Maharashtra Board Class 12 Biology Sample Paper Set 5 with Solutions

Maharashtra State Board Class 12th Biology Sample Paper Set 5 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Biology Model Paper Set 5 with Solutions

Time : 3 Hrs.
Max. Marks : 70

General Instructions:

The question paper is divided into four sections.

1. Section A: Q. No. I 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. Pollination by birds is called as
(A) Ornithophily
(B) Anemophily
(C) Chiropterophily
(D) Entomophily
Answer:
(A) Ornithophily

ii. Rupturing of follicles and discharge of ova is known as ___.
(A) capacitation
(B) gestation
(C) ovulation
(D) copulation
Answer:
(C) Ovulation

iii. In a fully turgid cell, DPD is
(A) always one
(B) zero
(C) always negative
(D) both (B) and (C)
Answer:
(B) zero

iv. Which of the following bacteria are NOT involved in the process of nitrification?
(A) Nitrosomonas
(B) Nitrosococcus
(C) Nitrobacter
(D) Bacillus spp.
Answer:
(D) Bacillus spp.

v. ____ is a sound producing organ.
(A) Larynx
(B) Pharynx
(C) Tonsils
(D) Trachea
Answer:
(A) Larynx

vi. Coughing and sneezing are controlled by ____.
(A) Cerebrum
(B) Cerebellum
(C) Medulla oblongata
(D) Pons Varolli
Answer:
(C) Medulla oblongata

vii. ____ Which of the following diseases can be contracted by droplet infection?
(A) Malaria
(B) Chicken pox
(C) Pneumonia
(D) Rabies
Answer:
(C) Pneumonia

viii. ___ is a method in which crops are bred for having higher levels of vitamins, minerals and fats i.e. better nutritive value.
(A) Biofortification
(B) Bioinformatics
(C) Bioremediation
(D) Biomagnification
Answer:
(A) Biofortification

ix. Diethylcarbamazine is the drug useful in treatment of ____.
(A) Malaria
(B) Ascariasis
(C) Amoebiasis
(D) Filariasis
Answer:
(D) Filariasis

x. The muscular structure that separates the thoracic and abdominal cavity is ___.
(A) pleura
(B) diaphragm
(C) trachea
(D) epithelium
Answer:
(B) diaphragm

Question 2.
Answer the following: – [8 Marks]

i. Define clones.
Answer:
Morphologically and genetically identical individuals produced by asexual reproduction are called as clones.

ii. If variation occurs in a population by chance alone and not by natural selection and brings change in frequencies of an allele. What is it called?
Answer:
If the variation in a population occurs by chance alone and not by natural selection and brings about a change in frequencies of an allele, it is called as genetic drift.

iii. How we can calculate the length of DNA molecule?
Answer:
The length of DNA can be calculated by multiplying the number of base pairs with the distance between two consecutive base pairs.

iv. Define apical dominance.
Answer:
In most of the higher plants, growing apical bud inhibits the growth of lateral buds. This is called as apical dominance.

v. Define guttation.
Answer:
The loss of water in the form of liquid is called guttation.

vi. Name the organ which prevents the entry of food into the trachea while eating.
Answer:
Epiglottis prevents the following the entry of food into the trachea while eating.

vii. What is ecological hierarchy?
Answer:
The ecological grouping of organisms is known as ecological hierarchy.

viii. Identify the last three stages of secondary succession in a forest ecosystem after fire.
Forest → Forest fire → Severe forest fire → Total nudation of forest land → Terrestrial Plants → a. → b. → c.
Answer;
a. Shrubs,
b. Wood land stage,
c. Forest stage

Section – B

Attempt any EIGHT of the following questions:

Question 3.
Describe the two types of dichogamy.
Answer:
Dichogamy is divided into two types:

1. Protandry: In this type, anthers mature first, but the stigma of the same flower is not receptive at that time. e.g. in the disc florets of sunflower.
2. Protogyny: In this type, stigma of carpel matures earlier than anthers of the same flower, e.g. Gloriosa.

Question 4.
Observe the given pedigree chart and answer the following questions.

i. Identify whether the trait is sex linked or autosomal.
ii. Give an example of a trait in human beings which shows such a pattern of inheritance.
Answer:
i. The trait represented in given pedigree is sex linked trait.
ii. Haemophilia, colour blindness are the examples of sex linked traits in humans.

Question 5.
i. Define the term codogen.
ii. What will be the base sequence on the template strand of DNA, which codes for methionine?
Answer:
i. Codogen: It is the smallest possible sequence (triplet) of nucleotides present on DNA strand which can specify one particular amino acid.
ii. The base sequence on the template strand of DNA that codes for methionine is 3′-TAC-5′ because the mRNA codon for methionine is AUG, which is complementary to the template strand of DNA.

Question 6.
Enlist any four factors responsible for genetic variations.
Answer:
Factors responsible for genetic variations are:

1. Gene mutation
2. Genetic recombination
3. Gene flow
4. Genetic drift
5. Chromosomal aberrations [Any four factors]

Question 7.
Write any four properties of water which makes it a significant molecule that connects physical world with biological processes.
Answer:
Properties of water:

1. Water is in the liquid form at room temperature and is the best solvent for most of the solutes.
2. In pure form, it is inert inorganic compound with neutral pH. Due to this, water is the best transporting medium for dissolved minerals and food molecules.
3. It is best aqueous medium for all biochemical reactions occurring in the cells.
4. It is an essential raw material for photosynthesis.
5. Water has high specific heat, high heat of vaporization and high heat of fusion. Due to this, it acts as thermal buffer.
6. Water molecules have good adhesive and cohesive forces of attraction.
7. Due to high surface tension and high adhesive and cohesive force, it can easily rise in the capillaries.
   These properties of water make it a significant molecule that connects physical world with biological processes.

Question 8.
Write a short note on Donnan equilibrium.
Answer:
Donnan equilibrium:

1. It is based on the assumption that certain negatively charged ions (i.e. anions), after their entry into the cell, become fixed on the inner side of the cell membrane and cannot diffuse outside through the cell membrane.
2. Therefore, additional mobile cations are required to balance these fixed anions.
3. Concentration of cations becomes more due to accumulation.
4. The process continues till cations and anions become equal both inside and outside the cell membrane.
5. This kind of passive absorption of anions/ cations from cell exterior against their own concentration gradient in order to neutralize the effect of cations/ anions, is called Donnan equilibrium.

Question 9.
Distinguish between active immunity and passive immunity.
Answer:

Active Immunity	Passive immunity
i. When resistance is developed by individuals as a result of an antigenic stimulus it is called as active immunity.	When ready-made antibodies are directly given to protect body against foreign agents, immunity is called as ‘Passive immunity’.
ii. The types of active immunity are natural acquired active immunity and artificial acquired active immunity.	The types of passive immunity are natural acquired passive immunity and artificially acquired passive immunity.
iii. It has no side effects.	It may cause reaction.
iv. It provides relief only after long period.	It provides immediate relief.
v. It is long lasting immunity. e.g. Polio vaccine, CG vaccine, etc.	It is short-lived immunity. Rabies vaccine, maternal antibodies, etc.

Question 10.
Write the objectives considered for breeding programs used in biofortification.
Answer:
Following objectives are considered for breeding programs used in biofortification:

1. Increasing protein content and quality
2. Increasing oil content and quality
3. Increasing vitamin content
4. Increasing micronutrient content and quality

Question 11.
Draw neat and labelled diagram of steps involved in PCR.
Answer:

Question 12.
Give reason: Adaptations in living being with different environments serve as basis for diversity.
Answer:

1. The diversity is with respect to size (microscopic to macroscopic), shape, colour, form, mode of nutrition, type of habitat, reproduction, motility, duration of life cycle span, etc.
2. This is due to the attempt of living beings to accommodate with different environmental conditions (like climatic, edaphic, topographic, geographic, etc.) or situations, solely for their survival and perpetuation.
3. In doing so, living organisms adapt themselves to overcome different situations and thus develop distinct but different features and that has led to the diversity in them.
4. The diversity in features becomes infused in the life cycle.
   Hence, adaptations in living being with different environments serve as basis for diversity.

Question 13.
Complete the following table.

Answer:

Question 14.
How Indian culture and traditions helped in bio-diversity conservation?
Answer:

1. Indian culture and traditions are always connected with nature, and rituals are laid down to protect biodiversity.
2. In many cultures, stretches of forests were set aside and protected in the name of Almighty, which are called sacred groves.
3. Such sacred groves are found in Khasi and Jaintia hills in Meghalaya, Western ghat regions of Maharashtra and Karnataka, Aravalli hills of Rajasthan and Bastar, and Chanda and Sarguja areas of Madhya Pradesh.
4. Sacred groves serve the only chance of survival for some endangered varieties of animal and plant species. Tribals do not allow to cut even a single branch of tree from sacred grove.

Section – C

Attempt any EIGHT of the following questions:

Question 15.
Give a brief account on syphilis with respect to the following points:
i. Causative agent
ii. Incubation period
iii. Site of infection
iv. Symptoms
v. Preventive measures
vi. Treatment
Answer:
i. Causative agent: Treponema pallidum (Bacteria)
ii. Incubation period: 3-4 weeks
iii. Site of infection: Mucous membrane in genital, rectal and oral region.
iv. Symptoms: Primary lesion called chancre at the site of infection. Chancre is formed on external genitalia, skin rashes and mild fever, inflamed joints, loss of hair. Paralysis, Degenerative changes occur in the heart and brain.
v. Preventive measures: Education about sex practices, sex hyqiene, avoidinq sex with unknown partner or multiple partners, using condom during coitus.
vi. Treatment: Antibiotic-Penicillin

Question 16.
Explain any three characteristics of genetic code.
Answer:
Genetic code of DNA has following characteristics:

1. Genetic code is a triplet code: Sequence of three consecutive bases constitutes a codon, which specifies one particular amino acid. Base sequence in a codon is always in 5′ → 3′ direction. In every living organism genetic code is a triplet code.
2. Genetic code has distinct polarity: Genetic code shows definite polarity i.e. direction. It is always read in 5′ → 3′ direction and not in 3′ → 5′ direction. Otherwise message will change e.g. 5′ AUG 3′
3. Genetic code is non-overlapping: Code is non-overlapping i.e. each single base is a part of only one codon. Adjacent codons do not overlap.
4. Genetic code is commaless: There is no gap or punctuation mark between successive/ consecutive codons.
5. Genetic code has degeneracy: Usually single amino acid is encoded by single codon. However, some amino acids are encoded by more than one codon, e.g. Cysteine has two codons, while isoleucine has three codons. This is called degeneracy of the code. Degeneracy of the code is explained by Wobble hypothesis. Here, the first two bases in different codons are identical but the third one, varies.
6. Genetic code is non-ambiguous: Specific amino acid is encoded by a particular codon. Alternatively, two different amino acids will never be encoded by the same codon.
7. Codon and anticodon: Codon is a part of DNA e.g. AUG is codon. It is always represented as 5′ AUG 3′. Anticodon is a part of tRNA. It is always represented as 3’UAC 5′.
8. Initiation codon and termination codon: AUG is always an initiation codon in any and every mRNA. AUG codes for amino acid methionine. Out of 64 codons, three codons viz. UAA, UAG and UGA are termination codons which terminate/ stop the process of elongation of polypeptide chain, as they do not code for any amino acid.
9. Genetic code is universal: In most of the living organisms the specific codon specifies same amino acid. e.g. Codon AUG always specifies amino acid methionine.

Question 17.
Which type of natural selection is represented in the following graph? Explain in detail.

Answer:
The given graph represents ‘Stabilizinq selection or Balancing selection’.

1. In this type of selection more individuals of a population acquire a mean character value.
2. It tends to favour the intermediate forms and eliminate both the phenotypic extremes. e.g. More number of infants with intermediate weight survive better as compared to those who are over weight or under-weight.
3. Stabilizing selection reduces variations.
4. It does not lead to evolutionary change but tends to maintain phenotypic stability within population, therefore, it is described as stabilizing selection.
5. Genetically, stabilizing selection represents a situation where a population has adapted to its environment.

Question 18.
Define turgor pressure (T.P) and wall pressure (W.P). How T.P is related to DPD and W.P?
Answer:

1. Turgor pressure (T.P) is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.
2. Cell wall being thick and rigid, exerts a counter pressure on the cell sap. This is called Wall pressure (W. P).
3. In a fully turgid cell, DPD is zero.
4. In a fully turgid cell, T. P. = W. P, but operating in opposite direction.

Question 19.
Complete the given chart representing stages of development and explain the concept of development.

Answer:
i. A: Cell division;
B: Cell elongation
C: Senescence

ii. Development:
a. It refers to the ordered or progressive changes in shape, form and degree of complexity.
b. It includes all the changes occurring in sequence from the germination of seed upto the senescence or death during life cycle of plants.
c. Thus development includes growth, morphogenesis, maturation and senescence.

Question 20.
i. Describe the structure of haemoglobin.
ii. What happens after the process of haemolysis?
Answer:
i. Structure of haemoglobin:
a. Each molecule of haemoglobin is a protein-iron complex.
b. It consists of four polypeptide (globin) chains 2 alpha and 2 beta chains. An iron-porphyrin (haem) group is attached to each chain and all four chains are bound together.

ii. After haemolysis, haemoglobin is broken down. Its globin part is broken to recycle the amino acids. Iron of haem group is stored as ferritin in the liver and porphyrin group of haem is converted into green pigment biliverdin and then into red-orange coloured bilirubin. These pigments (mainly bilirubin) are added to bile and finally removed out of body alonq with faeces.

Question 21.
Describe the T.S. of spinal cord.
Answer:

1. i. The spinal cord is dorso-ventrally flattened due to the presence of deep, narrow posterior fissure and shallow, broad anterior fissure. The fissures divide the spinal cord incompletely into a right and left side.
2. The fissures divide the grey matter into six horns, namely dorsal, lateral and ventral horns while the white matter is divisible into 6 columns or funiculi, namely dorsal, lateral and ventral funiculi.
3. The H-shaped or butterfly shaped grey matter is on the inner side, while the white matter is on the outer side.
4. The dorsal and ventral horns extend out of the spinal cord as dorsal root and ventral root of spinal cord respectively.
5. The dorsal root is connected to the dorsal root ganglion (lies just outside and lateral to the spinal cord). It has an aggregation/collection of unipolar sensory neurons.
6. A central canal can be seen in the centre.
7. The association or inter-neurons lie inside the grey matter. They receive signal from the sensory nerve, integrate it and direct the response towards the motor neurons lying towards the ventral horn. The lateral horns have neurons of autonomic nervous system (ANS). The nerves arisinq from these neurons emerge out from the ventral root of spinal nerve.
8. The white matter consists mainly of bundles of myelinated nerve fibre called ascending and descending tracts. The ascending tracts conduct sensory impulses from spinal cord to the brain and these lie in the dorsal column/funiculi. The descending tracts conduct motor impulses from brain to the lateral and ventral funiculi of spinal cord.

Question 22.
Complete the following table and explain plant breeding for developing resistance to insect pest.

Crop	Variety	Insect pest
Brassica	Pusa Gaurav	(A)
(B)	Pusa Sem 2, Pusa Sem 3	Jassids, Aphids, and fruit borer
Okra	(C)	Shoot and fruit borer

Answer:

Crop	Variety	Insect pest
Brassica	Pusa Gaurav	Aphids
Flat bean	Pusa Sem, Pusa Sem 3	Jassids, Aphids and fruit borer
Okra	Pusa Sawani, Pusa A – 4	Shoot and fruit borer

i. Insects being herbivores, incur heavy loss in the quantity and quality of crops.

ii. Resistance in crops can be developed by following ways:
a. Development of morphological characters like hairy leaves in cotton and wheat develop vector resistance from jassids and cereal leaf beetle, respectively.
b. Solid stem in wheat leads to resistance to stem borers.
c. Biochemical characters provide resistance to insects and pests. For example, the high aspartic acid, and low nitrogen and sugar content in maize, lead to resistance against stem borers.
d. The nectar-less cotton having smooth leaves develop resistance against bollworms.

Question 23.
Explain the mechanism underlying the insect resistance in plants produced using rDNA technology.
Answer:
The mechanism underlying the insect resistance in plants produced using rDNA technology is as follows:

1. Insect resistant plants contain either a gene from B. thuringiensis or the cowpea trypsin inhibitor gene.
2. The gene called cry gene present in B. thuringiensis produces a protein that forms crystalline inclusions in bacterial spores.
3. When ingested by a susceptible insect, a combination of high pH and the enzyme proteinase of the insect’s mid gut, processes them hydrolytically to release the core toxic fragments.
4. The effect of these fragments is seen within minutes of ingestion, beginning with mid gut paralysis and ending with disruption of mid gut cells of insect.
5. Bt toxin activity has been against many species of insects within the orders of Lepidoptera, Diptera, and Coleoptera.
6. Similarly, the gene of a-amylase inhibitor (aAl-Pv) has been isolated from adzuki bean (Phaseolus vulgaris) and transferred to tobacco and this gene works against pests like Zabrotes subfasciatus and Callosobruchus chinensis.

Question 24.
Describe any three important characteristics of a population.
Answer:
Some important characteristics of population are:

i. Population density: Population density tells us the number of individuals present per unit space, in a given time.
OR
Density of a population is the total number of individuals in that population present per unit area at a specific time.

ii. Natality: Natality is the birth rate of a population.

iii. Mortality: Mortality is the death rate of a population.

iv. Age distribution and Age pyramids:
a. A population consists of individuals with different ages. The entire population is divided into three age groups – pre-reproductive (0-14 years), reproductive (age 15-44 years), post reproductive (45-85+years). The relative proportion of individuals of various age groups in the population is referred to as age structure of the population.
b. If the age distribution (percent individuals of a given age or age group) is plotted for the population, the resulting structure is called as age pyramid.

v. Sex Ratio: Sex ratio is the ratio of the number of individuals of one sex to that of the other sex.

Question 25.
Elaborate on the following disorders of nervous system.
i. Parkinson’s disease
ii. Alzheimer’s disease
Answer:
i. Parkinson’s disease:
a. Degeneration of dopamine-producing neurons in the CNS causes Parkinson’s disease.
b. Symptoms are tremors,stiffness and difficulty in walking, balance and co-ordination.
c. The symptoms develop gradually over the years.

ii. Alzheimer’s disease:
a. It is the most common form of dementia.
b. Its incidence increases with the age, showing the loss of cognitive functioning – thinking, and remembering, reasoning and behavioural abilities to such an extent that it interferes with the person’s daily life and activities.
c. It occurs due to loss of cholinergic and other neurons in the CNS, accumulation of amyloid proteins.
d. There is no cure for Alzheimer’s, but treatment slows down the progression of the disease and may improve the quality of life.

Question 26.
Explain in detail Gross Primary Productivity and Net Primary Productivity (NPP):
Answer:
Primary productivity is divided into two types i.e. Gross Primary Productivity and Net Primary Productivity.

i. Gross Primary Productivity (GPP):
a. It is the rate of production of organic matter during photosynthesis.
b. Plants use considerable amount of GPP to carry out their own respiration.

ii. Net Primary Productivity (NPP):
a. It is the available biomass for the consumption to heterotrophs like herbivores, carnivores and decomposers.
b. The gross primary productivity (GPP) minus respiratory losses (R) constitutes the net primary productivity (NPP).
GPP – R = NPP
c. The annual net primary productivity of the whole biosphere is approximately 170 billion tons (dry weight) of organic matter.
d. Of the 170 billion tons, the productivity of the ocean is only 55 billion tons and rest is from land ecosystem.

Section – D

Attempt any THREE of the following questions:

Question 27.
i. Identify the types of pollination in the given plants.

Answer:
a. Figure A represents maize plant which shows anemophilous flowers. Thus it shows wind pollination (Anemophily).
b. Figure B represents male and female plant of Vallisneria. It shows epihydrophily type of water pollination (Hydrophily).

ii. Write a short note on entomophily.
Answer:
Entomophily:
a. Pollination carried out by insects is called as entomophily.
b. Entomophilous flowers show following adaptations:

1. They are large, showy and often brightly coloured.
2. The flowers produce sweet odour (smell) and have nectar glands.
3. The stigma is rough due to presence of hair or is sticky due to mucilaginous secretion.
4. The pollen grains are spiny and surrounded by a yellow sticky substance called pollenkitt.

c. Entomophily commonly occurs in Rose, Jasmine, Cestrum, Salvia, etc.

Question 28.
With the help of suitable diagram describe the permanent methods of birth control to avoid pregnancy.
Answer:
Permanent Methods of birth control to avoid pregnancy:

1. The permanent birth control method in men is called vasectomy and in women it is called tubectomy.
2. These are surgical methods, also called sterilization.
3. In vasectomy a small part of the vas deferens is tied and cut.
4. In tubectomy, a small part of the fallopian tube is tied and cut.
5. This blocks gamete transport and prevent pregnancy.

Question 29.
Sketch and label the heart wall and pericardium. Explain the structure of heart wall.
Answer:

Structure of heart wall:

1. It shows three layers, viz. epicardium, myocardium and endocordium.
2. The outer layer is the epicardium. middle myocordium and inner endocardium.
3. Epicardium is thin and formed of a single layer of flat squamous epithelium resting on basement membrane.
4. Myocardium is the middle thick layer formed of cardiac muscles.
5. Endocardium is a single thin layer formed of squamous epithelium.
6. The epicardium and endocardium are protective in function whereas myocardium is responsible for contraction and relaxation of heart.

Question 30.
Identify the types of neuroglial cells (i-iv) and explain each one of them.

Answer:

1. Ependymal cells: They are found in CNS. These cells form single layer of squamous or columnar cells, often ciliated epithelial cells lining the ventricles or brain cavities and central canal of spinal cord. They are mainly responsible for production and probably also for circulation of CSF in brain ventricles and central canal.
2. Oligodendrocytes: They are found in central nervous system. These cells have few branches. Oligodendrocytes mainly form myelin sheath around the central axons, which constitutes the white matter of CNS.
3. Satellite cells: These cells are found in PNS. They support the functions of neurons.
4. Astrocytes: These are star shaped and the most abundant glial cells of CNS. They have varied roles in the brain- secretion and absorption of neural transmitter, maintenance of blood-brain barrier (BBB) and regulation of the transmission of electrical impulses with the brain.

Question 31.
Study and identify cross I and cross II and explain the experiment.

Answer:
The given Cross I and Cross II explains Morgan’s experiment of linkage and recombination.

1. Morgan carried out dihybrid crosses in Drosophila to study sex-linked genes.
2. He crossed yellow bodied, white eyed female with brown bodied red eyed males.
3. He inter-crossed their F1 progeny.
4. He observed that the two genes did not segregate independently of each other and he got the 9 : 3 : 3 : 1 ratio in the F2 generation.
5. Morgan and his team knew that genes were located on the X chromosome. They concluded that when two qenes in a dihybrid cross were located on the same chromosome, the parental gene combinations were much higher than non-parental type. This occurs due to the physical association or linkage of two genes.
6. He also concluded that when genes are grouped on the same chromosome, some genes were tightly linked, while others were loosely linked.
7. Tightly linked genes show very few recombinations (1.3%).
8. When genes are loosely linked i.e. present far away from each other on chromosome, they show more (higher) recombinations (37.2 %).
9. The genes for yellow body and white eye were strongly linked and showed only 1.3% recombination.
10. White bodied and miniature wings showed 37.2% recombination.

Maharashtra Board Class 12 Biology Previous Year Question Papers