HSC Biology Board Question Paper March 2023 - Maharashtra Board

Board Question Paper: March 2023

Subject: Biology | Max. Marks: 70 | Time: 3 Hrs.

Maharashtra State Board of Secondary and Higher Secondary Education

SECTION - A

Q.1. Multiple Choice Questions

Instructions: Select and write the correct answer along with its alphabet.

i. Histones are rich in _______.
  • (A) Lysine and Arginine
  • (B) Leucine and Methionine
  • (C) Serine and Leucine
  • (D) Phenyl alanine and Lysine
Answer: (A) Lysine and Arginine
ii. How many mitotic divisions take place during the formation of a female gametophyte from a functional megaspore?
  • (A) One
  • (B) Two
  • (C) Three
  • (D) Four
Answer: (C) Three

Explanation: The functional megaspore undergoes 3 successive free nuclear mitotic divisions to form an 8-nucleate embryo sac.

iii. Which of the following is the only gaseous plant growth regulator?
  • (A) ABA
  • (B) Cytokinin
  • (C) Ethylene
  • (D) Gibberellin
Answer: (C) Ethylene
iv. The pH of nutrient medium for plant tissue culture is in the range of _______.
  • (A) 2 to 4.2
  • (B) 5 to 5.8
  • (C) 7 to 7.5
  • (D) 8 to 9.5
Answer: (B) 5 to 5.8
v. Rivet Popper Hypothesis is an analogy to explain the significance of _______.
  • (A) Biodiversity
  • (B) natality
  • (C) sex-ratio
  • (D) age distribution ratio
Answer: (A) Biodiversity
vi. Which of the following group shows ZW-ZZ type of sex determination?
  • (A) Pigeon, Parrot, Sparrow
  • (B) Parrot, Bat, Fowl
  • (C) Bat, Fowl, Crow
  • (D) Sparrow, Fowl, Cat
Answer: (A) Pigeon, Parrot, Sparrow

Explanation: ZW-ZZ type is found in birds.

vii. In Hamburger’s phenomenon, _______.
  • (A) Cl– diffuse into WBCs
  • (B) Cl– diffuse into RBCs
  • (C) Na+ diffuse into RBCs
  • (D) Na+ diffuse into WBCs
Answer: (B) Cl– diffuse into RBCs
viii. Calcium and Phosphate ions are balanced between blood and other tissues by ______.
  • (A) Thymosin and Parathormone
  • (B) Calcitonin and Somatostatin
  • (C) Collip’s hormone and Calcitonin
  • (D) Calcitonin and Thymosin
Answer: (C) Collip’s hormone and Calcitonin

Note: Collip's hormone is another name for Parathormone (PTH).

ix. Identify the INCORRECT statement.
  • (A) In a flaccid cell, T.P. is zero
  • (B) In a turgid cell, DPD is zero
  • (C) In a fully turgid cell, TP = OP
  • (D) Water potential of pure water is negative
Answer: (D) Water potential of pure water is negative

Explanation: The water potential of pure water is zero (at standard temperature and pressure).

x. Which of the following is a hormone releasing contraceptive?
  • (A) Cu-T
  • (B) Cu-7
  • (C) Multiload-375
  • (D) LNG-20
Answer: (D) LNG-20

HSC Biology

Q.2. Answer the following questions (VSA)

i. Which disease is caused by HPV?
Answer: Genital warts (or Cervical cancer).
ii. Which device is used to clean both dust and gases from polluted air?
Answer: Scrubber (Specifically, a wet scrubber can remove water-soluble gases like SO₂ and also particulate matter/dust).
iii. Mention the name of sterile animal produced by intergeneric hybridisation.
Answer: Mule (Cross between male Donkey and female Horse).
iv. Give the name of first transgenic plant.
Answer: Tobacco.
v. A child has low BMR, delayed puberty and mental retardation. Identify the disease.
Answer: Cretinism (Congenital Hypothyroidism).
vi. Identify ‘A’ in the given graph of population growth:
[Graph showing Sigmoid Curve: Lag phase -> 'A' -> Stationary]
Answer: Exponential phase (or Log phase).
vii. Complete the following box with reference to symptoms of mineral deficiency:
Abscission Pre-mature fall of flowers, fruits and leaves
? Appearance of green and non-green patches on leaves
Answer: Mottling (or Mosaic).
viii. Give an example of plant having both kidney and dumb-bell shaped guard cells in stomata.
Answer: This question likely asks for examples of plants for both types.
Kidney shaped: Sunflower (Dicot).
Dumb-bell shaped: Maize (Monocot).
SECTION - B

Attempt any EIGHT of the following questions

Q.3. Define the terms:
a. Gross Primary Productivity
b. Net Primary Productivity
Answer: a. Gross Primary Productivity (GPP): It is the rate of production of organic matter (biomass) during photosynthesis. It represents the total solar energy trapped by the plant.

b. Net Primary Productivity (NPP): It is the amount of biomass or organic matter available for consumption by heterotrophs (herbivores and decomposers). It is calculated as GPP minus respiration losses (R). (NPP = GPP - R).
Q.4. Draw a neat diagram of thyroid gland and label thyroid follicle, follicular cells and blood capillaries.
Answer:
[Diagram of Histology of Thyroid Gland]
Labels required:
1. Thyroid Follicle (The circular structures)
2. Follicular Cells (Cuboidal epithelium lining the follicle)
3. Blood Capillaries (In the connective tissue/stroma between follicles)
(Also: Colloid inside the follicle)
Q.5. i. Give reason – ABA is also known as antitranspirant.
ii. Explain the role of chlorophyllase enzyme in banana.
Answer: i. ABA (Abscisic Acid) is called an antitranspirant because, during water stress or drought conditions, it induces the closure of stomata to reduce the rate of transpiration and prevent water loss.

ii. In bananas, the enzyme chlorophyllase degrades the green pigment chlorophyll. This degradation results in the unmasking of other pigments (like carotenoids/xanthophylls), causing the peel to turn yellow during ripening.
Q.6. Complete the chart showing human proteins produced by rDNA technology to treat human diseases and re-write.
Disorders/diseases Recombinant Proteins
? Erythropoietin
Asthma ?
? Tissue plasminogen activator
Emphysema ?
Answer: 1. Disorder for Erythropoietin: Anemia.
2. Protein for Asthma: DNase (Note: DNase is primarily used for Cystic Fibrosis to treat mucus accumulation, but is often the expected answer in this context for respiratory disorders).
3. Disorder for Tissue plasminogen activator: Heart Attack / Myocardial Infarction.
4. Protein for Emphysema: Alpha-1 antitrypsin.
Q.7. i. Define – Imbibition.
ii. Explain how imbibition helps root hairs in adsorption of water.
Answer: i. Imbibition: It is the adsorption of water by hydrophilic colloids (solid particles) resulting in an increase in volume without forming a solution.

ii. Role in Root Hairs: The cell wall of root hairs is made up of pectic compounds and cellulose, which are hydrophilic colloids. These substances imbibe water from the surrounding soil, facilitating the initial adsorption and movement of water into the root hair cell.
Q.8. Draw a neat diagram of the conducting system of human heart and label AV node, Bundle of His and Purkinje fibres.
Answer:
[Diagram of Human Heart Conducting System]
Labels required:
1. AV Node (Atrio-Ventricular Node) - at the base of right atrium.
2. Bundle of His (AV Bundle) - arising from AV node towards septum.
3. Purkinje Fibres - spreading into the ventricular walls.
Q.9. Distinguish between heterochromatin and euchromatin with reference to staining property and activity.
Answer:
Feature Heterochromatin Euchromatin
Staining Property It stains strongly and appears dark. It stains lightly and appears light.
Genomic Activity It is genetically inactive (transcriptionally inert). It is genetically active (transcriptionally active).
Q.10. Complete the following chart regarding energy flow in an Ecosystem and re-write:
Answer:
Role Example
Primary Consumer Herbivores
Primary Producer Green Plants / Algae
Tertiary Consumer (or Top Carnivore) Man, Lion
Secondary consumer Small Carnivores (e.g., Wolf, Frog)
Q.11. i. What is biofortification?
ii. Mention one example each of fortification with reference to –
a. Amino acid content
b. Vitamin-C content
Answer: i. Biofortification: It is the method of breeding crops with higher levels of vitamins, minerals, or higher protein and healthier fats to improve public health.

ii. Examples:
a. Amino acid content: Maize hybrids (Shakti and Rattan) rich in Lysine and Tryptophan.
b. Vitamin-C content: Bitter gourd, Tomato, Mustard, Bathua (enriched varieties).
Q.12. Differentiate between X-chromosome and Y-chromosome with reference to –
i. length of non-homologous regions
ii. type as per position of centromere.
Answer: i. Length of non-homologous regions: The X-chromosome has a longer (larger) non-homologous region containing more genes, while the Y-chromosome has a shorter (smaller) non-homologous region.

ii. Type as per position of centromere: The X-chromosome is Sub-metacentric, while the Y-chromosome is Acrocentric.
Q.13. Define the terms:
i. Genetic drift
ii. Homologous organs
Answer: i. Genetic drift: It refers to the random change in allele frequency occurring by chance in a small population.

ii. Homologous organs: Organs that have the same structural origin and basic plan but may perform different functions (e.g., forelimbs of human, bat, and whale). They indicate divergent evolution.
Q.14. i. What is ex-situ conservation?
ii. Mention any two places where the ex-situ conservation is undertaken.
Answer: i. Ex-situ conservation: It is the conservation of threatened animals and plants outside their natural habitat in special settings where they can be protected and cared for.

ii. Places: Zoological parks, Botanical gardens, Seed banks, or Cryopreservation centers.
SECTION - C

Attempt any EIGHT of the following questions

Q.15. i. Define – Incomplete dominance.
ii. If a red flowered Mirabilis jalapa plant is crossed with a white flowered plant, what will be the phenotypic ratio in F2 generation? Show it by a chart.
Answer: i. Incomplete dominance: It is a phenomenon where neither of the two alleles of a gene is dominant over the other, resulting in a heterozygous phenotype that is intermediate between the two homozygous phenotypes.

ii. Chart for F2 Generation:
P generation: Red (RR) × White (rr)
F1 generation: All Pink (Rr)
Selfing F1: Pink (Rr) × Pink (Rr)
F2 Ratio:
- 1 Red (RR)
- 2 Pink (Rr)
- 1 White (rr)
Phenotypic Ratio: 1 : 2 : 1 (Red : Pink : White).
Q.16. i. Differentiate between sympathetic and parasympathetic nervous system with reference to the following:
a. Pre and post ganglionic nerve fibres.
b. Effect on heart beat.
ii. Give reason – All spinal nerves are of mixed type.
Answer: i. Differences:
a. Nerve Fibres: In Sympathetic, pre-ganglionic fibres are short and post-ganglionic are long. In Parasympathetic, pre-ganglionic fibres are long and post-ganglionic are short.
b. Heart beat: Sympathetic system increases the heart beat. Parasympathetic system decreases the heart beat.

ii. Reason: All spinal nerves are formed by the union of a dorsal sensory root (carrying sensory impulses) and a ventral motor root (carrying motor impulses). Thus, every spinal nerve contains both sensory and motor fibres, making them mixed nerves.
Q.17. i. Draw a suitable diagram of replication of eukaryotic DNA and label any three parts.
ii. How many amino acids will be there in the polypeptide chain formed on the following mRNA?
5' GCCACAUGGAGAUGACGACAAAAUUUUACUAGAAAA 3'
Answer: i. Diagram:
[Diagram of Replication Fork]
Labels: Leading Strand, Lagging Strand, Okazaki Fragments, RNA Primer, Template Strands.

ii. Amino Acid Count:
The given mRNA sequence is:
5' GCC ACA UGG AGA UGA CGA CAA AAU UUU ACU AGA AAA 3'

Logic: Translation typically starts at the start codon (AUG) and ends at a stop codon. However, in this sequence, the 5th triplet is UGA, which is a Stop codon.
- If we read triplets from the 5' end: GCC, ACA, UGG, AGA, UGA (Stop). This produces 4 amino acids.
- Note: Biologically, translation requires an AUG start codon. If we scan for the first AUG, it appears at the 6th nucleotide index, but board problems of this type often test the recognition of the stop codon in the reading frame provided by the spacing.
Final Answer: 4 amino acids (Translation stops at the UGA stop codon).
Q.18. Describe the steps in breathing.
Answer: Breathing (Ventilation) involves two main steps:
1. Inspiration (Inhalation): An active process where atmospheric air enters the lungs.
  • The diaphragm muscles contract, making the diaphragm flat and pushing it downwards.
  • The external intercostal muscles contract, pulling the ribs and sternum upwards and outwards.
  • This increases the volume of the thoracic cavity, decreasing the intra-pulmonary pressure. Air rushes in from high pressure (atmosphere) to low pressure (lungs).
2. Expiration (Exhalation): A passive process where air leaves the lungs.
  • The diaphragm relaxes and becomes dome-shaped.
  • The external intercostal muscles relax, returning ribs and sternum to their original position.
  • This decreases the thoracic volume, increasing the intra-pulmonary pressure. Air is expelled out.
Q.19. i. What is spermatogenesis?
ii. Draw a neat and labelled diagram of spermatogenesis.
Answer: i. Spermatogenesis: It is the process of formation of haploid male gametes (sperms/spermatozoa) from the diploid germinal epithelium (spermatogonia) of the testis.

ii. Diagram:
[Diagram of Spermatogenesis]
Structure to show:
- Spermatogonia (2n) -> Mitosis
- Primary Spermatocyte (2n) -> Meiosis I
- Secondary Spermatocytes (n)
- Spermatids (n) -> Spermiogenesis
- Spermatozoa (Sperms) (n)
Q.20. i. What is a connecting link?
ii. Which fossil animal is considered as the connecting link between reptiles and birds? Give any one character of each class found in it.
Answer: i. Connecting link: An organism (living or fossil) that possesses characters of two different groups of organisms, indicating their evolutionary relationship, is called a connecting link.

ii. Fossil Animal: Archaeopteryx.
- Reptilian Character: Presence of teeth in jaws, long tail with vertebrae, or claws on wings.
- Avian (Bird) Character: Presence of feathers covering the body, modification of forelimbs into wings, or presence of a beak.
Q.21. Complete the following chart regarding population interaction and re-write:
Sr. No. Name of interaction Interaction between
1 ? Plasmodium and Man
2 ? Leopard and Lion
3 ? Clown fish and Sea-anemone
Answer: 1. Parasitism
2. Competition
3. Commensalism
Q.22. i. What is composition of bio-gas?
ii. Mention any four benefits of bio-gas.
Answer: i. Composition: Biogas mainly consists of Methane (CH₄) [50-70%], Carbon dioxide (CO₂) [30-40%], and traces of Hydrogen (H₂), Nitrogen (N₂), and Hydrogen Sulphide (H₂S).

ii. Benefits: 1. It provides a cheap and safe fuel for cooking and lighting. 2. It is an eco-friendly and pollution-free energy source. 3. The leftover sludge is a rich fertilizer (manure). 4. It helps in effective waste management and sanitation.
Q.23. i. Give reason – Water acts as thermal buffer.
ii. Draw a neat and proportionate diagram of root hair and label mitochondria, nucleus and vacuole.
Answer: i. Reason: Water has a high specific heat capacity, meaning it requires a large amount of energy to change its temperature. This property allows it to absorb or release heat with minimal temperature change, thus acting as a thermal buffer to maintain constant body temperature in organisms.

ii. Diagram:
[Diagram of Root Hair Cell]
Labels required:
- Nucleus (Peripheral, pushed by vacuole)
- Vacuole (Large central)
- Mitochondria (In the cytoplasm)
Q.24. Explain three main functions of free antibodies produced by B-lymphocytes.
Answer: The three main functions of antibodies are: 1. Agglutination: Antibodies bind to antigens (like bacteria) forming clumps, immobilizing them for easier destruction by phagocytes. 2. Opsonization: Antibodies coat the surface of pathogens (bacteria), making them more appetizing and easier for phagocytes (macrophages) to engulf and destroy. 3. Neutralization: Antibodies bind to toxins or viruses, neutralizing their harmful effects and preventing them from entering host cells.
Q.25. i. Following are the diagrams of entry of pollen tube into ovule. Identify the type A and B.
ii. Give any four points of significance of double fertilization.
[Image A: Pollen tube enters through integuments/funicle side]
[Image B: Pollen tube enters through micropyle at the top]
Answer: i. Identification:
A: Mesogamy (Entry through integuments).
B: Porogamy (Entry through micropyle).

ii. Significance of Double Fertilization: 1. It involves the fusion of one male gamete with the egg to form a diploid Zygote, which develops into an embryo. 2. It involves the fusion of the second male gamete with polar nuclei to form the triploid Primary Endosperm Nucleus (PEN), which develops into nutritive endosperm. 3. It restores the diploid condition of the lifecycle. 4. It ensures that the nutritive tissue (endosperm) is formed only if fertilization is successful, preventing wastage of energy.
Q.26. i. Name the hormone which is responsible for apical dominance.
ii. A farmer wants to remove broad-leaved weeds from the jowar plantation in his field. Suggest any plant hormone to remove such weeds.
iii. Mention any two applications of cytokinin.
Answer: i. Hormone for apical dominance: Auxin (IAA).

ii. Hormone for weed removal: Synthetic Auxins like 2,4-D (2,4-Dichlorophenoxyacetic acid). It acts as a selective herbicide for broad-leaved weeds.

iii. Applications of Cytokinin: 1. Promotion of cell division (Cytokinesis). 2. Delaying senescence (aging) of leaves (Richmond-Lang effect). (Other options: Breaking seed dormancy, inducing shoot formation in tissue culture).
SECTION - D

Attempt any THREE of the following questions

Q.27. i. What is blood pressure?
ii. Give the name of the instrument which is used to measure the blood pressure.
iii. Differentiate between an artery and a vein with reference to lumen and thickness of wall.
Answer: i. Blood Pressure: It is the lateral pressure exerted by the flowing blood on the walls of the blood vessels (mainly arteries).

ii. Instrument: Sphygmomanometer.

iii. Difference between Artery and Vein:
Feature Artery Vein
Thickness of wall Thick, muscular, and elastic wall. Thin and less muscular wall.
Lumen Narrow lumen (cavity). Wide lumen (cavity).
Q.28. i. Describe any three adaptations in anemophilous flowers. Mention any one example.
ii. Describe any three adaptations in hydrophilous flowers. Mention any one example.
Answer: i. Anemophilous (Wind-pollinated) Flowers:
Adaptations: 1. Flowers are small, inconspicuous, colorless, without nectar and fragrance. 2. Pollen grains are light in weight, dry, and produced in large numbers to ensure pollination. 3. Stigma is feathery to trap pollen, and stamens have versatile anthers to release pollen easily.
Example: Maize, Wheat, Grasses.

ii. Hydrophilous (Water-pollinated) Flowers:
Adaptations: 1. Flowers are small, inconspicuous, and without nectar or fragrance. 2. Floral parts and pollen grains are often coated with mucilage to prevent wetting. 3. Pollen grains are long and needle-like (in Hypohydrophily) or float on the surface (in Epihydrophily).
Example: Vallisneria (Epihydrophily) or Zostera (Hypohydrophily).
Q.29. i. What is polymerase chain reaction (PCR)?
ii. Describe three steps involved in mechanism of PCR.
Answer: i. PCR: Polymerase Chain Reaction is an in vitro technique used to amplify (make multiple copies of) a specific segment of DNA in a very short time.

ii. Steps in PCR: 1. Denaturation (90-98°C): The double-stranded DNA is heated to high temperature to break the hydrogen bonds, separating the two strands. 2. Annealing (40-60°C): The temperature is lowered to allow two specific oligonucleotide primers to bind (anneal) to the complementary sequences on the single-stranded DNA templates. 3. Extension / Polymerization (70-75°C): The thermostable enzyme Taq polymerase adds nucleotides to the primers, synthesizing new DNA strands complementary to the templates.
Q.30. i. Give any four significances of fertilization in human.
ii. Mention the names of any two organs each derived from ectoderm and mesoderm.
Answer: i. Significance of Fertilization: 1. It restores the diploid number of chromosomes (2n) in the zygote. 2. It combines characters from both parents, leading to genetic variation. 3. It determines the sex of the offspring (XX or XY). 4. It stimulates the secondary oocyte to complete meiosis II and initiates cleavage (development).

ii. Derivatives: - From Ectoderm: Epidermis of skin, Nervous system (Brain/Spinal cord), Retina of eye, Enamel of teeth. - From Mesoderm: Dermis of skin, Muscles, Bones/Cartilage, Heart/Blood vascular system, Kidneys.
Q.31. i. Give any two functions of cerebellum.
ii. Write the names of any four motor cranial nerves with their appropriate serial number.
iii. Which hormones stimulate liver for glycogenesis and glucogenolysis?
Answer: i. Functions of Cerebellum: 1. Maintains equilibrium, posture, and balance of the body. 2. Coordinates voluntary muscle movements (walking, running) and regulates muscle tone.

ii. Motor Cranial Nerves: - III : Oculomotor - IV : Trochlear - VI : Abducens - XI : Spinal Accessory - XII : Hypoglossal
(Select any four).

iii. Hormones: - For Glycogenesis (Glucose to Glycogen): Insulin. - For Glucogenolysis (Glycogen to Glucose): Glucagon.
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