C3 and Dark reaction Questions in English

(B) The biosynthetic phase of photosynthesis,also known as the dark reaction or the Calvin cycle,involves the fixation of $CO_2$ into sugars.
This process was elucidated by Melvin Calvin and his colleagues using radioactive $^{14}C$ in algal photosynthesis studies.
Therefore,Melvin Calvin is credited with the discovery of the biosynthetic phase of photosynthesis.

(D) $RuBP$ stands for Ribulose$-1,5-$bisphosphate,which is a $5$-carbon ketose sugar (pentose sugar).
$RuBisCO$ stands for Ribulose$-1,5-$bisphosphate carboxylase-oxygenase,which is the enzyme responsible for the primary carboxylation step in the Calvin cycle.
Therefore,$RuBP$ is a pentose sugar and $RuBisCO$ is an enzyme.

(A) In the $C_3$ cycle (Calvin cycle),the enzyme $RuBisCO$ (Ribulose$-1,5-$bisphosphate carboxylase-oxygenase) catalyzes the carboxylation of $RuBP$ (Ribulose$-1,5-$bisphosphate).
When $RuBisCO$ acts as a carboxylase,it reacts $RuBP$ with $CO_2$ and $H_2O$ to produce $2$ molecules of $3$-phosphoglycerate ($3$-$PGA$).
This is the primary step of carbon fixation in $C_3$ plants.

(B) In the $C_3$ cycle (Calvin cycle),the primary carboxylation reaction involves the fixation of $CO_2$ by $RuBP$ to form two molecules of $3$-phosphoglyceric acid $(PGA)$,which is the first stable product $(P = PGA)$.
In the $C_4$ cycle (Hatch-Slack pathway),the primary carboxylation occurs in the mesophyll cells where $PEP$ combines with $CO_2$ to form oxaloacetic acid $(OAA)$,which is a $4$-carbon compound and the first stable product $(Q = OAA)$.

(D) The Calvin cycle consists of three main stages: carboxylation,reduction,and regeneration.
To produce one molecule of glucose $(C_6H_{12}O_6)$,the cycle must turn $6$ times because each turn fixes one molecule of $CO_2$.
In the reduction phase,for every $CO_2$ molecule fixed,$2 \, ATP$ are used for phosphorylation and $2 \, NADPH$ are used for reduction.
Since $6$ turns are required for one glucose molecule,the total requirement for the reduction phase is $6 \times 2 = 12 \, ATP$ and $6 \times 2 = 12 \, NADPH$.

(D) In the Calvin cycle,the conversion of $1,3$-bisphosphoglycerate to $3$-phosphoglycerate is a step where a phosphate group is removed from the molecule. This process is catalyzed by the enzyme phosphoglycerate kinase. Since a phosphate group is removed,the reaction is termed as dephosphorylation. This step also results in the formation of $ATP$ from $ADP$.

(C) The Calvin cycle consists of three main stages:
$1$. Carboxylation: The fixation of $CO_2$ into a stable organic intermediate,where $CO_2$ is utilized for the carboxylation of $RuBP$.
$2$. Reduction: $A$ series of reactions that lead to the formation of glucose,involving the utilization of $ATP$ and $NADPH$ for the reduction of $3-PGA$.
$3$. Regeneration: The formation of the $CO_2$ acceptor molecule $(RuBP)$ again,which is crucial for the cycle to continue.
Therefore,the correct sequence is Carboxylation $\rightarrow$ Reduction $\rightarrow$ Regeneration.

(A) The Calvin cycle consists of three main stages: Carboxylation,Reduction,and Regeneration.
$1$. Carboxylation: $CO_2$ combines with a $5$-carbon sugar called Ribulose $1,5$-bisphosphate $(RuBP)$ (represented by $P$) to form a $3$-carbon acid,$3$-phosphoglycerate ($3$-$PGA$) (represented by $R$).
$2$. Reduction: $3$-$PGA$ is converted into Triose phosphate (represented by $Q$) using $ATP$ and $NADPH$.
$3$. Regeneration: Triose phosphate is used to regenerate $RuBP$ to continue the cycle.
Therefore,$P$ is Pentose sugar $(RuBP)$,$R$ is $3$-carbon acid ($3$-$PGA$),and $Q$ is Triose sugar (Triose phosphate).

(A) In the Calvin cycle,the fixation of one molecule of $CO_2$ requires $3\,ATP$ and $2\,NADPH$ molecules.
For the fixation of three $CO_2$ molecules,the calculation is as follows:
$ATP$ required = $3 \times 3 = 9\,ATP$
$NADPH$ required = $3 \times 2 = 6\,NADPH$
Therefore,to fix three $CO_2$ molecules,$9\,ATP$ and $6\,NADPH$ are required.

(A) In $C_3$ plants,the entire process of photosynthesis occurs within the mesophyll cells.
The light-dependent reaction (light reaction) takes place in the thylakoid membranes (grana) of the chloroplasts.
The Calvin cycle (dark reaction) takes place in the stroma of the same chloroplasts within the mesophyll cells.
Therefore,both processes are localized in the mesophyll cells of $C_3$ plants.

(A) In the Calvin cycle,the reduction phase involves the conversion of $3$-phosphoglycerate ($3$-$PGA$) to glyceraldehyde $3$-phosphate (triose phosphate).
For the fixation of one molecule of $CO_2$,the reduction phase requires $1$ molecule of $ATP$ (for phosphorylation) and $1$ molecule of $NADPH$ (for reduction).
However,the overall stoichiometry for the production of one molecule of triose phosphate from $3$ molecules of $CO_2$ requires $6$ $ATP$ and $6$ $NADPH$ in the reduction phase.
Therefore,for the fixation of a single $CO_2$ molecule,the requirement is $1$ $ATP$ and $1$ $NADPH$ specifically for the reduction step.

(C) For every $CO_2$ molecule entering the Calvin cycle,$3$ molecules of $ATP$ and $2$ molecules of $NADPH_2$ are required.
To synthesize one molecule of glucose $(C_6H_{12}O_6)$,$6$ turns of the Calvin cycle are necessary because each turn fixes one carbon atom.
Therefore,the total requirement is calculated as:
$6 \times (3\,ATP + 2\,NADPH_2) = 18\,ATP$ and $12\,NADPH_2$.

(B) The dark reaction of photosynthesis,also known as the Calvin cycle or light-independent reaction,takes place in the stroma of the chloroplast.
It does not directly require light,chlorophyll,or the light-harvesting apparatus.
Instead,it utilizes the products of the light-dependent reactions to fix carbon dioxide into glucose.
The essential requirements for the dark reaction are:
$1$. $CO_2$: Acts as the carbon source for the synthesis of sugars.
$2$. $ATP$: Provides the necessary energy for the reduction process.
$3$. $NADPH$: Acts as a reducing agent to convert $3-PGA$ to $G3P$.
Therefore,$C, D,$ and $E$ are the correct requirements.

(C) In the Calvin cycle,the fixation of $1$ molecule of $CO_2$ involves three stages: carboxylation,reduction,and regeneration.
During the reduction phase,$2$ molecules of $ATP$ and $2$ molecules of $NADPH$ are used for the reduction of $1$ molecule of $3-PGA$ to $G3P$.
Additionally,$1$ molecule of $ATP$ is required during the regeneration phase to convert $RuMP$ back into $RuBP$.
Therefore,for the fixation of $1$ molecule of $CO_2$,a total of $3$ molecules of $ATP$ and $2$ molecules of $NADPH$ are consumed.

(D) $RuBisCO$ (Ribulose$-1,5-$bisphosphate carboxylase-oxygenase) is the most abundant enzyme on Earth.
It acts as a carboxylase in the Calvin cycle,where it catalyzes the carboxylation of $RuBP$ (Ribulose$-1,5-$bisphosphate) by adding $CO_2$ to it.
Option $A$ is incorrect because $RuBisCO$ is light-dependent and active during the day.
Option $B$ is incorrect because although it can bind to both,it has a much higher affinity for $CO_2$ than for $O_2$ under normal physiological conditions.
Option $C$ is incorrect because the photolysis of water is catalyzed by the oxygen-evolving complex associated with Photosystem $II$,not $RuBisCO$.
Therefore,the correct statement is that it catalyzes the carboxylation of $RuBP$.

(D) The Calvin cycle consists of three main stages: Carboxylation,Reduction,and Regeneration.
$1$. Carboxylation: $\text{CO}_2$ is fixed to $\text{RuBP}$ by the enzyme $\text{RuBisCO}$. No $\text{ATP}$ is used here.
$2$. Reduction: This stage utilizes both $\text{ATP}$ and $\text{NADPH}$ to convert $3$-phosphoglycerate into glyceraldehyde-$3$-phosphate.
$3$. Regeneration: This stage requires $\text{ATP}$ to regenerate $\text{RuBP}$ from triose phosphates so that the cycle can continue.
Therefore,both the Reduction and Regeneration stages utilize $\text{ATP}$.

(A) In the $C_3$-cycle (Calvin cycle),the synthesis of $1$ molecule of glucose $(C_6H_{12}O_6)$ requires $18$ $\text{ATP}$ and $12$ $\text{NADPH}$.
To form $3$ molecules of glucose,we multiply the requirements by $3$.
Total $\text{ATP}$ required $= 3 \times 18 = 54$ $\text{ATP}$.
Therefore,the correct option is $A$.

(A) The biosynthetic phase of photosynthesis is also known as the dark reaction or the Calvin cycle.
This phase does not directly require light but depends on the products of the light-dependent reaction ($ATP$ and $NADPH$).
Key processes in the biosynthetic phase include:
$1$. Use of $ATP$ (energy source).
$2$. Use of $NADPH + H^{+}$ (reducing agent).
$3$. Reduction of $CO_2$ to form carbohydrates (sugars).
Options $(A)$,$(C)$,and $(E)$ are correct components of this phase.
Option $(B)$ (Synthesis of $NADPH + H^{+}$) and $(D)$ (Release of oxygen) occur during the light-dependent phase (photochemical phase).
Therefore,the correct combination is $A, C$ and $E$ only.

(C) Statement $I$ is correct because the biosynthetic phase of photosynthesis,where $CO_2$ is fixed into sugars,is traditionally referred to as the dark reaction.
Statement $II$ is incorrect because these reactions do not require darkness; they occur in the presence of light. They are called 'dark reactions' simply because they do not directly require light energy (photons) to drive the enzymatic steps,but they are indirectly dependent on the products of the light-dependent reactions ($ATP$ and $NADPH$). Therefore,they can occur during the day as well.

(A) The Calvin cycle consists of three main stages: Carboxylation,Reduction,and Regeneration.
$1$. Carboxylation: In this stage,$CO_2$ is fixed by $\text{RuBP}$ with the help of the enzyme $\text{RuBisCO}$. This step does not require $\text{ATP}$ or $\text{NADPH}$.
$2$. Reduction: This stage involves the utilization of $2$ molecules of $\text{ATP}$ and $2$ molecules of $\text{NADPH}$ for the reduction of each molecule of $3$-phosphoglycerate ($3$-$PGA$) to glyceraldehyde $3$-phosphate.
$3$. Regeneration: This stage requires $1$ molecule of $\text{ATP}$ for the phosphorylation of ribulose $5$-phosphate to regenerate $\text{RuBP}$.
Therefore,the Carboxylation stage is the only one that does not involve the dephosphorylation of $\text{ATP}$.

(A) The synthesis of one molecule of glucose $(C_6H_{12}O_6)$ via the Calvin cycle requires $18$ molecules of $ATP$ and $12$ molecules of $NADPH_2$.
To synthesize $10$ molecules of glucose,we multiply the requirement per glucose molecule by $10$.
Calculation: $12 \text{ molecules of } NADPH_2 \times 10 \text{ glucose molecules} = 120 \text{ molecules of } NADPH_2$.
Therefore,$120$ molecules of $NADPH_2$ are required.

(D) The correct answer is $D$.
The Calvin cycle (also known as the $C_3$ cycle) is the light-independent reaction of photosynthesis.
For the synthesis of one molecule of glucose $(C_6H_{12}O_6)$,the cycle must turn six times.
The overall stoichiometry for the production of one glucose molecule is:
$6 CO_2 + 18 ATP + 12 NADPH + 12 H^+ \rightarrow 1 C_6H_{12}O_6 + 18 ADP + 18 Pi + 12 NADP^+ + 6 H_2O$.
Thus,the net outcome includes $1$ glucose,$18 ADP$,and $12 NADP^+$.

(A) The correct option is $A$.
In the Calvin cycle (dark reaction),the process consists of three main stages:
$1$. Carboxylation $(A)$: $CO_2$ is fixed by $RuBP$ to form an unstable compound which breaks down into $PGA$.
$2$. Reduction ($B$ and $C$): This involves two steps:
- Phosphorylation $(B)$: $PGA$ is converted to $1,3$-bisphosphoglycerate using $ATP$.
- Reduction $(C)$: $1,3$-bisphosphoglycerate is reduced to $G3P$ (or $3-PGA$ aldehyde) using $NADPH$.
$3$. Regeneration $(D)$: $RuBP$ is regenerated from $RuMP$ using $ATP$ to continue the cycle.

(A) The incorrect statement is $A$.
In the Calvin cycle,$18$ molecules of $ATP$ are not synthesized during carbon fixation; rather,$ATP$ is consumed.
To produce one molecule of glucose ($6$ carbons),the Calvin cycle must turn $6$ times.
Each turn requires $3$ $ATP$ and $2$ $NADPH$ molecules.
Therefore,a total of $18$ $ATP$ and $12$ $NADPH$ are consumed,not synthesized,during the process.
Option $B$ is correct as $NADPH$ is used for the reduction of $1,3$-bisphosphoglycerate to glyceraldehyde-$3$-phosphate.
Option $C$ is correct as $RuBisCO$ is the enzyme responsible for the carboxylation of $RuBP$.
Option $D$ is correct as $3$-phosphoglycerate ($3$-$PGA$) is the first stable product of the Calvin cycle.

(D) The correct answer is $D$.
In the process of photosynthesis,the carbon source is $CO_2$,which is incorporated into the glucose molecule during the Calvin cycle.
Since the plant is supplied with the radioactive isotope $^{14}CO_2$,the carbon atoms in the resulting glucose molecule will be radioactive (labelled).
Oxygen released during photosynthesis is derived from the photolysis of water $(H_2O)$,not from $CO_2$.
Therefore,the oxygen released will be normal $(^{16}O)$ and not labelled.

(B) In the Calvin cycle,the enzyme RuBisCO (RuBP carboxylase-oxygenase) catalyzes the carboxylation step.
This enzyme facilitates the fixation of atmospheric $CO_2$ into a $5$-carbon sugar called Ribulose $1,5$-bisphosphate (RuBP).

(B) For the synthesis of one molecule of glucose $(C_6H_{12}O_6)$ via the Calvin cycle,$6$ turns are required.
Each turn fixes one $CO_2$ molecule,consuming $3$ $ATP$ and $2$ $NADPH$.
Therefore,for $6$ $CO_2$ molecules,the total requirement is:
$6 \times 3 = 18$ $ATP$
$6 \times 2 = 12$ $NADPH$
Thus,$18$ $ATP$ and $12$ $NADPH$ molecules are required to produce one molecule of glucose.