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HSC Maharashtra Board Class 12 Biology Question Paper Solutions July 2023

Maharashtra Board Class 12 Biology Question Paper Solutions July 2023
Question Paper Page No. 1 Question Paper Page No. 2 Question Paper Page No. 3 Question Paper Page No. 4 Question Paper Page No. 5 Question Paper Page No. 6 Question Paper Page No. 7 Question Paper Page No. 8 Maharashtra Board Resources

Board Question Paper: July 2023

BIOLOGY

Time: 3 Hrs. Max. Marks: 70

General Instructions:
  1. The question paper is divided into four sections.
  2. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying one mark each.
    • For each multiple choice type of question, it is mandatory to write the correct answer along with its alphabet, e.g., (A) ……. / (B) ……. / (C) ……. / (D) ……. etc. No mark/s shall be given if ONLY the correct answer or alphabet of the correct answer is written.
    • In case of MCQ, evaluation will be done for the first attempt only.
    • Q. No. 2 Contains Eight very short answer type of questions carrying one mark each.
  3. Section B: Q. No. 3 to 14 are short answer type of questions carrying two marks each. (Attempt any Eight)
  4. Section C: Q. No. 15 to 26 are short answer type of questions carrying three marks each. (Attempt any Eight)
  5. Section D: Q. No. 27 to 31 are long answer type of questions carrying four marks each. (Attempt any Three)
  6. Begin the answer of each section on a new page.
SECTION – A

Q.1. Select the correct alternatives and write the answers: [10]

i. If members of two populations have difference in the structure of reproductive organs, then this type of isolation is called _______.

  • (A) ethological
  • (B) seasonal
  • (C) mechanical
  • (D) habitat
Answer: (C) mechanical

ii. The primary precursor of Indole-3-Acetic acid is _______.

  • (A) Tryptophan
  • (B) Phenyl alanine
  • (C) Mevalonic acid
  • (D) Methionine
Answer: (A) Tryptophan

iii. If only one DNA molecule is subjected to PCR and the time required for each cycle is three minutes, then after five cycles, how many DNA molecules are obtained?

  • (A) 10
  • (B) 15
  • (C) 32
  • (D) 64
Answer: (C) 32
Explanation: Number of DNA molecules = \(2^n\), where \(n\) is the number of cycles. \(2^5 = 32\).

iv. The specific gravity of CSF is _______.

  • (A) 1.005
  • (B) 1.02
  • (C) 1.502
  • (D) 1.81
Answer: (A) 1.005

v. Cardiac output of a person is 5400 ml and heart rate 72 per min. What will be his stroke volume?

  • (A) 65 ml
  • (B) 74 ml
  • (C) 75 ml
  • (D) 78 ml
Answer: (C) 75 ml
Explanation: Stroke Volume (SV) = Cardiac Output (CO) / Heart Rate (HR) = 5400 / 72 = 75 ml.

vi. Detritus food chain starts from _______.

  • (A) producers
  • (B) dead organic matter
  • (C) parasite
  • (D) photosynthesis
Answer: (B) dead organic matter

vii. Plants absorb _______ water.

  • (A) gravitational
  • (B) capillary
  • (C) combined
  • (D) hygroscopic
Answer: (B) capillary

viii. The organisms having tolerance for wide range of salinity are called _______.

  • (A) stenothermal
  • (B) euryhaline
  • (C) stenohaline
  • (D) eurythermal
Answer: (B) euryhaline

ix. Hisardale is a new breed of sheep developed by crossing _______.

  • (A) Bikaneri ram and Marino ewe
  • (B) Bikaneri ram and Bikaneri ewe
  • (C) Marino ram and Bikaneri ewe
  • (D) Marino ewe and Marino ram
Answer: (C) Marino ram and Bikaneri ewe

x. Perforins are secreted by _______.

  • (A) Helper T-cells
  • (B) Cytotoxic T-cells
  • (C) Suppressor T-cells
  • (D) Memory T-cells
Answer: (B) Cytotoxic T-cells

HSC Biology

Q.2. Answer the following questions: [8]

i. Mention the chromosome number having the mutated gene for β-Thalassemia.

Answer: The mutated gene for β-Thalassemia is located on chromosome 11.

ii. Which organ produces calcitriol?

Answer: The Kidney produces calcitriol.

iii. If the megaspore mother cell has 26 chromosomes, what will be the total number of chromosomes in endosperm of the same plant?

Answer: Megaspore Mother Cell (2n) = 26, so haploid (n) = 13.
Endosperm is triploid (3n).
Total chromosomes = 3 × 13 = 39 chromosomes.

iv. Define the term − Facilitated diffusion.

Answer: Facilitated diffusion is the passive transport of substances across the cell membrane from a region of higher concentration to lower concentration with the help of specialized carrier proteins or channels, without the expenditure of metabolic energy (ATP).

v. Give reason − Energy pyramid is always upright.

Answer: The energy pyramid is always upright because energy flows from one trophic level to the next with a significant loss of energy as heat at each step (according to the 10% law). Therefore, the amount of energy decreases at successive trophic levels.

vi. What will be the base sequence on the template strand of DNA, which codes for methionine?

Answer: Methionine is coded by the mRNA codon AUG. The template strand of DNA is complementary to the mRNA codon. Therefore, the base sequence on the template strand is TAC.

vii. Deficiency of which element causes Brown heart disease in plants?

Answer: Deficiency of Boron causes Brown heart disease in plants.

viii. Where are the cells of Rauber situated in the blastocyst of human embryo?

Answer: The cells of the trophoblast which are in contact with the inner cell mass (embryonal knob) in the blastocyst are called the cells of Rauber.
SECTION – B

Attempt any EIGHT of the following questions: [16]

Q.3. i. Give one example each of:

a. Autosomal dominant traits

b. Autosomal recessive traits

Answer: a. Autosomal dominant traits: Widow's peak / Huntington's disease.
b. Autosomal recessive traits: Phenylketonuria (PKU) / Cystic fibrosis / Sickle cell anaemia.

ii. If a carrier woman marries a colorblind man, what will be the phenotype of their progeny? Show in the form of a chart.

Answer: Parents: Carrier Woman (\(X^C X^c\)) × Colorblind Man (\(X^c Y\))

Gametes: (\(X^C\), \(X^c\)) and (\(X^c\), \(Y\))

Progeny Chart:
Gametes \(X^c\) \(Y\)
\(X^C\) \(X^C X^c\) (Carrier Daughter - Normal Vision) \(X^C Y\) (Normal Son)
\(X^c\) \(X^c X^c\) (Colorblind Daughter) \(X^c Y\) (Colorblind Son)
Phenotypic Ratio: 1 Normal daughter (carrier) : 1 Colorblind daughter : 1 Normal son : 1 Colorblind son.

Q.4. Sketch the appropriate diagrams showing following chromosomal aberrations:

i. deletion

ii. inversion

Answer:
Diagram showing Deletion and Inversion chromosomal aberrations [Please draw a diagram showing a chromosome segment being lost (Deletion) and a segment rotating 180 degrees (Inversion).]
i. Deletion: A segment of the chromosome (e.g., segment D) is lost.
ii. Inversion: A segment of the chromosome breaks, rotates 180°, and rejoins (e.g., B-C-D becomes D-C-B).

Q.5. Observe the following diagram of double circulation and identify A, B, C and D:

Pulmonary veins
Superior and Inferior vena cava
Heart
Right atrium ↓ Right ventricle
Left atrium ↓ Left ventricle
Right Ventricle → [ A ][ D ] → Pulmonary veins
Left Ventricle → [ B ][ C ] → Vena Cava
Answer:
  • A: Pulmonary artery
  • B: Dorsal Aorta (Systemic Aorta)
  • C: Body organs / Systemic capillaries
  • D: Lungs (Pulmonary capillaries)

Q.6. i. Select the names of fresh-water fishes from the given list:

Sardinella, Rastrelliger, Cirrhina, Harpadon, Labeo

Answer: The fresh-water fishes are: Cirrhina and Labeo.

ii. Write the economic importance of Lac (Any Two).

Answer: 1. Lac is used in the manufacture of bangles, toys, and woodwork.
2. It is used in the preparation of inks, polishes, and sealing wax.

Q.7. i. Define the term − Ecological succession.

Answer: The gradual and predictable change in the species composition of a given area is called ecological succession.

ii. What is the reason of eutrophication?

Answer: Eutrophication is caused by the nutrient enrichment of water bodies, particularly with nitrogen and phosphorus (from agricultural runoff, sewage, etc.), which leads to excessive growth of algae (algal bloom) and subsequent depletion of oxygen.

Q.8. i. Mention any two ill-effects of UV-rays on human beings.

Answer: 1. It causes skin cancer (melanoma).
2. It causes inflammation of the cornea (snow blindness) or cataract.

ii. Give significance of Ecosan.

Answer: Ecological sanitation (Ecosan) is a sustainable approach for handling human excreta using dry composting toilets. It transforms human waste into a natural fertilizer, thereby recycling nutrients and preventing water pollution.

Q.9. What are oral vaccines? Enlist the benefits of oral vaccines.

Answer: Oral vaccines are immunizing agents administered through the mouth.
Benefits:
1. They are easy to administer and painless (no needles required).
2. They are cost-effective and suitable for mass immunization campaigns (e.g., Pulse Polio).
3. They stimulate mucosal immunity (IgA production) in the gut.

Q.10. What is vernalization? Give the advantages of vernalization.

Answer: Vernalization: It is the dependence of certain plants on exposure to low temperatures, either quantitatively or qualitatively, to induce flowering.
Advantages:
1. It prevents precocious reproductive development late in the growing season.
2. It enables plants to have sufficient time to reach maturity.
3. It can induce early flowering in some plants.

Q.11. Enlist the causes of biodiversity losses.

Answer: The causes of biodiversity losses ("The Evil Quartet") are:
1. Habitat loss and fragmentation.
2. Over-exploitation.
3. Alien species invasions.
4. Co-extinctions.

Q.12. Give role of hormones Relaxin and Inhibin.

Answer: Relaxin: It is secreted by the ovary (corpus luteum) and placenta. It relaxes the pubic symphysis and dilates the cervix to facilitate the birth of the child (parturition).
Inhibin: It is secreted by Sertoli cells in males and Granulosa cells in females. It inhibits the secretion of FSH (Follicle Stimulating Hormone) from the pituitary gland to regulate spermatogenesis or follicular development.

Q.13. Sketch and label the diagram of a stoma showing kidney − shaped guard cells.

Answer:
[Students are expected to draw a diagram of a stomatal apparatus]
Labels to include:
- Kidney-shaped Guard cells
- Stomatal pore (aperture)
- Chloroplasts (inside guard cells)
- Subsidiary cells (surrounding guard cells)
- Epidermal cells

Q.14. Explain in brief the process of southern blotting and hybridization in DNA fingerprinting.

Answer: Southern Blotting: The DNA fragments separated by gel electrophoresis are transferred from the gel to a synthetic membrane, such as nitrocellulose or nylon membrane. This process preserves the arrangement of DNA fragments.
Hybridization: The membrane containing the DNA fragments is exposed to radioactive DNA probes (single-stranded DNA sequences complementary to the VNTRs). These probes bind (hybridize) to specific complementary DNA sequences on the membrane. The hybridized fragments can then be detected using autoradiography.
SECTION – C

Attempt any EIGHT of the following questions: [24]

Q.15. i. Define − Palaeontology.

Answer: Palaeontology is the branch of biology that deals with the study of fossils.

ii. Give any four points of significance of palaeontology.

Answer: 1. It provides direct evidence of evolution.
2. It helps in constructing the phylogeny (evolutionary history) of organisms.
3. It helps in the study of extinct organisms and their habits.
4. It connects links between different groups of organisms (e.g., Archaeopteryx connects reptiles and birds).
5. It helps in the study of the geological time scale.

Q.16. i. What is pollination?

Answer: Pollination is the transfer of pollen grains from the anther to the stigma of a flower.

ii. Differentiate between Anaemophily and Entomophily with reference to:

  • a. pollinating agent
  • b. stigma
  • c. nectar
  • d. fragrance
Answer:
Feature Anaemophily (Wind Pollination) Entomophily (Insect Pollination)
a. Pollinating agent Wind Insects
b. Stigma Large, feathery, and exposed to trap pollen. Sticky, often situated deep within the flower.
c. Nectar Absent Present (produced to attract insects)
d. Fragrance Absent Present (often sweet-smelling)

Q.17. i. Differentiate between hypotonic and hypertonic solutions.

Answer: Hypotonic solution: A solution having a lower concentration of solutes (and higher water potential) compared to the cell sap.
Hypertonic solution: A solution having a higher concentration of solutes (and lower water potential) compared to the cell sap.

ii. Mention the effect of exo-osmosis and endo-osmosis on shape of the cell.

Answer: Exo-osmosis: Causes the cell to shrink (plasmolysis in plant cells, crenation in animal cells).
Endo-osmosis: Causes the cell to swell and become turgid.

iii. Give one difference between symplast and apoplast pathway.

Answer: Symplast pathway: Water moves through the living parts of the cell (protoplasm) connected by plasmodesmata. It is slower.
Apoplast pathway: Water moves through the non-living parts (cell walls and intercellular spaces). It is faster.

Q.18. i. Mention the control measures to prevent ascariasis.

Answer: 1. Maintaining personal hygiene (washing hands before eating).
2. Washing vegetables and fruits thoroughly before consumption.
3. Proper disposal of human excreta and preventing soil pollution.

ii. With appropriate terms, complete the following chart and rewrite it.

Answer:
Sr. No. Name of disease Name of pathogen
a. Amoebiasis (Amoebic Dysentery) Entamoeba histolytica
b. Typhoid Salmonella typhi
c. Filariasis (Elephantiasis) Wuchereria bancrofti
d. Malaria Plasmodium species

Q.19. i. Define − Adaptation.

Answer: Adaptation is any attribute of the organism (morphological, physiological, or behavioural) that enables the organism to survive and reproduce in its habitat.

ii. Explain any two adaptations in Opuntia and Seal.

Answer: Opuntia (Desert Plant): 1. Leaves are modified into spines to reduce water loss through transpiration.
2. Stems are modified into flattened, green structures (phylloclades) to perform photosynthesis and store water.
Seal (Aquatic Mammal in cold climate): 1. They possess a thick layer of fat (blubber) below their skin that acts as an insulator to reduce the loss of body heat.

Q.20. Match the disease resistant varieties given in Column-I with the crops in Column-II and rewrite:

Answer:
Column-I (Variety) Column-II (Crop)
i. Himgiri b. Wheat
ii. Pusa shubhra c. Cauliflower
iii. Pusa sadabahar a. Chilli

Q.21. i. Explain the qualitative and quantitative aspects of growth phenomenon.

Answer: Quantitative aspect: Growth involves an increase in parameters like mass, volume, surface area, number of cells, or dry weight, which can be measured.
Qualitative aspect: Growth involves development, differentiation, and change in the form or function of cells and tissues leading to maturity.

ii. Explain the phase of cell maturation.

Answer: In the phase of cell maturation (differentiation), the cells that have elongated attain their maximum size. They undergo structural and physiological differentiation to perform specific functions. The cell walls become thickened, and the protoplasm undergoes modifications.

Q.22. i. What is co-dominance?

Answer: Co-dominance is a condition where both alleles of a gene pair in a heterozygote are fully expressed, with neither one being dominant or recessive to the other.

ii. If a red colored female cattle is crossed with a white male cattle, what will be the appearance of progeny in F2 generation? Show the genotypes with the help of a chart.

Answer: Parents: Red (RR) × White (WW)
F1 Generation: All Roan (RW)
F2 Generation (Selfing F1): RW × RW

Chart:
Gametes R W
R RR (Red) RW (Roan)
W RW (Roan) WW (White)
Appearance of Progeny in F2:
Red : Roan : White = 1 : 2 : 1.

Q.23. i. Give any two involuntary vital functions of medulla oblongata.

Answer: 1. Regulation of heart beat (Cardiac center).
2. Regulation of respiration (Respiratory center).
(Other functions: Vasomotor activities, peristalsis).

ii. Mention two functions of spinal cord.

Answer: 1. Conduction of sensory and motor impulses to and from the brain.
2. Acting as a center for spinal reflexes.

Q.24. i. Define − Transcription.

Answer: Transcription is the process of copying genetic information from one strand of the DNA into RNA.

ii. Write anticodons for the following triplet codons:

AUG, GAG, CUA, CCU

Answer:
  • AUG → UAC
  • GAG → CUC
  • CUA → GAU
  • CCU → GGA

Q.25. i. Mention the position of the following in human heart:

a. Eustachian valve

b. Bicuspid valve

Answer: a. Eustachian valve: Guards the opening of the Inferior Vena Cava into the right atrium.
b. Bicuspid valve (Mitral valve): Located between the left atrium and the left ventricle.

ii. Differentiate between open and closed circulation with reference to:

a. blood pressure

b. exchange of material

Answer: a. Blood pressure: In open circulation, blood flows at low pressure. In closed circulation, blood flows at high pressure.
b. Exchange of material: In open circulation, exchange occurs directly between blood and cells/tissues. In closed circulation, exchange occurs through the walls of capillaries.

Q.26. Draw a neat and proportionate diagram of Graafian follicle and label oocyte and antrum. Explain its structure in brief.

Answer:
[Students are expected to draw the Graafian Follicle]
Labels: Secondary Oocyte, Antrum, Theca externa, Theca interna, Membrana Granulosa, Cumulus oophorus, Zona pellucida.
Structure Explanation:
The Graafian follicle is a mature ovarian follicle. It consists of an outer protective layer called Theca externa and an inner vascular layer called Theca interna. Inside this, there are follicular cells forming the Membrana Granulosa. A large fluid-filled cavity called the Antrum (filled with liquor folliculi) is present. The secondary oocyte is situated eccentrically and is surrounded by a group of cells called Cumulus oophorus (or Discus proligerus). The oocyte is covered by a non-cellular layer called Zona pellucida.
SECTION – D

Attempt any THREE of the following questions: [12]

Q.27. i. What is placenta?

Answer: The placenta is a temporary structural and functional unit formed by the intimate connection between the foetal and maternal tissues, which facilitates the supply of oxygen and nutrients to the embryo and the removal of carbon dioxide and excretory wastes from it.

ii. Give reason − Placenta is considered as a temporary endocrine gland.

Answer: The placenta produces several hormones necessary for the maintenance of pregnancy, such as Human Chorionic Gonadotropin (hCG), Human Placental Lactogen (hPL), estrogens, progesterone, and relaxin. Since it secretes hormones directly into the blood, it acts as a temporary endocrine gland.

iii. Give significance of hCG.

Answer: Human Chorionic Gonadotropin (hCG) maintains the corpus luteum and stimulates it to secrete progesterone, which is essential for maintaining the endometrium and thus the pregnancy. Its presence in urine is the basis for pregnancy tests.

Q.28. Describe in brief the structural and hormonal changes during ovarian cycle.

Answer: The ovarian cycle consists of the following phases:

1. Menstrual Phase (Days 1-5):
If fertilization does not occur, the corpus luteum degenerates. Progesterone and estrogen levels fall. This triggers the breakdown of the endometrium.
2. Follicular Phase (Days 5-13):
Under the influence of FSH (Follicle Stimulating Hormone) from the pituitary, primary follicles in the ovary develop into mature Graafian follicles. The developing follicles secrete Estrogen, which helps in the regeneration of the endometrium.
3. Ovulatory Phase (Day 14):
Estrogen levels peak, stimulating a surge in LH (Luteinizing Hormone). This LH surge causes the rupture of the Graafian follicle and the release of the secondary oocyte (ovulation).
4. Luteal Phase (Days 15-28):
Under the influence of LH, the remaining cells of the ruptured follicle transform into the Corpus Luteum. The Corpus Luteum secretes large amounts of Progesterone (and some estrogen), which maintains the endometrium for implantation. If fertilization does not occur, the Corpus Luteum degenerates into a white scar called Corpus Albicans.

Q.29. i. Name the formed elements which are useful in blood coagulation. Give its normal number per cubic milimeter (mm³) in human blood.

Answer: The formed elements useful in blood coagulation are Platelets (Thrombocytes).
Normal number: 1.5 to 3.5 lakhs / mm³ (150,000 to 350,000 / mm³).

ii. Comment on the shape and secretion of the above mentioned formed elements.

Answer: Shape: They are cell fragments, irregular or rounded in shape, and non-nucleated.
Secretion: They release clotting factors like Thromboplastin (Platelet factor 3).

iii. Explain in brief the mechanism of blood coagulation.

Answer: 1. At the site of injury, platelets rupture and release Thromboplastin.
2. Thromboplastin initiates the formation of the enzyme complex Prothrombinase (or Thrombokinase).
3. Prothrombinase converts inactive Prothrombin into active Thrombin in the presence of Calcium ions (\(Ca^{++}\)).
4. Thrombin converts soluble Fibrinogen protein into insoluble Fibrin threads.
5. These fibrin threads form a mesh that traps blood cells, forming a clot (coagulum).

Q.30. i. Give full form of the cloning vectors BAC and YAC.

Answer: BAC: Bacterial Artificial Chromosome.
YAC: Yeast Artificial Chromosome.

ii. Write the appropriate palindrome for Eco RI and indicate by an arrow its recognition sequence.

Answer: Sequence:
5' — G ↓ A A T T C — 3'
3' — C T T A A ↑ G — 5'
(The arrows indicate the cut site between G and A).

iii. Give any four uses of gene therapy.

Answer: 1. Treatment of genetic disorders like SCID (Severe Combined Immunodeficiency) caused by ADA deficiency.
2. Treatment of Cystic Fibrosis.
3. Treatment of Haemophilia.
4. Treatment of certain cancers (using suicide genes).
5. Treatment of Parkinson’s disease.

Q.31. Explain any four contrivances to prevent self pollination in plants with an appropriate example of each type.

Answer: 1. Unisexuality (Dicliny): Plants produce unisexual flowers (either male or female). This prevents self-pollination. Example: Papaya (Dioecious), Maize (Monoecious).

2. Dichogamy: In bisexual flowers, anthers and stigma mature at different times.
- Protandry: Anthers mature first. Example: Sunflower, Disc florets.
- Protogyny: Stigma matures first. Example: Michelia, Gloriosa.

3. Prepotency: Pollen grains of another flower germinate more rapidly over the stigma than the pollen grains from the same flower. Example: Apple.

4. Heterostyly (Heteromorphy): Plants produce flowers with different lengths of styles and stamens, preventing pollen from reaching the stigma of the same flower. Example: Primula (Primrose).

5. Herkogamy: Physical barriers prevent self-pollination. Example: Calotropis (Pentangular stigma positioned above anthers).

6. Self-sterility (Self-incompatibility): Pollen grains fail to germinate on the stigma of the same flower due to genetic mechanisms. Example: Tobacco, Thea.