Students can access the CBSE Sample Papers for Class 11 Biology with Solutions and marking scheme Set 4 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Biology Set 4 with Solutions
Time : 3 Hours
Maximum Marks : 70
General Instructions:
- All questions are compulsory.
- The question paper has five sections and 33 questions. All questions are compulsory.
- Section-A has 16 questions ofl mark each; Section-B has 5 questions of 2 marks each; Section- C has 7 questions of 3 marks each; Section- Dhas 2 case-based questions of 4 marks each; and Section-E has 3 questions of 5 marks each.
- There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
- Wherever necessary, neat and properly labeled diagrams should be drawn.
Section – A
Question 1.
Which pigment acts directly to convert light energy to chemical energy? 1
(A) Chlorophyll a
(B) Chlorophyll b
(C) Xanthophyll
Answer:
Option (A) is correct.
Explanation: Chlorophyll ‘a’ is the pigment, which acts directly to convert light energy to chemical energy.
Question 2.
The ultimate electron acceptor of respiration in an aerobic organism is: 1
(A) Cytochrome
(B) Oxygen
(C) Hydrogen
(D) Carotenoid
Answer:
Option (B) is correct.
Explanation: The electron transport chain catalyses an electron flow from NADH (FADH2) to oxygen, which is the final electron acceptor of the respiratory process. It produces water in the mitochondrial matrix.
Question 3.
Match the following and mark the correct option. 1
Column I | Column II |
(a) Fast muscle fibres | (i) Myoglobin |
(b) Slow muscle fibres | (ii) Lactic acid |
(c) Actin filament | (iii) Contractile unit |
(d) Sarcomere | (iv) I-band |
(A) (a) – (i), (b) – (ii), (c) – (iv), (d) – (iii)
(B) (a) – (ii),(b) – (i),(C) – (iii),(d) – (iv)
(C) (a) – (ii), (b) – (i), (c) – (iv), (d) – (iii)
(D) (a) – (iii), (b) – (ii), (c) – (iv), (d) – (i)
Answer:
Option (C) is correct.
Explanation: The ultimate unit of contraction is also known to be the sarcomere. It is composed filaments of actin and myosin. The isotropic component of the sarcomere is formed by actin filaments, which are also referred to as thin filaments.
Few fast muscle fibres are ones that are present in the type of runners that runs for a little period of time. They have high glycogen content and lactic acid in them because of anaerobic respiration and muscle fatigue.
Slow muscle fibre is found in those runners that need to have long run. They also have a huge numbers of mitochondria and myoglobin because they need continuous ATP production or a long period of time.
Question 4.
Difference between virus and viroid is: 1
(A) Absence of protein coat in viroid but present in virus.
(B) Presence of low molecular weight RNA in virus but absent in viroid.
(C) Both (A) and (B)
(D) None of the above.
Answer:
Option (A) is correct.
Explanation: A virus is made up of nucleic acid (DNA or RNA) and a protein coat that is encapsulating the whole nucleic acid.
A viroid is made up of a short strand of circular, single-stranded RNA that has no protein coating.
Question 5.
Splitting of water is associated with: 1
(A) Photosystem I.
(B) Lumen of thylakoid.
(C) Both Photosystem I and II.
(D) Inner surface of the thylakoid membrane.
Answer:
Option (D) is correct.
Explanation: Inside the thylakoids- light reaction occurs, particularly those in the grana region. It involves two distinct types of reactions: the production of assimilation power and photosynthesis of water. Water splitting into hydrogen and oxygen in the illuminated chloroplasts is a phenomena known as photolysis or photocatalysis. The PS II, which is physically situated on the inner surface of the thylakoid membrane, is linked to the water-splitting complex.
Question 6.
The affect of apical dominance can be overcome by which of the following hormone? 1
(A) IAA
(B) Ethylene
(C) Gibberellin
(D) Cytokinin
Answer:
Option (D) is correct.
Explanation: Nutrients travel preferentially in the direction where cytokinin is present. When applied to lateral buds, they promote their growth even in the presence of an apical bud. As a result, they function in antagonism to auxin, which encourages apical dominance. That’s why cytokinin can overcome the apical dominance which is caused by auxins.
Question 7.
Which of the following dyes is best suited for staining chromosomes? 1
(A) Basic Fuchsin
(B) Safranin
(C) Methylene blue
(D) Carmine
Answer:
Option (D) is correct.
Explanation:
(A) Basic fuchsin is a crystallised green dye, which is used for staining tissue sections.
(B) Safranin is the stain for gram-negative bacteria, mucin, etc.,
(C) Methylene blue is used for microscopical examination of DNA or RNA.
(D) Carminestain is used for chromosomes, nuclei, and other structures.
Question 8.
Carefully study the diagram of the human respiratory system with labels (i), (ii), (iii) and (iv). Select the option which gives correct identification and main function and for characteris. 1
(A) (i) Trachea: It is supported by bony rings for conducting inspired air.
(B) (ii) Ribs: When we breathe out, ribs are lifted.
(C) (iii) Alveoli: Thin-walled sac like structures for exchange of gases.
(D) (iv) Diaphragm: It is pulled up when we breathe in.
Answer:
Option (C) is correct.
Explanation: The alveoli in our lungs are small sac- like structures where the exchange of gases takes place.
Question 9.
Apples are generally wrapped in waxed paper to: 1
(A) Prevent sunlight for changing its colour.
(B) Prevent aerobic respiration by checking the entry of O2.
(C) Prevent ethylene formation due to injury.
(D) Make the apples look attractive.
Answer:
Option (B) is correct.
Explanation:Apples respire because of the lenticels on their skin. After harvest, they are wrapped in wax paper to stop the apples from overripening and respiring.
Question 10.
Which stages of cell division do the following figures represent respectively? 1
(A) Metaphase – Telophase
(B) Late Anaphase – Prophase
(C) Telophase – Metaphase
(D) Prophase – Anaphase
Answer:
Option (B) is correct.
Explanation:- The figure A represent Late Anaphase whereas figure B represents Prophase stage of cell division.
Question 11.
ABA acts antagonistic to: 1
(A) Ethylene
(B) Gibberellic add
(C) Cytokinin
(D) IAA
Answer:
Option (B) is correct.
Explanation: Gibberellins are antagonist to the plant hormone ABA (Abscisic acid).
Question 12.
Identify the difference between chlorophyll ‘a’ and chlorophyll ‘b’. 1
Wavelength of light in nanometres (nm)
(A) Chlorophyll ‘a’ (CHQ), Chlorophyll ‘b’ (CH3)
(B) Chlorophyll ‘a’ (C2H5), Chlorophyll V (CH3)
(C) Chlorophyll ‘a’ (CH3), Chlorophyll ‘b’ (CHO)
(D) Chlorophyll ‘a’ (CH3), Chlorophyll ‘b’ (C2H5)
Answer:
Option (C) is correct.
Explanation: Chlorophyll ‘a’ has methyl (-CH3) group and chlorophyll ‘b’ has an aldehyde (-CHO) group.
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer the questions selecting the appropriate option given below.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true but R is false.
(d) A is false but R is true.
Question 13.
Assertion (A): There are only seven categories in taxonomy. 1
Reason (R): Others are called intermediate categories.
Answer:
Option (B) is correct.
Explanation:The biological classification follows a hierarchical order, that includes seven categories: kingdom, division or phylum, class, order, family, genus and species. Intermediate or sub-categories are generally added to the hierarchy of seven categories for finer distinction whenever necessary.
Question 14.
Assertion (A): Cell is the basic unit of life. 1
Reason (R): Robert Hooke discovered the cell.
Answer:
Option (B) is correct.
Explanation: Cell is the basic unit of life according to the cell theory. Cell was discovered by Robert Hooke for the first time.
Question 15.
Assertion (A): There exists a clear division of labour within the chloroplast. 1
Reason (R): The membrane system in chloroplast is responsible for the light reaction and dark reaction.
Answer:
Option (A) is correct.
Explanation: There is a clear division of labour within the chloroplast. The membrane system is responsible for the light reaction (trapping light energy and synthesis of ATP and NADPH) while dark reaction i.e., enzymatic reactions for the reduction of carbon dioxide into sugars using ATP and NADPH takes place in the stroma.
Question 16.
Assertion (A): The PNS comprises of all the nerves of the body associated with CNS. 1
Reason (R): PNS is the site of information processing and control.
Answer:
Option (C) is correct.
Explanation: The PNS comprises of all the nerves of the body associated with the CNS while the brain and the spinal cord are the site of information processing and control.
Section – B
Question 17.
What are trichomes? Write any two functions of it? 2
Answer:
Trichomes are hairs which are often found as epidermal outgrowths. The hairs of aerial parts may be unicellular or multicellular, branched or unbranched.
Function:
- They enclose stationary air and protect the plant organ against water loss and sudden fluctuation in atmospheric temperature.
- In some plants, the hairs are glandular which secrete essential oil and provide characteristic odours to the plant. e.g., Citrus, Ocimum, etc.
Question 18.
The diagram below shows the structure of ribosome. 2
Explain the structure of ribosome ?
Answer:
Structure of ribosomes-
- These are granular organelles which are not enclosed by any membrane.
- These are found either in a free state in the cytoplasm or attached to the endoplasmic reticulum.
- The ribosomes found in the mitochondrial matrix and chloroplast stroma are of 70S type.
- Ribosomes are commonly called as cell engines as they are the sites of protein synthesis.
- Under electron microscope, they appear as spherical bodies of about 150A to 250A in diameter. Ribosomes are composed of roughly equal amount of protein and RNA.
- The ribosomes consist of two unequal-sized units arranged on top of each other. The eukaryotic cells have large ribosomes than prokaryotic cells.
- The prokaryotic ribosomes are 70S while the eukaryotic ribosomes are 80S type. 70S ribosomes have two sub-units 50S and 30S while 80S ribosome has larger 60S and smaller 40S sub-units.
Question 19.
Given graph represents the absorption spectra of the photosynthesis pigments chlorophyll a, chlorophyll b and beta carotenoids. 2
Wavelength of light in nanometres (nm)
What is absorption spectrum? How it is studied?
Answer:
The curve showing the amount of different wavelengths of light absorbed by a substance is called the absorption spectrum.
It is studied with the help of spectrophotometer.
Question 20.
On the basis of your observations. 2
(a) If a patient’s ECG revealed an abnormally long delay between the P wave and the QRS deflection, what would that suggest?
(b) What is the importance of ECG?
Answer:
(a) If a patient s ECG revealed an abnormally long delay between P wave and the QRS deflection, it suggests that there is a delay of conduction from the atria to the ventricles, hence the stimulation from SA nodes are conducting stimuli very slowly to the ventricles.
(b) The importance of ECG is that it gives accurate information about the heart. It is of great diagnostic value in cardiac diseases.
Question 21.
Algae are divided into three main classes: Chlorophyceae, phaeophyceae and rhodophyceae. The presence or absence of pigments is the main basis of the classification of algae. Study the table and fill the blanks. 2
Classes | Common Name | Major Pigments |
(i) ……………………. | Green algae | Chlorophyll a, b |
Phaeophyceae | Brown algae | (ii) …………………………. |
(iii) …………………….. | (iv) ……………….. | Chlorophyll a, d, phycoerythrin |
OR
What do you mean by metagenesis? Give one example of animal that shows metagenesis.
Answer:
(i) Chlorophyceae.
(ii) Chlorophyll a, c, fucoxanthin.
(iii) Rhodophyceae.
(iv) Red algae.
OR
Certain cnidarians exhibit both body forms (polyp and medusa), exhibit alternation of generation, i.e., polyps produce medusae asexually and medusae form the polyps sexually. This phenomenon is called metagenesis. Example: Obelia.
Section – C
Question 22.
The figure given below shows the cellular structure of slime mould. 3
(a) Write a short note on slime moulds.
(b) What is the main body in slime moulds?
Answer:
(a)
- Slime moulds are both plant and animal-like.
- They are plant like in the production of spores during reproduction and animal like in the mode of nutrition.
- Their somatic structures consist of wall-less, multinucleate mobile mass of protoplasm called plasmodium.
- They absorb nutrients directly from the substratum.
- The reproductive stage consists of sporangia and spores which formed after meiosis.
(b) Thalloid body of a slime moulds is formed of slimy protoplasm, which consists of a large number of nuclei and is known as plasmodium.
Question 23.
What are thaliophyta ? How many groups are included in it ? 3
Answer:
Thallophyta possesses the simplest plants which possess undifferentiated or thallus-like forms. Vascular tissues are absent. Differentiation of true roots, stems and leaves is also absent.
Thallophyta includes three groups- algae, fungi and bryophyta.
Question 24.
(a) What are plasmids? 3
(b) Describe the role of plasmid in bacteria.
Answer:
(a) Plasmids are self-replicating, extra-chromosomal segments of double stranded, circular, naked DNA.
(b) Role of plasmid in bacteria is:
- They are used to transfer the information like antibiotic resistance, sex factor, fertility factor etc., from one cell to another.
- They are also used as vector in genetic engineering.
Question 25.
(a) Identify the type of placentation in the given diagram. 3
(b) How can you differentiate between free central and axile placentation?
Answer:
(a) The given diagram is of marginal placentation. In this, one or two alternate rows of ovules occur longitudinally along the ridge in the wall of the ovary in the area of fusion of two margins. A true placenta is believed to be absent. Ovary is unilocular, e.g., Pea, Acacia.
(b) When the ovules are borne on central axis and septa are absent, as in Dianthus and Primrose, the placentation is called free central.When the placenta is axial and the ovules are attached to it in a multilocular ovary, the placentation is said to be axile, as in China rose, tomato and lemon.
Question 26.
List any three salient features of Watson-Crick model of DNA. 3
Answer:
Three salient features of Watson and Crick model are:
- DNA molecule consists of two polynucleotide chains, which are coiled like a rope in helical fashion. So, the two strands form a ‘double helix’.
- The two strands of a polynucleotide are anti parallel i.e., run in the opposite direction. The backbone is formed by the sugar-phosphate sugar chain with the nitrogenous bases projected inside.
- Each DNA helix has alternate minor and major groove having width of 12 Å and 22 Å respectively.
- The distance between two consecutive spirals is 34 Å. Thus, between two consecutive spirals, 10 nucleotides can be adjusted. (Any three)
Question 27.
(a) How does air play a vital role in the production of voice ? 3
(b) On what factors the quality of sound depends ?
(c) Where does exchange of gas occur ?
OR
(a) Briefly explain :
(i) External respiration
(ii) Internal respiration
What is pulmonary respiration?
Answer:
(a) Sound is produced by the vocal cords. When expired air is passed through the true vocal cords under pressure from the lungs, the vocal cords are set into vibration which results in the production of sound.
(b) Factors: The pitch of a sound is determined by the tension on the vocal cords-the greater the tension, the higher the pitch.
The quality of voice depends on the resonators above the larynx, namely the pharynx, mouth and para-nasal sinuses.
(c) Exchange of gaseous occur in lung alveoli and tissue cells.
OR
(a) (i) External respiration: It is simply the intake of oxygen from the surrounding medium (air or water) and giving out of carbon dioxide into that surrounding medium.
(ii) Internal respiration: This involves three steps:
(a) uptake of oxygen by tissue cells.
(b) oxidation of food inside of the cells by oxidising enzymes.
(c) elimination of carbon dioxide from tissues.
(b) Respiration by lungs is termed as pulmonary respiration.
Question 28.
Inspiratory muscles and expiratory muscles play a very important role in respiratory process. Explain their role. 3
Answer:
The contraction of the external intercostal muscles and diaphragm leads to an increase in volume of thoracic cavity. An increase in pulmonary volume decreases the intra-pulmonary pressure to less than atmospheric pressure. Thus, forcing air into the lungs from outside. This is an inspiration. Relaxation of the diaphragm and the inter-costal muscles returns the diaphragm and the sternum to their normal positions and reduce the thoracic volume and thereby the pulmonary volume. This lead to an increase in intra-pulmonary pressure in comparison to the atmospheric pressure. This causes expiration.
Section – D
Question 29.
The structure below shows the cytoplasmic streaming. 4
(a) What is cytoplasmic streaming ?
(b) Which organelle helps in this?
(c) How many types of streaming are there ?
OR
Why cytoplasmic streaming is present in eukaryotes?
Answer:
(a) The autonomic movement of the matrix of cytoplasm of eukaryotic cells is called cytoplasmic or protoplasmic streaming or cyclosis.
(b) Cyclosis is caused by the activity of microfilaments.
(c) These are of two types-
(i) Rotation
(ii) Circulation.
In rotation, the cytoplasmic matrix moves in one direction around a central vacuole. In circulation, the matrix moves in different directions around different vacuoles of the cell.
OR
Cytoplasmic streaming, also called cyclosis, transports nutrients, enzymes, and larger particles within cells, enhances the exchange of materials between organelles, as well as between cells. In some unicellular organisms, such as amoeba, it provides the mechanism for cell locomotion.
Question 30.
Haemoglobin is a red coloured iron containing pigment present in the RBCs. Oxygen can bind with haemoglobin in a reversible manner to form oxyhaemoglobin. Similarly CO2 is carried bv haemoglobin as carbamino-haemoglobin. The oxygen dissociation curve is highly useful in studying the effect of factors like pCO2/ H+ concentration, etc., on binding of O2 with haemoglobin. In the alveoli, there is high pO2, low pCO2, lesser H+ concentration and lower temperature. These factors are all favourable for the formation of HbO2, whereas in the tissues, where low pO2, high pCO2, high H+ concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from the oxyhaemoglobin. 4
(a) Which pigment is formed when O2 binds with haemoglobin?
(b) Which pigment is formed when CO2 binds with haemoglobin?
(c) Name the factors favourable for the formation HbCL.
OR
Based on the diagram, explain what will be the pO2 and pCO2 in the atmospheric air compared to those in the alveolar air?
Answer:
(a) Oxyhaemoglobin pigment is formed when O2 binds with haemoglobin.
(b) Carbamino-haemoglobin pigment is formed when CO2 binds with haemoglobin.
(c) The factors favourable for the formation HbO2 are High pO2, low pCO2, lesser H+ concentration and lower temperature.
OR
pO2 higher, pCO2 lesser. The partial pressure of oxygen in atmospheric air is more than that of oxygen in alveolar air. In atmospheric air pO2 is about 159 mm Hg. In alveolar air, it is about 104 mm Hg. The pCO2 in atmospheric air is lesser than that of pCO2 in alveolar air. In atmospheric air, pCO2 is about 0.3 mm Hg and in alveolar air it is about 40 mm Hg.
Section – E
Question 31.
(a) Name the structures involved in the protection of the brain. 5
(b) Write short note on-
(i) Cerebrum
(ii) Diencephalon
(c) Where is the hunger centre located in human brain?
OR
Distinguish between
(a) Afferent neurons and efferent neurons.
(b) Impulse conduction in a myelinated nerve fibre and unmyelinated nerve fibre.
(c) Explain the terms given below.
(i) Cranial nerves.
(ii) Spinal nerves.
Answer:
(a) The brain is protected by the skull. Inside the skull, it is well protected by cranial meninges that are made up of an outer layer called dura mater, a thin middle layer called arachnoid, and an inner layer called pia mater.
(b)
(i) Cerebrum: Cerebrum is the largest and most prominent part of the brain. It consists of:
- Motor area: Controls the voluntary movements of muscles. It is found in the posterior parts of frontal lobe.
- Sensory area: Found in parietal lobe (temperature, touch, pressure, pain, taste etc.), occipital lobe (vision) and temporal lobes (hearing and smell).
- Association area: Seen in frontal lobe. It is neither clearly sensory nor motor in function. It is responsible for intersensory associations, memory and communication.
- Integrated activities of different centres of the cerebral cortex control intelligence, memory, judgment, learning, thinking and articulate speech.
(ii) Diencephalon: It includes thalamus and Hypothalamus.
- Thalamus: Thalamus is the main centre of coordination for sensory and motor signalling. It is wrapped by cerebrum. It acts as relay station for sensory and motor impulses between cerebrum and other parts of the brain.
- Hypothalamus: Hypothalamus regulates temperature, thirst, hunger and emotions. It also controls pituitary gland. It controls sleep, wakefulness, food intake, blood pressure, heart rate, etc.
(c) The hunger centre located in human brain is hypothalamus.
OR
(a)
S. No | Afferent neuron | Efferent neuron |
(i) | They conduct impulses from the receptors to the CNS. | They conduct impulses from CNS to the effectors. |
(ii) | They are sensory in nature. | They are motor in nature. |
(b)
S. No | Impulse conduction in a myelinated nerve fibre | Impulse conduction in an non-myelinated nerve fibre |
(i) | In a myelinated nerve fibre, the action potential is conducted from one node to another. | In a non-myelinated nerve fibre, the action potential is not conducted from node to node. It is carried along the whole length of the nerve fibre. |
(ii) | The conduction of impulses is faster. | The conduction of impulses is slower. |
(c) (i) Cranial nerves- Cranial nerves arises from different parts of the brain. There are 12 pairs of cranial nerves present in man. Cranial nerves are sensory-motor and mixed.
(ii) Spinal nerves- A pair of spinal nerves arises from each segment of the spinal cord. There are 31 pairs of spinal nerves present in man. All spinal nerves are mixed nerves.
Question 32.
What are the criteria used by Whittaker for delimiting the five kingdom classification? What are the advantages and
disadvantages of this classification? 5
OR
Describe the various salient features of Protista. Name the major groups of this kingdom.
Answer:
He has used the following three criteria for delimiting the different kingdom:
- Complexity of cell structure: prokaryotic and eukaryotic.
- Complexity of body structure or level of structural organisation: unicellular and multicellular.
- Mode of nutrition: autotrophic and heterotrophic.
Advantages of this classification are :
- It exhibits phylogenetic relationship between diverse groups.
- It justifies the position of algae, fungi, higher plants and animals. (Any two)
- Animal and plant kingdoms are more homogenous than the earlier two-kingdom classification.
Disadvantages of this classification are :
(i) Protists show great diversity.
(ii) Three higher kingdoms are polyphyletic.
(iii) Viruses are neither prokaryotes nor eukaryotes. They do not belong to any kingdom. (Any two)
OR
Features of protista are:
- They are single celled, colonial, filamentous eukaryotes.
- They grow in humid and moist environments.
- Some are photosynthetic but some are non-photosynthetic.
- Some forms are like animals whereas some are like plants.
- They have membrane bound cell organelles.
- Examples are protozoa, slime moulds, Euglenoids, Chrysophytes, protistan algae such as diatoms, dinoflagellates or phytoplanktons, etc.
- The protozoan are unicellular (single celled) heterotrophs. Euglena is autotroph.
- Slime moulds are plant-like or animal like. Their somatic body is called plasmodium (acellular, multinucleate, mobile mass of protoplasm lacking the cell wall). (Any three)
The major groups of this kingdom are:
- Photosynthetic protist or protistan algae.
- Consumer decomposer protists-Slime moulds.
- Protozoan protists.
Question 33.
Based on position of calyx, corolla and androecium with respect to ovary on thalamus, the flowers are described as
hypogynous flower, epigynous flower and Perigynous flower. State differences between the three on the basis of
aspects given in the table. 5
Aspects | Hypogynous flower | Epigynous flower | Perigynous flower |
Ovary position | |||
Floral parts | |||
Ovary relationship | |||
Floral symmetry | |||
Examples |
OR
The essential functions of roots are anchorage and absorption of water and minerals in the terrestrial plant. What functions are associated with the roots of aquatic plants. There are certain characteristics provided in the table. Based on them, state the differences between the two.
Characteristics | Roots of aquatic plants | Roots of terrestrial plants |
Oxygen requirement | ||
Absorption of nutrients | ||
Functions | ||
Vascular bundles | ||
Anchorage | ||
Structure | ||
Aerial adaptations | ||
Examples |
Answer:
Aspects | Hypogynous flower | Epigynous flower | Perigynous flower |
Ovary position | Superior ovary (above the attachment of other parts). | Inferior ovary (below the attachment of other parts). | Partially inferior ovary (partially below the attachment of other parts). |
Floral parts | Sepals, petals, and stamens are attached below the ovary. | Sepals, petals, and stamens are attached above the ovary. | Sepals, petals, and stamens attached at the same level as the ovary. |
Ovary
relationship |
Free from floral parts (not fused to them). | Fused with floral parts (attached to them). | Partially fused with floral parts. |
Floral
symmetry |
Actinomorphic (radially symmetrical) | Zygomorphic (bilaterally symmetrical) | Actinomorphic (radially symmetrical) |
Examples | Rose, Lily, Sunflower | Apple, Cherry, Strawberry | Rosemary, Cherry blossom |
OR
Characteristics | Roots of aquatic plants | Roots of terrestrial plants |
Oxygen requirement | Adapted for low oxygen availability. | Adapted for sufficient oxygen availability. |
Absorption of nutrients | Absorb nutrients from the water column or sediment. | Absorb nutrients from the soil. |
Functions | Modified roots carry out the function of photosynthesis, food storage and exchange of gases. | Roots help in anchorage and absorption of nutrients from soil. |
Vascular bundles | Vascular bundles are poorly developed. | Vascular bundles are well developed. |
Anchorage | Provide stability in water currents. | Anchor plants in the soil and provide stability. |
Structure | Often fine, feathery, or filamentous. | Generally thicker and more robust. |
Aerial adaptations | Absent or limited aerial roots. | Present, such as prop roots or aerial roots. |
Examples | Water lilies, Water hyacinths, Pondweeds. | Oak trees, Grasses, Rose bushes. |