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CBSE Class 11 Biology — NCERT Solutions

CBSE Class 11 Biology NCERT solutions, chapter by chapter — 212 textbook questions solved across 19 chapters. Follows the CBSE syllabus.

About these solutions

212 NCERT textbook questions for CBSE Class 11 Biology, solved step by step across 19 chapters. Each chapter page has every exercise: half the solutions are open to read and the rest are free with a Super Tutor account.

1

The Living World

10 questions solved

  • EXERCISES — The Living World · 10 questions
Q1.Why are living organisms classified?

Given/Context: There are millions of living organisms on Earth, varying enormously in size, shape, colour, habitat, and physiology.

Reason for Classification:

Living organisms are classified for the following reasons:

  1. To ease study: It is impossible to study each organism individually. Classification allows us to study a group of organisms with common characteristics together, making the study manageable.
  1. To understand diversity: Classification helps us understand the enormous diversity of life on Earth in a systematic and organised manner.
  1. To establish relationships: It helps in understanding the evolutionary and phylogenetic relationships among different organisms.
  1. Practical utility: Taxonomic studies are useful in agriculture, forestry, industry, and for knowing our bio-resources. For example, identifying a plant as medicinal or harmful.
  1. To avoid confusion: Each organism is given a unique scientific name, preventing confusion caused by different local/common names.
  1. Predictive value: Once an organism is classified, we can predict many of its characteristics based on the group it belongs to.

Conclusion: Classification is essential to bring order to the vast diversity of living organisms and to facilitate their systematic study.

Q2.Why are the classification systems changing every now and then?

Given/Context: Classification systems have evolved from the two-kingdom system (Linnaeus) to the five-kingdom system (Whittaker) and continue to be revised.

Reasons for changing classification systems:

  1. New discoveries: As new organisms are discovered, existing classification systems may not accommodate them properly, necessitating revision.
  1. Advancement in technology: Development of new tools such as electron microscopy, biochemical analysis, and molecular techniques (DNA sequencing, genomics) provides new data that may alter our understanding of relationships among organisms.
  1. Better understanding of evolutionary relationships: As knowledge of phylogeny (evolutionary history) improves, organisms may be reclassified to reflect true evolutionary relationships more accurately.
  1. Re-evaluation of characteristics: Characteristics once considered important for classification may later be found to be less significant, while new, more reliable characters are identified.
  1. Molecular phylogenetics: Comparison of nucleic acid sequences and protein structures has revealed relationships not apparent from morphology alone, leading to reclassification.

Conclusion: Classification is a dynamic and ever-evolving science. As our knowledge and technology advance, classification systems are updated to reflect the most accurate understanding of the living world.

All 10 The Living World solutions
2

Biological Classification

12 questions solved

  • Exercises · 12 questions
Q1.Discuss how classification systems have undergone several changes over a period of time?

Given/Concept: The history of biological classification spans from ancient times to the modern five-kingdom system.

Answer:

Step 1 – Aristotle's Classification (earliest attempt):
Aristotle was the first to attempt a scientific classification. He classified plants into trees, shrubs, and herbs based on morphology, and animals into those with red blood and those without.

Step 2 – Two-Kingdom Classification (Linnaeus):
Carolus Linnaeus (18th century) classified all living organisms into two kingdoms:

  • Kingdom Plantae (plants)
  • Kingdom Animalia (animals)

Limitation: It did not distinguish between eukaryotes and prokaryotes, unicellular and multicellular organisms, or photosynthetic and non-photosynthetic organisms.

Step 3 – Three-Kingdom Classification (Ernst Haeckel, 1866):
Haeckel proposed a third kingdom Protista to include unicellular organisms, separating them from plants and animals.

Step 4 – Four-Kingdom Classification (Copeland, 1956):
Copeland added Kingdom Monera for prokaryotes (bacteria and blue-green algae), recognising the fundamental difference between prokaryotic and eukaryotic cell organisation.

Step 5 – Five-Kingdom Classification (R.H. Whittaker, 1969):
Whittaker proposed five kingdoms based on:

  1. Cell structure (prokaryotic vs. eukaryotic)
  2. Body organisation (unicellular vs. multicellular)
  3. Mode of nutrition (autotrophic vs. heterotrophic)
  4. Phylogenetic relationships

The five kingdoms are:

KingdomKey Features
MoneraProkaryotes, e.g., bacteria
ProtistaUnicellular eukaryotes
FungiHeterotrophic, saprophytic eukaryotes
PlantaeAutotrophic, multicellular eukaryotes
AnimaliaHeterotrophic, multicellular eukaryotes

Step 6 – Six-Kingdom / Three-Domain System (Carl Woese, 1990):
Based on ribosomal RNA analysis, Woese proposed separating Monera into Archaebacteria (Archaea) and Eubacteria (Bacteria), giving a six-kingdom or three-domain system.

Conclusion: Classification systems have evolved from simple morphology-based two-kingdom systems to complex, phylogeny-based multi-kingdom systems, reflecting advances in cell biology, biochemistry, and molecular biology.

Q2.State two economically important uses of: (a) heterotrophic bacteria (b) archaebacteria

Given/Concept: Bacteria have immense economic importance in industry, medicine, and agriculture.

(a) Two economically important uses of Heterotrophic Bacteria:

  1. In food industry: Lactobacillus and other lactic acid bacteria are used in the production of curd, cheese, and yoghurt. They ferment milk sugar (lactose) to lactic acid.
  1. In medicine/industry: Certain bacteria (e.g., Streptomyces) are used to produce antibiotics such as streptomycin. Bacteria are also used in the production of vitamins (e.g., Vitamin B12), enzymes, and in sewage treatment (decomposing organic waste).

(b) Two economically important uses of Archaebacteria:

  1. Biogas production: Methanogens (e.g., Methanobacterium) present in the gut of ruminants and in marshy areas produce methane (biogas) from organic matter. This biogas is used as a fuel source.
  1. Biotechnology applications: Thermophilic archaebacteria (e.g., Thermus aquaticus) produce heat-stable enzymes such as Taq polymerase, which is essential in the Polymerase Chain Reaction (PCR) technique used in genetic engineering and diagnostics.
All 12 Biological Classification solutions
3

Plant Kingdom

11 questions solved

  • Exercises · 11 questions
Q1.What is the basis of classification of algae?

Given: We need to identify the basis on which algae are classified.

Concept: Algae are classified based on two main criteria:

  1. Type of pigment (photosynthetic pigments) present in the cell.
  2. Type of stored food (reserve food material).

Classification:

ClassPigmentsStored Food
Chlorophyceae (Green algae)Chlorophyll a, b; carotenoidsStarch
Phaeophyceae (Brown algae)Chlorophyll a, c; fucoxanthinMannitol, laminarin
Rhodophyceae (Red algae)Chlorophyll a, d; phycoerythrinFloridean starch

Additional bases include the nature of the cell wall, flagellation of reproductive cells, and the type of sexual reproduction (isogamy, anisogamy, or oogamy).

Conclusion: The primary basis of classification of algae is the type of photosynthetic pigments and the nature of stored food material.

Q2.When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?

Given: We need to identify the site and time of meiosis (reduction division) in five different plant groups.

Concept: Reduction division (meiosis) occurs during spore formation (sporogenesis) in all embryophytes. The location differs based on the dominant generation.

(i) Liverwort:

  • When: During spore formation.
  • Where: In the sporophyte (capsule), specifically in the spore mother cells (sporocytes) inside the capsule (sporangium). Meiosis produces haploid spores.

(ii) Moss:

  • When: During spore formation.
  • Where: In the capsule of the sporophyte, in the spore mother cells present in the sporangium. Meiosis produces haploid spores.

(iii) Fern (Pteridophyte):

  • When: During spore formation.
  • Where: In the sporangia borne on the leaves (sporophylls) of the sporophyte. Meiosis occurs in the spore mother cells within the sporangia.

(iv) Gymnosperm:

  • When: During the formation of microspores and megaspores.
  • Where:
  • In microsporangia (on microsporophylls of male cones) — meiosis in microspore mother cells produces haploid microspores (pollen grains).
  • In megasporangia/ovules (on megasporophylls of female cones) — meiosis in megaspore mother cells produces haploid megaspores.

(v) Angiosperm:

  • When: During the formation of microspores and megaspores.
  • Where:
  • In anthers (microsporangia) — meiosis in microspore mother cells produces haploid pollen grains.
  • In ovules (megasporangia inside the ovary) — meiosis in megaspore mother cells produces haploid megaspores.

Summary Table:

PlantSite of Meiosis
LiverwortCapsule (sporangium) of sporophyte
MossCapsule (sporangium) of sporophyte
FernSporangia on sporophyte
GymnospermMicrosporangia & megasporangia (ovule)
AngiospermAnther (microsporangia) & ovule (megasporangia)
All 11 Plant Kingdom solutions
4

Animal Kingdom

15 questions solved

  • EXERCISES — Animal Kingdom (Class 11 Biology) · 15 questions
Q1.What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?

Given/Concept: Classification requires common fundamental features such as level of organisation, symmetry, coelom, segmentation, notochord, etc.

Answer:
If common fundamental features are not taken into account, the following difficulties would arise:

  1. No systematic grouping: Animals could not be placed into logical, hierarchical groups. Every animal would have to be treated as a separate entity.
  2. Confusion due to superficial similarities: Animals that look alike but are fundamentally different (e.g., a whale and a fish) might be grouped together incorrectly.
  3. Inability to identify evolutionary relationships: Phylogenetic (evolutionary) relationships among animals could not be established.
  4. Lack of universal system: Different scientists would use different criteria, making communication and comparison impossible.
  5. Difficulty in studying millions of species: With over a million animal species, without common features it would be impossible to study, remember, or compare them systematically.
  6. Misidentification: Animals with convergent evolution (similar adaptations but different ancestry) would be misclassified.

Conclusion: Common fundamental features provide the scientific basis for a natural, logical, and universally accepted classification system.

All 15 Animal Kingdom solutions
5
  • Exercises · 10 questions
Q1.How is a pinnately compound leaf different from a palmately compound leaf?

Given: Two types of compound leaves — pinnately compound and palmately compound.

Concept: In a compound leaf, the lamina (leaf blade) is divided into several leaflets. The manner in which leaflets are arranged distinguishes the two types.

FeaturePinnately Compound LeafPalmately Compound Leaf
DefinitionThe leaflets (pinnae) are arranged on both sides of a common axis called the rachis (an extension of the petiole).The leaflets are attached at a single point — the tip of the petiole — and spread out like fingers of a hand.
RachisPresent (midrib-like axis).Absent; leaflets arise from one point.
Arrangement of leafletsAlong the rachis on either side.Radiate from the tip of the petiole.
ExamplesNeem (Azadirachta indica), Rose, Cassia.Silk cotton (Bombax), Hemp (Cannabis), Cleome.

Conclusion: The key difference is the presence of a rachis in pinnately compound leaves, whereas in palmately compound leaves all leaflets arise from a common point at the tip of the petiole.

All 10 Morphology of Flowering Plants solutions
6

Anatomy of Flowering Plants

7 questions solved

  • Exercises · 7 questions
Q1.Draw illustrations to bring out the anatomical difference between (a) Monocot root and Dicot root (b) Monocot stem and Dicot stem

Given: We need to highlight anatomical differences through illustrations and descriptions.


(a) Monocot Root vs. Dicot Root

FeatureMonocot RootDicot Root
Number of xylem bundles (protoxylem)Many (polyarch; 6 or more)Few (2–6; diarch to hexarch)
PithLarge and well-developedSmall or absent
CortexBroadBroad
EndodermisProminent with Casparian stripsProminent with Casparian strips
PericycleSingle-layeredSingle-layered (gives rise to lateral roots and vascular cambium)
Secondary growthAbsentPresent
Conjunctive tissueParenchymatousParenchymatous (later becomes cambium)

Illustration description (T.S. Monocot Root):
From outside inward — Epidermis (with root hairs) → Cortex (multilayered parenchyma) → Endodermis (with Casparian strips) → Pericycle (single layer) → Many alternating xylem and phloem bundles arranged in a ring → Large central pith.

Illustration description (T.S. Dicot Root):
From outside inward — Epidermis → Cortex → Endodermis → Pericycle → 2–6 xylem bundles (exarch) alternating with phloem bundles → Small or no pith.


(b) Monocot Stem vs. Dicot Stem

FeatureMonocot StemDicot Stem
EpidermisSingle layer, cuticle presentSingle layer, cuticle present
HypodermisSclerenchymatousCollenchymatous
Ground tissueNot differentiated into cortex and pithDifferentiated into cortex, endodermis, pericycle and pith
Vascular bundlesScattered throughout ground tissueArranged in a ring
Bundle sheathSclerenchymatous (present)Absent (or parenchymatous)
Bundle typeConjoint, collateral, closed (no cambium)Conjoint, collateral, open (cambium present)
PithAbsent (not distinct)Large, distinct
Secondary growthAbsentPresent

Illustration description (T.S. Monocot Stem — e.g., maize):
Epidermis → Sclerenchymatous hypodermis → Undifferentiated ground tissue with vascular bundles scattered (more numerous and smaller towards periphery, larger towards centre) → Each vascular bundle oval-shaped, surrounded by sclerenchymatous bundle sheath, with phloem (including water-containing cavity) and xylem (Y-shaped metaxylem).

Illustration description (T.S. Dicot Stem — e.g., sunflower):
Epidermis (with trichomes) → Collenchymatous hypodermis → Cortex (parenchyma) → Endodermis (starch sheath) → Pericycle (sclerenchyma + parenchyma) → Ring of vascular bundles (each with xylem below, cambium in middle, phloem above) → Large central pith.

Conclusion: The key differences lie in the arrangement of vascular bundles (scattered vs. ringed), type of hypodermis, presence/absence of cambium, and differentiation of ground tissue.

All 7 Anatomy of Flowering Plants solutions
  • Exercises · 2 questions
Q1.Draw a neat diagram of digestive system of frog.

Given: We need to draw and label the digestive system of a frog (Rana tigrina).

Concept: The digestive system of a frog consists of the alimentary canal and the associated digestive glands.

Diagram Description (to be drawn in the exam):

Draw a neat, well-labelled diagram showing the following parts in order:

Alimentary Canal (in sequence):

  1. Mouth / Buccal cavity — wide opening at the anterior end of the head
  2. Pharynx — short region behind the buccal cavity
  3. Oesophagus — short, wide tube connecting pharynx to stomach
  4. Stomach — J-shaped, muscular sac; divided into cardiac (anterior) and pyloric (posterior) parts
  5. Small Intestine (Duodenum + Ileum) — duodenum is the first part (U-shaped loop); ileum is the long, coiled part
  6. Large Intestine (Rectum) — short, wide tube
  7. Cloaca — common chamber opening to the exterior via the cloacal aperture

Associated Digestive Glands (to be shown with ducts opening into the alimentary canal):

  1. Liver — large, bilobed gland; secretes bile stored in the gall bladder; bile duct opens into the duodenum
  2. Pancreas — lies in the loop of the duodenum; secretes pancreatic juice; pancreatic duct opens into the duodenum

Labels to include in the diagram:

  • Oesophagus
  • Stomach
  • Liver
  • Gall bladder
  • Bile duct
  • Pancreas
  • Pancreatic duct
  • Duodenum
  • Ileum
  • Large intestine / Rectum
  • Cloaca
  • Cloacal aperture

Note for students: The diagram should be drawn in pencil, with all parts clearly labelled using straight label lines. The liver is the largest gland and should be shown on the right side of the stomach.

All 2 Structural Organisation in Animals solutions
8

Cell : The Unit of Life

14 questions solved

  • EXERCISES — Cell: The Unit of Life · 14 questions
Q1.Which of the following is not correct?
(a) Robert Brown discovered the cell.
(b) Schleiden and Schwann formulated the cell theory.
(c) Virchow explained that cells are formed from pre-existing cells.
(d) A unicellular organism carries out its life activities within a single cell.

Correct Answer: (a) Robert Brown discovered the cell.

Justification:
Robert Brown discovered the nucleus (1831), not the cell. The cell was first discovered and described by Robert Hooke in 1665 when he observed cork slices under a microscope and called the box-like structures 'cells'.

  • Option (b) is correct: Schleiden (1838) and Schwann (1839) together formulated the cell theory.
  • Option (c) is correct: Rudolf Virchow (1855) proposed Omnis cellula e cellula — cells arise from pre-existing cells.
  • Option (d) is correct: A unicellular organism (e.g., Amoeba, Paramecium) performs all life functions within a single cell.
All 14 Cell : The Unit of Life solutions
9

Biomolecules

11 questions solved

  • EXERCISES — Biomolecules (Class 11 Biology) · 11 questions
Q1.What are macromolecules? Give examples.

Given/Concept: Macromolecules are large molecular weight biomolecules (generally >10,000 Da) found in living systems. They are polymers made up of repeating monomeric units.

Answer:
Macromolecules are very large molecules with high molecular weights, formed by the polymerisation of smaller units called monomers. They are found in the acid-insoluble fraction of living tissues.

Types and Examples:

MacromoleculeMonomer Units
ProteinsAmino acids
Nucleic acids (DNA, RNA)Nucleotides
Polysaccharides (starch, glycogen, cellulose)Monosaccharides (e.g., glucose)

Note: Lipids are also found in the macromolecular fraction because of their association with membranes, although they are not strictly polymers.

Conclusion: Thus, the three principal classes of true macromolecules in living systems are proteins, nucleic acids, and polysaccharides.

All 11 Biomolecules solutions
10

Cell Cycle and Cell Division

16 questions solved

  • EXERCISES — Cell Cycle and Cell Division · 16 questions
Q1.What is the average cell cycle span for a mammalian cell?

Given/Concept: The cell cycle is the sequence of events from one cell division to the next.

Answer: The average cell cycle span for a mammalian cell is approximately 24 hours.

  • G1G_1 phase: ~11 hours
  • S phase: ~8 hours
  • G2G_2 phase: ~4 hours
  • M phase (Mitosis): ~1 hour

Thus, the total duration ≈ 24 hours. However, this duration can vary greatly depending on the cell type and physiological conditions.

All 16 Cell Cycle and Cell Division solutions
  • EXERCISES — Photosynthesis in Higher Plants · 9 questions
Q1.By looking at a plant externally, can you tell whether a plant is C₃ or C₄? Why and how?

Given/Concept: C₃ and C₄ plants differ in their photosynthetic pathways, but we need to assess whether external morphology alone can distinguish them.

Answer: No, it is generally not possible to distinguish a C₃ plant from a C₄ plant simply by looking at it externally. There are no definitive external morphological features that reliably indicate which photosynthetic pathway a plant uses.

Reasoning:

  • Both C₃ and C₄ plants can look similar in terms of leaf shape, size, colour, and overall plant architecture.
  • However, there is a broad ecological/habitat clue: C₄ plants are often found in tropical, high-temperature, high-light-intensity environments (e.g., sugarcane, maize, sorghum), while C₃ plants are more common in temperate, cooler environments (e.g., wheat, rice, pea).
  • This habitat association is only a rough indicator and not a definitive external test.

Conclusion: External appearance alone cannot confirm whether a plant is C₃ or C₄. Internal anatomical examination (Kranz anatomy) is required for a definitive answer.

All 9 Photosynthesis in Higher Plants solutions
12

Respiration in Plants

16 questions solved

  • Exercises · 16 questions
Q1(a).Respiration and Combustion

Respiration is the step-wise oxidation of food inside the cell that releases energy and traps it in ATP. Combustion is a rapid, uncontrolled burning of a substance, usually outside the cell, in which energy is released mostly as heat.

Differences

  • Respiration occurs inside living cells and is controlled by enzymes.
  • Combustion occurs outside living cells and is not enzyme-controlled.
  • In respiration, energy is released slowly in steps and stored in ATP.
  • In combustion, energy is released suddenly and mainly as heat and light.
  • Respiration occurs at normal biological temperatures; combustion usually needs ignition.
  • Respiration may be aerobic or anaerobic; combustion requires oxygen and is a complete burning process.

So, respiration is a controlled biological process, while combustion is an uncontrolled chemical burning process.

All 16 Respiration in Plants solutions
13

Plant Growth and Development

10 questions solved

  • EXERCISES — Plant Growth and Development · 10 questions
Q1.Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.

Growth: Growth is an irreversible permanent increase in size, shape, number, volume or dry weight of a cell, organ or whole organism. It involves an increase in protoplasmic material and is expressed in parameters such as length, area, volume, cell number, etc.

Differentiation: It is the process by which cells derived from the meristem (which are initially similar) undergo structural and functional changes to become specialised for performing specific functions. For example, cells differentiate to form xylem vessels, sieve tubes, etc.

Development: Development is the sum total of all changes that an organism goes through during its life cycle — from germination of seed to senescence. It includes both growth and differentiation:
Development=Growth+Differentiation\text{Development} = \text{Growth} + \text{Differentiation}

Dedifferentiation: It is the process by which living differentiated cells that have lost the capacity to divide regain the capacity to divide under certain conditions. For example, formation of meristems (interfascicular cambium, cork cambium) from differentiated parenchyma cells.

Redifferentiation: After dedifferentiation, when the cells that have regained the capacity to divide lose it again and mature to perform specific functions, the process is called redifferentiation. For example, secondary xylem and phloem formed from vascular cambium.

Determinate Growth: Growth that ceases after reaching a certain size or stage is called determinate (or limited/closed) growth. It is characteristic of most animals and leaves, flowers, and fruits in plants.

Meristem: Meristems are regions of active cell division in plants. They are composed of undifferentiated, actively dividing cells. Based on position, they are classified as:

  • Apical meristem (at root and shoot tips)
  • Intercalary meristem (at internodes/leaf bases)
  • Lateral meristem (vascular cambium, cork cambium)

Growth Rate: Growth rate is the increase in growth per unit time. It can be expressed as arithmetic or geometric growth rate:
Growth Rate=Increase in parameter (size, number, etc.)Time\text{Growth Rate} = \frac{\text{Increase in parameter (size, number, etc.)}}{\text{Time}}
It can be absolute (total increase) or relative (increase per unit of existing material per unit time).

All 10 Plant Growth and Development solutions
14
  • Exercises · 14 questions
Q1.Define vital capacity. What is its significance?

Given / Concept: Vital capacity is a pulmonary volume measured using a spirometer.

Definition: Vital Capacity (VC) is the maximum volume of air a person can exhale after a maximum inhalation (or vice versa). It is the sum of:
VC=IRV+TV+ERVVC = IRV + TV + ERV
where IRV = Inspiratory Reserve Volume (~2500 mL), TV = Tidal Volume (~500 mL), ERV = Expiratory Reserve Volume (~1000 mL).

Thus, VC≈2500+500+1000=4000 mL (approximately 3.5 – 4.5 L)VC \approx 2500 + 500 + 1000 = 4000 \text{ mL (approximately 3.5 – 4.5 L)}.

Significance:

  1. It is of great clinical significance as it indicates the functional capacity of the lungs.
  2. It reflects the overall health of the respiratory system — a reduced VC indicates restrictive or obstructive lung diseases (e.g., fibrosis, emphysema).
  3. Athletes and trained individuals have a higher VC, indicating better respiratory efficiency.
  4. It helps physicians assess the extent of lung damage and monitor recovery.

Conclusion: Vital capacity is the maximum usable volume of air in the lungs and serves as an important diagnostic indicator of respiratory health.

All 14 Breathing and Exchange of Gases solutions
15

Body Fluids and Circulation

14 questions solved

  • EXERCISES — Body Fluids and Circulation · 14 questions
Q1.Name the components of the formed elements in the blood and mention one major function of each of them.

Given: Formed elements are the cellular components of blood.

Components and their major functions:

  1. Red Blood Cells (RBCs / Erythrocytes):
  • They contain haemoglobin and are responsible for the transport of respiratory gases (O₂ and CO₂).
  1. White Blood Cells (WBCs / Leucocytes): These are of two main types:
  • Granulocytes (Neutrophils, Eosinophils, Basophils)
  • Agranulocytes (Lymphocytes, Monocytes)
  • Major function: Defence against infections and foreign substances (immune response).
  • Neutrophils are phagocytic; Eosinophils resist infections and are associated with allergic reactions; Basophils are involved in inflammatory responses; Lymphocytes produce antibodies; Monocytes are phagocytic.
  1. Platelets (Thrombocytes):
  • Major function: Blood coagulation (clotting) — they release substances necessary for the clotting of blood at the site of injury, thereby preventing excessive blood loss.

Summary Table:

Formed ElementMajor Function
RBCsTransport of O₂ and CO₂
WBCsDefence / Immunity
PlateletsBlood coagulation
All 14 Body Fluids and Circulation solutions
  • EXERCISES — Excretory Products and their Elimination · 12 questions
Q1.Define Glomerular Filtration Rate (GFR).

Given / Concept: GFR refers to the volume of filtrate formed by both kidneys per minute.

Definition: Glomerular Filtration Rate (GFR) is defined as the amount of filtrate formed by the glomeruli of both kidneys per minute.

Key value: In a normal healthy adult, the GFR is approximately 125 mL per minute, which means about 180 litres of filtrate is produced per day.

This filtration is driven by the glomerular capillary blood pressure and is a non-selective process — all small molecules (water, glucose, amino acids, urea, ions) pass through, while large proteins and blood cells are retained.

All 12 Excretory Products and their Elimination solutions
17

Locomotion and Movement

10 questions solved

  • Exercises · 10 questions
Q1.Draw the diagram of a sarcomere of skeletal muscle showing different regions.

Given: We need to represent the structural unit of a myofibril — the sarcomere.

Description of the diagram:

A sarcomere is the functional unit of a myofibril, bounded on either side by Z lines (Z discs).

∣ ⁣ ⁣ ⁣I-band∣A-band⏟H-zone in centre∣I-band⏟Sarcomere ⁣ ⁣ ⁣∣⏟Z lineZ line\underbrace{|\!\!\!\underbrace{\quad I\text{-band}\quad|\underbrace{\quad\quad A\text{-band}\quad\quad}_{\text{H-zone in centre}}|\quad I\text{-band}\quad}_{\text{Sarcomere}}\!\!\!|}_{Z\text{ line}\qquad\qquad\qquad\qquad\qquad\qquad Z\text{ line}}

Regions of a sarcomere:

  1. Z line (Z disc): Boundary of each sarcomere. Thin (actin) filaments are anchored here.
  2. I band (Isotropic band): Light band on either side of the Z line. Contains only thin filaments (actin). It is bisected by the Z line.
  3. A band (Anisotropic band): Dark central band. Contains thick filaments (myosin) throughout, and thin filaments overlap at the periphery of this band.
  4. H zone: Central lighter region within the A band where only thick filaments (myosin) are present (no overlap with actin).
  5. M line: A thin line at the centre of the H zone that holds the thick filaments in position.

Key filaments:

  • Thick filaments: Myosin — present in the A band.
  • Thin filaments: Actin — present in the I band and extend into the A band.

During contraction: The I band and H zone shorten; the A band length remains constant.

All 10 Locomotion and Movement solutions
18
  • Exercises · 10 questions
Q1.Briefly describe the structure of the Brain.

Given: The human brain is the central organ of the neural system enclosed within the skull.

Structure of the Human Brain:

The human brain is protected by the bony cranium (skull) and is covered by three meningeal layers: dura mater (outermost), arachnoid (middle), and pia mater (innermost). The brain is divided into three major regions:


I. Forebrain:
It consists of:

  • Cerebrum: The largest part of the brain. It is longitudinally divided into two cerebral hemispheres connected by the corpus callosum. The outer layer is the cerebral cortex (grey matter), which is highly folded into gyri (ridges) and sulci (grooves), increasing surface area. The inner region contains white matter. The cerebrum is divided into four lobes: frontal, parietal, temporal, and occipital. It controls voluntary movements, memory, intelligence, and sensory perception.
  • Thalamus: Acts as a relay centre for sensory and motor signals to and from the cerebral cortex.
  • Hypothalamus: Controls body temperature, hunger, thirst, sleep, and regulates the pituitary gland. It forms the floor of the diencephalon.
  • Limbic System: Formed by inner parts of cerebral hemispheres and associated deep structures. It is concerned with olfaction, autonomic responses, sexual behaviour, emotional reactions, and motivation.

II. Midbrain:

  • Located between the forebrain and hindbrain.
  • The dorsal portion has four rounded lobes called corpora quadrigemina (two superior and two inferior colliculi).
  • It receives and integrates visual, tactile, and auditory inputs.
  • The midbrain and hindbrain together form the brain stem.

III. Hindbrain:
It consists of:

  • Pons: Contains fibre tracts that interconnect different regions of the brain. It also helps in regulating respiration.
  • Cerebellum: Has a highly convoluted surface. It integrates information from the semicircular canals of the ear and the auditory system. It coordinates voluntary movements, maintains posture and balance.
  • Medulla Oblongata: Connects the brain to the spinal cord. It contains vital centres that control respiration, cardiovascular reflexes, and gastric secretions.

Conclusion: The brain is a highly complex organ that integrates and coordinates all body functions through its three major divisions.

All 10 Neural Control and Coordination solutions
  • Exercises · 9 questions
Q1.Define the following:
(a) Exocrine gland
(b) Endocrine gland
(c) Hormone

(a) Exocrine gland:
Exocrine glands are glands that pour their secretions into a duct (duct glands). These ducts open either onto a body surface or into a body cavity. Examples: salivary glands, sweat glands, liver (bile secretion), pancreas (exocrine part secreting digestive enzymes).

(b) Endocrine gland:
Endocrine glands are ductless glands that secrete their products (hormones) directly into the blood (or lymph). Because they lack ducts, they are also called ductless glands. Examples: pituitary gland, thyroid gland, adrenal gland, pancreas (islets of Langerhans).

(c) Hormone:
Hormones are non-nutrient chemical messengers produced in trace amounts by specialised tissues (endocrine glands or cells). They are secreted directly into the blood and are transported to distant target organs/tissues where they produce specific physiological effects. They act as intercellular messengers and are generally inactivated after their action. Examples: insulin, thyroxine, adrenaline.

All 9 Chemical Coordination and Integration solutions

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