System Physiology Plant MCQ Quiz - Objective Question with Answer for System Physiology Plant - Download Free PDF

The correct answer is C and D

Concept:

• The interconversion of fructose 6-phosphate to fructose 1,6-bisphosphate is a crucial step in the glycolytic pathway, a central metabolic process in plants. This step is tightly regulated to ensure proper energy production and metabolic balance.
• Two key enzymes involved in this step are phosphofructokinase (PFK) and pyrophosphate-dependent phosphofructokinase (PPi-PFK), each having distinct roles and regulatory mechanisms.

Explanation:

Statement C: Cytosolic fructose 1,6-bisphosphatase is strongly inhibited by fructose 2,6-bisphosphate.

• This inhibition is a critical regulatory mechanism in plants, ensuring that glycolysis and gluconeogenesis do not occur simultaneously in the cytosol.
• Fructose 2,6-bisphosphate acts as a potent allosteric inhibitor of cytosolic fructose 1,6-bisphosphatase, favoring glycolysis when energy is required.

Statement D: Pyrophosphate-dependent phosphofructokinase catalyzes a reversible reaction of interconversion of fructose 6-phosphate to fructose 1,6-bisphosphate.

• PPi-PFK is unique to plants and some microorganisms. Unlike ATP-dependent PFK, PPi-PFK uses pyrophosphate (PPi) as the phosphoryl donor.
• This enzyme allows the reaction to be reversible, providing flexibility in metabolic regulation under varying conditions such as energy scarcity.

Incorrect Options:

Statement A: Phosphofructokinase catalyzes the C6 phosphorylation of fructose 6-phosphate.

• This statement is incorrect because Phosphofructokinase (PFK), specifically Phosphofructokinase-1 (PFK-1), catalyzes the C1 phosphorylation of fructose 6-phosphate (F6P) to form fructose 1,6-bisphosphate (FBP).

The reaction is Fructose 6-phosphate + ATP → Fructose 1,6-bisphosphate + ADP

Statement B: Plastid phosphofructokinase is activated by Pi, while cytosolic phosphofructokinase is activated by phosphoenolpyruvate.

• This statement is incorrect. Plastid phosphofructokinase is activated by Pi, but cytosolic phosphofructokinase is inhibited by phosphoenolpyruvate, not activated.
• Phosphoenolpyruvate typically acts as a feedback inhibitor, signaling high energy status and reducing glycolytic flux.

The correct answer is A and D

Concept:

• Phosphoenolpyruvate carboxylase (PEPCase) is a critical enzyme in plants that participate in the fixation of CO₂ during photosynthesis.
• PEPCase is particularly significant in C₄ and CAM (Crassulacean Acid Metabolism) plants because of its role in adapting plants to different environmental conditions.
• The regulation of PEPCase activity in C₄ and CAM plants is achieved through phosphorylation by PEPCase kinase, which modifies the enzyme's sensitivity to inhibitors like malate.

Statement A: Light activates PEPCase kinase in C₄ plants.

• This statement is correct.
• In C₄ plants, PEPCase kinase is activated in the presence of light. This activation leads to the phosphorylation of PEPCase, which enhances its activity during the daytime for efficient CO₂ fixation.

Statement D: Phosphorylated PEPCase is less sensitive to malate.

• This statement is also correct.
• Phosphorylation of PEPCase reduces its sensitivity to malate, a feedback inhibitor. This allows the enzyme to continue functioning efficiently even when malate levels are high, which is critical for photosynthetic processes in C₄ and CAM plants.

Other Options:

Statement B: Phosphorylation inactivates PEPCase in C₄ plants while it activates the enzyme in CAM plants.

• This statement is incorrect.
• Phosphorylation activates PEPCase in both C₄ and CAM plants by reducing its sensitivity to malate, enabling it to function efficiently during CO₂ fixation.

Statement C: PEPCase kinase gets activated by light in CAM plants.

• This statement is incorrect.
• In CAM plants, PEPCase kinase activation is controlled by the circadian rhythm and not directly by light. CAM plants typically fix CO₂ at night, and PEPCase activity is regulated accordingly.

The correct answer is C and D only

Concept:

Polar auxin transport is a fundamental mechanism in plants that regulates growth and development. Auxins are a class of plant hormones responsible for processes such as cell elongation, root development, and tropic responses. This transport is directional and occurs from the shoot apex to the root apex. The polar nature of auxin transport is facilitated by specific cellular mechanisms involving active transport and protein carriers.

• Auxin transport occurs via specialized pathways that ensure the hormone's movement in a controlled and polarized manner.
• Protein carriers such as PIN and AUX/LAX proteins play a significant role in mediating this transport across plant cell plasma membranes.

Explanation:

• Statement C: Polar auxin transport is specific for active auxins, including both natural ones such as indole-3-acetic acid (IAA) and synthetic auxins like 2,4-D. This specificity is crucial for the regulation of plant growth processes. Thus, this statement is correct.
• Statement D: The process is mediated by protein carriers located on the plasma membrane, primarily PIN proteins and AUX/LAX transporters, which actively move auxins in a polar manner. 

Incorrect Statements:

• Statement A: Polar auxin transport does not proceed via the symplast (cytoplasmic continuity between cells). Instead, it involves active transport across the plasma membrane mediated by protein carriers. Therefore, this statement is incorrect.
• Statement B: The velocity of polar auxin transport is slower than the phloem translocation rates. Phloem translocation is a much faster process, driven by pressure flow mechanisms. Thus, this statement is incorrect.

The correct answer is A = Low light; B = Darkness; C = High light

Explanation:

• Chloroplasts in plant cells can move within cells to adapt to different light conditions. This process is known as chloroplast movement or chloroplast relocation.
• Under low light conditions, chloroplasts distribute across the surface of the cell to maximize the absorption of light for photosynthesis.
• In darkness, chloroplasts tend to remain in a scattered, resting position as no light stimulus is present to drive active movement.
• Under high light intensity, chloroplasts move to the sides of the cell walls (along the periphery) to avoid damage caused by excessive light, a phenomenon called photoprotection.
  • A = Low light: Under low light, chloroplasts spread across the surface of the palisade cells to maximize light capture for photosynthesis. 
  • B = Darkness: In the absence of light, chloroplasts remain scattered and stationary within the cell. 
  • C = High light: At high light intensities, chloroplasts move to the peripheral sides of the cells to minimize exposure and prevent photodamage. 

Fig: Schematic diagram of chloroplast distribution patterns in Arabidopsis palisade cells in response to different light intensities. (A) Under low light conditions, chloroplasts optimize light absorption by accumulating at the upper and lower sides of palisade cells. (B) Under high light conditions, chloroplasts avoid sunlight by migrating to the side walls of palisade cells. (C) Chloroplasts move to the bottom of the cell in darkness (Source: Plant Physiology and Development by Taiz and Zeiger VI Edition)

The correct answer is (A) - (II), (B) - (IV), (C) - (I), (D) - (III)

Explanation:

• Phylloclade (A) - Euphorbia (II):
  • Phylloclades are flattened or cylindrical, photosynthetic stems that resemble leaves.
  • They help plants like Euphorbia and Opuntia survive in arid conditions by reducing water loss.
  • In Euphorbia, phylloclades are a common feature, enabling the plant to perform photosynthesis while conserving water.
• Cladode (B) - Opuntia (IV):
  • Cladodes are flattened, photosynthetic stems that function like leaves, commonly found in cacti such as Opuntia.
  • Opuntia, a desert plant, uses cladodes to store water and perform photosynthesis efficiently in hot, dry environments.
• Phyllode (C) - Acacia (I):
  • Phyllodes are flattened petioles that take over the function of leaves. They are a characteristic feature of Acacia species.
  • In Acacia, true leaves are reduced to minimize water loss, while phyllodes perform photosynthesis and adapt the plant to arid climates.
• Inflated Petiole (D) - Pistia (III):
  • Inflated petioles are air-filled structures that provide buoyancy to floating aquatic plants like Pistia (water lettuce).
  • These modified petioles help Pistia float on water surfaces, facilitating efficient gas exchange and sunlight absorption for photosynthesis.

The correct answer is B, C and E.

Explanation:

A. The receptors for plant steroid hormones are found in the nucleus, similar to animal steroid hormones: This statement is incorrect. Unlike animal steroid hormone receptors that are often nuclear receptors, plant steroid hormone receptors, such as those for brassinosteroids, are usually found on the cell membrane. The well-known receptor for brassinosteroids in plants is BRI1, a membrane-bound receptor kinase.
B. There are multiple pathways for the plant steroid hormone biosynthesis involving cytochrome P450 class of enzymes: This is correct. The biosynthesis of brassinosteroids, a major class of plant steroid hormones, involves multiple pathways, and several cytochrome P450 enzymes play essential roles in these pathways.
C. The first plant steroid hormone was isolated from male gametophytes: This is correct. The first brassinosteroid, brassinolide, was identified from pollen (male gametophytes) of rapeseed (Brassica napus).
D. Plants deficient for the steroid hormone brassinosteroid show underproliferation of phloem and overproliferation of xylem cells: This is incorrect. Plants deficient in brassinosteroids exhibit abnormal vascular development, characterized by reduced phloem tissue and increased xylem tissue.
E. Castasterone is a plant steroid hormone abundant in the vegetative tissues of the plant: This is correct. Castasterone is one of the active brassinosteroids and is indeed abundant in the vegetative tissues of plants.

Conclusion: Based on the analysis, the correct statements about plant steroid hormones are: B, C, and E

Explanation:

Whenever a ray of light falls on a substance, then either it reflects the light or absorbs it.

The color which is reflected by the substance appears to be its color to the human eye. All other colors are absorbed by the substance and are not visible to the eye.

Hence, when a plant with green leaves is kept in a dark room with only green light on, then it will reflect all the green light and appears brighter than the surrounding.

∴ The plant will appear brighter than the surroundings.

 The correct answer is A, D, and E.

Explanation:

Crassulacean Acid Metabolism (CAM) is an adaptation in certain plants, allowing them to conserve water by opening their stomata at night to capture CO₂ and close them during the day. 

Statement A: "The HCO₃⁻ concentration is enriched in the cytosol during the night by the CO₂ coming from the external atmosphere through the open stomata and the mitochondrial respiration."

• This is correct. At night, CAM plants open their stomata to take in CO₂, which is converted to bicarbonate (HCO₃⁻) in the cytosol. CO₂ is also generated from mitochondrial respiration.

Statement B: "Oxaloacetate produced by the action of PEPCase is stored in the vacuole during dark."

• This is incorrect. Oxaloacetate is not stored directly in the vacuole. Instead, it is rapidly converted into malate, which is stored in the vacuole during the night.

Statement C: "During light, oxaloacetate produces malate that provides CO₂ for the Calvin Benson cycle in the chloroplast."

• This is incorrect. During the day, malate (stored during the night) is decarboxylated to release CO₂, which then enters the Calvin-Benson cycle in the chloroplast. Oxaloacetate itself is not directly involved in this process during the day.

Statement D: "During dark, phosphoenolpyruvate is produced by the breakdown of starch present in the chloroplast."

• This is correct. During the night, starch is broken down to produce phosphoenolpyruvate (PEP), which is used by PEP carboxylase (PEPCase) to fix CO₂ into oxaloacetate.

Statement E: "CAM is a mechanism of concentrating CO₂ around Rubisco by keeping stomata closed during the day."

• This is correct. CAM plants concentrate CO₂ around Rubisco during the day by decarboxylating malate when the stomata are closed, thus reducing water loss.

Key Points

• HCO₃⁻ is enriched in the cytosol at night when the stomata are open.
• Malate, not oxaloacetate, is stored in the vacuole during the night and provides CO₂ during the day for the Calvin-Benson cycle.
• Phosphoenolpyruvate (PEP) is produced at night by the breakdown of starch.
• CAM plants close their stomata during the day, concentrating CO₂ around Rubisco to reduce water loss.

Fig. CAM Plant

The correct answer is Option 2

Explanation:

In the ABCDE model of flower development, five classes of genes (A, B, C, D, and E) work together in various combinations to specify the four floral whorls: sepals, petals, stamens, and carpels. These genes interact as follows:

• A class genes: control the formation of sepals.
• A and B class genes: together control the formation of petals.
• B and C class genes: together control the formation of stamens (male organs).
• C class genes: control the formation of carpels (female organs).
• D class genes: are involved in ovule development.
• E class genes: are required for floral organ identity in all whorls.

In unisexual flowers, only male or female reproductive organs are present. The genes controlling the stamens (male organs) or carpels (female organs) must be altered or repressed to form such flowers.

Step-by-step process:

1. If the B and C class genes, responsible for stamen development, are suppressed or absent, the resulting flower will lack male reproductive structures, leading to a female (carpellate) flower.
2. If the C class genes, responsible for carpel formation, are absent, the flower will lack female reproductive organs, resulting in a male (staminate) flower.

Key Points

• The ABCDE model describes the development of floral organs based on specific gene combinations.
• To generate unisexual flowers, modifications must occur in the B or C class gene functions, resulting in either male or female reproductive structures only.
• The ABC model has been gradually expanded to include class D- and E-function genes, which are necessary for the ovules and the definition of the floral whorls, respectively.
• The class D genes (e.g., SEEDSTICK, and SHATTERPROOF) specify the identity of the ovule
• Class E genes (e.g., SEPALLATA), expressed in the entire floral meristem, & are necessary

Image of the ABCDE Model:

The correct answer is Reducing sugars.

Explanation:

1. Reducing Sugars: These sugars, such as glucose and fructose, have free aldehyde or ketone groups that can participate in oxidation-reduction reactions. While they can be transported in the phloem, they are less stable over long distances due to potential for oxidation and involvement in Maillard reactions. As a result, they are not the primary form of carbon transported in the phloem.

2. Mannitol: This is a sugar alcohol that can be transported in the phloem and is stable over long distances. It is used by some plants as an osmotic agent and can help with water retention, making it suitable for transport.

3. Galactosyl-sucrose Oligosaccharides: These are forms of non-reducing sugars and are commonly found in the phloem sap. They are stable and effective for long-distance transport of carbon because they do not have reactive groups that can lead to degradation.

4. Non-reducing Sugars: These sugars, like sucrose, are preferred for long-distance transport in phloem because they are stable and do not readily participate in reactions that can lead to loss of carbon or energy.

Conclusion:

While reducing sugars can be transported, they are not ideal for long-distance transport due to their instability. In contrast, non-reducing sugars and stable compounds like mannitol and oligosaccharides are more suited for this function.

Thus, the correct answer is Reducing sugars.

The correct answer is Root water potential falls below leaf water potential at night.

Concept:

• Water potential is a measure of the potential energy of water in a system and determines the direction of water movement. Water moves from areas of higher water potential (less negative) to areas of lower water potential (more negative).

• In a well-watered plant, water generally moves from the soil (higher water potential) through the roots (moderate water potential) to the leaves (lowest water potential) due to transpiration. During a drought (like when a plant is not watered for 7 days), water potential throughout the plant becomes more negative, but the relative gradient remains the same.

Explanation:

1. Pre-dawn leaf water potential declines over the 7 days:

   • As the plant experiences water stress, leaf water potential will become more negative over time because of reduced water availability, especially at pre-dawn when transpiration is minimal. This scenario is likely to happen.

2. Leaf water potential shows a diurnal cycle of highs and lows:

   • Diurnal changes in leaf water potential are typical, with lower water potential (more negative) during the day due to transpiration and higher (less negative) water potential at night when transpiration slows. This is a common observation, so this is also likely to happen.

3. Root water potential fluctuates between day and night:

   • Water potential in the roots can fluctuate between day and night because the plant experiences changes in water uptake and soil moisture availability throughout the day, especially under water stress. So, this is likely to happen

4. Root water potential falls below leaf water potential at night:

   • This is least likely to happen because water moves from roots to leaves due to the gradient created by transpiration. Even under drought stress, the root water potential typically remains higher (less negative) than leaf water potential to allow for water flow upward. At night, when transpiration decreases, the gradient between root and leaf water potential diminishes but does not reverse.

Thus, the least likely scenario is root water potential falling below leaf water potential at night.

The correct match is  A - III, B - I, C - II, D - IV.

Explanation:

Abscisic acid (ABA) is a key hormone involved in plant responses to environmental stresses, particularly water stress. The biosynthesis and degradation of ABA involve several important enzymes, each playing a unique role in the pathway.

Step-by-step matching of enzymes and their functions:

1. 9-cis-epoxycarotenoid dioxygenase: This enzyme is involved in the cleavage of carotenoids, leading to the production of xanthoxin, a precursor of ABA. Therefore, it matches with the function of "Xanthoxin production."
2. Cytochrome P450 monooxygenase (CYP707A3): This enzyme is part of the oxidative pathway of ABA catabolism, contributing to the degradation of ABA.
3. ABA glucosyltransferase: This enzyme catalyzes the conjugation of ABA to glucose, resulting in a sugar-conjugated form of ABA.
   Match: C - II
4. β-glucosidase: This enzyme hydrolyzes ABA-glucose conjugates, releasing active ABA from its sugar-bound form.
   Match: D - IV

Fig: Abscisic acid (ABA) biosynthesis pathway.

ABA Biosynthesis pathway

• De novo ABA biosynthesis begins in plastid, catalyzed by several enzymes ZEP  (zeaxanthin epoxidase); VDE (violaxanthin de-epoxidase); ABA4 (ABA-deficient 4), NCED, nine-cis-apoxicarotenoid-dioxygenase) that convert zeaxanthin into xanthoxin.
• After transport into the cytosol, xanthoxin is subjected to oxidation steps mediated by three enzymes ABA2 (ABA-deficient 2); AAO3 (abscisic aldehyde oxidase) and ABA3 (ABA-deficient 3) giving rise to ABA.
• Abscisic acid can be modified by CYP707A enzymes that yield phaseic acid (PA). The latter is the substrate of the PA reductase (PAR) that forms dihydrophaseic acid.
• Alternatively, ABA is glucosylated by ABA glucosyltransferase (GT) into ABA-GT, which can be transported into the vacuole and the endoplasmic reticulum (ER).
• Abscisic acid–GT can be converted back into ABA by the glycosyl hydrolase (BG1)

Key Points:

• 9-cis-epoxycarotenoid dioxygenase: Responsible for xanthoxin production, a key precursor in ABA biosynthesis.
• Cytochrome P450 monooxygenase (CYP707A3): Involved in ABA catabolism and its degradation pathway.
• ABA glucosyltransferase: Produces a sugar-conjugated form of ABA, aiding in its storage and regulation.
• β-glucosidase: Releases active ABA from its sugar-conjugated form, reactivating the hormone when needed.

The correct answer is Malate and Aspartate

Explanation:

In C4 photosynthesis, plants use a pathway that minimizes photorespiration by fixing CO₂ into a four-carbon compound before it enters the Calvin cycle. The four-carbon intermediates that play a significant role in this process include Malate and Aspartate.

Steps in C4 Photosynthesis:

• CO2 Capture in Mesophyll Cells: The initial step in C4 photosynthesis involves the capture of CO2 in the mesophyll cells, where it reacts with phosphoenolpyruvate (PEP) to form oxaloacetate (OAA) in a reaction catalyzed by the enzyme PEP carboxylase. This reaction is the primary carboxylation step in C4 photosynthesis and occurs in the mesophyll cells.
• Conversion of OAA: The OAA can then follow different pathways, depending on the type of C4 plant. In general, OAA is either converted into malate by the enzyme malate dehydrogenase or into aspartate. Malate and aspartate serve as transport molecules to carry the fixed CO2 to the bundle sheath cells.
• Decarboxylation in Bundle Sheath Cells: Once in the bundle sheath cells, malate or aspartate undergoes decarboxylation to release CO2. The released CO2 is then fixed again by the Calvin cycle, operating in the bundle sheath cells.

Fig: A simplified schematic representation of C4 photosynthetic metabolite flow highlighting interconnectivity between bundle sheath and mesophyll metabolism.

Other Options:

• Alanine is a three-carbon amino acid, not a four-carbon compound involved in C4 photosynthesis.
• Phosphoenolpyruvate (PEP) is a three-carbon molecule, not a four-carbon intermediate.
• Pyruvate is a three-carbon molecules, not four-carbon intermediates in C4 photosynthesis.

The correct answer is Conversion of PFR to PR

Explanation:

Phytochromes are light-sensitive proteins found in plants that regulate various physiological processes in response to light. Phytochromes exist in two interconvertible forms:

• P_R (Phytochrome Red): Absorbs red light (~660 nm) and is converted to the active form P_FR.
• P_FR (Phytochrome Far-Red): Absorbs far-red light (~730 nm) and can either revert back to P_R in far-red light or undergo dark reversion in the absence of light.

Dark Reversion:

• Dark Reversion refers to the spontaneous conversion of P_FR (the active form) back to P_R (the inactive form) in the absence of light. This process occurs in the dark and serves to reset the phytochrome system in plants.

Conversion of P_R to P_FR: This happens in response to red light, not in the dark.
Export of P_FR from cytosol to nucleus: P_FR can move to the nucleus for signaling, but this is not related to dark reversion.
Export of P_R from cytosol to nucleus: P_R does not play a role in nuclear signaling and this does not describe dark reversion.

The correct answer is Option 2 i.e.A,B and E

Concept:

• Photoperiodism refers to the ability of the plant to measure the length of the photoperiod. 
• It was reported by W.W.Garner and H.A.Allard in a variety of Maryland Mammoth 
• The period of photoperiodism is responsible for many physiological events such as initiation of flowering, seed germination, dormancy, asexual reproduction, etc and they are triggered by a photoperiod of a certain length. 
• This specific length of the photoperiod is referred to as the critical day length. It varies from species to species. 
• Plants can be classified based on their photoperiod responses based on flowering 

1. Long day plant - this plants flowers only when the day length is longer than the critical day length. Their flowering is accelerated by longer day length. 
2. Short day plant - This plant flowers only when the day length is less than the critical day length. Their flowering is accelerated by shorter day length. 
3. Day-neutral plant -This plant flowers in intermediate day length. It remains in a vegetative state when the days are either too long or either too short. 

Explanation: 

• Plants measure the duration of the dark period. 
• If the night or dark time is interrupted by light, then it will interrupt flowering.
• Light treatment A -  This treatment will promote flowering in SDP because they require a short day to flower but it will not cause LDP to flower cause they require more daylight to flower. 
• So, the responses shown for light treatment A are correct.
• Light treatment B -  This treatment will not promote flowering in SDP because they require a short day to flower but it will cause LDP to flower cause they require more daylight to flower. 
• So, the responses shown for light treatment B are correct. 
• Light treatment C - This treatment will not promote flowering in SDP because the dark period of the plant is interrupted by light which will further not promote flowering in short-day plants. The LDP will flower here because the light interruption will send a signal for the flower. 
• So, the responses shown for light treatment C are incorrect. 
• Light treatment D - This treatment will not promote flowing in SDP because a short-day plant requires a short day to flower whereas it will cause LDP to flower because the day length of the flower is longer than the critical day length. 
• So, the responses shown for light treatment D is incorrect.
• Light treatment E - This treatment will not promote flowering in SDP as the length of the dark period is very short while it will promote flowering in LDP because the dark length is less than the critical day length.
• So, the responses shown for light treatment E are correct. 

Hence, the correct answer is Option 2.