Plant Physiology

Step 1: Statement (a) is correct. The striated (banded) appearance of a sarcomere is due to differences in the thickness, size, and density of actin (thin) and myosin (thick) filaments.
Step 2: Statement (b) is correct. The A band corresponds to the region occupied by thick (myosin) filaments.
Step 3: Statement (c) is correct. The A band includes:

M line (center),
H zone (region with only thick filaments),
Zones of overlap between thick and thin filaments.

Step 4: Statement (d) is correct.

A band is anisotropic (dark band),
I band is isotropic (light band).

Step 5: Since all statements (a), (b), (c), and (d) are correct, there are no incorrect statements.

Step 1: At physiological pH ( ), amino acids may carry positive, negative, or neutral charges depending on their side chains (R groups).
Step 2: Aspartate and glutamate contain a carboxyl (-COO ) group in their R chains.
Step 3: At pH 7.0, these carboxyl groups are deprotonated, giving the R groups a negative charge.
Step 4: The other options are incorrect because:

Arginine and histidine have positively charged or basic side chains.
Cysteine and methionine are neutral.
Proline and valine are non-polar and neutral.

Step 1: In the light reaction of photosynthesis, Photosystem II (PS-II) absorbs light energy using chlorophyll \textit{a}.
Step 2: When chlorophyll molecules of PS-II become excited, they release high-energy electrons.
Step 3: The first stable electron acceptor in PS-II is pheophytin.
Step 4: From pheophytin, electrons are passed to plastoquinone and then through the cytochrome complex.

Step 1: The complete oxidation of glucose occurs in four main stages:

Glycolysis
Conversion of pyruvic acid to acetyl CoA
Krebs cycle
Electron Transport Chain (ETC)

Step 2: Glycolysis and Krebs cycle produce only a small number of ATP molecules by substrate-level phosphorylation.
Step 3: Most of the reduced coenzymes (NADH and FADH ) are formed during glycolysis, pyruvate oxidation, and Krebs cycle.
Step 4: These reduced coenzymes donate electrons to the Electron Transport Chain, where a large proton gradient is created across the inner mitochondrial membrane.
Step 5: ATP synthase uses this proton gradient to synthesize the maximum number of ATP molecules from ADP by oxidative phosphorylation.

Step 1: Immunosuppressive agents are substances that reduce or suppress the immune response.
Step 2: During organ transplantation, the recipient’s immune system may recognize the transplanted organ as foreign and try to reject it.
Step 3: Cyclosporine A is a bioactive compound produced by the fungus \textit{Trichoderma polysporum}.
Step 4: It selectively suppresses T-lymphocyte activity, thereby preventing organ rejection in transplant patients.
Step 5: The other options are incorrect:

Acetic acid and ethanol are not immunosuppressants.
Pectinase is an enzyme used in fruit juice clarification.

Step 1: Essential elements are those nutrients that are required by plants for normal growth and completion of their life cycle.
Step 2: Iron, zinc, and potassium are essential plant nutrients involved in various physiological processes.
Step 3: Iodine is not required for plant growth and is therefore not considered an essential element for plants. Hence, the correct answer is iodine.

Step 1: ATP stands for adenosine triphosphate.
Step 2: Structurally, ATP is composed of a nitrogenous base (adenine), a sugar (ribose), and three phosphate groups.
Step 3: This structure classifies ATP as a nucleotide.
Step 4: ATP functions as the energy currency of the cell but is not an enzyme, vitamin, or nucleic acid. Hence, ATP is a nucleotide.

Step 1: Guard cells surround the stomatal pore present on the leaf surface.
Step 2: By opening and closing the stomata, guard cells regulate the exchange of gases and the loss of water vapour.
Step 3: This regulated loss of water vapour from the plant body is known as transpiration. Hence, guard cells help in transpiration.

Step 1: Coconut milk is the liquid endosperm of coconut and is rich in growth-promoting substances.
Step 2: The active growth factor present in coconut milk is cytokinin.
Step 3: Cytokinins promote cell division (cytokinesis), delay senescence, and stimulate growth in plant tissues. Hence, coconut milk factor is cytokinin.

Step 1: Transpiration is the loss of water vapour from the aerial parts of plants, mainly through stomata.
Step 2: At low atmospheric pressure, the diffusion of water vapour from the leaf surface to the surrounding air becomes faster.
Step 3: Faster diffusion of water vapour increases the rate of transpiration. Hence, due to low atmospheric pressure, the rate of transpiration increases.

Step 1: Lipid molecules consist mainly of long hydrocarbon chains, which are non-polar in nature.
Step 2: Water is a polar solvent and dissolves substances that are polar or charged.
Step 3: Since lipids are non-polar and hydrophobic, they do not dissolve in water. Therefore, lipids are water insoluble because they are hydrophobic.

Step 1: Quantasomes are functional units of photosynthesis located on the thylakoid membranes of chloroplasts.
Step 2: Each quantasome consists of a group of chlorophyll molecules along with accessory pigments and proteins.
Step 3: A quantasome contains approximately 250 chlorophyll molecules, which work together to capture light energy efficiently. Therefore, the correct answer is 250 chlorophyll molecules.

Step 1: Trace elements are required by plants in very small quantities for normal growth and development.
Step 2: Cobalt is a micronutrient essential for nitrogen fixation in leguminous plants.
Step 3: The radioactive isotope cobalt-60 is widely used in cancer radiotherapy for treating malignant tumors. Hence, the correct answer is cobalt.

Plants that can prepare and store their own food using simple inorganic substances such as carbon dioxide, water, and minerals exhibit autotrophic nutrition. Step 1: Autotrophs synthesize their own food by the process of photosynthesis using sunlight, carbon dioxide, and water.
Step 2: During photosynthesis, plants convert inorganic raw materials into organic food (like glucose), which can then be stored for later use.
Step 3: Heterotrophs depend on other organisms for food and hence cannot prepare food from inorganic substances. Therefore, the given quality belongs to autotrophs.

Step 1: Monosaccharides are classified as aldoses or ketoses based on the functional group present.
Step 2: Ketose sugars contain a ketone group, while aldose sugars contain an aldehyde group.
Step 3: Fructose contains a ketone functional group and is therefore a ketose sugar.
Step 4: Galactose, mannose, and glucose are aldose sugars. Hence, fructose is the correct answer.

Step 1: Water potential ( ) is a measure of the free energy of water.
Step 2: By convention, the water potential of pure water at atmospheric pressure and standard temperature is taken as zero.
Step 3: Solutions have lower (more negative) water potential compared to pure water. Hence, the water potential of pure water at atmospheric pressure is zero bar.

Step 1: Fat soluble vitamins (A, D, E, and K) are soluble in organic solvents such as alcohol and fats.
Step 2: These vitamins are generally derived from isoprene (propene-related) units, making them lipid-like in nature.
Step 3: Fat soluble vitamins are stored in the liver and adipose tissues, unlike water soluble vitamins which are excreted easily. Hence, all the given statements are correct.

Step 1: Reducing sugars are those sugars which have a free aldehyde or ketone group capable of acting as a reducing agent.
Step 2: Maltose, lactose, and glucose possess a free reducing group and hence are reducing sugars.
Step 3: In sucrose, the glycosidic bond involves both the reducing ends of glucose and fructose, leaving no free reducing group. Therefore, sucrose is a non-reducing sugar.

Step 1: Soil water exists in different forms such as gravitational, capillary, and hygroscopic water.
Step 2: Capillary water is held in the small pores of soil particles and is easily available for absorption by plant roots.
Step 3: Gravitational water drains away quickly and is not readily available, while hygroscopic water is tightly bound to soil particles and cannot be absorbed. Hence, the water readily available to plants is capillary water.

Step 1: Epithelial cells show polarity with distinct apical, lateral, and basal surfaces.
Step 2: The basal lamina provides positional cues and signals that induce and maintain this cellular polarity.
Step 3: Proteins like vinculin and occludin are junctional proteins involved in cell adhesion and tight junctions, but they do not induce polarity.
Step 4: The extracellular matrix provides support, but the basal lamina specifically induces epithelial polarity. Hence, the induction source of cellular polarity in epithelial cells is the basal lamina.

Step 1: Understand what a complex fertilizer is. A complex fertilizer is one that:
contains two or more primary plant nutrients (N, P, K),
provides these nutrients in chemical combination within a single compound.
Step 2: Analyse each option.
Potassium sulphate — supplies only potassium (K); single nutrient fertilizer.
Calcium ammonium nitrate — supplies mainly nitrogen; considered a mixed fertilizer, not complex.
Triple super phosphate — supplies phosphorus only; single nutrient fertilizer.
Urea ammonium phosphate — supplies both nitrogen (N) and phosphorus (P) in a single compound; hence a complex fertilizer.
Step 3: Select the correct option. Only urea ammonium phosphate fulfills the criteria of a complex fertilizer.
Hence, the correct answer is

Step 1: Identify what hop flowers are. Hop flowers are obtained from the plant \textit{Humulus lupulus}. They contain resins and essential oils that impart bitterness, aroma, and preservative properties.
Step 2: Recall their industrial use. Hop flowers are widely used in the brewing industry:
They give beer its characteristic bitter taste.
They improve flavor and aroma.
They act as natural preservatives.
Step 3: Eliminate incorrect options.
Gluconic acid production — involves oxidation of glucose by bacteria.
Vinegar production — involves acetic acid bacteria.
Alcohol production — primarily carried out by yeast; hops are not used to produce alcohol.
Hence, the correct answer is

Step 1: Define parthenocarpy. Parthenocarpy is the development of fruit without fertilization.
Step 2: Understand the consequence of no fertilization. Since fertilization does not occur:
ovules do not develop into seeds,
the ovary still develops into a fruit. Thus, the fruit formed is seedless.
Step 3: Eliminate incorrect options.
(A) Seed fruit — incorrect, no seeds are formed.
(C) No fruit — incorrect, fruit does develop.
(D) Seed formation — incorrect, fertilization is required for seed formation.
Hence, the correct answer is

Step 1: Recall the structure and function of grana. Grana are stacks of thylakoid membranes present in the chloroplast. They are the site of the light-dependent reactions of photosynthesis.
Step 2: Identify the process occurring in grana. During the light reaction:
Light energy is absorbed by chlorophyll,
Energy is used to synthesize ATP from ADP and inorganic phosphate ( ). This process is called photophosphorylation.
Step 3: Eliminate incorrect options.
Phosphorylation: General term, not specific to light reactions.
Oxidative phosphorylation: Occurs in mitochondria during respiration.
Photolysis: Splitting of water into , , and electrons.
Hence, the correct answer is

Step 1: Understand the phenomenon described. Garden pea tendrils coil around a support in response to touch. The movement involves directional growth toward the stimulus.
Step 2: Recall definitions of the given terms.
Thermotaxis: Movement of organisms in response to temperature.
Thigmotaxis: Directional movement of \emph{entire organisms} in response to touch.
Thigmotropism: Directional growth movement of a plant part in response to touch.
Thigmonasty: Non-directional movement in response to touch (e.g., closing of Mimosa leaves).
Step 3: Identify the correct category. Since tendril coiling:
is a growth response,
is directional,
occurs in plants due to touch, it is classified as thigmotropism.
Hence, the correct answer is

Step 1: Recall common dietary sources of Vitamin C. Vitamin C (ascorbic acid) is abundantly found in citrus fruits.
Step 2: Analyse each option.
Rice: Mainly a source of carbohydrates, negligible Vitamin C.
Milk: Contains proteins and calcium, very little Vitamin C.
Egg: Rich in proteins and fats, not a significant source of Vitamin C.
Lemon: A citrus fruit, rich in Vitamin C.
Step 3: Select the correct option. Among the given options, lemon is the richest source of Vitamin C.
Hence, the correct answer is

Step 1: Understand the given condition. The system consists of:
Sugar solution on one side,
Pure water on the other side,
A semipermeable membrane separating them.
Step 2: Recall the definition of osmosis. Osmosis is the movement of solvent (water) molecules through a semipermeable membrane from a region of higher water potential (dilute solution or pure water) to a region of lower water potential (concentrated solution).
Step 3: Eliminate other options.
Diffusion: Occurs without a semipermeable membrane.
Imbibition: Absorption of water by colloids or solids.
Translocation: Transport of food materials in plants.
Step 4: Identify the correct phenomenon. Since the key requirement—semipermeable membrane—is present, the phenomenon is osmosis.
Hence, the correct answer is