12th Biology Question Paper 2024 Maharashtra Board Pdf

Maharashtra State Board Class 12th Biology Question Paper 2024 with Solutions Answers Pdf Download.

Class 12 Biology Question Paper 2024 Maharashtra State Board with Solutions

Time : 3 Hrs
Marks : 70

General Instructions: The question paper is divided into four sections.

1. Section A: Q. No.1 contains Ten multiple choice type of questions carrying One mark each. Evaluation will be done for the first attempt only.
   Q.No. 2 Contains Eight every short answer type of questions carrying one mark each.
2. Section B: Q.No. 3 to 14 are short answer type of questions carrying two marks each.
   (Attempt any Eight)
3. Section C: Q.No. 15 to 26 are Short answer type of questions carrying three marks each. (Attempt any Eight)
4. Section D: Q.No. 27 to 31 are long answer type of questions carrying four marks each. (Attempt any Three)
5. Begin the answer of each section on a new page.

Time: 3 Hrs.
Total Marks: 70

Section – A

Question 1.
Select the correct alternatives and write the answers: (10)

(i) Identify the growth hormone in plants which causes inhibitory effect.
(a) Cytokinins
(b) Abscissic acid
(c) Gibberellin
(d) Ethylene
Answer:
(b) Abscissic acid

Explanation:
It is also called stress hormone. It acts as anti-transpirant.

(ii) Which one of the following is not a part of lac operon?
(a) Promoter
(b) Regulator
(c) Inducer
(d) Operator
Answer:
(c) Inducer

Explanation:
Inducer acts as switch off and switch on mechanism for operon.

(iii) In absence of fertilization, corpus luteum degenerates into ………
(a) tunica albugenia
(b) membrana granulosa
(c) zona pellucida
(d) corpus albicans
Answer:
(d) corpus albicans.

Explanation:
As fertilisation has not taken place, corpus luteum will not secrete progesterone which acts as pregnancy hormone. Hence, it will degenerate into corpus albicans.

(iv) Which of the following divides nasal cavity?
(a) Hyaline cartilage
(b) Mesethmoicfcartilage
(c) Ligamentum arteriosum
(d) Laryngopharynx
Answer:
(b) Mesethmoid cartilage.

Explanation:
The nose has a pair of slit like openings called external nares or nostrils for entry of air into the nasol cavity. The nasal cavity is divisible into right and left nosal chambers by a mesethmoid cartilage.

(v) Which of the following is caused by unsterilized needle?
(a) Elephantiasis
(b) AIDS
(c) Malaria
(d) Dengue
Answer:
(b) AIDS

Explanation:
AIDS is caused by HIV virus.
This virus can transmit through sharing syrings, needles and blood transfusion.

(vi) Recognition sequence of restriction enzymes are generally ……. nucleotides long.
(a) 2 to 4
(b) 4 to 8
(c) 8 to 10
(d) 14 to 18
Answer:
(b) 4 to 8

Explanation:
Restriction enzymes can give proper nick at 4 to 8 nudeotide sequence which can lead to either blunt or cohesive ends.

(vii) Which of the following types require pollinator but result is genetically similar to autogamy?
(a) Geitonogamy
(b) Xenogamy
(c) Apogamy
(d) Cleistogamy
Answer:
(a) Geitonogamy

Explanation:
Geitonogamy is functionally similar to cross-pollination as it involves pollinating agent, but it cannot bring about genetic variations because transfer of pollen grain to stigma of different flower on same plant and is only of ecological significance e.g., cucurbita maxima.

(viii) Which one of the following does not evolve further?
(a) Climax community
(b) Primary Succession
(c) Pioneer Species
(d) Seral Community
Answer:
(a) Climax community

Explanation:
It is the most stable community, as the final stage of ecological succession where balance is achieved between species composition.

(ix) Identify the appropriate term for the number of births under ideal conditions:
(a) Absolute mortality
(b) Realized natality
(c) Realized mortality
(d) Absolute natality
Answer:
(d) Absolute natality

Explanation:
It is no. of births under ideal conditions with no competition abundance of resources etc.

(x) Observe the graph and select correct option:

(a) Line ‘A’ represents S-CA²
(b) Ling ‘B’ represents log C = log A + Z log S
(c) Line A represents S = CA²
(d) Line B represents log S = log Z + C log A
Answer:
(b) Line ‘B’ represents log C = log A + Z log S

Explanation:
S = CA^Z
S = Species richness
A = Area
C = Constant
Z = Regression Constant.

Question 2.
Answer the following questions: (8)

(i) What are vestigeal organs?
Answer:
Vestigeal organs (Rudimentary organs) : They are imperfectly developed and non functional, degenerate structures which were functional in ancestors.
e.g: Nictitating membrane, wisdom teeth, coccyx (tail bone), Vermiform appendix and the caecum.

(ii) Expand the term ZIFT.
Answer:
ZIFT: Zygote Intra-fallopian transfer.

(iii) Give the name of endocrine gland which is prominent at birth but gets gradually atrophied in adult stage.
Answer:
Thymus gland. It is a prominent gland at birth but gets gradually atrophied in the adult, so it is called temporary gland.

(iv) What is the full form of IAA?
Answer:
Indole-3-Acetic – Acid.

(v) Give the name of microbial source of antibiotic Chloromycetin.
Answer:
Streptomyces venezuelac.

(vi) Which cells of islets of Langerhans produce a hormone insulin?
Answer:
β (Beta) cells.

(vii) How many meiotic divisions are required for the formation of 300 seeds in angiosperm?
Answer:
To produce 300 seeds in angiosperms, 300 meiotic divisions are required, one for each seed.

(viii) Explain the term Emigration.
Answer:
Emigration : It is the number of individuals of the population, who leave a habitat or place with the intent of moving to a different place permanently.

Section – B

Attempt any EIGHT of the following questions: (16)

Question 3.
What are the reasons for the success of Mendel?
Answer:
Reasons for Mendel’s Success

• Experiments carefully planned and involved lange samples.
• Recorded and expressed his results as ratios.
• Contrasting characters can be easily recognized.
• 7 different characters in pea plant controlled by a single factor each. The factors are located on separate chromosomes and these factors are transmitted from generation to generation.

Question 4.
Arrange the following steps of DNA fingerprinting in correct sequence:
(a) Gel electrophoresis
(b) Isolation of DNA
(c) Southern blotting
(d) Restriction digestion
Answer:

Question 5.
Distinguish between human sperm and ovum.
Answer:

Human Sperm	Ovum
1. Sperm is Male gamete.	1. Ovum is the female gdmete.
2. Sperms are produced in testes.	2. Ova are produced in ovaries.
3. Four sperms are produced from one spermatogonium.	3. Only one ovum is produced from one oogonium.
4. It has very small amount of cytoplasm.	4. It has large amount of cytoplasm called ooplasm.

Question 6.
Enlist the uses of gene therapy.
Answer:
Uses of Gene therapy.

• Replacing of missing or defective genes.
• Deliver genes that speed the destruction of cancer cells.
• Supply genes to revert cancer cells to normal cells.
• Supply of gene for impairing viral replication.
• Provide genes that promote or impede the growth of new tissue.
• Deliver genes that stimulate the healing of damaged tissue.

Question 7.
Define the following terms:
(a) Gene flow
(b) Chromosomal aberrations
Answer:
(a) Gene flow : It is movement of genes from one population to another population.
(b) Chromosomal aberrations : The structural, morphological change in chromosome due to rearrangement in single or multiple chromosomes.

Question 8.
What are the significances of double fertilization?
Answer:
Significance of Double fertilization.

• It is a unique feature of angiosperms. It ensures that the parent invests a seed with food store, only if the egg is fertilised.
• The Diploid Zygote develops into an embryo, which consequently develops into a new plant.
• The triploid PEN (3N) develops into nutritive endosperm tissue.
• It restores diploid condition by fusion of haploid male gamete with haploid female gamete.
• Avoid polyembryony.

Question 9.
Identify and define ‘A’ and ‘B’ in relation to uptake of water by the root:

Answer:
A → Symplast pathway : Water passes across from one living cell to another living cell through plasmodesmata.
B → Apoplast pathway : Water passes across the root through the cell-wall and the intercellular spaces of cortical cells of root.

Question 10.
Describe mutualism.
Answer:
Mutualism is a type of interaction between two living organisms in which both are equally benefited and dependent on each other for survive.

For example, lichen is a mutualistic relationship between a fungus and algae. Algae provide food to fungus obtained from photosynthesis. The fungus provides anchoring and protection to the algae.

Question 11.
Explain factors affecting water absorption.
Answer:
Factors affecting water absorption.

• Presence of capillary water is essential.
• Rate of water absorption is maximum at soil temperature between 20 to 30° C.
• High concentration of solutes in soil, reduce the rate of absorption of water.
• Poorly aerated soil shows poor absorption rate.
• Increased transpiration accelerate the rate of absorption of water in the irrigated soil.

Question 12.
What is differentiation and redifferentiation?
Answer:
Differentiation : Maturation of cells derived from apical meristem of root and shoot. A permanent change occur in structure and function of cells.
Redifferentiation : Cells produced by dedifferentiation lose the capacity to divide and mature to perform specific functions are called redifferentiation.
For example : Secondary xylem and Secondary phloem.

Question 13.
Select and rewrite appropriate disorder of respiratory system with the given symptoms;
[sinusitis, emphysema, silicosis and asbestosis, laryngitis]
(a) Breakdown of alveoli, shortness of breath.
(b) Inflammation of the sinuses mucous discharge.
(c) Inflammation of larynx, vocal cord, sore throat, hoarseness of voice, mucous build up and cough.
(d) Inflammation of fibrosis, lung damage.
Answer:
(a) Emphysema
(b) Sinusitis
(c) Laryngitis
(d) Occupational Respiratory disorders silicosis, asbestoiss.

Question 14.
Explain the steps involved in preliminary treatment of sewage.
Answer:
Preliminary treatment:

(i) Screening : Sewage and waste water contains plenty of suspended floating materials, course and solid particles along with dissolved substance. The suspended objects are filtered and removed.
Sewage is passed through screens or Nets.

(ii) Grit chamber : After screening, filtered sewage is then passed into series of grit chambers. These chambers contain large stone (pebbles) and brick-ballest. Course particles settle down by gravity and removed.

Section – C

Attempt any EIGHT of the following questions: (24)

Question 15.
Give the different steps involved in formation of m-RNA from hn-RNA.
Answer:
Formation of mRNA from hn RNA.
In eukaryotes RNA transcribed from DNA is called primary transcripts.
Splicing : Primary transcript is non-functional and contains both exons and introns. During processing introns are removed.
Capping : Methylated guanosine triphosphate is added to 5′ end of hnRNA like a cap.
Tailing : Polyadenylation takes place, when hnRNA is fully processed, it is known as mRNA.
For translation, mRNA is transported out of nucleus through nuclearpore.

Question 16.
What is reproductive isolation? Describe any two types each of pre-mating and post-mating isolating mechanism.
Answer:
Reproductive isolation : It occurs due to change in genetic material, gene pool and structure of genital organs. It prevents interbreeding between populations.

Types (A): Pre mating or Pre-zygotic

1. Habital isolation/Ecological isolation : Members living in the same geographic region but occupy separate habitats so that potential mates do not meet.
2. Seasonal or Temporal Isolation : Members of a population living in the same geographic region, but are sexually mature at different years or different time of the years.

(B) Post mating/Post zygotic barriers

1. Zygote Mortality : Egg is fertilized but zygote dies due to one or the other reason.
2. Gamete Mortality : Gametes have a limited life span. Due to one or the other reason, if the union of the two gametes does not occur in the given time, it results in gamete mortality.

Question 17.
Explain unique features of acquired immunity.
Answer:
Unique features of acquired immunity

1. Specificity : It can produce a specific antibody or T. lymphocytes against a particular antigen / pathogen.
2. Diversity : It can recognise a vast variety of diverse pathogens or foreign molecules.\
3. Discrimination between self, and non-self : Differentiates between own body cells (self) and foreign (non-self) molecules.
4. Memory Immune System builds an (imprint) memory after first encounter. As a result, a second encounter with some pathogen brings quick and strong immune response.

Question 18.
Name and describe hormones secreted by ovaries.
Answer:
Hormones secreted by Ovary

1. Estrogeon : It is secreted by developing follicles. Estradio is the main estrogen.
   It is responsible for development of secondary sexual characters in females.
2. Progesterone: It is secreted by corpus luteum of the ovary after ovulation. It is essential for thickening of uterine endometrium. Development of Mammary glands inhibits uterine contraction during pregnancy.
3. Relaxin : Relaxes the Cervix of the pregnant female and ligaments of pelvic girdle for easy birth of young one.
4. Inhibin: Inhibits FSH-and GnRH production.
5. Activin : It increases FSH binding and FSH induced aromatisation.

Question 19.
Explain different steps involved in PCR technique.
Answer:
Steps in PCR

Step (i): Denaturation
Reaction mixture is heated to temperature (90 – 98 °C) to separate two strands of sample DNA.

Step (ii): Annealing.
Mixture is alLowed to cool upto 40 – 60°C that permits pairing of the primer to DNA-complementary stands.

Step (iii): Polymerisation :
Temperature (70 – 75°C) allows thermostable Taq DNA polymerase to use SS-DNA as template and adds nucleotide. This is called primer extension.

Question 20.

Identify A, B, and C from the above diagrams and give their functions.
Answer:
(a) Neutrophils : Neutrophils have fine granules that stain with neutral dyes, make up about 70% of total WBCs, and have a multi-lobed nucleus. They are able to move in an amoeboid manner and perform phagocytosis, playing a crucial role in destroying pathogens.

(b) Eosinophils: Eosinophils contain lysosomal granules that are stained tp red colour with acidic stains like eosin. Eosinophils are about 1 – 3 % of total WBCs. The nucleus is bilobed. They destroy antigen-antibody complex by phagocytosis. Their number increases in allergic condition and they show antihistaminic property. They are also responsible for detoxification as they produce antitoxins.

(c) Monocytes: Monocytes are the largest of all the WBCs. Its nucleus is large and bean or kidney shaped. They form 3-5% of WBCs. Monocytes are actively motile and give rise to macrophages. They are mainly phagocytic and destroy the bacteria and dead or damaged tissue by phagocytosis.

Question 21.
What are the limitations of root pressure theory?
Answer:
Limitations of Root Pressure Theory.

• Not applicable to plants taller than 20 meters.
• Ascent of sap occurs in absence of root system.
• Root pressure value is almost, zero in taller Gymnosperm trees.
• In actively transpiring plant, no root pressure is developed.
• Xylem sap under normal condition is under tension.

Question 22.
Explain green house effect with reference to gases responsible for it and their sources.
Answer:
Greenhouse effect: Responsible for heating of earth’s surface and atmosphere.
Gases responsible : CO₂, CH₄, CFC, N₂O (Nitrous oxide) and water vapours.
Sources of Green House Gases : Burning of fossil fuels, agricultural wastes, automobiles increases the levels of CO₂.
Biogas plants, paddy fields, cattle sheds add CH₄ to atmosphere. CFC’s are emitted by fire extinguishers and air conditioners.

Question 23.
Describe physiological effects and applications of ethylene.
Answer:
Physiological Effects and Applications of Ethylene.

• Promotes ripening of fruits like banana, apple and mangoes.
  Initiation of lateral roots.
• Breaking of dormancy.
• Accelerates abscission activity.
• Inhibits growth of lateral buds and causes apical dominance and retards flowering.
• Enhancement of process of senescence.
• Causes epinasty.
• Increases activity of chlorophyllase enzyme causing degreening effect in banana and citrus.

Question 24.
Give the name and type of I, IV and VII cranial nerves.
Answer:

Sr.No.	Name	Type
I	Factory	Sensory
IV	Pathetic	Motor
VII	Facial	Mixed

Question 25.
Describe pyramid of energy with the help of diagram.
Answer:
Pyramid of energy:

It is always upright. It can never be inverted.
Energy flows from one trophic level to next level as heat at each step.
In smaller food chains, more energy is available than in the longer food chains.

Question 26.
What is lac? Enlist economic importance of Lac.
Answer:
Lac is a resin-like substance produced by the dermal glands of the female lac insect, Trachardia lacca. The insect feeds on the succulent twigs of certain plants and secretes a pink- coloured resin, which hardens upon exposure to air, forming lac.

Economic Importance of lac:

1. It is used in the preparation of sealing wax, paints, varnish, electrical goods.
2. It is used in the preparation of bracelets, buttons, toys and in filling hollow gold ornaments.
3. It is used in artificial leather and pottery.
4. It is also used in gramophone industry.
5. It is used for coating fruits and vegetables.

Section – D

Attempt any THREE of the following questions: (12)

Question 27.
Describe histological structure of Testis with well labelled diagram.
Answer:

Histology of Testis:

1. Externally, the testis is covered by three layers. These are:

A. Tunica vaginalis: it is the outermost incomplete peritoneal covering made up of connective tissue and epithelium.

B. Tunica albuginea: It is the middle layer formed by collagenous connective tissue.

C. Tunica vasculosa/vascularis: It is the innermost layer. It is a thin and membranous layer.
Fibers from tunica albuginea divide each testis into about 200-300 testicular lobules Seminiferous tubule is lined by cuboidal germinal epithelial cells and few large pyramidal cells called Sertoli or sustentacular cells.

Stages of spermatogenesis is:

Question 28.
What are chromosomal disorders? Describe Turner’s syndrome and Klinefelter’s syndrome.
Answer:
Chromosomal disorders : are caused due to absence or excess of one or more chromosomes or their disarrangement.

Turner’s Syndrome. (X Monosomy / XO females)

• Sex chromosomal disorder.
• Non – disjunction of chromosomes.
• 44 autosomes + XO.
• Phenotypically females.
• Short Statue and webbed neck.
• Lower posterior hair line, broad shield shaped chest.
• Poorly developed ovaries and breast.
• Low intelligence.

Klinefelter’s Syndrome (XXY)

• Extra X chromosome.
• Geno type – 44 + XXY
• Feminised males.
• Non- disjunction of X chromosome during meiosis.
• Voice pitch is harsh.
• Underdeveloped testes.
• Tall with long arms, feminine development, Gynae- comastia, no spermatogenesis. Individuals are sterile.

Question 29.
Describe nervous system in planaria with well labelled diagram.
Answer:
Nervous System in Planaria (flatworm)

• Most primitive animal with a CNS (Central nervous system) located on ventral side of body.
• It consists of mass of cerebral or cephalic ganglion appearing like an inverted U-shaped brain.
• These lie in the anterior or head region and from each ganglion arise 9-branches towards the outer side.
• Ventrally from below the ganglion arise a pair of VNC (Ventral Nerve Cords) or Long Nerve Cords.
• These are interconnected to each other by transfer nerve or Commissure in a ladder like manner.
• Peripheral nerve plexus arising laterally from VNC.
• PNS include sensory cells arranged in lateral cords in the body.
• Eyes are located on dorsal side of the brain.
• Single Sensory cells scattered in the body.

Question 30.
Explain following terms:
(a) Grafting
(b) Apomixis
(c) Polyembryony
(d) Parthenocarpy
Answer:
Grafting: Two plants are joined to grow as one plant.
In this method, part of stem containing more than one bud (scion) is joined onto rooted plant stock is called grafting.
Bud grafting: Only one bud is joined on stock.
Eg : Apple, Pear, Rose etc.

Apomixis : Formation of embryo (s) through asexual method of reproduction without formation of gametes.
Parthenocarpy : Condition in which fruit is developed without process of fertilization. The fruit resembles the normally produced fruit but it is seedless.
Eg: Pineapple, Banyan, Papaya.
Polyembryony : Development of more than, one embryos, inside the seed.

Types of Polyembryony.
Adventive: Embryo develops directly from diploid cell of nucellus and integuments.
Cleavage : Zygote proembryo sometimes divides into many parts. Each part develops into embryo.

Question 31.
Interpret the given diagrams A and B. Enlist the changes occuring during inspiration and expiration.

Answer:
Diagrams A and B represent the human respiratory system during the processes of inspiration and expiration.
In diagram A, we observe the following changes typical of inspiration:

1. During inspiration, the diaphragm, a dome-shaped muscle beneath the lungs, contracts and flattens out. This movement increases the vertical dimension of the thoracic cavity.
2. Simultaneously, the external intercostal muscles (located between the ribs) contract, pulling the ribcage upward and outward. This expansion increases the front-to-back and side-to-side dimensions of the thoracic cavity.
3. As the thoracic cavity’s volume increases, the pressure inside the lungs falls below the atmospheric pressure

(Boyle’s law). This pressure differential causes air to rush into the lungs, filling them with oxygen, which is crucial for our cells’ metabolic processes.
In diagram B, which illustrates expiration, the opposite occurs:

1. During expiration, the diaphragm relaxes, resuming its dome shape, which decreases the vertical space in the thoracic cavity.

2. The internal intercostal muscles may contract to actively pull the ribcage down and in (during forceful expiration), while the external intercoastal relax. However, in normal, relaxed breathing, expiration is mostly a passive process without significant muscular effort.

3. The decrease in the thoracic cavity’s volume leads to an increase in pressure inside the lungs. When this pressure surpasses the atmospheric pressure, it pushes air out of the lungs, expelling carbon dioxide, a waste product of cellular metabolism.

These changes in volume and pressure in the thoracic cavity are critical for the process of breathing, as they enable the flow of air in and out of the lungs.

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