Maharashtra Board Class 12 Biology Sample Paper Set 4 with Solutions

Maharashtra State Board Class 12th Biology Sample Paper Set 4 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Biology Model Paper Set 4 with Solutions

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 very 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.

Section – A

Question 1.
Select and write the correct answer: [10 Marks]

i. The supporting cells that produce myelin sheath in the PNS are ____.
(A) Oligodendrocytes
(B) Satellite cells
(C) Astrocytes
(D) Schwann cells
Answer:
(D) Schwann cells

ii. In human female, the fertilized egg gets implanted in uterus after about ___.
(A) 7 days of fertilization
(B) 30 days of fertilization
(C) two months of fertilization
(D) 3 weeks of fertilization
Answer:
(A) 7 days of fertilization

iii. Water absorption takes place through
(A) lateral roots
(B) root cap
(C) root hair
(D) primary root
Answer:
(C) root hair

iv. Epiphytic plants like orchids absorb water vapours from air with the help of epiphytic roots having special tissue called ____.
(A) sporogenous tissue
(B) sporophytic tissue
(C) sclerenchymatous tissue
(D) velamen
Answer:
(D) velamen

v. In grafting, the part of stem containing more than one bud called ____ is joined onto a rooted plant.
(A) stock
(B) cutting
(C) scion
(D) clone
Answer:
(C) scion

vi. Infectious stage of Plasmodium is
(A) Trophozoite
(B) Sporozoite
(C) Cryptozoite
(D) Metacercaria
Answer:
(B) Sporozoite

vii. What will be the parents blood groups if the blood group of a child is AB?
(A) A and O
(B) B and O
(C) AB and O
(D) A and AB
Answer:
(D) A and AB

viii. Antibody producing plasma cells are derived from
(A) Memory T-cells
(B) Suppressor T-cells
(C) Helper T-cells
(D) B-lymphocytes
Answer:
(D) B-lymphocytes

ix. The micro-organism used in the production of acetic acid is
(A) Aspergillus niger
(B) Rhizopus arrhizus
(C) Neurospora gossypii
(D) Acetobacter aceti
Answer:
(D) Acetobacter aceti

x. The bacterium which causes a plant disease called crown gall is ____.
(A) Helicobacter pylori
(B) Agrobacterium tumefaciens
(C) Thermophilus aquaticus
(D) Bacillus thuringiensis
Answer:
(B) Agrobacterium tumefaciens

Question 2.
Answer the following: (8 Marks)

i. Write the function of Corpus luteum.
Answer:
Corpus luteum releases progesterone, small amounts of estrogen and inhibin. It is thus, essential for establishing and maintaining pregnancy.

ii. Write the name of the small molecule required to initiate / start the process of synthesis of new complementary strand during replication of DNA.
Answer:
RNA primer is required to initiate the synthesis of new complementary strand during DNA replication.

iii. What is gravitational water?
Answer:
The water which percolates deep in the soil, due to the gravity is called ‘gravitational water’.

iv. What is genetic drift?
Answer:
Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift.

v. Name the instrument used for the measurement of blood pressure.
Answer:
The instrument used for the measurement of blood pressure is sphygmomanometer.

vi. What are transgenic animals?
Answer:
Transgenic animals are those animals which have their DNA manipulated to possess and express foreign gene.

vii. Define Stratification.
Answer:
Stratification is the vertical distribution of different species of plants and animals which occupies different levels.

viii. What is the function of International Union for Conservation of Nature and Natural Resources (IUCN)?
Answer:
The function of International Union for Conservation of Nature and Natural Resources (IUCN) is to maintain a Red Data Book also known as Red List, where conservation status of plant and animal species is recorded.

Section – B

Attempt any EIGHT of the following questions:

Question 3.
Write a short note on colostrum.
Answer:
Colostrum:

1. Colostrum is sticky and yellow fluid secreted by the mammary glands soon after child birth.
2. It contains proteins, lactose and mother’s antibodies e.g. IgA.
3. The fat content in colostrum is low.
4. The antibodies present in it helps in developing resistance for the new born baby at a time when its own immune response is not fully developed.

Question 4.
Why law of segregation is also called the law of purity of gametes?
Answer:

1. A diploid organism contains two factors for each trait in its diploid cells and the factors segregate during the formation of gametes.
2. The two alleles (contrasting characters) do not mix, alter or dilute each other and the gametes formed are ‘pure’ for the characters which they carry.
3. A gamete may carry either dominant or recessive factor but not both. Hence, this law is also called the law of purity of gametes.

Question 5.
Write a short note on cloverleaf structure of tRNA.
Answer:

1. Cloverleaf structure of tRNA possesses an anticodon loop that has bases complementary to the codon. It is called anticodon.
2. It shows amino acid acceptor end (3′ end) having unpaired CCA bases (i.e. amino acid binding site) to which amino acid binds.
3. For every amino acid, there is specific tRNA.
4. Initiator tRNA is specific for methionine.
5. There are no t-RNAs for stop codons.

Question 6.
What is adaptive radiation? Explain with suitable example.
Answer:
i. The process of evolution which results in transformation of original species to many different varieties, is called adaptive radiation.

ii. Darwin’s Finches is one of the best example of adaptive radiation.
During his visit to Galapagos Islands, Charles Darwin also noticed a variety of small birds. These birds are now called Darwin’s finches.

Darwin concluded that the American main land species of the bird was the original one from which they migrated to the different islands of Galapagos.

These birds adapted to the different environmental conditions of these islands. From original seed eating features, many other forms with altered beaks evolved into insectivorous features.

iii. Another example of adaptive radiation is Australian Marsupials. In Australia, there are many marsupial mammals who evolved from common ancestor.

iv. Adaptive radiation leads to divergent evolution.

Question 7.
What are the limitations of root pressure theory?
Answer:
Limitations to root pressure theory:

1. It is not applicable to plants taller than 20 meters.
2. Ascent of sap can also occur even in the absence of root system.
3. Root pressure value is almost nearly zero in taller gymnosperm trees.
4. In actively transpiring plants, no root pressure is developed.
5. Xylem sap under normal condition is under tension i.e. it shows negative hydrostatic pressure or high osmotic pressure.

Thus, root pressure is not the sole mechanism explaining the ascent of sap in all plants of varying heights.

Question 8.
In the given figure representing stages during embryo development showing arithmetic and geometric growth, identify phase ‘X’ and ‘Y’.

Answer:
In the given figure representing stages during embryo development showing arithmetic and geometric growth;
X: Geometric phase
Y: Arithmetic phase

Question 9.
What is addiction? How do people get addicted?
Answer:

1. Addiction is a psychological attachment to certain effects such as euphoria and a temporary feeling of well-being associated with drugs and alcohol.
2. These feelings drive people to take them even when these are not needed, or even when their use becomes self-destructive.
3. With repeated use of drugs, the tolerance level of the receptors present in our body increases.
4. Consequently the receptors respond only to higher doses of drugs or alcohol leading to greater intake and addiction.

Question 10.
Following diagram represents L.S. of Azolla leaf. Identify the labels (A), (B) (C), (D) and redraw the diagram.

Answer:
(A) Photosynthetic zone
(B) Anabaena
(C) Cavity
(D) Dorsal lobe

Question 11.
Enlist any four properties of transgenic plants.
Answer:
Following are the properties of transgenic plants:

1. Insect pest resistance
2. Improved nutritional qualities
3. Modified post harvesting characteristics
4. Transgenic plants act as factories/bioreactors
5. Edible vaccine production

Question 12.
Define population and community.
Answer:
i. Population:
Organisms of same kind inhabiting a geographical area constitute population.
OR
Individuals live in groups in a well-defined geographical area, share or compete for similar resources, potentially interbreed and thus form a population.
OR
Population is defined as a group of individuals of a species occupying a definite geographic area at a given time.

ii. Community: Several populations of different species in a particular area constitute community that interact with one another in several ways.

Question 13.
i. Auxin is called as a growth regulator. Justify.
ii. Name the standard bioassay method used for auxins.
Answer:
i. a. Plant growth regulators are organic compounds that inhibit, promote or modify the growth.
b. Auxins influence various aspects of growth, especially cell elongation. Hence, it is called a growth regulator.
ii. Avena Curvature Test is the standard bioassay method for auxins.

Question 14.
Distinguish between Habitat and Niche.
Answer:

Habitat	Niche
i. A habitat is an area, where a species lives and interact with the other factors and prosper.	Niche not only describes the position of a species in an environment but also describes the functional role played by an organism.
ii. Habitat deals with effects of temperature, rainfall and other abiotic factors.	Niche deals with the flow of energy from one organism to another.
iii. Habitat supports numerous species at a time.	Niche supports a single species at a time.
iv. Habitat is a physical place.	Niche is an activity or role performed by organisms.
v. Habitat is not species specific.	Niche is species specific.

Section – C

Attempt any EIGHT of the following questions: [24 Marks]

Question 15.
Name the following and give example wherever possible.
i. Female reproductive whorl of flower.
ii. Individual member of gynoecium is called as.
iii. Flower in which gynoecium possesses many free carpels is called as
iv. Flower in which gynoecium possesses many fused carpels is called as
v. Ovary with single ovule.
vi. Ovary with many ovules.
Answer:
i. Gynoecium (Pistil)
ii. Carpel (megasporophyll)
iii. Apocarpous (e.g. Michelia)
iv. Syncarpous (e.q. Brinjal)
v. Uniovulate (e.q. paddy, wheat and mango)
vi. Multiovulate (e.g. tomato and lady’s finger)

Question 16.
In the light of Griffith’s experiment, explain the action of two strains of Diplococcus pneumoniae and give his conclusion.
Answer:
i. In 1928, Frederick Griffith performed an experiment with Streptococcus pneumoniae (bacterium responsible for causing pneumonia).

ii. There are two strains of the bacteria:
a. S-strain whose cells produce a capsule of polysaccharides (capsulated). This strain is virulent (pathogenic) and causes pneumonia.
b. R-strain whose cells lack a capsule (non-capsulated) and is avirulent or non-pathogenic.

iii. Griffith performed his experiment by injecting the above strains of bacteria into mice and found the following results:
a. R-strain (non-virulent) bacteria were injected into mice, the mice suffered no illness because R-strain was non-pathogenic.
b. S-strain (virulent) bacteria were injected into mice, the mice developed pneumonia and finally died.
c. Then, Griffith injected heat killed S-strain bacteria into mice and they survived.
d. A mixture of living R-strain and heat killed S-strain were injected into mice, the mice developed pneumonia and died.

iv. He concluded that some genetic factor (transforming principle) from heat killed S-strain cells transformed live R-strain cells into live S-strain and produced the disease.

v. Thus, Griffith’s experiment helped to identify the transformation of aenetic material.

Question 17.
i. Define isolating mechanism.
ii. Describe geographical isolation.
Answer:
i. A barrier which prevents gene flow or exchange of genes between isolated populations, is called isolating mechanism.

ii. Geographical Isolation:

It is also called as physical isolation. It occurs when an original population is divided into two or more groups by geographical barriers such as rivers, oceans, mountains, glaciers, etc. These barriers prevent interbreeding between isolated groups.

The separated groups are exposed to different kinds of environmental factors and they acquire new traits by mutations.

The separated populations develop distinct gene pool and they do not interbreed.
Thus, new species have been formed by geographical isolation, e.g. Darwin’s Finches.

Question 18.
Root pressure is not the sole mechanism explaining the ascent of sap in all plants of varying heights. Give reason.
Answer:
Although, ascent of sap takes place due to root pressure, there are certain limitations to root pressure theory:

1. It is not applicable to Dlants taller than 20 meters.
2. Ascent of sap can also occur even in the absence of root system.
3. Root pressure value is almost nearly zero in taller gymnosperm trees.
4. In actively transpiring plants, no root pressure is developed.
5. Xylem sap under normal condition is under tension i.e. it shows negative hydrostatic pressure or high osmotic pressure.

Thus, root pressure is not the sole mechanism explaining the ascent of sap in all plants of varying hight.

Question 19.
Explain Absolute and Relative Growth Rate by comparing the growth of two leaves per day in the pictures given below.

Answer:

1. In the given figure, two leaves ‘A’ and ‘B’ are of different sizes but show some absolute increase in area in a given time.
2. Both leaves grow and increase their area by 5cm² /day to produce A’ and B’ leaves. Thus their absolute growth rate is same i.e. 5cm²/ day.
3. Leaf A of 5cm² in size grows 5cm²/ day thus, its RGR would be 100%.
4. Whereas leaf B of 50cm² in size grows 5cm²/ day thus, its RGI would be 10%.
5. Thus, leaf A shows more Relative Growth late as compared to leaf B.

Question 20.
Complete the following flow diagram of double circulation:

Answer:

Question 21.
Describe the chemical nature of hormones.
Answer:
Chemical nature of hormones:

1. Amines: These are simple amines. Catecholamines secreted by adrenal medulla, epinephrine and non-epinephrine and melatonin from pineal gland. Some are modified from amino acids.
   e.g. Thyroxine.
2. Peptide harmones: These hormones consist of long or short chains of amino acids.
   e.g. Hormones of hypothalamus oxytocin, ADH, GnRH.
3. Protein hormone: Insulin, glucagon TSH, FSH, LTH, GH, relaxin.
4. Fatty acid derivatives: Prostaglandin
5. Steroid hormones; These hormones are lipid soluble and derived from cholesterol and other steroids. Action of these hormones is concerned with long lasting responses. e. g. Oestrogen, testosterone, al dosterone.
6. Gas: NO (Nitric Oxide)

Question 22.
Draw a flowchart representing the sewage treatment.
Answer:

Question 23.
Identify labels (A), (B), and (C) in the following diagram representing the plasmid cloning vector.

Answer:
(A) On (Origin of replication)
(B) (ampr) Ampicillin resistance gene
(C) Polylinker

Question 24.
Label the different stages (i to vi) of hydrarch succession of plants.

Answer:
i. Phytoplankton stage
ii. Submerged plant stage
iii. Submerged and free floating plant stage
iv. Reed-swamp stage
v. Marsh-meadow stage
vi. Scrub stage

Question 25.
Enlist the functions of CSF.
Answer:
Functions of CSF:

1. The meninges and CSF act as a shock absorber and protect the brain and spinal cord from mechanical injuries.
2. It also maintains constant pressure inside cranium.
3. It helps in exchange of nutrients and wastes between blood and brain tissue.
4. It helps in the supply of oxygen to the brain.
5. It protects the brain from desiccation.

Question 26.
Explain the Rivet Popper Hypothesis.
Answer:
i. Paul Ehrlich, an ecologist from Stanford gave an analogy to explain significance of diversity. It is called Rivet Popper Hypothesis.

ii. He compared aeroplane to ecosystem and the species as rivets that keep all parts of the aeroplane together. There are thousands of rivets needed to hold all the parts of the aeroplane together.

iii. If each passenger decides to pop even one rivet, initially, not much of the turbulence will be experienced but slowly, as number of popped rivets will increase, there will be a serious threat to the safety of the aeroplane. Also, which rivets are removed will also matter. Suppose, rivets at key positions such as the ones that bind the wings to the body of the aeroplane, are removed, situation will become serious.

iv. Thus, we can say that relationship between diversity and well-being of ecosystem is not linear. But, it is certain that loss of species may not pose threat to the ecosystem only initially. However, loss of key species will certainly cause threat in very short span of time. It will affect food chains, food web, energy flow, natural cycles, etc. In short, it will affect the balance of ecosystem.

Section – D

Attempt any THREE of the following questions: [ 12 Marks]

Question 27.
Complete the given flow chart and explain each type with the help of neat and labelled diagram.

Answer:
A: Endosperm; B: Nuclear Type; C: Cellular Type; D: Helobial Type
i. Nuclear Type:

a. It is the most common type of endosperm found in 161 angiospermic families.

b. In this, the primary endosperm nucleus repeatedly divides mitotically without wall formation. Due to which large number of free nuclei are formed.
c. A big central vacuole appears in the centre of cell which pushes the nuclei towards the periphery.
d. Later, wall formation occurs between the nuclei, hence multicellular endosperm is formed.
e. But in several cases cell wall formation remains incomplete, e.g. wheat, sunflower and coconut.
f. Coconut has multicellular endosperm in the outer part and free nuclear as well as vacuolated endosperm in the centre.

ii. Cellular Type:

a. In some plants, division of triploid primary endospermic nucleus is immediately followed by wall formation.

b. So that the endosperm is cellular right from the beginning.
c. It is mostly observed in 72 families of dicots as in members – Balsam, Petunia, Adoxa, etc.

iii. Helobial Type:

a. It occurs in the order Helobiales of monocotyledons.
b. In this case, first division of primary endosperm nucleus is followed by a transverse wall, which divides the cell unequally.
c. The smaller cell is called chalazal cell and larger cell is the micropylar cell.
d. The nuclei in each cell divide by free nuclear divisions and then walls develop between nuclei in micropylar chamber.
e. It is intermediate between cellular and nuclear type endosperm e.q. Asphodelus.

Question 28.
Identify the given diagram and explain it’s histological structure.

Answer:
The given diagram is the histological structure of ovary.
The histological structure of the ovary shows the different stages of development of the oocyte in the ovary. Each ovary is a compact structure differentiated into a central part called medulla and the outer part called cortex.

i. Medulla: The stroma or loose connective tissue of the medulla has blood vessels, lymph vessels, and nerve fibres.

ii. Cortex

a. The cortex is covered externally by a layer of germinal epithelium.
b. The outer cortex is more compact, granular and shows large number of tiny masses of cells called ovarian follicles.
c. Ovarian follicles are formed from the immature ova originating from cells of the dorsal endoderm of the yolk sac.
d. The cells migrate to the gonadal ridge during embryonic development and divide mitotically.
Now these cells are called oogonia.
e. As the oogonia continue to grow in size they are surrounded by a layer of granulosa cells. This assembly forms the rudiments of the ovarian follicles.

iii. The ovary has more than two million primordial follicles in it.

iv. Each primordial follicle has, at its center a large primary oocyte (2n) surrounded by a single layer of flat follicular cells.

v. The primary oocyte starts with its meiotic division but gets arrested at meiosis I.

vi. A primordial follicle starts to grow, developing into primary follicles.

vii. Each primary follicle has multi-layered cuboidal follicular cells.

viii. The stroma cells add theca over the follicle, which then changes into a secondary follicle.

ix. The secondary follicle grows into the Graafian follicle by addition of more follicular cells.

x. The egg is released from the Graafian follicle during ovulation and the remaining part of the follicle changes into a temporary endocrine gland called corpus luteum.

xi. If fertilization does not take place the corpus luteum degenerates into a white scar called corpus albicans.

Question 29.
Identify (i), (ii), (iii) and (iv) and redraw the following chart representing the sex determination in honey bees. Name and explain in detail the type of reproduction observed in honey bees.

Answer:

1. In honey bees, chromosomal mechanism of sex determination is of haplo-diploid type.
2. In this type, sex of individual is determined by the number of set of chromosomes received.
3. Females are diploid (2n=32) and males are haploid (n=16).
4. The female produces haploid eggs (n=16) by meiosis and male produces haploid sperms (n=16) by mitosis.
5. If the egg is fertilized by sperm, the zygote develops into a diploid female (2n=32) (queen and worker) and unfertilised egg develops into haploid male (n=16) (Drone) by way of parthenogenesis.
6. The diploid female gets differentiated into either worker or queen bee depending on the food they consume during their development.
7. Diploid larvae which get royal jelly as food develops into queen (fertile female) and other develops into workers (sterile females).

Question 30.
Explain the structure of pituitary gland with the help of neat and labelled diagram.
Answer:
Pituitary gland: It is the smallest pea sized, reddish green coloured gland that controls almost all other endocrine glands.

Structure: The pituitary gland shows two lobes called anterior lobe (adenohypophysis) and posterior lobe (neurohypophysis).

i. Adenohypophysis: It is an outgrowth from the roof of buccal cavity. This outgrowth is called Rathke’s pouch. It grows upwards towards the brain. It is the larger lobe of pituitary gland. It is a highly cellular and vascular part of pituitary gland. It contains various types of epitheloid secretory cells, acidophils, basophils, chromatophores.

It is further divided into three parts – Pars distalis, pars tuberalis and pars intermedia. Pars intermedia is poorly developed in human beings. It is a small reduced pact dying in the cleft between the anterior and posterior lobe. It secretes Melanocyte Stimulating Hormone (MSH) in some lower vertebrates.

MSH stimulates the dispersion of melanin granules in melanocytes and is responsible for skin pigmentation.

ii. Neurohypophysis: The neurohypophysis grows as a downward extension of hypothalamus. Neurohypophysis is connected directly to the hypothalamus by axon fibres.
It is composed of three parts – Pars nervosa/ neural lobe, infundibulum and median eminence.

The pars nervosa acts as storage area for the secretions of hypothalamus. It stores and releases the hormones – oxytocin and vasopressin.

Question 31.
i. Identify the labels P, Q, R and S.

Answer:
P- Breathing control centers (pons and medulla),
Q – Phrenic nerves,
R – Intercostal nerves,
S – O₂ Sensors in artery (aortic arch)

ii. Explain the Hering Breuer reflex.
Answer:
a. During inspiration, when the lungs expand to a critical point, the stretch receptors are stimulated and impulses are sent along the vagus nerves to the expiratory centre.

It then sends out inhibitory impulses to the inspiratory center. The inspiratory muscles relax and expiration follows. As air leaves the lungs during expiration, the lungs are deflated and the stretch receptors are no longer stimulated. Thus, the inspiratory centre is no longer inhibited and a new respiration begins. These events are called the Herinci-Breuer reflex.

b. The Hering-Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

c. The respiratory centre has connections with the cerebral cortex which means the patterns of breathing can be changed voluntarily. Voluntary control is protective because it enables us to prevent water or irritating gases from entering the lungs. But the ability to stop breathing is also limited by the build-up of carbon dioxide in the blood.

Maharashtra Board Class 12 Biology Previous Year Question Papers