Maharashtra Board Class 12 Biology Sample Paper Set 3 with Solutions

Maharashtra State Board Class 12th Biology Sample Paper Set 3 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Biology Model Paper Set 3 with Solutions

Max. Marks : 70
Time : 3 Hrs.

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. When pollen tube enters ovule through integuments it is called as
(A) syngamy
(B) porogamy
(C) chalazogamy
(D) mesogamy
Answer:
(D) mesogamy

ii. A colour blind man marries a woman, who is homozygous for normal colour vision, the probability of their son being colour blind is –
(A) 0%
(B) 25%
(C) 50%
(D) 100%
Answer:
(A) 0%

iii. Which of the following represents frequency of homozygous recessive individual in Hardy – Weinberg equation?
(A) p²
(B) pq
(C) q²
(D) 2pq
Answer:
(C) q²

iv. What is the minimum number of plasma membranes that oxygen has to diffuse across to pass from air in the alveolus to haemoglobin inside a RBC?
(A) Two
(B) Three
(C) Four
(D) Five
Answer:
Answer:
(D) Five

v. Abscisic acid controls
(A) cell division
(B) leaf fall and dormancy
(C) shoot elongation
(D) cell elongation and wall formation
Answer:
(B) leaf fall and dormancy

vi. Hormones thyroxine, adrenaline and nor adrenaline are formed from .
(A) Glycine
(B) Arginine
(C) Ornithine
(D) Tyrosine
Answer:
(D) Tyrosine

vii. Removal of large pieces of floating debris, oily substances, etc. during sewage treatment is called
(A) primary treatment
(B) secondary treatment
(C) final treatment
(D) amplification
Answer:
(A) primary treatment

viii. Which of the following is responsible for activation of protoxin to active Bt toxin of the Bacillus thuringiensis in boll worm?
(A) Alkaline pH of gut
(B) Acidic pH of stomach
(C) Body temperature
(D) Moist surface of midgut
Answer:
(A) Alkaline pH of gut

ix. The entire sequence of communities that changes successively in a given area is called ____.
(A) xerarch succession
(B) sere
(C) pioneer community
(D) climax community
Answer:
(B) sere

x. Which of the following refers to ‘Terror of Bengal?
(A) Algal bloom
(B) Water hyacinth
(C) Increased BOD
(D) Eutrophication
Answer:
(B) Water hyacinth

Question 2.
Answer the following: [8 Marks]

i. What is artificial hybridization?
Answer:
Artificial hybridization is the process in which only desired pollen grains are used for pollination and fertilization.

ii. How does the onset of puberty differ for males and females?
Answer:
The onset of puberty usually occurs in males at the age 12-15 years and in females at the age 10-14 years.

iii. Due to which properties water acts as a thermal buffer?
Answer:
Water has hiqh specific heat, hiqh heat of vaporization and hiqh heat of fusion. Due to this, it acts as thermal buffer.

iv. Give the name of meristem primarily responsible for increasing length of internodes.
Answer:
Intercalary meristem is primarily responsible for increasing length of internodes.

v. Observe the given diagram of lymphatic system. Identify the labels.

Answer:
a – Tonsil,
b – Spleen

vi. Name the type of hormones binding to DNA and alter gene expression.
Answer:
Steroids
[Note: Steroid hormone receptors are proteins that have a binding site for a particular steroid molecule. Their response elements are DNA sequences that are bound by the complex of the steroid bound to its receptor.]

vii. Identify bX in the given figure.

Answer:
In the given figure’X’represents tube nucleus.

viii. What is sex ratio?
Answer:
Sex ratio is the ratio of the number of individuals of one sex to that of the other sex.

Section – B

Attempt any EIGHT of the following questions:

Question 3.
Describe the given picture with respect to pollination.

Answer:

1. The given picture shows lever mechanism or turn-pipe mechanism in Salvia.
2. Salvia is an entomophilous plant, thus pollination occurs through insects.
3. Salvia plants have special adaptations for the insect visitor to help in cross pollination.

Question 4.
Explain the statement: Test cross is back cross but back cross is not necessarily a test cross.
Answer:

1. In back cross F₁ generation can be crossed with either dominant or recessive parent.
2. But in test cross, F₁ qeneration is crossed with recessive parent only.
3. Thus, in back cross if F₁ generation is crossed with recessive parent it will be a test cross, but if F₁ generation is crossed with a dominant parent it will not be a test cross.

Therefore, test cross is back cross but back cross is not necessarily a test cross.

Question 5.
Differentiate between X and Y chromosomes.
Answer:

X chromosome	Y Chromosome
i. It is metacentric, hence appears X shaped.	It is acrocentric, hence appears Y shaped.
ii. It is longer than ‘Y’ chromosomes.	It is shorter than ‘X’ chromosomes.
iii. It contains large amount of euchromatin and small amount of heterochromatin.	It contains large amount of heterochromatin and small amount of euchromatin.
iv. It is found in both male and females.	It is found only in males.
v. Non-homologous part of X chromosome shows more genes than Y chromosome.	Non-homologous part of Y chromosome contains few genes as compared to X chromosome.
vi. X – linked genes are present on X chromosome.	Y — linked genes (Holandric genes) are present on Y chromosome.
vii. Genes present on X chromosome show crisscross inheritance.	Genes present on Y chromosome show straight inheritance.

Question 6.
Complete the flow chart on central dogma.

Answer:
(i) – Replication;
(ii) – Transcription;
(iii) – Translation;
(iv) – Protein

Question 7.
Draw neat and labelled diagram of root showing various zones.
Answer:

Question 8.
Identify the type of growth shown in the given figures.

Answer:
Figure ‘A’ represents Arithmetic growth and figure ‘B’ represents Geometric growth.

Question 9.
Mention the causes, symptoms and treatment of emphysema.
Answer:
Emphysema: It is characterized by destruction of walls of alveoli. People suffering from emphysema should quit smoking, avoid polluted water
Symptoms: Shortness of breath
Causes: Smoking, air pollution
Treatment: Administer oxygen

Question 10.
Give reason – Injury to medulla oblongata may prove fatal.
Answer:

1. Medulla oblongata is a part of brain stem.
2. It controls involuntary vital functions like heartbeat, respiration, vasomotor activities and peristalsis.
3. It also controls non vital reflex activities like coughing, sneezing, swallowing, vomiting, yawning etc.
4. Thus, damage or injury to medulla oblongata may disrupt these vital functions.

Therefore, injury to medulla oblongata may prove fatal.

Question 11.
i. What is a carcinogen?
ii. Name one chemical carcinogen with its target tissue.
Answer:
Factors that are known to cause cancer are called carcinogens.

Carcinogen	Organ affected
i. Soot	Skin, lungs
ii. Coal tar (3, 4 benzpyrene)	Skin, lungs
iii. Cigarette smoke (N-nitrosodimethylamine)	Lungs
iv. Cadmium oxide	Prostate gland
v. Aflatoxin (a metabolite of Aspergillus flavus, a mould)	Liver
vi. 2 – naphthylamine and 4 – aminobiphenyl	Urinary bladder
vii. Mustard gas	Lungs
viii. Nickel and chromium	Lungs
ix. Asbestos	Lungs
x. biethyistilbestrol (bES)	Vagina
xi. Vinyichloride (VC)	Liver

Question 12.
Write any four advantages of biofertilizers.
Answer:
Advantages of Biofertilizers:

1. Low cost and can be used by marginal farmers.
2. Free from pollution hazards.
3. Increase soil fertility.
4. BGA as biofertilizers secret growth promoting substances, organic acids, proteins and vitamins.
5. Azotobacter supply nitrogen and antibiotics in the soil.
6. Biofertilizers increase physico-chemical properties of soil- like texture, structure, pH, water holding capacity of soil by providing nutrients and organic matter. [Any four advantages]

Question 13.
Write the information on Ascariasis with respect to the following points:
i. Diagnosis
ii. Treatment
Answer:
i. Diagnosis: Diagnosis can be done by microscopic examination of the stool. [1 Mark]
ii. Treatment: Anti-helminthic drugs like Piperazine, Mebendazole, Levamisole, and Pyrantel are effective against Ascaris lumbricoides.

Question 14.
Write the applications of transgenic mice.
Answer:
Applications of transgenic mice:

1. Transgenic mice that have been modified using a particular oncogene (cancer causing gene) and thus developed a certain type of cancer, is useful to answer questions concerning the relationship between oncogenes and cancer development.
2. Theoretically, such animals can also be used for research into cancer treatment and prevention of malignancy.
3. In laboratory one such transgenic mouse model for the investigation of the breast cancer was developed. The oncogenes myc and ras were analysed to find out if they lead to breast cancer in mice transformed with these genes.

Section – C

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

Question 15.
Draw neat and labelled diagram representing development of female gametophyte.
Answer:

Question 16.
i. Identify A, B, C, D, in the given diagram of human reproductive system:

Answer:
A: Testis, B: Vas deferens, C: Seminal vesicle, D: Prostate gland [2 Marks]

ii. What is cryptorchidism?
Answer:
Failure of testis to descend into scrotum is called cryptorchidism.

Question 17.
Satish is a colorblind boy. His mother has normal vision but his maternal grandfather is colourblind. His father and maternal grandmother have normal vision. Explain the pattern of inheritance with a suitable chart.
Answer:

Pattern of inheritance:
Criss-cross inheritance: The inheritance of characters from the father to his grandson through his daughter is called criss-cross inheritance. In the given situation, since the father has normal vision, the mother must be a carrier.
Also, as the maternal grandmother has normal vision but the maternal grandfather is colour blind, the mother (Satish’s mother) is a carrier.

Question 18.
Explain the concept of natural selection by taking industrial melanism as one example.
Answer:

1. Natural selection encourages those genes or traits that assure highest degree of adaptive efficiency between population and its environment.
2. Industrial melanism is one of the best examples for natural selection.
3. In Great Britain, before industrialization (1845) grey white winged moths (Biston betularia) were more in number than black-winged moth (Biston carbonaria).
4. These moths are nocturnal and during day time they rest on tree trunk.
5. White-winged moths were camouflaged (hide in the background) well with the lichen covered trees that helped them to escape from the predatory birds.
6. However, the black-winged moth resting on lichen covered tree trunks were easy victims for the predatory birds and their number was reduced.
7. During industrial revolution, large number of industries came up in Great Britain.
8. The industries released black sooty smoke that covered and killed the lichens growing on tree and turn the tree black due to pollution.
9. This change became an advantage to the black-winged moths that camouflaged well with the black tree trunks and their number increased.
10. The white-winqed moths however became victims to predatory birds due to which their number reduced.

Thus, natural selection has resulted in the establishment of a phenotypic trait in the changing environmental conditions.

Question 19.
With the help of a neat, labelled diagram describe the structure of root hair.
Answer:
Structure of root hair:

1. Root hair is cytoplasmic extension (prolongation) of epiblema cell.
2. Each root hair may be approximately 1 to 10 mm long and tube like structure.
3. It is colourless, unbranched, short-lived (ephemeral) and very delicate.
4. It has a large central vacuole surrounded by thin film of cytoplasm, plasma membrane and thin cell wall, which is two layered.
5. Outer layer is composed of pectin and inner layer is made up of cellulose.
6. Cell wall of a root hair is freely permeable but plasma membrane is selectively permeable.

Question 20.
i. What is cardiac output?
ii. A man’s pulse rate is 70 and cardiac output is 5500 cm³. Find the stroke volume.
Answer:
i. Cardiac output is the volume of blood pumped out per minute.
ii. Cardiac output = Stroke volume × Heart rate
∴ 5500 = Stroke volume × 70
Stroke volume = 5500/ 70 = 78.57 mL

Question 21.
What would be expected to happen if:
i. GA₃ is applied to rice seedlings
ii. a rotten fruit is kept with unripe fruits
iii. a research student forgets to add cytokinin to the culture medium.
Answer:
i. If GA₃ (Gibberellic acid) is applied to rice seedlings, the rice seedlings will show increase in height due to internode elongation.
ii. If a rotten fruit gets mixed with unripe fruits, then ethylene gas released from the rotten fruits will speed up the ripeninq process of the unripe fruits.
iii. If a research student forgets to add cytokinin to the culture medium, then it will affect the cell division, differentiation and growth of culture.

Question 22.
Draw a flow chart of various steps involved in tissue culture.
Answer:
Steps involved in tissue culture:

Question 23.
i. Identify ‘P’, ‘Q’, ‘R’ and ‘S’ in the given structure.

Answer:
P – Cochlea,
Q – Auditory canal,
R – Semicircular ducts,
S – Cochlear nerve

ii. How does the inner ear control the body balance?
Answer:
Vestibular apparatus is composed of three semi-circular canals filled with endolymph and the utriculo-saccular region with the otolith organ. All three semi-circular canals lie in different planes at right angle to each other. The base of each of the canal has an ampulla in which there is a sensory spot called crista. The cristae help in maintaining equilibrium.

Question 24.
With the help of suitable diagram describe the logistic population growth curve.
Answer:

1. Resources like food and space are not always unlimited. They may be plenty in the beginning; but as the population density increases, competition for those resources starts, resulting in slowdown in the rate at which the original population was growing. This results in logistic or sigmoid growth curve.
2. Competition between individuals for limited resources will weed out the ‘weaker’ ones. Only the ‘fittest’ individuals will survive and reproduce.
3. A given habitat has enough resources to support a maximum possible number, beyond which no further growth is possible. This limit can be called as nature’s carrying capacity (K) for that species in that habitat.
4. A population growing in a habitat with limited resources show initially a lag phase, followed by phases of acceleration and deceleration and finally an asymptote, when the population density reaches the carrying capacity.
5. A plot of population density (N) in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth.
6. Since resources for growth of most animal populations, are finite and become limiting sooner or later, the logistic growth model is considered a more realistic one.

Question 25.
What are the types of ecological succession based on water availability?
Answer:

1. The amount of water determines the pattern of ecological succession in the given area.
2. Depending on the availability of water, there are two types of succession – hydrarch succession (hydrosere) and xerarch succession (xerosere).
3. Hydrarch succession takes place in wetter areas and the successional series progress from hydric to the mesic conditions.
4. Xerarch succession takes place in dry areas and the series progress for xeric to mesic conditions.
5. Hence, both hydrarch and xerarch successions lead to medium water conditions (mesic) – neither too dry (xeric) nor too wet (hydric).

Question 26.
i. Define Biomagnification.
ii. Draw flow chart showing biomagnification of DDT in the food chain.
Answer:
i. The phenomenon throuqh which certain pollutants qet accumulated in tissues in increasing concentration along the food chains (successive trophic levels) is called Biological Magnification or Biomagnification.

ii. Biomagnification of DDT:

Section – D

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

Question 27.
Observe the diagram and answer the questions given below.

Answer:
i. ‘X’ represents the placenta:

a. Placenta is a flattened, discoidal organ in the uterus of a pregnant woman.
b. It is a temporary structural and functional connection between foetal and maternal circulation.
c. It is attached to the wall of the uterus and to the baby’s umbilical cord.
d. Placenta is the only organ, which is formed of tissues from two different individuals- the mother and the foetus.
e. Part of the placenta contributed by the foetus is foetal placenta. It is called as the chorionic villi.
f. The other part which is rich in blood supply shared by the mother. It is a part of uterine wall, termed as maternal placenta. So, human placenta is called haemochorial.
g. The umbilical cord is formed of three blood vessels. Of these three blood vessels, two are small arteries which carry blood towards the placenta and one is a large vein which returns blood to the foetus. [2 Marks]

ii. Functions of the placenta:

a. The placenta facilitates the supply of oxygen, nutrients, hormones, antibodies to the foetus and also removal of carbon dioxide and excretory wastes produced by the foetus.

b. The placenta also acts as an endocrine tissue:
It produces hormones like hCG, proqesterone, oestroqen and relaxin. Relaxin is secreted by the placenta in the later phase of pregnancy.

Level of hCG increases upto the end of first trimester and then it declines.
By the end of first trimester progesterone is produced by placenta.
These hormones are required for foetal growth and maintenance of pregnancy.

Question 28.
On a crime scene in a hotel, forensic investigators found the blood spots and hairs near the dead body of a person. Police arrest the suspected individuals. Which technique would be applied by forensic investigators to identify and confirm the criminal? Explain the technique in detail.
Answer:
The technique used by forensic investigators to identify and confirm the criminal is DNA finqerprinting.

The steps of the DNA fingerprinting technique are as follows:

i. Isolation of DNA: The DNA must be isolated from the cells or tissues of the body (host). Only small amount of tissue like blood, hair roots, skin, etc. is required.

ii. Restriction digestion: The isolated DNA is treated with restriction enzymes.
The restriction enzymes cut the DNA into small fragments having variable lengths.
This phenomenon is called Restriction Fragment Length Polymorphism (RFLP).

iii. Gel electrophoresis: The DNA samples are loaded for agarose gel electrophoresis under an electric influence.
The DNA fragments, which are negatively charged move to the positive pole.
The movement of these fragments depends on length of the fragments. This results in formation of bands. dsDNA splits into ssDNA by alkali treatment.

iv. Southern blotting: It is a technique used for detecting specific DNA. The separated DNA fragments are transferred to a nylon membrane or a nitrocellulose filter paper by placing it over the gel and soaking them with filter paper overnight.

v. Selection of DNA probe: A known sequence of single- stranded DNA is prepared. It is called DNA Probe. DNA Probe is obtained from organisms or prepared by cDNA preparation method. The DNA probe is labelled with radioactive isotopes.

vi. Hybridization: Probe DNA is added to the nitrocellulose filter paper containing host DNA. The single-stranded DNA probe pairs with the complementary base sequence of the host DNA strand. As a result, DNA-DNA hybrids are formed on the nitrocellulose filter paper. Remaining single stranded DNA probe fragments are washed off.

vii. Photography: The nitrocellulose filter paper is photographed on an X-ray film by autoradiography. The film is analysed to determine the presence of hybrid DNA.

Question 29.
Describe the conducting system of human heart.
Answer:
Conducting system of Heart:

i. The human heart is myogenic i.e. the heart is capable of generating a cardiac contraction independent of nervous input. It also shows auto rhythmicity due to the presence of specialized muscles known as nodal tissues. These nodal tissues are distributed in the heart.

ii. SA node or sinoatrial node: It is present in the upper right corner of the right atrium. It lies at the base of opening of superior vena cava. It acts as a pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction. SA node passes the contraction to the left ventricle and also to the AV node.

iii. AV node or atrio-ventricular node: It is present in the right atrial wall near the base of inter-atrial septum.
It acts as pace setter of heart.

iv. Bundle of His/ Tawara: Branches start from AV node and pass through inter- ventricular septum. Bundle of His forms two branches, the right and left bundles, one for each ventricle. These branches form network in ventricular walls and these are called Purkinje fibers.

Bundle of His and Purkinje fibers spread impulses in ventricles. As a result, both the ventricles contract simultaneously.

Question 30.
To produce insulin on industrial scale, human gene for insulin is isolated and cloned in a host bacterium. Then the final product is extracted and purified. Explain this in detail.
Answer:
The steps involved in gene cloning are as follows:

i. Isolation of DNA (gene) from the donor organism:
a. The desire gene to be cloned is obtained from the source organism (donor).
b. Initially the cells of the donor organism are sheared with the blender and treated with suitable detergent.
c. Genetic material from the donor is isolated and purified using several techniques.
d. Isolated DNA can be spooled on to a glass rod.

ii. Cutting of desired gene:
a. Isolated purified DNA is then cleaved by using restriction enzymes i.e. restriction endonucleases.
b. These enzymes cleave DNA at restriction sites and break the DNA into fragments.
c. There are several types of restriction endonucleases.
d. Cleaved DNA fragments have cohesive, sticky, staggered ends or blunt ends.
e. From cleaved DNA fragments, a fragment containing desired gene is isolated and selected for cloning. This is now called foreign DNA or passenger DNA.
f. A desired gene can also be obtained directly from genomic library or cDNA library.

iii. Insertion of desired foreign gene into a cloning vector (vehicle DNA):
a. The foreign DNA or passenger DNA is now inserted into a cloning vector or vehicle DNA.
b. The most commonly used clonina vectors are plasmids of bacteria and the bacteriophage viruses like lambda phage and M13.
c. The most commonly used plasmid is pBR322.
d. Plasmids are isolated from the vector organisms i.e. bacterium.
e. By using same restriction enzyme (which is used in the isolation of the desired gene from the donor), plasmid i.e. vector DNA is cleaved.
f. Now by using enzyme DNA ligase, foreign DNA is inserted/ integrated into the vector DNA.
g. The combination of vector DNA and foreign DNA is now called Recombinant DNA or Chimeric DNA and the technology is referred to as rDNA technology.

iv. Transfer of rDNA into suitable competent host or cloning organism:
a. Finally the recombinant DNA is transferred for expression into a competent host cell which is usually a bacterium.
b. Host cell takes up naked rDNA by process of ‘transformation’ and incorporates into its own chromosomal DNA which finally expresses the trait controlled by passenger DNA.
c. The transfer of rDNA into a bacterial cell is assisted by divalent Ca⁺⁺.
d. The cloning organisms used in plant biotechnology are E. coli and Agrobacterium tumefaciens.
e. The host/ competent cell which has taken up rDNA is now called transformed cell.
f. Foreign DNA can also be transferred directly into the naked cell or protoplast of the competent host cell, without using vector.
g. This is done by using techniques like electroporation, microinjection, lipofection, shot gun, ultra-sonification, biolistic method, etc. But in plant biotechnology the transformation is throuqh Ti plasmids of A. tumefaciens.

v. Selection of the transformed host cell:
a. The transformation process generates a mixed population of transformed (recombinant) and non-transformed (non-recombinant) host cells.
b. For isolation of recombinant cell from non-recombinant cell, marker gene of plasmid vector is employed.
c. For example, pBR322 plasmid vector contains different marker gene (Ampicillin resistant gene and Tetracycline resistant gene).
d. When PstI restriction enzyme is used, it knocks out Ampicillin resistant gene from the plasmid, so that the recombinant cell becomes sensitive to Ampicillin.

vi. Multiplication of transformed host cell:
a. Once transformed, host cells are separated by the screening process.
b. In this step the transformed host cells are introduced into fresh culture media.
c. At this stage the host cells multiply along with the replication of the recombinant DNA carried by them.

vii. Expression of the gene to obtain the desired product:
a. The next step involves the production of desired products like alcohol, enzymes, antibiotics, etc.
b. Finally the desired product is separated and purified through downstream processing using suitable bioreactor.

Question 31.
Identify X, Y and Z and explain the process shown in the diagram below.

Answer:

i. The transmission of the nerve impulse along the long nerve fibre/ axon tube is a result of electrical changes across the neuronal membrane during conduction of an impulse.

ii. Each neuron has a charged cellular membrane with a voltage which is different on the
outer and inner side of the membrane.

iii. The resting nerve remains in a polarised state where there is +’ve charge towards the outside and -‘ve charge towards the inside. It is established by maintaining an excess of Na⁺ on the outer side, whereas on the inner side there is an excess of K⁺ along with large negatively charge protein molecules and nucleic acid.

iv. Application of stimulus on the resting nerve results in the increased permeability of the membrane for Na⁺. Na⁺ rush into the axon from the extracellular fluid (ECF) and bring about depolarisation.

v. The membrane potential changes from -70 mV (resting potential) to about +30 mV to +60 mV (action potential).

vi. Depolarisation is now triggered in the next part while the initial part itself starts undergoing repolarisation.

vii. After a short interval (refractory period), the large number of Na⁺ on the inside causes a drop in permeability of membrane to Na⁺ and simultaneously increasing the permeability to K⁺ by opening the K⁺ voltage gates and slowly closing the Na⁺ gates. This is known as repolarisation.

viii. The K⁺ ions diffuse out of the axon and the axoplasm inside of the membrane becomes negatively charged as compared to the ECF which becomes positively charged.

ix. The process of producing a wave of stimulation, causing depolarisation and repolarisation is repeated continuously upto the end of the axon terminal. This is a self-propagating process.

Maharashtra Board Class 12 Biology Previous Year Question Papers