4$ plants have higher water use efficiency than

$C_3$ plants.
(c) Photorespiration is minimised in

$C_4$ pathway.
(d) Conversion of oxaloacetate to malate occurs in bundle sheath cells.

0%
a and b
0%
a, b and c
0%
b and c
0%
b, c and d.

Explanation

A) Kranz anatomy is an essential feature of

${ C }_{ 4 }$ plants as it enables the Hatch and Slack pathway. So, 'a' is correct.

${ C }_{ 4 }$ plants have improved leaf and plant water use efficiency. So, 'b' is correct.
B) 'a' and 'b' are correct. Photorespiration is overcome in

${ C }_{ 4 }$ plants. So, 'c' is correct.

C) 'a' is correct in addition to 'b' and 'c'.
D) 'd' is incorrect as conversion of oxaloacetate to malate occurs in mesophyll cells.

So the correct answer is 'a, b and c'.

Which is associated with electron transport in photosynthesis?

0%
Sodium
0%
Potassium
0%
Iron
0%
Cobalt

Explanation

Iron is the most important metal involved in thylakoid reactions mainly electron transport. It is the main constituent of many electron carriers. It helps in electron transport from water to NADP+ in PS II. Cytochrome b

$_6$ -f also contains Iron element, PS I and ferredoxin is also iron associated protein.

In photosrespiration, glycine passes from

0%
Chloroplast to peroxisome
0%
Chloroplast to mitochondrion
0%
Peroxisome to mitochondrion
0%
Mitochondrion to peroxisome

Photorespiration is a characteristic feature of plants having

0%
$C_4$ cycle
0%
$C_3$ cycle
0%
Aerobic respiration
0%
None of the above

Explanation

Photorespiration is a characteristic feature of plants having

$C_{3}$ cycle. This process does not occur in

$C_{4}$ plants. It is a process which involves loss of fixed carbon as

$CO_{2}$ in plants in the presence of light. It is initiated in chloroplasts. This process does not produce ATP or NADPH and is a wasteful process. Photo-respiratory reactions are operated in chloroplasts, mitochondria, and peroxisomes. Loss of

$CO_{2}$ occurs in mitochondria. 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. It decreases photosynthetic output.

Which is not a

$C_3$ plant?

0%
Amaranth
0%
Rice
0%
Wheat
0%
Potato

Explanation

Rice, wheat, and potato all are C

$_3$ plants but amaranthus is only the C

$_4$ plant because amaranthus is found in tropical and subtropical regions, these plants show Kranz anatomy in which vascular bundles in leaves are surrounded by a layer of bundle sheath cells that contain large number of chloroplasts i.e., dimorphic. The chloroplasts of bundle cells are larger in size and are arranged centripetally while,the mesophyll cells have normal type of chloroplasts.
So, the correct option is A.

During photorespiration, which compounds are formed having 2C and 3C respectively in peroxisome?

0%
Glycolate, glycine
0%
Serine, glycine
0%
Glycine, glycerate
0%
Phosphoglycerate, glycolate

The form of carbon used for carboxylation of phosphoenol pyruvate in

$C_4$ plants is

0%
$H_2CO_3$
0%
$C_2H_4$
0%
$CH_4$
0%
$HCO_3^-$ .

Explanation

A) The ionised form of

$H_{ 2 }{ CO }_{ 3 }$ is used. It cannot be used directly.
B) Ethane is not used in the carboxylation of phosphoenolpyruvate.

C) Methane is not used in the carboxylation of phosphoenolpyruvate.
D)

${ HCO }_{ 3 }^{ - }$ is used for the carboxylation of phosphoenol pyruvate in

${ C }_{ 4 }$ plants.
So the correct answer is '

${ HCO }_{ 3 }^{ - }$ '.

Number of ATP molecules formed from 8 molecules of water due to noncyclic electron transport and subsequent photophosphorylation is

0%
8
0%
4
0%
16
0%
12

ATP synthesis in cell requires

0%
$H^+$ gradient across the membrane
0%
$K^+$ gradient across the membrane
0%
$PO^{3-}_4$ gradient across the membrane
0%
$Ca^{2+}$ gradient across the membrane

Explanation

A) A

${ H }^{ + }$ gradient or proton pump is developed in the lumen of the thylakoid which helps in ATP synthesis.

${ K }^{ + }$ does not play a role in ATP synthesis.

${ PO }_{ 4 }^{ 3- }$ does not play a role in ATP synthesis.

${ Ca }^{ 2+ }$ is involved in the photolysis of water.

So the correct answer is '

${ H }^{ + }$ gradient across the membrane'.

Statement a. Photorespiration decreases photosynthetic output.
Statement b. In photorespiratory pathway, neither ATP nor NADPH is produced

0%
Both the statements a and b are correct.
0%
Both the statements a and b are wrong.
0%
Statement a is correct and statement b is wrong.
0%
Statement b is correct and statement a is wrong.

Explanation

Photorespiration is a process which involves loss of fixed carbon as

$O_{2}$ 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. It decreases photosynthetic output.

Oxygen released in photosynthesis is formed during

0%
Oxidative phosphorylation
0%
Cyclic photophosphorylation
0%
Noncyclic photophosphorylation
0%
Carbon assimilation during dark reactions

Explanation

Non-cyclic phosphorylation involves both Photosystem I and Photosystem II. These two photosystems work in series, first PS II and then PS I. The two photosystems are connected through an electron transport chain.

$NADPH$ is synthesized by this kind of electron flow. Electrons are transferred from the PS II and then passed on to the pigments of PS I. Then, electrons get excited and transferred to another acceptor. These electrons are then moved downhill again to a molecule of

$NADP^{+}$ . The addition of these electrons results in the formation of

$NADPH$ . The electron released from PS II by

$P_{680}$ must be replaced. This is achieved by electrons available due to the splitting of water. The water splitting complex is associated with PS II, which itself is physically located on the inner side of the membrane of the thylakoid. Water is split into

$H^{+}$ ,

$[O]$ and electrons. The protons and oxygen formed by splitting of water are released within the lumen of the thylakoids. Oxygen released in photosynthesis is formed during this process.

$2H_{2}O$

$\longrightarrow$

$4H^{+}$ +

$O_{2}$ +

$4e^{-}$

Choose the correct statement.

0%
Stroma lamella membrane lacks PS II and PS I.
0%
When PS I is functional, electron flow in a noncyclic way.
0%
ATPase enzyme consists of $F_0$ and $F_1$ units.
0%
NADP reductase is a part of PS II.

Explanation

Correct Option: C

Explanation:

ATP synthase is an enzyme that helps in ATP synthesis.
The enzyme ATP synthase has two parts $F_0$ and $F_1$ . $F_0$ is the part that transports protons across the membrane as a transmembrane channel while configuration changes in $F_1$ activate the enzyme.
They phosphorylate ADP i.e., add one more phosphorus group to the ADP and converts ADP(adenosine diphosphate) to ATP (adenosine triphosphate).
The proton gradient developed across the membrane is the driving force of ATP synthase.

The correct sequence of involvement of the following in non-cyclic electron transport
(i) PC (ii) PQ (iii) Pheo (d) Fd is

0%
ii, i, iii, iv
0%
iii, ii, iv, i
0%
iv, i, ii, iii
0%
iii, ii, i, iv

Explanation

In electron transport process the excited electron from PS II to Pheophytin is accepted by primary acceptor quinone, which sends them to an electron transport system consisting of plastoquinone(PQ), cytochrome complex and plastocyanin(PC).

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

Hence, the correct answer is 'iii,ii,i,iv'

1167856_666050_ans_7be8c40535574c1bb7608683076404be.jpg

With reference to three Calvin cycles, which of the given options is correct for the following questions?
I. How many gross PGAL molecules are produced?
II. Total, how many ATP molecules are required for synthesis of PGAL molecules?
III. Total, how many NADPH

$_2$ molecules are required for the synthesis of obtained PGAL molecules?

0%
I= 3 PGAL, II = 3 ATP, III = 3 $_2$
0%
I= 6 PGAL, II = 6 ATP, III = 6 $_2$
0%
I= 18 PGAL, II = 18 ATP, III = 18 $_2$
0%
I= 9 PGAL, II = 9 ATP, III = 9 $_2$

Explanation

In one Calvin cycle, 1

$CO_2$ (1 carbon) combines with 1 molecule of Ribulose bisphosphate (5 carbon) to form 2 molecules of 3-phosphoglycerates (3 carbon each). The 2 molecules of 3-phosphoglycerates (PGA) in a couple of steps are converted to 2 molecules of Glyceraldehyde-3- phosphate (PGAL - 3 carbon). In this entire process of formation 2 molecules of PGAL, a total of 2 ATP and 2 NADPH are utilized. So, for 3 cycles, net PGAL will be 6, ATP and NADPH consumption will allso be 6. Thus, the correct option is B.

A plant in your garden avoids photorespiratory loss, has improved water use efficiency, shows high rate of photosynthesis at high temperatures and has improved efficiency of nitrogen utilisation. In which of the following physiological groups would you assign this plant

0%
Nitrogen fixer
0%
$C_3$
0%
$C_4$
0%
CAM.

Explanation

This plant is a

$C_{4}$ plant.

$C_{4}$ plants are special as they have a special type of leaf anatomy called Kranz Anatomy. They can tolerate higher temperatures, they show a response to high intensities of light, they lack a wasteful process called photorespiration and they have improved efficiency of nitrogen utilization. Most of the plants that are adapted to dry tropical regions have the

$C_{4}$ pathway and are known as

$C_{4}$ plants. In these plants, double fixation of carbon dioxide occurs.

So, the correct answer is 'C4'.

DCPIP (dichlorophenol indophenol) is

0%
Inhibitor of electron transport
0%
Herbicide
0%
Inhibitor of photosynthesis
0%
Blue coloured hill oxidant which becomes colourless on being reduced.

In photosynthesis, light independent reactions take place at

0%
Thylakoid lumen
0%
Photosystem I
0%
Photosystem III
0%
Stromal matrix

Choose the correct option for the chloroplast of bundle sheath from the following

0%
They show granal organisation
0%
They do not show granal organisation
0%
They do not possess thylakoids
0%
They possess thylakoid and granal organisation

Explanation

C $_4$ plants have Kranz leaf anatomy. The vascular bundles in these leaves are surrounded by a layer of bundle sheath cells that contain a large number of chloroplasts.

The chloroplasts in these leaves are dimorphic:

Bundle sheath- Chloroplast larger in size and arranged centripetally. It has unstacked thylakoids called agranal chloroplast. The cells perform C $_3$ cycle.

Mesophyll cells- Normal type of chloroplast. The cells perform C $_4$ cycle.

Hence, the correct answer is 'They do not show granal organisation'.

How many molecules of ATP and NADPH are required in formation of two molecules of glucose? How many Calvin cycles are required?

0%
24 ATP, 36 NADPH, 12 Calvin cycles
0%
18 ATP, 12 NADPH, 6 Calvin cycles
0%
36 ATP, 24 NADPH, 6 Calvin cycles
0%
36 ATP, 24 NADPH, 12 Calvin cycles

Photo-respiratory reactions are operated in

0%
Chloroplasts, ribosomes and peroxisomes
0%
Chloroplasts, mitochondria and peroxisomes
0%
Mitochondria, peroxisomes and lysosomes
0%
Mitochondria, chloroplasts and ribosomes

Explanation

Photo-respiratory reactions occur during the process of photorespiration. It is a process which involves loss of fixed carbon as

$O_{2}$ 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. There is neither the synthesis of sugar nor of ATP. 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.

Which of the option is correct for photorespiration?

0%
In peroxisome, glycerate forms phosphoglycolate
0%
In chloroplast, glycerate forms glycine
0%
In mitochondrion, glycine forms serine
0%
In bundle sheath, serine forms glycine.

Explanation

The phosphoglycolate which is formed in the choloroplast (as a result of affinity of RuBisCo to the

$O_2$ ) is transferred to the peroxisome in the form of glycolate. The Glycolate in peroxisomes is converted to glycine through some intermediates and then transferred to the mitochondria. In the mitochondria the glycine is converted to serine and transferred back to peroxisome. Thus, the correct option is C

In which cells of leaf, pyruvate is converted to PEP in C

$_4$ pathway?

0%
Epidermal cells
0%
Mesophyll cells
0%
Bundle sheath cells
0%
Guard cells

Explanation

In C

$_4$ plants, initial fixation of CO

$_2$ occurs in mesophyll cells. The primary acceptor of CO

$_2$ is phosphoenolpyruvate (PEP). CO

$_2$ is again fixed inside bundle sheath cells through the Calvin cycle. The pyruvate is sent back to mesophyll cells where it is changed to phosphoenolpyruvate (PEP).

So, the correct answer is 'Mesophyll cells'.

The process which makes major difference between

$C_3$ and

$C_4$ plants is

0%
Respiration
0%
Glycolysis
0%
Calvin cycle
0%
Photorespiration

Explanation

Photorespiration: It is inhibition of photosynthesis by high O $_2$ level. It was termed as Warburg effect. However, the inhibition was observed mostly in C $_3$ plants. The C $_4$ plants were hardly affected by varying O $_2$ concentrations. This is because of the spatial difference between the carbon dioxide fixation and the commencement of the C $_3$ cycle in the C $_4$ plants. The former occurs in the mesophyll cells, the latter occurs in the bundle-sheath cells.

Hence, the correct answer is 'Photorespiration'.

How many molecules of glycine is required to release one

$CO_2$ molecule in photorespiration?

0%
One
0%
Two
0%
Three
0%
Four

Explanation

Photorespiration is a wasteful process which occurs due to attachment of oxygen at the active site of enzyme RuBisCO instead of carbon dioxide. Phosphoglycolate is formed in place of 3-phosphoglycerate. It takes place in the chloroplasts, leaf peroxisomes and mitochondria. Two molecules of glycine combine in mitochondria during photorespiration and form serine and release one CO

$_2$ .

So, the correct answer is option B.

$_2$ acceptor in C4 plants is

0%
PGA
0%
PEP
0%
RuBP
0%
OAA

Explanation

PEP (phosphoenol pyruvate) is the CO

$_2$ acceptor in C4 plants. The molecule accepts the CO

$_2$ and allows the fixation of it. This results in the fixation of the gas at lower concentrations as RUBISCO does not bind

$CO_2$ at lower concentrations.

During photorespiration, in which of the following NAD is reduced?

0%
Mitochondria and peroxisome
0%
Chloroplast and peroxisome
0%
Chloroplast and mitochondria
0%
Mitochondria

Explanation

In Mitochondria, the decarboxylation of Glycine in the presence of glycine decarboxylase and also deamination releases

$CO_2$ and

$NH_3$ resulting in formation of serine. Furthermore, NAD is reduced to NADH. The serine thus formed enters the peroxisome.

How many Calvin cycles are required to produce 5 molecules of glucose?

0%
60
0%
15
0%
30
0%
90

Explanation

Glucose is a hexose sugar. It's one molecule contains six carbons. As the Calvin cycle takes in only one carbon (as

$CO_2$ ) at a time, six turns of this cycle will be required to produce one molecule of glucose (6C). Hence, for producing a net gain of 5 molecules of glucose 30 molecules of carbon are required. Hence 30 turns of the Calvin cycle are needed. Thus the correct answer is option C.

TCA cycle is named after

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Embden
0%
Calvin
0%
Krebs
0%
Emerson

During light reaction of photosynthesis, which of the following phenomenon is observed during cyclic phosphorylation as well as non-cyclic phosphorylation?

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Release of O $_2$
0%
Formation of ATP
0%
Formation of NADPH
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Involvement of PS-I and PS-lI pigment systems

$C_4$ plants, chloroplasts also occur in

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Bundle sheath cells
0%
Guard cells
0%
Spongy parenchyma
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Epidermis

Explanation

$C_4$ plants have a special type of leaf anatomy called Kranz anatomy.

$C_4$ pathway requires two types of cells i.e., mesophyll cells and bundle sheath cells. Bundle sheath cells are the particularly large cells around the vascular bundles. In

$C_4$ plants, chloroplasts also occur in bundle sheath cells. They have a large number of dimorphic chloroplasts.

Chemiosmotic hypothesis of ATP synthesis in chloroplasts is based on

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Accumulation of $K^+$ ions
0%
Proton gradient
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Accumulation of $Na^+$ ions
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Membrane potential

Chloroplast dimorphism is a characteristic feature of

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Plants with Calvin cycle
0%
$C_4$ plants
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All plants
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Only in algae.

Explanation

Dimorphic chloroplasts are the chloroplasts which are concentrated into two types of cells having different size and structure. The dimorphic chloroplast is present in

$C_4$ plants because of the presence of special leaf anatomy called Kranz anatomy. They have two types of cells i.e., mesophyll cells and bundle sheath cells, which are required for

$C_4$ pathway. The particularly large cells around the vascular bundles of

$C_4$ plants are called bundle sheath cells, these cells may form several layers around the vascular bundles, they are characterised by having a large number of dimorphic chloroplasts, grana are absent, thick walls impervious to gaseous exchange and no intercellular spaces. Mesophyll cells have dimorphic chloroplast having grana. Chloroplast dimorphism is characteristic feature of

$C_4$ plants, e.g. Amaranthus, Sugarcane, Maize, etc.

The process which makes major difference between

$C_3$ and

$C_4$ plants is

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Respiration
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Glycolysis
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Calvin cycle
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Photorespiration

Explanation

Photorespiration is a process which involves loss of fixed carbon as

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

$O_{2}$ does 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. There is neither the synthesis of sugar nor of ATP. This process makes the major difference between

$C_{3}$ and

$C_{4}$ plants.

$C_{4}$ plants, photorespiration does not occur. This is because these plants have a mechanism that increases the concentration of

$CO_{2}$ at the enzyme site. Thus, the productivity and yields are better in

$C_{4}$ plants as compared to

$C_{3}$ plants.

Reason behind negligible/ no rate of photorespiration in

$C_{4}$ plant is _________________.

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High level of internal $CO_{2}$
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Agranal chloroplast of Bundle sheath cell
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Low level of internal $O_{2}$ aroung RuBisco
0%
More than one option correct

CBSE Questions for Class 11 Medical Biology Photosynthesis In Higher Plants Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Medical Biology Quiz Questions and Answers

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