MCQ Questions for Class 11 Medical Biology Photosynthesis In Higher Plants Quiz 17

Which group of plant are not benefited by

$CO_2$ fertilisation effect?

0%
Plant requiring 3 ATP per $CO_2$ fixation
0%
Plant with kranz anatomy
0%
Plant having high rate of photorespiration
0%
Plant having single carboxylation

Explanation

$_{2}$ fertilization refers to the increase in the concentration of the carbon dioxide due to the anthropogenic activities. The plants that perform the C

$_{3}$ cycle are benefitted by this process as increase in CO

$_{2}$ will not allow the oxygen to combine with the enzyme RuBisCo and so, the wasteful process of the photorespiration could be avoided. The C

$_{4}$ plants already have the mechanism to use the maximum CO

$_{2}$ because of the Kranz anatomy, hence they will not be benefited. The C

$_{3}$ plants use 3ATP per CO

$_{2}$ fixation and the C

$_{4}$ plants use 5ATP per CO

$_{2}$ fixation.

Hence, the correct answer is 'Plant with Kranz anatomy'

Number of cell organelles involved in photorespiration are

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

Dark reaction involves utilisation of

$CO_2$ in producing

$C_6H_12O_6(Glucose)$ . Why is this phase called dark reaction?

0%
The dark reaction is not dependent on light
0%
It occurs simultaneously with the light reaction
0%
In this process, ATP and NADP molecules formed in light reaction enter in to dark reaction
0%
All of the above.

During phenomenon of photorespiration the oxygen consuming steps occurs at

0%
Chloroplast and mitochondria
0%
Mitochondria and peroxisome
0%
Peroxisome and chloroplast
0%
Only chloroplast

Explanation

The initial oxygenation reaction by RuBiSCO occurs in the chloroplast. This is where O

$_{2}$ is consumed by photorespiration. The 2-phosphoglycolate that is formed is dephosphorylated and exported as glycolate to the peroxisome. The glycolate is then converted to glyoxylate. This reaction consumes O

$_{2}$ and generates hydrogen peroxide.

So, the correct option is 'Peroxisome and chloroplast'.

$C_4$ plants additional ATP.s consumed in

0%
Reduction of PGA
0%
Regeneration of RUBP
0%
Regeneration of PEP
0%
Regeneration of OAA

Amount of assimilatory power required to produce 2 molecules of sucrose in a

$C_4$ plant is

0%
30 ATPand 12 NADPH
0%
60 ATP and 24 NADPH
0%
36 ATP and 24 NADPH
0%
120 ATP and 48 NADPH

Explanation

As we know that for fixing of 1 CO

$_2$ molecule in C

$_4$ plants: 5ATPs and 2NAPDH molecules are required.

So, for formation of one molecule of sucrose(C

$_1$

$_2$ H

$_2$

$_2$ O

$_1$

$_1$ ) which consist of 12 Carbon atoms- 12 CO

$_2$ molecules are needed to be fixed. So, number of ATP and NADPH required are 60 and 24 respectively.

For formation of 2 molecules of sucrose in C

$_4$ plants 60 x 2 ATP and 24 x 2 NADPH will be required.

So, the correct answer is '120 ATP and 48 NADPH'.

Among the following the strongest oxidant is

0%
$P680^+$
0%
$H_2O$
0%
$P700^+$
0%
$P700$

How many quanta of light are required per NADPH formation?

0%
8 quanta
0%
16 quanta
0%
4 quanta
0%
2 quanta

Explanation

$\textbf{Correct Option C}$

$\textbf{Solution}$

$\bullet$ Quanta of light means the tiniest particle of the light present in the nature. Light quanta is also called photons.

$\bullet$ During photosynthesis, energy from the sunlight is captured by the leaves in the form of photons which are then converted into chemical energy and stored for future use.

$\bullet$ The stored energy is generally in the form of ATP and NADPH. For the formation of two molecules of ATP, NADPH there is a requirement of 8 photons.

$\bullet$ So requirement of quantity of light per NADPH formation is

8/2= 4 quanta of light.

$\bullet$ So the correct option is C.

Product (s) of

$Z-scheme$ which is/are utilise in calvin cycle, is/are:

0%
Only $ATP$
0%
$ATP,NADPH$ and $O_2$
0%
$ATP$ and $NADPH$
0%
Only $O_2$

Primary carboxylating enzyme in

$C_4$ plant is found in __________________.

0%
Chloroplast of mesophyll cells
0%
Cytoplasm of mesophyll cell
0%
chloroplast of bundle sheath cells
0%
Cytoplasm of bundle sheath cells

Explanation

Primary carboxylating enzyme in

${ C }_{ 4 }$ plant is found in Chloroplast of mesophyll cell. In

${ C }_{ 4 }$ plants, initial fixation of carbon dioxide occurs in mesophyll cells. PEP is the primary acceptor of carbon dioxide which combines with carbon dioxide in the presence of PEPcase to form oxaloacetate.

Why does the total amount of

$CO_2$ in the atmosphere remains constant even though respiration liberates

$CO_2$ ?

0%
$CO_2$ acts as a buffer
0%
$CO_2$ is converted to carbohydrates in photosynthesis
0%
$CO_2$ forms carbonic acid
0%
$CO_2$ undergoes cleavage during photosynthesis.

Explanation

Most organisms release or exhale

$CO_2$ during respiration. This

$CO_2$ present in the environment is utilized by plants in the process of photosynthesis with the final product being a carbohydrate. This is how the amount of

$CO_2$ in the atmosphere is kept under check.

So, the correct answer is '

$CO_2$ is converted to carbohydrates in photosynthesis'.

How many ATP equivalents are required to produce one glucose in

$C_4$ plants?

0%
$66ATP$
0%
$54 ATP$
0%
$18 ATP$
0%
$93ATP$

Explanation

In C4 plants, for production of one Glucose molecule 12 ATP and 18 energy NADPH

$_2$ are required. One NADPH

$_2$ is equivalent to 3 ATPs. Therefore a total of 54 equivalents of ATP will be required for the production of one glucose unit.

So, the correct answer is '54 ATP'.

How many turns of Calvin cycles are required for the production of one molecule of glucose and fructose respectively?

0%
$6, 6$
0%
$6, 12$
0%
$12, 3$
0%
$12, 6$

Photorespiration takes place in?

0%
Chloroplast
0%
Mitochondria
0%
Peroxisome
0%
All of the above

Explanation

(A) Correct answer (D)

(B) Explanation of correct answer:

Photorespiration is a process in $C_3$ plants that inhibits the rate of photosynthesis in the presence of high oxygen concentration in light.
In the presence of high concentration of oxygen, $C_3$ plants start releasing carbon dioxide in light during an extra input of oxygen.
It is an enzymatic process that involves specialised reactions which take place in three organelles in plants namely, chloroplasts, mitochondria and peroxisomes.
In the presence of low carbon dioxide concentration, enzyme RUBISCO acts as an oxygenase and splits a molecule of RUBP into 2-phosphoglycolic acid and 3- phosphoglyceric acid.

How many plants are

$C_4$ plant?

Amaranthus, Sugarcane, Maize, Sorghum, Opuntia, Euphorbia, Wheat, Rice.

0%
$8$
0%
$6$
0%
$5$
0%
$4$

Explanation

A C $_4$ plant is a plant that cycles carbon dioxide into four-carbon sugar compounds to enter into the calvin cycle.
These plants are very efficient in hot, dry climates and make a lot of energy.
Amaranthus, Sugarcane, Maize, Sorghum, Euphorbia are C $_4$ plants.
So, the correct answer is option C.

The correct sequence of flow of electrons in Z-scheme during photosynthesis is?

0%
$H_2O \rightarrow PS I \rightarrow ETS \rightarrow PS II \rightarrow NAD^+$
0%
$PS II \rightarrow H_2O \rightarrow ETS \rightarrow PS I \rightarrow NAD^+$
0%
$H_2O \rightarrow PS II \rightarrow ETS \rightarrow PS I \rightarrow NADP^+$
0%
$ETS \rightarrow PS II \rightarrow PS I \rightarrow H_2O$

Number of ATP and NADPH

$+$ H

$^+$ required to synthesise one glucose in Sorghum respectively.

0%
$18, 12$
0%
$3, 2$
0%
$5, 2$
0%
$30, 12$

CBSE Questions for Class 11 Medical Biology Photosynthesis In Higher Plants Quiz 17 - 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 17 - MCQExams.com

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Practice Class 11 Medical Biology Quiz Questions and Answers

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