Text Book Back Questions and Answers - Chapter 13 Photosynthesis 11th Biology Botany Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Photosynthesis
Text Book Back Questions and Answers
Questions 1.
Assertion (A): Increase in Proton gradient inside lumen responsible for ATP synthesis.
Reason (R): Oxygen evolving complex of PS I located on thylakoid membrane facing Stroma, releases $\mathrm{H}+$ ions.
(a) Both Assertion and Reason are True.
(b) Assertion is True and Reason is False.
(c) Reason is True and Assertion is False.
(d) Both Assertion and Reason are False.
Answer:
(a) Both Assertion and Reason are True.
Question 2.
Which chlorophyll molecule does not have a phytol tail?
(a) $\mathrm{Chl}-\mathrm{a}$
(b) $\mathrm{Chl}-\mathrm{b}$
(c) $\mathrm{Chl}-\mathrm{c}$
(d) $\mathrm{Chl}-\mathrm{d}$
Answer:
(c) $\mathrm{Chl}-\mathrm{c}$
Question 3.
The correct sequence of flow of electrons in the light reaction is:
(a) PS II, plastoquinone, cytochrome, PSI, ferredoxin.
(b) PS I, plastoquinone, cytochrome, PS II ferredoxin.
(c) PS II, ferredoxin, plastoquinone, cytochrome, PS I.
(d) PS I, plastoquinone, cytochrome, PS II, ferredoxin.
Answer:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.
Question 4.
For every $\mathrm{CO}_2$ molecule entering the $\mathrm{C}_3$ cycle, the number of ATP & NADPH required:
(a) 2 ATP $+2 \mathrm{NADPH}$
(b) 2 ATP + 3 NADPH
(c) 3 ATP + 2 NADPH
(d) 3 ATP + 3 NADPH
Answer:
(c) 3 ATP $+2 \mathrm{NADPH}$.
Question 5.
Identify true statement regarding light reaction of photosynthesis?
(a) Splitting of water molecule is associate with PS I.
(b) PS I and PS II involved in the formation of $\mathrm{NDPH}+\mathrm{H}^{+}$.
(c) The reaction center of $\mathrm{PS} \mathrm{I}$ is Chlorophyll a with absorption peak at $680 \mathrm{~nm}$.
(d) The reaction center of PS II is Chlorophyll a with absorption peak at 700 $\mathrm{nm}$.
Answer:
(b) PS I and PS II involved in the formation of $\mathrm{NDPH}+\mathrm{H}^{+}$.
Question 6.
Two groups (A \& B) of bean plants of similar size and same leaf area were placed in identical conditions. Group A was exposed to light of wavelength $400-450 \mathrm{~nm}$ and Group B to light of wavelength of $500-550 \mathrm{~nm}$. Compare the photosynthetic rate of the 2 groups giving reasons.
Answer:
The rate of photosynthesis in group $A$ bean plants is more than what is found in Group B plants because the rate of absorption of light is more at the wavelength is less beyond the wavelength of $500-550 \mathrm{~nm}$.
Question 7.
A tree is believed to be releasing oxygen during night time. Do you believe the truthfulness of this statement? Justify your answer by giving reasons?
Answer:
Yes, a tree is believed to be releasing $\mathrm{O}_2$ during night time because at night CAM plants fix $\mathrm{CO}_2$ with the help of phospho Enol Pyruvic acid and produce oxala acetic acid, which is converted into malic acid like $\mathrm{C}_4$ cycle.

Question 8.
Grasses have an adaptive mechanism to compensate. photorespiratory losses - Name and describe the mechanism.
Answer:
Rate of respiration is more in light than in dark. Photorespiration is the excess respiration taking place in photosynthetic cells due to absence of $\mathrm{CO}_2$ and increase of $\mathrm{O}_2$. This condition changes the carboxylase role of RUBISCO into oxygenase. $C_2$ Cycle takes place in chloroplast, peroxisome and mitochondria. RUBP is converted into PGA and a $2 C$ - compound phosphoglycolate by Rubisco enzyme in chloroplast. Since the first product is a $2 \mathrm{C}$ - compound, this cycle is known as $\mathrm{C}_2$ Cycle. Phosphoglycolate by loss of phosphate becomes glycolate.
Glycolate formed in chloroplast enters into peroxisome to form glyoxylate and hydrogen peroxide. Glyoxylate is converted into glycine and transferred into mitochondria. In mitochondria, two molecules of glycine combine to form serine. Serine enters into peroxisome to form hydroxy pyruvate. Hydroxy pyruvate with help of $\mathrm{NADH}+\mathrm{H}+$ becomes glyceric acid. Glyceric acid is cycled back to chloroplast utilising ATP and becomes Phosphoglyceric acid (PGA) and $v$ enters into the Calvin cycle (PCR cycle). Photorespiration does not yield any free energy in the form of ATP. Under certain conditions $50 \%$ of the photosynthetic potential is lost because of Photorespiration
Question 9.
In Botany class, teacher explains, Synthesis of one glucose requires 30 ATPs in $C_4$ plants and only 18 ATPs in $C_3$ plants. The same teacher explains $C_4$ plants are more advantageous than $C_3$ plants. Can you identify the reason for this contradiction?
Answer:
$\mathrm{C}_4$ plants requires 30 ATPs and $12 \mathrm{NADPH}+\mathrm{H}+$ to synthesize one glucose, but $\mathrm{C}_3$ plants require only 18 ATPs and $12 \mathrm{NADPH}+\mathrm{H}$ + to synthesize one glucose molecule. If then, how can you say $C_4$ plants are more advantageous? $C_4$ plants are more advantageous than $C_3$ plants because $C_4$ photosynthesis is advantages over $C_3$ plant, because $C_4$ photosynthesis avoids photorespiration and is thus potentially more efficient than $\mathrm{C}_3$ plants. Due to the absense of photorespiration, carbon di oxide compensation point for $C_4$ is lower than that of $\mathrm{C}_3$ plants.
Question 10.
When there is plenty of light and higher concentration of $\mathrm{O}_2$, what kind of pathway does the plant undergo? Analyse the reasons.
Answer:
The rate of photosynthesis decreases when there is an increase of oxygen concentration. This Inhibitory effect of oxygen was first discovered by Warburg (1920) using green algae, Chlorella.