Maharashtra Board Class 12 Biology Sample Paper Set 6 with Solutions

Maharashtra State Board Class 12th Biology Sample Paper Set 6 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Biology Model Paper Set 6 with Solutions

Time: 3 Hrs.
Max. Marks : 70

General Instructions:

The question papqr 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:

i. Which of the following component is not remobilized in plant?
(A) Phosphorous
(B) Sulphur
(C) Calcium
(D) Nitrogen
Answer:
(C) Calcium

ii. Attachment of embryo to the wall of the uterus is known as ____.
(A) fertilization
(B) gestation
(C) cleavage
(D) implantation
Answer:
(D) implantation

iii. In a cross performed between parents considering two contrasting characters yellow round and green wrinkled seed, in the F₂ generation progeny, the ratio of yellow and green will be ________.
(A) 9 : 3
(B) 12 : 4
(C) 4 : 1
(D) 4 : 12
Answer:
(B) 12 : 4

iv. In which of the following case ploidy level is NOT the same?
(A) Integuments and nucellus
(B) Root tip and shoot tip
(C) Secondary nucleus and endosperm
(D) Antipodals and Synergids
Answer:
(C) Secondary nucleus and endosperm

v. Gibberellins are synthesized from ___.
(A) aspartic acid
(B) glutamic acid
(C) phenolic acid
(D) mevalonic acid
Answer:
(D) mevalonic acid

vi. The pulmonary trunk and systemic aorta are connected by
(A) Chordae tendineae
(B) Columnae carneae
(C) Ligamentum arteriosum
(D) Purkinje fibres
Answer:
(C) Ligamentum arteriosum

vii. Thrombocytopenia means
(A) increased WBC count
(B) decreased WBC count
(C) increased platelet count
(D) decreased platelet count
Answer:
(D) decreased platelet count

viii. If parathyroid glands of man are removed, the specific result will be
(A) onset of aging
(B) disturbance of Ca⁺⁺
(C) onset of myxoedema
(D) elevation of blood pressure
Answer:
(B) disturbance of Ca⁺⁺

ix. The antiviral proteins released by a virus-infected cell are called ____.
(A) histamines
(B) interferons
(C) pyrogens
(D) allergens
Answer:
(B) interferons

x. Diapedesis is performed by
(A) erythrocytes
(B) thrombocytes
(C) adipocytes
(D) leucocytes
Answer:
(D) leucocytes

ix. The antiviral proteins released by a virus-infected cell are called ___.
(A) histamines
(B) interferons
(C) pyrogens
(D) allergens
Answer:
(B) interferons

Question 2.
Answer the following: [8 Marks]

i. What is microsporogenesis?
Answer:
It is a process in which each microspore mother cell divides meiotically to form tetrad of haploid microspores (pollen grains).
OR
Microsporogenesis is the process of formation of microspores from diploid microspore mother cell through meiotic cell division inside the microsporangia or pollen sacs.

ii. Define Gene frequency.
Answer:
The proportion of an allele in the gene pool, to the total number of alleles at a given locus, is called gene frequency.

iii. What is combined water?
Answer:
Water present in the form of hydrated oxides of silicon, aluminium, etc., is called ‘combined water’.

iv. Write the normal haemoglobin content in human male and female.
Answer:
The normal haemoglobin content for males is 14 – 17gm% of blood, whereas for normal female, it is 13-15gm% of blood.

v. What is the role of gibberellin in rosette plants?
Answer:
Gibberellin promotes bolting i.e. elongation of internodes just prior to flowering in plants with rosette habit e.g. beet, cabbage.

vi. What is germplasm collection?
Answer:
The entire collection having all the diverse alleles (i.e. variations) for all genes in a given crop is called germplasm collection.

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

viii. Name the Act passed by Government of India to safeguard water resources.
Answer:
Water (prevention and control of pollution) Act 1974.

Section – B

Attempt any EIGHT of the following questions:

Question 3.
Enlist the adaptations in chiropterophilous flowers.
Answer:
Adaptations in chiropterophilous flowers:

1. Flowers are dull coloured with strong fragrance.
2. They secrete abundant nectar.
3. Flowers produce large amount of edible pollen grains.
4. Chiropterophily is shown by plants like Antkocephalus (kadamb tree), Adansonia (Baobab tree), Kigelia (Sausage tree).

Question 4.
Which are the two major functions DNA has to perform to regulate and control the activities of the cell?
Answer:
As a carrier of genetic information, DNA has to perform two important functions:

i. Heterocatalytic function: When DNA directs the synthesis of chemical molecules other than itself, then such functions of DNA are called heterocatalytic functions.
e.g. Synthesis of RNA (transcription), synthesis of protein (translation), etc.

ii. Autocoftalytic function: When DNA directs the synthesis of DNA itself, then such function of DNA is called autocatalytic function, e.g. Replication

Question 5.
Write a short note on mutations.
Answer:
Mutations:

1. Sudden permanent heritable changes are called mutations.
2. Mutation can occur in the gene, in the chromosome and in the chromosome number.
3. Mutation that occurs within the single gene, is called point mutation or gene mutation or in a larger segment of genes by chromosomal aberrations. Both point mutations and chromosomal aberrations can alter gene frequency.
4. Mutation leads to the change in the phenotype of the organism, causing variation.

Question 6.
Identify the types of chromosomal aberrations in the following figures A, B, C, D:

Answer:
Chromosomal aberrations shown in the given figures are:
A: Deletion
B: Duplication
C: Inversion
D: Translocation

Question 7.
Complete the given chart.

Answer:
i. Apoplast pathway
ii. Vacuolar

Question 8.
Identify label ‘i’, ‘ii’, ‘iii’ and ‘iv’ in the given figure representing zones of growth phases in root.

Answer:
i. Zone of mature cells.
ii. Zone of cell differentiation.
iii. Zone of cell elongation.
iv. Zone of cell formation.

Question 9.
Identify the correct disease and route of transmission by the given symptoms.
i. Cough produces yellow or greenish sputum or phlegm.
ii. High fever
iii. Shortness of breath (Dyspnea)
iv. Chest pain during deep breath or coughing
v. Loss of appetite, fatigue, headaches, vomiting, joint pains and muscle aches.
Answer:
i. Name of the disease – pneumonia (1 Mark)
ii. Mode of transmission – Pneumonia mostly spreads by direct person to person contact. It can also spread via droplets released by infected person or even by using shared clothes and utensils.

Question 10.
i. Name the two basic techniques that help us to increase food production.
ii. Give any two examples of biofortification.
Answer:
i. Plant breeding and animal breeding are the two basic techniques that help us to increase food production. (1 Mark)

ii. Examples of biofortification are as follows:
a. Fortified Maize having twice the amount of amino acids- lysine and tryptophan.
b. Wheat -Atlas 66 has high protein content and Iron-fortified rice has 5 times more iron, ore developed.
c. Vegetable crops like carrot and spinach have more vitamin A and minerals.
d. Vitamin C enriched bitter gourd, tomato have been developed by IARI.

Question 11.
Complete the following table representing human proteins produced by rDNA technology to treat human diseases.

Disorder / Diseases / Health condition	Recombinant proteins
Cancer	(A)
Asthma	(B)
(C)	Factor IX
Dissolving bood clot	(D)

Answer:

Disorder / Diseases / Health condition	Recombinant proteins
Cancer	Interferons, tumour necrosis factor, interlukins, macrophage activating factor
Asthma	Interleukin – 1 receptor
Haemophilia B	Factor IX
Dissolving bood clot	Urokinase, Tissue plasminogen activator

Question 12.
Write a short note on primary succession of aquatic habitat.
Answer:
Primary succession of aquatic habitat:

1. Phytoplanktons are the small pioneers.
2. Phytoplanktons are then replaced by rooted submerged plants like Hydrilla, rooted floating angiosperm like lotus.
3. These are followed by free floating plants like Pistia, then reed swamp like Typha, marsh meadow like Cyperus, scrub like Alnus, and finally trees like Quercus.
4. The climax would be the forest.
5. As time passes the water body is converted into land.

Question 13.
Match the pairs and rewrite:

Recombinant Protein	It is use in or for
i. Factor VIII	a. Anaemia
ii. α1-antitrypsin	b. Cystic fibrosis
iii. DNase	c. Emphysema
iv. Erythropoietin	d. Haemophilia A

Answer:

Recombinant Protein	It is use in or for
i. Factor VIII	d. Haemophilia A
ii. α1-antitrypsin	c. Emphysema
iii. DNase	b. Cystic fibrosis
iv. Erythropoietin	d. Anaemia

Question 14.
How phosphorus causes eutrophication? How does it affect aquatic life?
Answer:

1. Phosphorus is always in short supply and hence acts as limiting factor for plant growth.
2. Sudden influx of phosphorus in the water bodies in the form of agricultural runoff or industrial effluents rich in phosphorus leads to eutrophication.
3. Eutrophication occurs due to overgrowth of algae at instance of high phosphorus dissolved in water. The overgrowth of algae kills or harms the aquatic life.

Section – C

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

Question 15.
What do you mean by Monozygotic, Dizygotic and Conjoined twins?
Answer:
i. When two embryos develop simultaneously in the same uterus and two offsprings are delivered from the same pregnancy, they are called as twins.

ii. The two major types of twins are:

a. Monozygotic twins: During early period of embryonic development (within 8 days of zygote formation), cells of the single embryo divide into two groups and develop as two separate embryos.
Such twins are genetically exactly similar to each other, hence their appearance as well as the gender is same.

b. Dizygotic twins: Sometimes, two oocytes are released from the ovary of a woman and both are fertilized by two separate sperms. As a result, two separate zygotes are formed. Embryos formed from these zygotes are separately implanted in the uterus and thus, dizygotic twins are formed. Dizygotic twins are genetically different and may be of same or different gender.

iii. In case of monozygotic twins, if embryonic cells are divided into two groups, 8 days after zygote formation they might form conjoined twins (Siamese twins).

Question 16.
Vidisha shows normal blood clotting but her mother is haemophilic. Suresh shows normal blood clotting but his father is haemophilic. If Suresh and Vidisha were to many, then find out the possible phenotypes of their offsprings.
Answer:
i. Vidisha’s mother is haemophilic that means her genetic constitution is X^hX^h. As Vidisha shows normal blood clotting that means her genetic constitution must be either X^HX^H or X^HX^h. But as Vidisha’s mother is haemophilic Vidisha had received one X^h chromosome (X chromosome carrying defective gene) and one normal X^H chromosome from her father. Therefore, she is carrier (XHXh) for haemophilia.

ii. Suresh’s father is haemophilic (X^hY) but Suresh shows normal blood clotting. Therefore, Suresh’s genetic constitution is X^HY (As males do not receive X chromosome from their father).

iii. If Suresh and Vidisha were to marry, following would be possible phenotypes of their offsprings:

Question 17.
Why the ratio in pleiotropy is 2 : 1? Explain it with example.
Answer:

1. When a single gene controls two (or more) different traits, it is called pleiotropic gene and this phenomenon is called pleiotropy or pleiotropism.
2. The ratio is 2:1 instead of 3:1.
3. According to Mendels principle of unit character, one gene (factor) controls one character (trait), but sometimes single gene produces two related or unrelated phenotypic expressions.
4. For example, the disease, sickle cell anaemia is caused by a gene Hb⁵. Normal or healthy gene is Hb^A and is dominant.
5. The carriers (heterozygotes – Hb^A/Hb^S) show signs of mild anaemia as their RBCs become sickle-shaped (half-moon shaped) in oxygen deficiency. They are said to have sickle-cell trait and are normal in normal conditions.
6. The homozygotes with recessive gene Hb^S however, die of fatal anaemia.
   
7. Thus, the gene for sickle-cell anaemia is lethal in homozygous condition and produces sickle cell trait in heterozygous carrier.
8. Two different expressions are produced by a single gene.
9. A marriage between two carriers will produce normal, earners and sickle-cell anaemic children in 1 : 2 : 1 ratio. But, sickle-cell anaemic who are homozygous for gene Hb^S will die, as Hbs is a lethal gene causing death of the bearer.

Thus the sickle cell anaemic die leaving carriers and normal in the ratio 2 : 1.

Question 18.
Explain in detail vertical and lateral translocation of food.
Answer:
The translocation of food occurs in vertical and lateral direction.

i. Vertical translocation:
a. In vertical (longitudinal) transport, food is translocated in downward direction from leaves (source) to stem and root (sink).
b. It also occurs in upward direction during germination of seed, bulbils, corm, etc.
c. Upward translocation also occurs from leaves to growing point of stem, to developing flowers and fruits situated near the ends of the branches of stem.

ii. Lateral translocation:
a. It occurs horizontally/laterally across the root and stem.
b. When food is translocated from phloem to pith, it is called radial translocation and from phloem to cortex, it is called tangential translocation.

[Note: Radial translocation is the lateral movement of organic solute. It occurs from the cells of pith to the cells of cortex and epidermis. The radial translocation occurs through the medullary rays.]

Question 19.
Write about the discovery of cytokinin. Add a note on natural and synthetic cytokinin. Biology.
Answer:

1. The first cytokinin was discovered by Skoog and Miller (1954) during investigation of nutritional requirements of callus tissue culture of Nicotiana tabacum (Tobacco).
2. They observed that the callus proliferated when the nutrient medium was supplemented with coconut milk and degraded sample of DNA (obtained from herring sperm). They named it as kinetin. Chemically kinetin are 6-furfurylamino purine.
3. First natural cytokinin was obtained from unripe maize grains by Letham et al. It is known as Zeatin.
4. Seven different types of cytokinins are recorded from plants.
5. Natural cytokinins are also reported from plants like Banana flowers, apple and tomato fruits, coconut milk, etc.
6. 6-benzyl adenine is a synthetic cytokinin hormone.

Question 20.
Draw neat and labelled diagram of internal structure of human heart.
Answer:

Question 21.
Describe any three properties of the nerve fibres.
Answer:
Properties of nerve fibers:

1. Excitability/Irritability: Nerve fibres have polarized membrane, thus they have the ability to perceive stimulus and enter into a state of activity.
2. Conductivity: It is ability of nerve to transmit impulses along the whole length of axon.
3. Stimulus: It is any detectable, physical, chemical, electrical change in the external or internal environment which brings about excitation in a nerve/muscle/organ/organism. A stimulus must have a minimum intensity called threshold stimulus, in order to be effective. Subliminal (weak) stimulus will have no effect while supraliminal (strong) stimulus will produce the same degree of impulse as the threshold stimulus.
4. Summation effect: A single subliminal stimulus will have no effect but when many such weak stimuli are given again and again they may produce an impulse due to summation of effects.
5. All or none law: The nerve will either conduct the impulse along its entire length or will not conduct the impulse at all. This occurs in case of subliminal or weak stimulus.
6. Refractory period: It is the time interval (about millisecond) during which a nerve fails to respond to a second stimulus even if it is strong.
7. Synaptic delay: The impulse takes about 0.3 to 0.5 milliseconds to cross a synapse. It is required for release of neurotransmitter from the axon terminal and excitation in the dendron of the next neuron.
8. Synaptic fatigue: The transmission of nerve impulse across the synapse stops temporarily due to depletion of the neurotransmitter.
9. Velocity: The rate of transmission of impulse is higher in long and thick nerves. It is higher in homeotherms than in poikilotherms. The velocity of transmission is higher in voluntary fibres (100 – 120 m/s in man) as compared to autonomic or involuntary nerves (10-20 m/s). In medullated nerve fibre, the velocity of transmission is higher as impulse has to jump from one node of Ranvier to the next.

Question 22.
Rewrite the following paragraph by writing most appropriate words in the blank places.

Improvement in oil content and oil quality of oil crops like soybean, oil palm, ___ and sunflower, have been achieved by transfer of ____. _____ is also a serious nutritional
problem, affecting an estimated 30% of the world population. For production of transgenic crops that will produce food rich in iron, an iron storage protein-___ is targeted. Ferritin is found in many animals, plants and ____. The genes for ferritin protein isolated from soybean and ___ have been transferred to rice.
Answer:
Improvement in oil content and oil quality of oil crops like soybean, oil palm, rapeseed and sunflower, have been achieved by transfer of ‘Arabidopsis genes’. Iron deficiency is also a serious nutritional problem, affecting an estimated 30% of the world population. For production of transgenic crops that will produce food rich in iron, an iron storage protein-ferritin is targeted. Ferritin is found in many animals, plants and bacteria. The genes for ferritin protein isolated from soybean and Phaseolus have been transferred to rice.

Question 23.
Explain the process of biogas production by giving reactions.
Answer:
Process of biogas production:
Anaerobic digestion involves three processes:

i. Hydrolysis or solubilization:
a. In initial stage raw material (cattle dung) is mixed with water in equal proportion to make slurry which is then fed into the digester.
b. Here anaerobic hydrolytic bacteria (e.g. Clostridium, Pseudomonas) hydrolyse carbohydrates into simple sugars, proteins into amino acids and lipids into fatty acids.

ii. Acidogenesis: In this stage, facultative anaerobic, acidogenic bacteria and obligate anaerobic organisms, convert simple organic material into acids like formic acid, acetic acid, H₂ and CO₂.

iii. Methanogenesis: This is last stage in which anaerobic Methanogenic bacteria like Methanobacterium, Methanococcus convert acetate, H₂ and CO₂ into Methane, CO₂ and H₂O and other products.

Question 24.
Make a concept map representing the parts of nervous system and brain.
Answer:

Question 25.
i. Which type of ecological interaction is shown in given pictures?
ii. Explain it by giving examples.

Answer:
i. The type of ecological interaction shown in the given pictures is Commensalism.

ii. Commensalism
a. This is the interaction in which one species benefits and the other is neither . harmed nor benefited.
b. An orchid growing as an epiphyte on a branch of mango tree, will get benefit while the mango tree derives no benefit.
c. The cattle egret and grazing cattle in close association, is a classic example of commensalism. Cattle egrets always forage close to cattle, as cattle move they flush out insects that might be difficult for the egrets to find and catch. Another example of commensalism is the interaction between sea anemone that has stinging tentacles and the clown fish that lives among them. The fish gets protection from predators which stay away from the stinging tentacles. The anemone does not appear to derive any benefit by hosting the clown fish.

Question 26.
Explain ex-situ conservation.
Answer:

1. Sometimes, when a species is critically endangered, special measures must be undertaken to protect it. It might be protected in captivity, as one of the measures of protection. This is called ex-situ conservation.
2. In this type of conservation, living beings are protected away from their natural habitats in special settings.
3. Wildlife safari parks, zoological parks and botanical gardens serve this purpose.
4. Animals which have decreased in number are allowed to breed in captivity in order to protect them.
   e.g. Crocodile bank of Chennai.
5. Seed banks are established to conserve wild varieties of food grains and vegetables.
6. Now a days, modern techniques like tissue culture, in vitro fertilization of eggs and cryopreservation (preservation at low temperature, -196°C) of gametes, are used to protect endangered species.

Section – D

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

Question 27.
With the help of neat and labelled diagram explain the T.S. of anther.
Answer:

i. Sporogenous tissue: Some hypodermal cells get transformed into archesporial cells.
The archesporial cell divides into an inner sporogenous cell and outer primary parietal cell.
Sporogenous cell forms sporogenous tissue.
Each cell of sporogenous tissue is capable of giving rise to a microspore tetrad.

ii. Anther wall: Parietal cell undergoes divisions to form anther wall layers. The anther wall is divided into four layers as follows:
a. Epidermis: It is the outermost protective layer made up of tabular (flattened) cells.
b. Endothecium: It is sub-epidermal layer made up of radially elongated cells with fibrous thickenings.
c. Middle layers: Inner to endothecium is middle layer made up of thin walled cells (1-2 layered), which may disintegrate in mature anther.
d. Tapetum: It is the inner most nutritive layer of anther wall. It immediately encloses the sporoqenous tissue (microspore mother cells).

Question 28.
Describe the process of replication of eukaryotic DNA.
Answer:
The process by which DNA duplicates to form identical copies is known as replication.
Semi-conservative method of replication:

1. After replication, each daughter DNA molecule has one old and other new strand.
2. As parental DNA is partly conserved in each daughter DNA, the process of replication is called semi-conservative.
3. The model of semi-conservative replication was proposed by Watson and Crick.
4. The semi-conservative model of DNA replication using heavy isotope of nitroqen N¹⁵ and E. coli, was experimentally verified by Meselson and Stahl (1958).

Mechanism of replication is as follows:

a. Activation of Nucleotides:

1. The four types of nucleotides of DNA i.e. dAMP, d£MP, dCMP and dTMP are present in the nucleoplasm.
2. They are activated by ATP in presence of an enzyme phosphorylase.
3. This results in the formation of deoxyribonucleotide triphosphates i.e. dATP, dGTP, dCTP and dTTP. This process is known as Phosphorylation.

b. Point of Origin or Initiation point:

1. Replication begins at specific point ‘O’ origin and terminates at point ‘T.
2. Origin is flanked by ‘T sites. The unit of DNA in which replication-occurs, is called replicon.
3. In prokaryotes, there is only one replicon however in eukaryotes, there are several replicons in tandem.
4. At the point ‘O’, enzyme endonuclease nicks one of the strands of DNA, temporarily.
5. The nick occurs in the sugar-phosphate back bone or the phosphodiester bond.

c. Unwinding of DNA molecule:

1. Enzyme DNA helicase breaks weak hydroqen bonds in the vicinity of ‘O’.
2. The strands of DNA separate and unwind. This unwinding is bidirectional and continues as Y shaped replication fork.
3. Each separated strand acts as template.
4. The two separated strands are prevented from recoiling (re-joining) by SSBP (Single strand binding proteins).
5. SSB proteins remain attached to both the separated strands for facilitating synthesis of new polynucleotide strands.

d. Replicatiftg fork:

1. The point formed due to unwinding and separation of two strands appears like a Y-shaped fork, called replicating/ replication fork.
2. The unwinding of strands imposes strain which is relieved by super-helix relaxing enzyme.

e. Synthesis of new strands:

1. Each separated strand acts as a mould or template for the synthesis of new complementary strand.
2. It requires a small RNA molecule, called RNA primer.
3. RNA primer attaches to the 3′ end of template strand and attracts complementary nucleotides from surrounding nucleoplasm.
4. These nucleotides bind to the complementary nucleotides on the template strand by forming hydrogen bonds (i.e. A = T or T = A; G = C or C = G).
5. The newly bound consecutive nucleotides get interconnected by phosphodiester bonds, forming a polynucleotide strand.
6. The synthesis of new complementary strand is catalysed by enzyme DNA polymerase.
7. The new complementary strand is always formed in 5‘→ 3′ direction.

f. Leading and Lagging strand:

1. The template strand with free 3′ end is called leading template and with free 5′ end is called lagging template.
2. The process of replication always starts at C-3 end of template strand and proceeds towards C-5 end.
3. As both the strands of the parental DNA are antiparallel, new strands are always formed in 5′ → 3′ direction.
4. One of the newly synthesized strand which develops continuously towards replicating fork is called leading strand.
5. Another new strand develops discontinuous away from the replicating fork and’ is called lagging strand.
6. Maturation of Okazaki fragments: DNA synthesis on lagging template takes place in the form of small fragments called as Okazaki fragments (named after scientist Okazaki).
7. Okazaki fragments are joined by enzyme DNA ligase.
8. RNA primers are removed by DNA polymerase and replaced by DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-a in eukaryotes.
9. Finally, DNA gyrase (topoisomerase) enzyme forms double helix to form daughter DNA molecules.

g. Formation of daughter DNA molecules:

1. At the end of the replication, two daughter DNA molecules are formed.
2. In each daughter DNA, one strand is parental and the other one is totally newly synthesized.
3. Thus, 50% is contributed by mother DNA. Hence, it is described as semiconservative replication.

[Note: ON A gyrase (topoisomerase) relieves torsional strain generated by DNA unwinding.]

Question 29.
With the help of a suitable diagram, describe the process of gastrulation.
Answer:

Gastrulation is the process of formation of ‘gastrula’ from the blastocyst.

In the gastrula stage, rate of cleavage or divisions slows but there are two important events that take place actively:

1. Differentiation of blastomeres: This process results in the formation of three germinal layers i.e. ectoderm, mesoderm and endoderm from the cells of the embryoblast.
2. Morphogenetic movements: These are different types of movements to reach their definite place in the embryo.
3. Development of extraembryonic membranes:

a. On about 8th day after fertilization, cells on the free end of inner cell mass called hypoblasts (primitive endoderm) become flattened, start dividing and grow downwards towards the blastocoel, cavity of blastocyst.
b. It grows within the blastocoel and forms a sac called yolk sac.
c. The remaining cell of the inner cell mass, in contact with cells of Rauber are called epiblasts (primary ectoderm).
d. Both the layers form a flat, bilaminar embryonal disc.
e. After the formation of endoderm the second layer to be differentiated is the ectoderm.
f. Cells of epiblast divide and re-divide and move in such a way that they enclose the amniotic cavity.
g. The floor of this cavity has the embryonal disc and the roof of amniotic cavity is lined by amniogenic cells (amnion forming cells).
h. Later, the amniogenic cells divide and re-divide to form the amnion.
i. Amnion is an extra embryonic membrane that surrounds and protects the embryo.
j. As a result of all these changes, the bilaminar embryonic disc is positioned in between amniotic cavity and Yolk sac.

iv. Process of gastrulation:
a. Gastrulation occurs about 15 days after fertilization, in which the bilaminar embryonic disc (consisting of hypoblast and epiblast) is transformed into trilaminar embryonic disc (consisting of three primary germ layers i.e. ectoderm, mesoderm and endoderm).
b. This transformation occurs by division, rearrangement and migration of cells of epiblast.
c. It begins with formation of primitive streak and a shallow groove on the surface is called primitive groove.
d. This streak progresses from posterior to anterior end of embryo.
e. Anterior end of primitive groove communicates with yolk sac by an aperture called blastopore (future anus).
f. Cells of the epiblast move inwards below the primitive streak and detach from the epiblast (invagination). Some of these invaginated cells displace the hypoblast forming the endoderm.
g. After formation of endoderm the remaining cells of the epiblast now result in the formation of the second layer, the ectoderm.
h. From the site of a primitive streak, a third layer of cells called mesoderm extends between ectoderm and endoderm. Anterior end of primitive groove communicates with yolk sac an aperture called blastopore
i. The embryonal disc now has differentiated into three layers ectoderm, mesoderm and endoderm.
j. Gastrulation is followed by organogenesis.

Question 30.
i. If the duration of the atrial systole is 0.1 sec and that of complete diastole is 0.4 sec, then how does one cardiac cycle complete in 0.8 sec?
ii. What is tachycardia?
Answer:
i. a. One cardiac cycle includes atrial systole, ventricular systole and joint/complete diastole.
b. The duration for atrial systole is 0.1 sec, duration for complete diastole is 0.4 sec, which means if one cardiac cycle completes in 0.8 sec then duration for ventricular systole is 0.3 sec.
c. Therefore, the duration of one cardiac cycle = Atrial systole + Ventricular systole + Complete diastole = 0.1 sec +0.3 sec + 0.4 sec = 0.8 sec
d. Also the relaxation period shortens as the heart beats faster whereas the durations of atrial systole and diastole shortens slightly. Hence, one cardiac cycle completes in 0.8 sec. [3 Marks]

ii. Pulse rate higher than normal i.e. above 100 beats / min is called tachycardia.

Question 31.
i. Identify the gland (x).

Answer:
i. The gland (X) given in the picture is thyroid gland. [1 Mark]

ii. Write an account of hormones secreted by gland (x).
Answer:
a. Thyroid gland is stimulated to secrete its hormones by thyroid stimulating hormone (TSH).
The two hormones secreted by the follicular cells are Thyroxine/tetra iodothyronine/ T₄ (four atoms of iodine) and Triiodothyronine or T₃ (three atoms of iodine).
b. Thyroxine is synthesized by attaching iodine to amino acid tyrosine by enzymatic action. The amino acid tyrosine molecule binds to iodine to produce Monoiodothyronine (T₁) or 2 atoms of iodine to produce Diodothyronine (T₂).
T₁ and T₂ molecules bind end to end to make colloidal mass inside, the follicle.
They are further metabolised to prepare T₃ and T₄.
c. Triidothyronine or T₃ is also secreted in small quantity. It is physiologically more active. Thyroid gland is the only gland that stores its hormones. T₃ and T₄ hormones are stored before secretion and are regulated by thyrotropin of pituitary gland by negative feed back mechanism.

Maharashtra Board Class 12 Biology Previous Year Question Papers