Class 11 Biology Archives

NCERT Solutions For Class 11 Biology Plant Growth and Development

July 30, 2018

NCERT Solutions For Class 11 Biology Plant Growth and Development

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development:

Section Name	Topic Name
15	Plant Growth and Development
15.1	Growth
15.2	Differentiation, Dedifferentiation and Redifferentiation
15.3	Development
15.4	Plant Growth Regulators
15.5	Photoperiodism
15.6	Vernalisation
15.7	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Solution: Growth is defined as a vital process which brings about an irreversible and permanent change in the shape, size, form, weight and volume of a cell, organ or whole organism, accompanied with increase in dry matter.
Differentiation is a localised qualitative change in size, biochemistry, structure and function of cells, tissues or organs, e.g., fibre, vessel, tracheid, sieve tube, mesophyll, leaf etc. Thus it is a change in form and physiological activity. It results in specialisation for particular functions.
Development may be defined as a process which includes growth, differentiation and maturation in a regular sequence in the life history of a cell, organ or organism viz., seed germination, growth, differentiation, flowering, seed formation and senescence. Dedifferentiation is the process by which the differentiated cells which have lost the ability to divide under certain circumstances, become meristematic and regain the divisibility. Redifferentiation is defined as maturation or differentiation of dedifferentiated cells to form cells which are unable to divide e.g., secondary xylem elements, cork cells etc., are formed by redifferentiation of secondary cambial cells.
Determinate growth is the ability of a cell, tissue or the organism to grow for a limited period of time. Meristem is a tissue consisting of unspecialised immature cells, possessing the power of continuous cell division and adding new cells to the body. Growth rate is defined as the increased growth per unit time.

2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Solution: A flowering plant consists of a number of organs viz., roots, stem, leaves, flowers, fruits etc. growing differently under different stages of life cycle. These plant organs require different parameters to demonstrate their growth. In plant organs like fruits, bulbs, corms etc. fresh weight is used for measuring their growth. In case of fruits, increase in volume, diameter etc., are also used as other parameters for the measurement of their growth. For flat organs like leaves, increase in surface area is used as the parameter. Stem and roots primarily grow in length and then in girth, thus increase in length and diameter are used for measuring their growth. Consequently, the flowering plants exhibit several parameters to demonstrate growth.

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3. Describe briefly
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
Solution: (a) Arithmetic growth: If the length of a plant organ is plotted against time it shows a linear curve, the growth is called arithmetic growth. In this growth, the rate of growth is constant and increase in growth occurs in arithmetic progression e.g., length of a plant is measured as 2,4, 6, 8,10,12 cms at a definite interval of 24 hrs. It is found in root or shoot elongating at constant rate. Arithmetic growth is expressed as L_t = L₀ + r_tHere, L_t= length after time t. L₀ = length at the beginning, r = growth rate.

(b) Geometric growth: Geometric growth is the growth where both the progeny cells following mitosis retain the ability to divide and continue to do so. It occurs in many higher plants and in unicellular organisms when grown in nutrient rich medium. Number of cells is initially small so that initial growth is slow which is called lag phase. Later on, there is rapid growth at exponential rate. It is called log or exponential phase.

(c) Sigmoid growth curve: Geometric growth cannot be sustained for long. Some cells die. Limited nutrient availability causes slowing down of growth. It leads to stationary phase. There may be actually a decline. Plotting the growth against time will give a typical sigmoid or S-curve.

S-curve of growth is typical of most living organisms in their natural environment. It also occurs in cells, tissues and organs of plants.
(d) Absolute growth rate is the measurement of total growth per unit time. Relative growth rate is growth per unit time per unit initial growth.
Growth in given time period/ Measurement at start of time period
Suppose two leaves have grown by 5 cm² in one day. Initial size of leaf A was 5 cm² while that of leaf B was 50 cm². Though their absolute growth is the same (5 cm²/day), relative rate of growth is faster in leaf A(5/5) because of initial small size than in leaf B(5/50).

4. List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/ horticultural applications of any one of them.
Solution: There are five main groups of natural plant growth regulators which are very much recognised as natural hormones in plants. These are:

Auxins
Gibberellins
Cytokinins
Abscisic acid
Ethylene

Discovery of auxin: In 1880, Charles Darwin and Francis Darwin worked with the coleoptile of canary grass (Phalaris sp.) and found the existence of a substance in coleoptile tip, which was able to recognise the light stimulus and leads to the bending of tip towards light. Boysen and Jensen (1910-1913) worked on Avena seedling and explained that the substances secreted in the tip are soluble in water (gelatin).
Paal (1919) reported that the substances secreted in the tip are translocated downwards and caused cell elongation in half portion which was on the dark side and hence bending was observed in opposite direction.
F.W. Went (1928) further refined this experiment and supported the observations of Paal. He was the first person to isolate and name these substances of tip as auxins (Greek Auxein – means ‘to grow’).
In 1931, Kogl and Haagen-Smith isolated
crystalline compounds from human urine.
These were named as auxin-a, auxin-b and
heteroauxin.

Physiological functions of auxins:

Auxins induce cambial cell divisions, shoot cell elongation and early differentiation of xylem and phloem in tissue culture experiments.
In general, auxins initiate rooting but inhibit the growth of roots. IBA is the most potent root initiator.
Auxins inhibit the growth of axillary buds (apical dominance) but enhance the size of carpel and hence earlier fruit formation.
Application of auxins retards the process of senescence (last degradative phase), the abscission of leaves, fruits, branches, etc.
Auxins induce feminisation, i.e., on male plant, female flowers are produced.

Agricultural/horticultural application of auxins:

Application of auxins like IAA, IBA, NAA induce rooting in stem cuttings of many plants. This method is widely used to multiply several economically useful plants.
Normally, auxins inhibit flowering however in litchi and pineapple, application of auxin promotes flowering thus used in orchards.
Auxin induces parthenocarpy in some plants including tomato, pepper, cucumber and Citrus, thus, produces seedless fruits of more economic value.
Auxins like 2, 4-D and 2, 4, 5-T are commercially used as weedicides, due to their low cost and greater chemical stability. They are selective herbicides (killing broad-leaved plants, but not grasses).
For checking premature fruit drop, auxins are applied which prevent the formation of abscission zone in the petiole or just below the fruit. Auxin regulates maturing fruit on the trees of apples, oranges and grape fruit. High doses of auxins can
cause fruit drop. Thus, heavy applications of synthetic auxins are used commercially to promote a coordinated abscission of various fruits to facilitate harvesting.
Auxin, produced in the apical bud, suppresses the development of lateral buds, i.e., apical dominance. Thus practically used in prolonging the dormancy period of potato tubers.
Naphthalene acetamide is used to prevent the lodging (excessive elongation and development of weak plants, specially in gramineae) or falling of crops.
Auxin (2,4-D) promotes callus formation in tissue culture. Complete plantlets are regenerated from callus tissue, using auxins and cytokinin which are then transplanted into the soil. Now-a-days, this is a widely practised method of propagation in the field of agriculture and horticulture.

5. What do you understand by photoperiodism and vernalisation? Describe their significance.
Solution: The physiological mechanism for flower-ing is controlled by two factors: photoperiod or light period, i.e., photoperiodism and low temperature, i.e., vernalisation. Photoperiodism is defined as the flowering response of a plant to relative lengths of light/ dark period. Significance of photoperiodism is as follows:

Photoperiodism determines the season in which a particular plant shall flower. For example, short day plants develop flowers in autumn-spring period (e.g., Dahlia, Xanthium) while long day plants produce flowers in summer (e.g., Amaranthus).
Knowledge of photoperiodic effect is useful in keeping some plants in vegetative growth (many vegetables) to obtain higher yield of tubers, rhizomes etc. or keep the plant in reproductive stage to yield more flowers and fruits.
A plant can be made to flower throughout the year by providing favourable photoperiod.
Helps the plant breeders in effective cross-breeding in plants.
Enable a plant to flower in different seasons.
Vernalisation is promotion or induction of flowering by exposing a plant to low temperature for some time. Significance of vernalisation is as follows :
(i) Crops can be grown earlier.
(ii)Plants can be grown in such regions where normally they do not grow.
(iii)Yield of the plant is increased.
(iv)Resistance to cold and frost is increased.
(v) Resistance to fungal diseases is increased.

6. Why is abscisic acid also known as stress hormone?
Solution: A fairly high concentration of abscisic acid (ABA) is found in leaves of plants growing under stress conditions, such as drought, flooding, injury, mineral deficiency etc. It is accompanied by loss of turgor and closure of stomata. When such plants are transferred to normal conditions, they regain normal turgor and ABA concentration decreases. Since the synthesis of ABA is accelerated under stress condition and the same is destroyed or inactivated when stress is relieved, it is also known as stress hormone.

7. ‘Both growth and differentiation in higher plants are open’. Comment.
Solution: Plant growth is generally indeterminate. Higher plants possess specific areas called meristems which take part in the formation of new cells. The body of plants is built on a modular fashion where structure is never complete because the tips (with apical meristem) “are open ended – always growing and forming new organs to replace the older or senescent ones. Growth is invariably associated with differentiation. The exact trigger for differentiation is also not known. Not only the growth of plants are open- ended, their differentiation is also open. The same apical meristem cells give rise to different types of cells at maturity, e.g., xylem, phloem, parenchyma, sclerenchyma fibres, collenchyma, etc. Thus, both the processes are indeterminate, unlimited and develop into
different structures at maturity i.e., both are open.

8. ‘Both a short day plant and a long day plant can produce flower simultaneously in a given place’. Explain.
Solution: A short day plant (SDP) flowers only when it receives a long dark period and short photoperiod, e.g., Xanthium, Dahlia etc. On the other hand, a long day plant (LDP) will flower only when it receives a long photoperiod and short dark period, e.g., wheat, oat etc. Thus critical photoperiod is that continuous duration of light which must not be exceeded in SDP and should always be exceeded in LDP in order to bring them to flower. Xanthium requires light for less than 15.6 hrs and Henbane requires light for more than 11 hrs. Xanthium (a SDP) and Henbane (DP) will flower simultaneously in light period between 11 to 15.6 hrs.

9. Which one of the plant growth regulators would you use if you are asked to
(a) induce rooting in a twig
(b) quickly ripen a fruit
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt’ a rosette plant
(f) induce immediate stomatal closure in leaves.
Solution: (a) Auxins like IBA, NAA.
(b) Ethylene
(c) Cytokinins
(d) Cytokinins
(e) Gibberellins
(f) Abscisic acid (ABA)

10. Would a defoliated plant respond to photo- periodic cycle? Why?
Solution: No, a defoliated plant would not respond to photoperiodic cycle because photoperiodic stimulus is picked up by the leaves only. Even one leaf or a part of it is sufficient for this purpose. For perception of photoperiodic cycle, there must be the presence of leaves under inductive photoperiod, so that, the hormone responsible for flowering can be produced.

11. What would be expected to happen if:
(a) GA₃ is applied to rice seedlings
(b) dividing cells stop differentiating
(c) a rotten fruit gets mixed with unripe fruits
(d) you forget to add cytokinin to the culture medium.
Solution:
(a) The coleoptile will elongate rapidly, as GA₃ helps in cell growth.
(b) The development of callus (mass of undifferentiated cells) will take place.
(c) The unripe fruits will ripe quickly because of the increased rate of respiration due to emission of ethylene from rotten fruit.
(d) Cell division will retard and shoot will not initiate from the callus.

NCERT Solutions For Class 11 Biology Respiration in Plants

July 30, 2018

NCERT Solutions For Class 11 Biology Respiration in Plants

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants:

Section Name	Topic Name
14	Respiration in Plants
14.1	Do Plants Breathe?
14.2	Glycolysis
14.3	Fermentation
14.4	Aerobic Respiration
14.5	The Respiratory Balance Sheet
14.6	Amphibolic Pathway
14.7	Respiratory Quotient
14.8	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Give the schematic representation of an overall view of Krebs’ cycle.
Solution:

2. Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’cycle
(c) Aerobic respiration and Fermentation
Solution: (a) Differences between respiration and combustion are as follows :

(b) Differences between glycolysis and Krebs’ cycle are as follows:

(C)Differences between aerobic respiration and fermentation are as follows:

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3. What are respiratory substrates? Name the most common respiratory substrate.
Solution: Respiratory substrates are those organic substances which are oxidised during respiration to liberate energy inside the living cells. The common respiratory substrates are carbohydrates, proteins, fats and organic acids. The most common respiratory substrate is glucose. It is a hexose monosaccharide.

4. Give the schematic representation of glycolysis.
Solution:

5. Explain ETS.
Solution: An electron transport chain or system (ETS) is a series of coenzymes and cytochromes that take part in the passage of electrons from
a chemical to its ultimate acceptor. Reduced coenzymes participate in electron transport chain. Electron transport takes place on cristae of mitochondria [oxysomes ( F₀ -F₁ , particles) found on the inner surface of the membrane of mitochondria]. NADH formed in glycolysis and citric acid cycle are oxidised by NADH dehydrogenase (complex I) and the electrons are transferred to ubiquinone. Ubiquinone also receives reducing equivalents via FADH2 through the activity of succinate dehydrogenase (complex II). The reduced ubiquinone is then oxidised by transfer of electrons of cytochrome c via cytochrome Fc, complex (complex III). Cytochrome c acts as a mobile carrier between complex III and complex IV. Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a₃and two copper centres. When the electrons are shunted over the carriers via complex I to IV in the electron transport chain, they are coupled to ATP synthetase (complex V) for the formation of ATP from ADP and Pi. Oxygen functions as the terminal acceptor of electrons and is reduced to water along with the hydrogen atoms. Reduced coenzymes (coenzyme I, II and FAD) do not combine directly with the molecular O₂. Only their hydrogen or electrons are transferred through various substances and finally reach O₂. The substances useful for the transfer of electron are called electron carriers. Only electrons are transferred through cytochromes (Cyt F₁ Cyt c,,C₂, a, a₃) and finally reach molecular O₂. Both cytochrome a and a3 form a system called cytochrome oxidase. Copper is also present in Cyt a₃ in addition to iron. The molecular oxygen that has accepted electrons now receives the protons that were liberated into the surrounding medium to give rise to a molecule of water. The liberated energy is utilised for the synthesis of ATP from ADP and Pi.

6. What are the main steps in aerobic respiration? Where does it take place?
Solution: Aerobic respiration is an enzymatically controlled release of energy in a stepwise catabolic process of complete oxidation of organic food into carbon dioxide and water with oxygen acting as terminal oxidant. It
occurs by two methods, common pathway and pentose phosphate pathway. Common pathway is known so because its first step, called glycolysis, is common to both aerobic and anaerobic modes of respiration. The common pathway of aerobic respiration consists of three steps – glycolysis, Krebs’ cycle and terminal oxidation. Aerobic respiration takes place within mitochondria. The final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria.

7. What are the assumptions made during the calculation of net gain of ATP?
Solution: It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:

There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
transferred into the mitochondria and undergoes oxidative phosphorylation.
None of the intermediates in the pathway are utilised to synthesise any other compound.
Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

But these kind of assumptions are not really valid in a living system; all pathway work simultaneously and do not take place one after another; substrates enter the pathways and are withdrawn from it as and when necessary; ATP is utilised as and when needed; enzymatic rates are controlled by multiple means. Hence, there can be a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

8. Distinguish between the following:
(a) Aerobic respiration and Anaerobic respira¬tion.
(b) Glycolysis and Fermentation.
(c) Glycolysis and Citric acid cycle.
Solution: (a) Differences between aerobic and anaerobic respiration are as follows:

(b) Differences between glycolysis and fermentation are as follows:

9. Discuss “The respiratory pathway is an amphibolic pathway”.
Solution: Amphibolic pathway is the one which is used for both breakdown (catabolism) and build-up (anabolism) reactions. Respiratory pathway is mainly a catabolic process which serves to run the living system by providing energy. The pathway produces a number of intermediates. Many of them are raw materials for building up both primary and secondary metabolites. Acetyl CoA is helpful not only in Krebs’ cycle but is also raw material for synthesis of fatty acids, steroids, terpenes, aromatic compounds and carotenoids, a-ketoglutarate is organic acid which forms glutamate (an important amino acid) on amination. OAA (Oxaloacetic acid) on amination produces asparate. Both aspartate and glutamate are components of proteins. Pyrimidines and alkaloids are other products. Succinyl CoA forms cytochromes and chlorophyll.
Hence, fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Hence, the respiratory pathway comes into the picture both during breakdown and synthesis of fatty acids. Similarly, during breakdown and synthesis of proteins too, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

10. Define RQ. What is its value for fats?
Solution: Respiratory quotient (RQ) is the ratio of the volume of carbon dioxide produced to the volume of oxygen consumed in respiration over a period of time. Its value can be one, zero, more than 1 or less than one.

Volume of C02 evolved Volume of 02 consumed
RQ is less than one when the respiratory substrate is either fat or protein.
C₅₇ H₁₀₄O₆ + 80 O₂-» 57 CO₂+ 52H₂O
RQ = 57CO₂/80O₂ = 0.71
RQ is about 0.7 for most of the common fats.

11. What is oxidative phosphorylation?
Solution: Oxidative phosphorylation is the synthesis of energy rich ATP molecules with the help of energy liberated during oxidation of reduced co-enzymes (NADH, FADH₂) produced in respiration. The enzyme required for this synthesis is called ATP synthase. It is considered to be the fifth complex of electron transport chain. ATP synthase is located in FT or head piece of F₀ -F₁ or elementary particles. The particles are present in the inner mitochondrial membrane. ATP synthase becomes active in ATP formation only where there is a proton gradient having higher concentration of H⁺ or protons on the F₀ side as compared to F x side (chemiosmotic hypothesis of Peter Mitchell).
Increased proton concentration is produced in the outer chamber or outer surface of inner mitochondrial membrane by the pushing of proton with the help of energy liberated by passage of electrons from one carrier to another. Transport of the electrons from NADH over ETC helps in pushing three pairs of protons to the outer chamber while two pairs of protons are sent outwardly during electron flow from FADH_2. The flow of protons through the F₀ channel induces F₁ particle to function as ATP-synthase. The energy of the proton gradient is used in attaching a phosphate radical to ADP by high energy bond. This produces ATP. Oxidation of one molecule of NADH₂ produces 3 ATP molecules while a similar oxidation of FADH₂ forms 2 ATP molecules.

12. What is the significance of step-wise release of energy in respiration?
Solution: The utility of step-wise release of energy in respiration are given as follows :
(i) There is a step-wise release of chemical bond energy which is very easily trapped in forming ATP molecules.
(ii) Cellular temperature is not allowed to rise.
(iii) Wastage of energy is reduced.
(iv) There are several intermediates which can be used in production of a number of biochemicals.
(v) Through their metabolic intermediates different substances can undergo respiratory catabolism.
(vi) Each step of respiration is controlled by its own enzyme. The activity of different enzymes can be enhanced or inhibited by specific compounds.
This helps in controlling the rate of respiration and the amount of energy liberated by it.

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NCERT Solutions For Class 11 Biology Structural Organisation in Animals

July 30, 2018

NCERT Solutions For Class 11 Biology Structural Organisation in Animals

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 7 Structural Organisation in Animals:

Section Name	Topic Name
7	Structural Organisation in Animals
7.1	Animal Tissues
7.2	Organ and Organ System
7.3	Earthworm
7.4	Cockroach
7.5	Frogs
7.6	Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1. Answer in one word or one line.
(i) Give the common name of Periplaneta americana.
(ii) How many spermathecae are found in earthworm?
(iii) What is the position of ovaries in cockroach?
(iv) How many segments are present in the abdomen of cockroach?
(v) Where do you find Malpighian tubules?
Solution: (i) Cockroach.
(ii) Four pairs.
(iii) In cockroach two large ovaries, lie laterally in the 2nd – 6th abdominal segments’.
(iv) Abdomen of cockroach consists of 10 segments.
(v) Malpighian tubules are present at the junction of midgut and hindgut in cockroach.

2. What are the following and where do you find them in animal body?
(a) Chondrocytes
(b) Axons.
(c) Ciliated epithelium
Solution: (a) Chondrocytes – Chondrocytes are the only cells found in cartilage. They are present in spaces called lacunae and they produce and maintain the matrix of cartilage. Bending ability of cartilage is due to chondrocytes. Cartilage is present at tip of nose, pinna of ear, epiglottis etc.
(b) Axon – Axon is one of the processes of neuron, which is the structural and functional unit of nervous system. The part of cyton – n’here axon arises is axon hillock and axon ends in group of branches called terminal arborizations. It conducts impulses away from the cyton. Neurons (nerve cells)
are present in brain and spinal cord.
(c) Ciliated epithelium – If the columnar or cuboidal cells bear cilia on their free surface they are called ciliated epithelium. Their function is to move particles or mucus in a specific direction over the epithelium. They are mainly present in the inner surface of hollow organs like bronchioles and Fallopian tube.

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3. Draw a labelled diagram of the reproductive organs of an earthworm.
Solution:

4. Answer the following.
(i) What is the function of nephridia?
(ii) How many types of nephridia are found in earthworm based on their location?
Solution: (i) Nephridia are excretory organs of earthworm, which perform the function of excretion and osmoregulation. Nephridia regulate the volume and composition of the body fluids. A nephridium is a coiled tubular and microscopic structure which starts out as a funnel that collects excess fluid from coelomic chamber. The funnel connects with a tubular wastes through a pore to the surface in the body wall or into the digestive tube.
(ii) In earthworm, nephridia are present in all segments except the first two. There are three types of nephridia on the basis of their location:
(a) Septal nephridia, present on both the sides of intersegmental septa from segment 15 to the last that open into intestine.
(b) Integumentary nephridia, attached to lining of the body wall of segment 3 to the last that open on the body surface and
(c) Pharyngeal nephridia, present as three paired tufts in the 4th, 5th and 6th segments.

5. Draw a labelled diagram of alimentary canal of a cockroach.
Solution:

6. What are the cellular components of blood?
Solution: Blood is a fluid connective tissue. It is composed of plasma (fluid) and blood cells (corpuscles). Cellular components of blood (blood corpuscles) constitute about 45% of blood volume.
Three types of blood cells are:
(i) Erythrocytes or red blood cells: They are most abundant blood cells. Normal RBC count is 5-5.5 million/mm³ in males and 4.5-5 million/mm³ in females) RBCs help in transport of gases and maintain blood pH.
(ii) Leucocytes or white blood cells: The normal WBC count is 5000-6000/mm³ of blood. They are involved in immune response of body and act as soldiers and scavangers.
(iii) Thrombocytes or blood platelets: There are about 2,50,000 platelets/mm³ of blood. They are involved in blood clotting.

7. Distinguish between the following:
(a) Prostomium and peristomium
(b) Septal nephridium and pharyngeal
Solution: (a) Differences between prostomium and peristomium are

(b) Differences between septal and pharyngeal nephridia are:

8. Mark the odd one in each series.
(a) Areolar tissue; blood; neuron; tendon
(b) RBC; WBC; platelets; cartilage
(c) Exocrine; endocrine; salivary gland; ligament
(d) Maxilla; mandible; labrum; antennae
(e) Protonema; mesothorax; metathorax; coxa.
Solution:
(a) Neuron: Areolar tissue, blood and tendon are connective tissues while neuron is a part a nervous tissue.
(b) Cartilage: RBC, WBC and platelets are parts of vascular connective tissue while cartilage is skeletal connective tissue.
(c) Ligament: Ligament is a connective tissue.
(d) Antennae: Maxilla, mandible and labrum are mouth parts of cockroach while antennae are sense organs.
(e) Protonema: Protonema is a filamentous juvenile stage in life cycle of Bryophytes, while mesothorax, metathorax and coxa are appendages of cockroach.

9. Match the terms in column I with those in column II.

Column I	Column II
(a) Compound epithelium (b) Compound eye (c) Septal nephridia (d) Open circulatory system (e) Typhlosole (f) Osteocytes (g) Genitalia	(i) Alimentry canal (ii) Cockroach (iii) Skin (iv) Mosaic vision (v) Earthworm (vi) Phallomere (vii) Bone

Solution: (a) – (iii), (b) – (iv), (c) – (v), (d) – (ii), (e) – (i), (f) – (vii), (g) – (vi)

10. Mention briefly about the circulatory system of earthworm.
Solution: Earthworm possesses a closed type of blood vascular system, as the blood flows through closed blood vessels. Blood is red in colour due to respiratory pigment haemoglobin. Prominent blood vessels in earthworm includes dorsal, ventral, sub- neural, lateral oesophageal and supra- oesophageal blood vessels. There are four pairs of tubular hearts, provided with valves. The anterior two pairs of hearts, known as lateral hearts lie in the 7th and 9th segments and connect the dorsal blood vessel with the ventral blood vessel. They receive blood from the dorsal blood vessel and convey it to the ventral blood vessel. The posterior two pairs of hearts are called latero-oesophageal hearts and are situated in the 12th and 13th segments. The latero-oesophageal hearts apart from connecting the dorsal and ventral blood vessels are also joined with the supra oesophageal blood vessel. Latero-oesophageal hearts carry blood from the dorsal vessel and the supra oesophageal vessel to the ventral blood vessel.Contractions keep blood circulating in one direction. Blood glands are present in the 4th, 5th and 6lh segments which produce blood cells and haemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature.

11. Describe various types of epithelial tissues with the help of labelled diagrams.
Solution: Epithelial tissue is a tissue made of one or more layers of compactly arranged cells that covers external surface and internal free surface of body organs and which is underlined by a basement membrane. The various types of epithelial tissue along with the diagram are given below:
(i) Simple epithelium : It is composed of single layer of cells which rest on basement membrane. Simple epithelium generally occurs over secretory and absorptive surfaces and forms lining of body cavities, ducts and tubes. Simple epithelium is of several types.
(a) Squamous epithelium: It consists of single layer of flat cells, tightly linked together and have centrally located oval or spherical nucleus. It is also called pavement epithelium. It is found in walls of blood vessels, air sacs of lungs, and lining of eye lens.
(b) Cuboidal epithelium: Cells of cuboidal epithelium are as tall as wide, with centrally placed nucleus. Its main functions are secretion and absorption. It lines sweat gland, thyroid follicles, salivary glands. Brush bordered cuboidal epithelium, i.e., cells having microvilli on their free surface lines proximal part of uriniferous tubule, pancreatic duct, testis and ovary.
(c) Columnar epithelium: Cells are with basally located nucleus. It helps in secretion and absorption. It occurs in lining of intestine, stomach, gall bladder.
(d) Ciliated epithelium: Free surface of columnar and cuboidal cells are covered with cilia. Cilia help in moving fluids, particles, mucus, etc. in a specific direction. It occurs in the inner surface of Fallopian tubules, nasal passage, bronchioles.

(e) Pseudostratified epithelium: It consists of single layer of cells but some cells are shorter than others. Due to difference in size of cells, the epithelium appears 2-3 layered. Pseudostratified columnar epithelium occurs in urethra and parotid salivary gland. Pseudostratified columnar ciliated epithelium (only larger cells ciliated) occurs in lining layer of nasal’ chambers, trachea and large bronchi. It helps in moving mucus and foreign particles.

(ii) Compoundepithelium/stratifiedepithelium: It is multilayered epithelium where cells of only the lowermost or basal layer are in contact with basement membrane. It provides protection against mechanical and chemical stresses and has limited role in secretion and absorption. It covers dry surface of skin, moist surface of buccal cavity, pharynx, etc. Different types of compound epithelium are:
(a) Stratified squamous epithelium: The cells of outer layer are flattened and squamous while the inner layers are cuboidal cells. It is of two types: Non- keratinised lining oesophagus, pharynx, buccal cavity, cornea, vagina and anal canal and keratinised (comified): forming epidermis of skin, hair, horn and nail.

(b) Stratified cuboidal epithelium: The outer layer of cuboidal cells and basal layer of columnar cells. It lines ducts of sweat glands, large salivary and pancreatic ducts.
(c) Stratified columnar epithelium: Both upper and basal layers are made of columnar cells, e.g., epiglottis covering, part of urethra.
(d) Stratified ciliated columnar epithelium: Outer layer consists of ciliated columnar cells and basal layer of columnar cells, e.g., larynx.
(iii) Transitional Epithelium: This is stratified epithelium which contains cuboidal or columnar shaped cells, which are thin and stretchable. No basement membrane is present as it would impede stretchability. It lines the inner surface of renal calyces, urinary bladder, ureter. Because of its t distribution, it is also called urothelium.

(iv) Glandular epithelium: It consists of specialised epithelial cells which synthesise intracellular macromolecules (protein in pancreas, lipids in adrenal glands, glycoprotein in salivary glands and all the three in mammary glands) and pour out the same in the form of a useful fluid secretion which is different from blood or any other extracellular fluid. Glands can be unicellular or multicellular on the basis of number of cells.
(a) Unicellular glands: Single-celled, e.g., goblet (mucous) cells of respiratory tract and alimentary canal.
(b) Multicellular glands: Consist of cluster of cells, e.g., Salivary glands.

On the basis of presence or absence of duct glands can be:
(a) Exocrine glands : These glands pour their secretion through a duct. They secrete milk, saliva, mucus, earwax. e.g., goblet cells, salivary glands, tear glands, gastric glands, intestinal glands.
(b) Endocrine glands: They are ductless glands, which pour their secretions into blood or lymph for reaching the target region. Their secretion is called hormone e.g., pituitary gland, thyroid gland, parathyroid glands, adrenal glands.
(c) Heterocrine glands: Both exocrine and endocrine, e.g., pancreas.
On basis of mode of secretion glands can be:
(a) Merocrine: Secretion is discharged
through diffusion, e g., goblet cells, sweat glands.
(b) Apocrine glands: Glandular secretion accumulates in the terminal part of the cell which is pinched off, e.g., mammary glands.
(c) Holocrine glands : The cell filled with secretory product disintegrates during discharge of the product, e.g., sebaceous gland.
(v) Modified epithelium : It is of following types:
(a) Germinal epithelium (generally cuboidal, produces gametes), (b) Glandular epithelium (columnar or cuboidal secretes chemicals and mucus), (c) Sensory epithelium or neuroepithelium. Epithelial cells having sensory hair on free surface and connected with nerve fibres on the other surface (generally columnar, receives and conveys stimuli), e.g, nasal epithelium, taste buds, retina, sensory spots of internal ear. (d) Pigmented epithelium – The cells possess melanin granules, e.g, retinal layer in contact with choroid of eye.

12. Distinguish between
(a) Simple epithelium and compound epithelium.
(b) Cardiac muscle and striated muscle.
(c) Dense regular and dense irregular connective tissues.
(d) Adipose and blood tissue.
(e) Simple gland and compound gland.
Solution:
(a) Differences between simple and compound epithelium are as follows:

(b) Differences between cardiac and striated muscles are as follows:

(c) Differences between dense regular and dense irregular connective tissues are as follows:

(d) Differences between adipose tissue and blood tissue are as follows:

(e) Differences between simple gland and compound gland are as follows:

13. Draw a neat diagram of digestive system of frog.
Solution:

14. Mention the function of the following:
(a) Ureters in frog
(b) Malpighian tubules
(c) Body wall in earthworm.
Solution:
(a) Ureters in frog: Ureter is a transparent duct which arise from outer portion of kidney. In the “male frogs, ureter acts as urinogenital duct which runs backwards from kidneys and opens into the cloaca. It carries both urine and spermatozoa from kidney to the cloaca. In female, ureter conducts only urine from kidneys to the cloaca.
(b) Malpighian tubules: Malpighian tubules are excretory organs present in cockroach. These are present at junction of mid gut and hindgut. These are fine, long, unbranched, yellowish and blind tubules and are 100-150 in number. They help in the removal of excretory products from haemolymph.
(c) Body wall in earthworm: It consists of cuticle, epidermis, muscular layer and parietal peritoneum.
(i) It maintains the characteristic shape of’ the body.
(ii) It protects the internal organs.
(iii) The cuticle prevents excessive evaporation.
(iv) It serves as an ideal respiratory organ.
(v) The receptor cells play a vital sensory function.
(vi) The albumen helps in the formation of cocoon. It also serves as a food for the developing earthworm inside the cocoon.
(vii) Setae and muscles are responsible for locomotion.
(viii) Excretory matter is passed out through nephridiopores.

Category: Class 11 Biology

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