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

The correct option is: 2

Explanation: In the context of phloem transport, pressure flow refers to the process by which sugars (primarily sucrose) and other nutrients are transported from the source (usually leaves or photosynthetic tissues) to the sink (areas where the sugars are used or stored, such as roots, fruits, or growing tissues). This process is driven by a difference in pressure between the source and sink regions.

1. In the source (where photosynthesis occurs), sugars are actively loaded into the sieve tubes of the phloem, reducing the water potential in the phloem, causing water to flow into the phloem via osmosis.
2. This creates a positive pressure in the sieve tubes at the source.
3. In the sink, sugars are unloaded from the phloem, and water is lost, increasing the water potential and creating a lower pressure.
4. The pressure difference between the source and sink regions causes the flow of the sugar-rich solution (phloem sap) from the high-pressure source to the low-pressure sink.

Other options:

• Movement of water into the sieve tube: While water does move into the sieve tube from surrounding tissues (due to osmotic pressure), this is a part of the process, but the "pressure flow" specifically refers to the movement of sugars, not just water.
• Diffusion of ions across membranes: This is a separate process, generally related to ion transport, and is not the driving force behind pressure flow in the phloem.
• Evaporation of water from the leaf: This process is related to transpiration, which affects water movement in the xylem but is not directly involved in the "pressure flow" in the phloem.

The correct option is: 

Explanation: 

• Root Hair (A):

  • Root hair cells actively absorb water from the soil through osmosis. As they take in water, they have a moderate solute concentration, but their main feature is the active water absorption. This results in a low DPD, as the cell's ability to take in water actively reduces the deficit between water inside and outside the cell.

• Mesophyll Cell (B):

  • Mesophyll cells are involved in photosynthesis and have a higher solute concentration (due to sugars and organic acids), leading to a high DPD. The high solute concentration causes a strong tendency for water to move into the cell, but the pressure potential is moderate, which results in a more balanced scenario for water absorption.

• Cortical Cell (C):

  • Cortical cells are located in the root cortex and are responsible for storing water and nutrients. They have a moderate DPD, as they contain solutes but not at the high concentration levels found in mesophyll cells. The pressure potential is relatively balanced between solute concentration and water absorption.

The correct option is: 2

Explanation:

Diffusion Pressure Deficit (DPD) is the difference between the diffusion pressure of pure water and the diffusion pressure of water inside the plant cell. It reflects the deficit in water pressure due to the presence of solutes inside the plant cell.

When a plant cell becomes turgid, the following occurs:

• The cell takes in water through osmosis due to the difference in water potential between the inside of the cell and its surroundings.
• As water enters, turgor pressure (the pressure exerted by the cell membrane against the cell wall) increases.
• This results in the water potential inside the cell becoming less negative (approaching zero) because the pressure from the cell wall balances the osmotic pressure.
• As a result, the DPD decreases. The osmotic potential is counteracted by the turgor pressure, which reduces the deficit in diffusion pressure.

Key Points

• DPD and Osmotic Potential: DPD is closely related to osmotic potential. As osmotic potential becomes more negative due to solutes in the cell, the DPD increases, but in a turgid cell, it decreases as the pressure from the cell wall balances the osmotic potential.

• Turgor Pressure and Cell Wall: The increase in turgor pressure during turgidity occurs because the plant cell membrane pushes against the rigid cell wall, preventing the cell from bursting despite the continuous influx of water.

• DPD and Water Potential: DPD is the difference between the water potential of the cell and the water potential of pure water. In a turgid cell, the water potential approaches zero due to the balance between turgor pressure and osmotic potential.

• Effect of High Solute Concentration on DPD: In a cell with high solute concentration (e.g., during plasmolysis), DPD increases because the water potential becomes more negative due to the osmotic pressure exerted by the solutes.

• Water Flow in Plant Cells: Water always moves from areas of high water potential to areas of low water potential. In a turgid cell, the movement of water slows as the turgor pressure builds up, balancing the osmotic pressure.

• Role of DPD in Water Movement: DPD helps in understanding the driving force for water movement in plants, influencing processes like osmosis, transpiration, and nutrient uptake.

The correct Option is: 2

Explanation: Water potential (Ψ_w) in a plant cell is influenced by both osmotic potential (Ψ_s) and turgor pressure (Ψ_p). The relationship is given by the equation:

•  Osmotic potential (Ψ_s): When solute concentration increases, the osmotic potential becomes more negative, because the presence of solutes reduces the potential energy of the water.
• Turgor pressure (Ψ_p): This does not change in this scenario (since pressure remains constant).

Thus, if solute concentration increases and pressure remains unchanged, osmotic potential becomes more negative, which leads to the overall water potential (Ψ_w) becoming more negative.

• The water potential becomes more positive: This is incorrect because an increase in solute concentration makes the water potential more negative.
• The osmotic potential becomes more positive: This is incorrect, as an increase in solute concentration makes osmotic potential more negative, not positive.
• The turgor pressure decreases in the cell: This is incorrect, as the turgor pressure does not change in this scenario.

The correct option is: 1

Explanation: Imbibition is the process by which dry seeds absorb water, leading to their swelling. The primary factor contributing to the swelling of dry seed tissues is the increase in turgor pressure as water enters the seed cells. This pressure helps in the expansion of cells, making the seed swell.

• Turgor pressure is the pressure exerted by the cell membrane against the cell wall, and when water enters the seed, it increases the internal pressure, causing the seed to swell.
• The osmotic potential driving water into the seed: Osmotic potential plays a role in water absorption, but turgor pressure is the direct cause of the swelling.
• The breakdown of starch to release stored sugars for water absorption: This process occurs later in seed germination, but it is not the main factor in the initial swelling during imbibition.
• The structural modification of cell membranes to allow water influx: While the permeability of the cell membrane is essential for water entry, it is the turgor pressure that directly leads to the swelling.

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.