Ch 1Living World
5-Mark Questions
Write the distinguishing features of Monera.
The kingdom Monera encompasses all prokaryotic organisms, which are characterized by the absence of a true nucleus and other membrane-bound organelles. These organisms are typically microscopic and include diverse forms such as Mycoplasma, bacteria, actinomycetes, and cyanobacteria. Monerans exhibit various modes of nutrition; some are autotrophic, performing photosynthesis or chemosynthesis, while others are heterotrophic, acting as decomposers, parasites, or symbionts. …
Why do farmers plant leguminous crops in crop rotations/mixed cropping?
Farmers plant leguminous crops in crop rotations and mixed cropping systems because these plants form symbiotic associations with Rhizobium bacteria in their root nodules. Rhizobium is a nitrogen-fixing bacterium that converts atmospheric nitrogen into nitrates and other nitrogen compounds that enrich the soil and increase its fertility. When leguminous crops such as pulses, beans, or clover are grown alternately with non-leguminous crops like paddy or wheat, the soil nitrogen content is naturally replenished without requiring synthetic nitrogen fertilizers. …
2-Mark Questions
Identify the correctly matched pair a. Actinomycete – a) Late blight b. Mycoplasma – b) lumpy jaw c. Bacteria – c) crown gall d. Fungi – d) sandal spike
The correctly matched pair is: a. Actinomycete – lumpy jaw, b. Mycoplasma – sandal spike, c. Bacteria – crown gall, d. Fungi – late blight. Lumpy jaw is a disease of cattle caused by Actinomycetes. Sandal spike disease affecting sandalwood trees is caused by Mycoplasma. Crown gall, a plant disease characterized by tumor-like growths, is caused by the bacterium Agrobacterium tumefaciens. Late blight of potato is a fungal disease caused by Phytophthora infestans.
Differentiate Homoiomerous and Heteromerous lichens.
Homoiomerous lichens have algal cells evenly and uniformly distributed throughout the entire thallus, mixed with fungal hyphae in a homogeneous manner without any distinct layering or organization. In contrast, heteromerous lichens display a distinct stratified structure with clearly defined layers: an upper cortex of fungal hyphae, a middle algal layer containing photosynthetic algal cells, and a lower cortex of fungal hyphae, with a medulla of loosely arranged hyphae in between. …
Bacterial chlorophyll is also known as a. Chlorophyll b. Bilirubin c. Chromatium d. Chioridin.
Bacterial chlorophyll is also known as Chloridin. Unlike the chlorophyll found in plants, bacterial chlorophyll is a type of bacteriochlorophyll that absorbs light at longer wavelengths, typically in the infrared spectrum. This allows photosynthetic bacteria to utilize light energy that is not accessible to plants. Chloridin is the specific term used for this type of photosynthetic pigment found in certain bacteria.
1-Mark Questions (MCQ)
Which one of the following statements about viruses is correct? a. Possess their own metabolic system b. They are the facultative parasites c. They contain DNA or RNA d. Enzymes are present
b. They are the facultative parasites
Ch 2Plant Kingdom
5-Mark Questions
Differentiate Haplontic and Diplontic life cycle.
The haplontic and diplontic life cycles represent two different patterns of alternation of generations in organisms. In the haplontic life cycle, the gametophyte phase is dominant and photosynthetically independent, representing the main vegetative body of the organism. The sporophyte phase is recessive and represented only by the zygote. When the zygote undergoes meiosis, haploidy is restored, and the haploid gametophyte develops again. Examples of organisms with haplontic life cycles include Volvox and Spirogyra. …
What do we infer from the term Pucnoxylic?
Secondary growth is also traced in gymnosperms, E.g. Cycas and Pinus. The wood may be compact with narrow medullary ray this condition known as Pycnoxlic seen in Pinus. It is opposite to Manoxylic condition which is seen in Cycas.
2-Mark Questions
What is plectostele – Give example.
Plectostele is a type of stele in which the xylem plates alternate with the phloem plates in a radial arrangement. This arrangement creates a characteristic pattern where vascular tissues are interspersed with one another rather than forming concentric rings. The plectostele is found in certain pteridophytes, particularly in club mosses. A classic example of a plant exhibiting plectostele is Lycopodium clavatum, commonly known as the club moss or stag's horn moss, where this vascular arrangement is clearly observable in cross-sections of the stem.
Gemmae formation is not traced in which three of the given four options a. Marchanlia b. Riella c. Ricciocarpus d. Anthoceros (i) ab & c (ii) be & d (iii) ab & d (iv) ac & d
The correct answer is (ii) be & d. Gemmae formation is not traced in Riella, Ricciocarpus, and Anthoceros. Gemmae are asexual reproductive structures found in certain bryophytes, particularly in liverworts like Marchantia, where they are produced in specialized cup-like structures called gemma cups. However, not all bryophytes produce gemmae. Riella and Ricciocarpus, which are also liverworts, do not form gemmae as a means of reproduction. Anthoceros, which is a hornwort, similarly does not produce gemmae. …
Find out the aquatic bryophytes of the following. a. Riella b. Ricciocarpus c. Riccia d. Bryopteris (i) a&c (ii) b&c (iii) c&d (iv) a&b
(iv) a&b
1-Mark Questions (MCQ)
Choose the correct statement regarding the common characters of Gymnosperm and Angiosperm only a. Pollen tube help in the transfer of male nucleus & fertilization is Siphonogamous b. Heterospory is of common occurrence c. Vessels are the chief water-conducting elements d. Pollination is by Anemophilous method only
a. Pollen tubes help in the transfer of the male nucleus & fertilization is Siphonogamous.
Ch 3Vegetative Morphology
5-Mark Questions
Write the similarities and differences between * Avicennia & trapa * Banyan & silk cotton * Fusiform and Napiform root
I. Avicennia & trapa Avicennia Trapa (water chestnut) Live in marshy leaves Live in aquatic habitat Has negatively geotrophic root known as respiratory roots-with pneumatophores help in exchange of gases Has photosynthetic or assimilatory roots – help in photosynthesis. II. Banyan & silk cotton Banyan Silk cotton Has pillar roots- grow vertically downward from the lateral branches to soil -to give additional support. Has broad plant-like outgrowths develop obliquely towards the base all around the trunk – to give support. III. …
How root climbers differ from stem climbers?
Root climbers and stem climbers differ significantly in their climbing mechanisms and structural adaptations. Root climbers, such as Piper betel and Piper nigrum, climb with the help of adventitious roots that arise from the nodes of the stem. These roots attach to the support and help the plant climb upward. In contrast, stem climbers, such as Ipomoea littoralis, possess no special climbing structures; instead, the stem itself coils around the support in a helical manner, allowing the plant to ascend. …
2-Mark Questions
Why lateral roots are endogenous?
Lateral roots are endogenous in origin because they arise from the pericycle, which is the innermost layer of the cortex located internal to the endodermis. Since lateral roots originate from internal tissues rather than from the epidermis or cortex, they are classified as endogenous. This internal origin distinguishes them from adventitious roots, which arise from non-root tissues such as stems or leaves.
Identify the Diagram and Label ABCD from the diagram
The question asks to identify a diagram and label its parts A, B, C, and D. Without the diagram, a specific answer cannot be provided. However, typically in botany, such diagrams might represent plant organs like roots, stems, or leaves, or their modifications. For example, if it were a diagram of a root, A might be the root cap, B the region of cell division, C the region of elongation, and D the region of maturation.
Among the given, Find out the odd man with reference to the fibrous root system. a. Eleusineeoracana b. Pennisetumamericanum c. Zingiferaoffieinalis d. Ficusbenahaliensis
Among the given options, Ficus benghalensis is the odd one out with reference to the fibrous root system. Eleusine coracana (finger millet), Pennisetum americanum (pearl millet), and Zingifera officinalis (ginger) all typically exhibit a fibrous root system, which is characteristic of monocotyledonous plants or plants with modified stems. Ficus benghalensis, commonly known as the Banyan tree, is a dicotyledonous plant and possesses a tap root system, which later develops into extensive prop roots for support, but its primary root system is not fibrous.
1-Mark Questions (MCQ)
The study about external features of an organism is known as …………….
(a) morphology
Ch 4Reproductive Morphology
5-Mark Questions
Explain different types of placentation with example
Placentation refers to the arrangement of placentae (the tissue to which ovules are attached) within the ovary. Understanding different types of placentation is crucial for plant classification and identification. Marginal placentation occurs when the placentae are located along the margin of a unicarpellate ovary, as seen in members of Fabaceae. Axile placentation is found in compound ovaries with septa, where the placentae arise from the central column or axis, characteristic of Hibiscus, tomato, and lemon. …
Attachment of Anther Description Example 1. Basiifixed …………………. Datura 2. …………………. The apex of filament is attached to the dorsal side of the anther Hibiscus 3. Versatile The filament is attached to the anther at the midpoint …………….. 4. ………….. The filament is continued from the base to the apex Ranunculus
The anther attachment to the filament varies based on the point of connection and the nature of the filament. Basifixed attachment occurs when the base of the anther is attached to the tip of the filament, as seen in Datura. Dorsifixed attachment is when the apex of the filament is attached to the dorsal side of the anther, characteristic of Hibiscus and many other plants. Versatile attachment describes the condition where the filament is attached to the anther at the midpoint, allowing the anther to swing freely, which is typical in grasses and many wind-pollinated flowers. …
2-Mark Questions
A true fruit is the one where a. only ovary of the flower develops into fruit. b. ovary and caly x of the flower develops into fruit. c. ovary, caly x, and thalamus of the flower develops into fruit. d. All floral whorls of the flower develops is to fruit.
The correct answer is a. only ovary of the flower develops into the fruit. A true fruit is defined as the mature ovary of the flower that develops after fertilization of the ovule. The ovary wall develops into the pericarp, which forms the fruit wall, while the ovules inside develop into seeds. In true fruits, only the ovary undergoes this transformation, whereas in accessory or false fruits, other floral parts such as the calyx, thalamus, or other floral whorls may also contribute to the fruit structure. …
Explain different type of fleshy fruit with suitable example
Fleshy fruits are fruits with a soft, succulent pericarp that is edible and contains seeds. There are several types of fleshy fruits classified based on their structure and development. Berry is a fleshy fruit developed from a single carpel with seeds embedded in the pulp, such as banana, tomato, and grape. Drupe is a fleshy fruit with a hard, woody endocarp (stone) surrounding the seed, with examples including mango, coconut, and peach. Pome is a fleshy fruit where the receptacle forms the edible part along with the ovary wall, as seen in apple and pear. …
Br, Ebrl, O7 p3+3 A(3) Go – is a. This F.D of male flower of musa b. The F.D of crotalaria juncea c. The F.D. of male flower of phyllanthus amaras d. The F.D of male flower of cocos nucifera
The floral diagram (F.D.) represented by Br, Ebrl, O7 p3+3 A(3) Go is the F.D. of the male flower of Phyllanthus amarus. This formula indicates the presence of bracts (Br) and bracteoles (Ebrl). The perianth (P) consists of two whorls of three tepals each (3+3), which are free. The androecium (A) has three stamens, which are free (3). The gynoecium (G) is superior (o) and has three carpels, which are fused (3), but in a male flower, the gynoecium is represented as rudimentary or absent, hence the 'o' indicating a superior ovary that is non-functional or absent in male flowers.
1-Mark Questions (MCQ)
Vexillary aestivation is characteristic of the family a. Fabaceae b. Asteraceae c. Solanaceae d. Brassieaceae
a. Fabaceae
Ch 5Taxonomy and Systematic Botany
5-Mark Questions
What is the role of national gardens in conserving biodiversity
National botanical gardens play a multi-faceted and crucial role in conserving biodiversity. Firstly, they serve as living museums, maintaining extensive collections of diverse plant species, including rare, endangered, and endemic ones, thereby acting as ex-situ conservation sites. These collections provide a genetic reservoir that can be vital for future restoration efforts or scientific research. Secondly, many botanical gardens have significant aesthetic value, attracting a large number of visitors. …
Where will you place the plants which contain two cotyledons with cup-shaped thalamus
Plants that contain two cotyledons are classified under the Class Dicotyledoneae. Within the classification system, particularly Bentham and Hooker's system, the presence of a cup-shaped thalamus is a distinctive feature used to further categorize dicotyledonous plants. Specifically, plants exhibiting a cup-shaped thalamus are placed under the Subclass Polypetalae, which is characterized by free petals. Furthermore, within Polypetalae, they belong to the Series Thalamiflorae. …
2-Mark Questions
The latest ICBN was held at a) Cambridge – England b) Leningrad – Russia c) Shenzhen – China d) Rio-de genero – America
The latest International Code of Nomenclature for algae, fungi, and plants (ICN, formerly ICBN) was held at Shenzhen, China. This congress, the 19th International Botanical Congress, took place in 2017 and resulted in significant updates and revisions to the rules governing the naming of these organisms. These congresses are crucial for maintaining stability and clarity in botanical nomenclature worldwide.
Phenology is the study a) Pollen grains structure b) Development of gametes c) Study of climate and weather on plants d) Study of functional aspects of plants
Phenology is the scientific study of periodic plant and animal life cycle events and how these are influenced by seasonal and interannual variations in climate and weather. It specifically focuses on the timing of biological events, such as flowering, leafing, fruiting, and dormancy in plants, and relates these timings to environmental factors like temperature, rainfall, and day length. This field is crucial for understanding the impacts of climate change on ecosystems, agricultural planning, and predicting ecological responses to environmental shifts. …
Find out the Correct Statements the given below. a) Scientific Names are treated as Latin regardless of their derivation b) Cryptogams include non-flowering plants c) Linnaeus system of classification is known as the Natural system of classification d) According to APG IV Monocots contain 10 orders and 37 families (I) a & b (II) b & c (III) c & d (IV) a & d
The correct statements are (a) Scientific Names are treated as Latin regardless of their derivation, and (b) Cryptogams include non-flowering plants. Therefore, option (I) a & b is the correct answer. Scientific names are indeed latinized to ensure universality and stability in nomenclature, making them understandable across different languages and regions. Cryptogams, such as algae, fungi, bryophytes, and pteridophytes, are characterized by their hidden reproductive organs and do not produce flowers or seeds, hence they are non-flowering plants. …
1-Mark Questions (MCQ)
Who is called the father of Botany?
(b) Theophrastus
Ch 6Cell The Unit of Life
5-Mark Questions
State the Protoplasm theory
Fischer in 1894 & Hardy ( 1899 ) Proposed the Colloidal theory of Protoplasm (the physical basis of life) It is a colloidal system with water, many biological import things, glucose, fatty acids, amino acids minerals, vitamins hormones & enzymes are seen. Homogenous -These solutes are soluble Heterogenous – Solutes are not soluble – This Forms the basis for its colloidal nature. Protoplasm occur in 2 states but interconvertible
Distinguish between Prokaryotes & Eukaryotes.
Prokaryotes and Eukaryotes represent two fundamental types of cells, differing significantly in their structural organization and complexity. Prokaryotic cells, typically 1-5 µm in size, lack a true nucleus; their genetic material is located in a region called the nucleoid, which is not enclosed by a nuclear membrane and does not contain a nucleolus. The DNA in prokaryotes is usually circular and is not associated with histone proteins. RNA and protein synthesis are coupled and occur simultaneously in the cytoplasm. Ribosomes are of the 70S type, composed of 50S and 30S subunits. …
2-Mark Questions
Which of the three, come under the system of the membrane in Eukaryotic cell a) Mitochondria b) Nuclear Membrane c) Golgi apparatus d) Endoplasmic reticulum (i) a, b & c (ii) b, c & d (iii) a, c & d (iv) a, b & d
The endomembrane system in eukaryotic cells is a collection of membranes and organelles that work together to modify, package, and transport lipids and proteins. This system includes the nuclear membrane, which encloses the nucleus; the endoplasmic reticulum, a network of interconnected membranes involved in protein and lipid synthesis; and the Golgi apparatus, which further processes and packages these molecules. …
The 60 s large subunit of Eukaryotes contain a) 23 s & 5 s – large subunit b) 16 s r RNA in large subunit c) 18 s r RNA in large subunit d) 28 s, 5-8 sand 5 s in large subunit
The 60S large ribosomal subunit in eukaryotes is a complex structure composed of ribosomal RNA (rRNA) and proteins. This large subunit specifically contains three distinct rRNA molecules: the 28S rRNA, the 5.8S rRNA, and the 5S rRNA. These rRNA molecules, along with numerous ribosomal proteins, are crucial for the catalytic activity and structural integrity of the ribosome, playing a vital role in protein synthesis. The other options provided contain incorrect rRNA components or sizes for the eukaryotic 60S subunit. Therefore, the correct answer is d) 28 s, 5-8 s and 5 s in large subunit.
Bring out the significance of Phase Contrast Microscope
The phase-contrast microscope is a specialized optical instrument that enables the observation of living cells, tissues, and cultured cells in their natural state without the need for staining or fixing. This microscope is particularly valuable for observing dynamic cellular processes such as mitosis, where the visualization of living chromosomes and spindle fiber movements is essential. …
1-Mark Questions (MCQ)
Scientist who named the unicellular particles as ‘animalcules’ …………….
(c) Antonie van Leeuwenhoek
Ch 7Cell Cycle
5-Mark Questions
Write any three significance of mitosis
Mitosis is a fundamental process of cell division that results in two daughter cells each having the same number and kind of chromosomes as the parent nucleus, typically for growth, repair, and asexual reproduction. One significant aspect of mitosis is that it produces an exact copy of the parent cell, meaning the daughter cells are genetically identical to the parent cell. This genetic stability is crucial as it ensures that all cells in an organism carry the same genetic information, which is vital for proper functioning and development. …
Differentiate between Mitosis and Meiosis.
Mitosis and meiosis are two distinct types of cell division with significant differences. Mitosis involves one division of the nucleus, whereas meiosis involves two successive divisions. In mitosis, the number of chromosomes in daughter cells remains the same as the parent cell (2n), while in meiosis, the chromosome number is halved in the daughter cells (n), producing haploid cells. …
2-Mark Questions
Un differentiated cells include a) Stem cells in animals b) Meristematic cells in plants c) RBC which carry out the transportation of oxygen d) Mesophyll cells which carry out photosynthesis (i) a & b (ii) c & d (iii) a & c (iv) b & c
The correct answer is (i) a and b. Undifferentiated cells are those that have not yet specialized and retain the ability to divide and differentiate into various cell types. Stem cells in animals are undifferentiated cells that can self-renew and differentiate into specialized cell types. Meristematic cells in plants are undifferentiated cells located in growing regions such as root tips and shoot apices that continuously divide to produce new cells for growth. …
In the S phase of cell cycle a) Amount of DNA doubles in each cell b) Amount of DNA remain same in each cell c) Chromosome number is increased d) Amount of DNA is reduced to half in each cell
The correct answer is a) Amount of DNA doubles in each cell. During the S phase (synthesis phase) of the cell cycle, DNA replication occurs. Each chromosome, which initially consists of a single chromatid, is replicated to form two identical sister chromatids joined at the centromere. Since the cell contains multiple chromosomes, and each chromosome is replicated, the total amount of DNA in the cell doubles. However, the chromosome number remains the same because sister chromatids are still counted as one chromosome until they separate during anaphase. …
During Anaphase (I) The daughter chromosome move to the opposite poles due to the shortening of the phragmoplast (II) due to the thickening of chromosomes (III) Shortening of microtubules (IV) Shortening of asters
The correct answer is (III) Shortening of microtubules. During anaphase, the separation and movement of chromosomes to opposite poles of the cell is primarily facilitated by the shortening of microtubules that form the spindle apparatus. The spindle microtubules attached to the kinetochores of chromosomes shorten, pulling the separated chromatids (now individual chromosomes) toward opposite poles. This shortening of microtubules is the primary mechanism responsible for chromosome movement during anaphase. …
1-Mark Questions (MCQ)
The correct sequence in cell cycle is a) S M G1 G2 b) S G1 G2 M c) G1 S G2 M d) M G1 G2 S
c) G1 S G2 M
Ch 8Biomolecules
5-Mark Questions
Explain the three types of Co-Factors.
Co-factors are non-protein organic or inorganic substances that assist enzymes in catalyzing biochemical reactions. There are three main types of co-factors. First, metal ions such as Mg2+, Ca2+, Zn2+, and Fe2+ act as co-factors by facilitating enzyme-substrate interactions and stabilizing enzyme structure. Second, prosthetic groups are organic co-factors that remain permanently or tightly bound to the enzyme protein. …
Choose the right answer a) Amylose – linear unbranched polymer of with 20% starch b) Amylopectin – a polymer with some 1,6 linkages that give it a linear structure c) Inulin – Polymer of galactose d) Amino acid – Here a basic structure of carbon linked to a basic amino group
The correct answer is d) Amino acid – Here a basic structure of carbon linked to a basic amino group. An amino acid is an organic compound with a characteristic structure consisting of a central carbon atom bonded to four different groups: an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and a variable side chain (R group). This basic structure is fundamental to all amino acids and allows them to link together through peptide bonds to form proteins. …
2-Mark Questions
Why do we call Glucose and Fructose Isomers -Discuss.
Glucose and fructose are classified as isomers because they possess the same molecular formula, C6H12O6, but differ in their structural arrangements. Although both contain six carbon atoms, twelve hydrogen atoms, and six oxygen atoms, the atoms are arranged differently in space. Glucose is an aldohexose with an aldehyde group at the first carbon, forming a six-membered pyranose ring in its cyclic form. Fructose is a ketohexose with a ketone group at the second carbon, forming a five-membered furanose ring in its cyclic form. …
Classify enzyme reactions:
Enzymes can be classified based on their location and site of action into two main categories. Extracellular enzymes are secreted outside the cell and function in external environments. These enzymes work in the extracellular space, such as in the digestive tract, where they catalyze the breakdown of complex food molecules. Examples include digestive enzymes such as amylase, which breaks down starch; protease, which breaks down proteins; and lipase, which breaks down fats. …
Draw the structural formula of 3 simple amino acids – Glycine, Alanine & Valine.
Amino acids with non-polar aliphatic R groups include glycine, alanine, valine, leucine, methionine, and isoleucine. Glycine has the simplest structure with a hydrogen atom as its R group, making it the only amino acid without a chiral center. Its structural formula shows the central carbon bonded to an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and another hydrogen atom as the R group. Alanine has a methyl group (CH3) as its R group attached to the central carbon, giving it a small non-polar side chain. …
1-Mark Questions (MCQ)
This has nothing to do with the structure of a. Cytosine b. Pyrimidine c. Adenine d. Thyamine
c. Adenine
Ch 9Tissue and Tissue System
5-Mark Questions
Explain sclereids with their types
Sclereids are dead cells with highly thickened and lignified cell walls that provide mechanical support and protection to plant tissues. These cells are characterized by their isodiametric shape, meaning they are roughly equal in all dimensions, though some sclereids may be somewhat elongated. The cell wall of sclereids is extremely thick due to extensive lignification, which is the deposition of lignin in the cell wall matrix, making them very hard and rigid. The lumen, or the internal cavity of the cell, is much reduced in size due to the thick cell wall deposition. …
What are sieve tubes? Explain.
Sieve tubes are long tube-like conducting elements in the phloem. These are formed from a series of cells called sieve tube elements. The sieve tube elements are arranged one above the other and form vertical sieve tube. The end wall contains a number of pores and it looks like a sieve. So it is called as sieve plate. The sieve elements show nacreous thickenings on their lateral walls. They may possess simple or compound sieve plates. The function of sieve tubes are believed to be controlled by campanion cells In mature sieve tube, Nucleus is absent. It contains a lining layer of cytoplasm. …
2-Mark Questions
When a leaf trace extends from a vascular bundle in a dicot stem, what would be the arrangement of vascular in the veins of the leaf? a) Xylem would be on top and the pholem on the bottom b) Pholem would be on the top and the xylem on the bottom c) Xylem would encircle the pholem d) Pholem would encircle the xylem
The correct answer is (a) Xylem would be on top and the phloem on the bottom. In dicot leaves, the vascular bundles are arranged with a characteristic pattern where the xylem is positioned on the upper (adaxial) surface facing the upper epidermis, while the phloem is positioned on the lower (abaxial) surface facing the lower epidermis. This arrangement is consistent with the vascular bundle organization in dicot stems, where the xylem is typically oriented toward the center and the phloem toward the outer region. …
Why the cells of sclerenchyma and tracheids become dead?
The cells of sclerenchyma and tracheids become dead because they lack protoplasm. During their development and maturation, these cells undergo programmed cell death where the protoplasm, including the nucleus, cytoplasm, and organelles, degenerates and is completely lost. This process occurs after the cell wall has been fully formed and lignified. The loss of protoplasm is essential for the functioning of these cells because it allows for the formation of large hollow lumens that facilitate the transport of water and minerals in the case of tracheids, and provides mechanical strength in the ca …
The tunica is: (a) the peripheral zone of shoot apex, that forms cortex (b) the inner zone of shoot apex, that forms stele (c) the peripheral zone of shoot apex, that forms the epidermis (d) the inner zone of shoot apex, that forms cortex and stele
(c) the peripheral zone of shoot apex, that forms the epidermis. The tunica is the outermost layer of the shoot apical meristem and consists of one or more layers of cells that divide anticlinally (perpendicular to the surface). Through these divisions, the tunica gives rise to the epidermis and contributes to the formation of the outer tissues of the shoot. This is distinct from the corpus, which is the inner zone responsible for forming the stele and cortex through various planes of cell division.
1-Mark Questions (MCQ)
Read the following sentences and identify the correctly matched sentences. i) In exarch condition, the protoxylem lies outside of the metaxylem. ii) In endarch condition, the protoxylem lies towards the centre. iii) In centrarach condition, metaxylem lies in the middle of the protoxylem iv) In mesarch condition, protoxylem lies in the middle of the metaxylem a) i, ii, and iii only b) ii, iii, and iv only c) i, ii, and iv only d) All of these
c) i, ii and iv only
Ch 10Secondary Growth
5-Mark Questions
Continuous state of dividing tissue is called meristem. In connection to this, what is the role is lateral meristem?
The secondary growth in dicots and gymnosperms is brought about by two lateral meristems. * Vascular cambium and * Cork cambium 1. Vascular cambium: The vascular cambium is the lateral meristem that produces the secondary vascular tissues, i.e.. secondary xylem and secondary phloem. Origin and Formation of Vascular Cambium: * A strip of vascular cambium originate from the procambium is present between xylem and phloem of the vascular bundle. This cambial strip is known as intrafascicular or fascicular cambium. …
A timer merchant bought 2 logs of wood from, a forest & named them A & B, The log A was 50 year old & B was 20 years old. Which log of wood will last longer for the merchant? Why?
In wood, the older it is, the stronger it becomes. Log A – Which was 50 years old is stronger and it will last longer. In a tree the central part of the wood will be darker in colour, dead in nature known as Heartwood or Duramen, and the outer sad wood is lighter in colour, living and conducting water. In the central Heartwood the conduction is blocked by the formation of tyloses from the nearby parenchyma cells, and dead. In the fully developed tyloses, starch crystals, resins, gums, oils tannins and coloured substances are found and it becomes very hard and durable. …
2-Mark Questions
Commercial cork is obtained from a) Oak b) Silver oak c) Pine d) Ficus
The correct answer is a) Oak. Commercial cork is obtained from the bark of the oak tree, specifically from Quercus suber, commonly known as the cork oak. The cork tissue, which is the outer protective layer of the bark produced by the phellogen, is harvested from these trees and used commercially for various purposes including bottle stoppers, insulation materials, and other products. The other options like silver oak, pine, and Ficus do not yield commercially viable cork.
Which of the statement is not correct? (a) In temperate regions, the cambium is very active in the winter season. (b) In temperate regions, the cambium is very active in the spring season. (c) In temperate regions, cambium is less active in the winter season. (d) In temperate regions earlywood is formed in the spring season.
The correct answer is (a) In temperate regions, the cambium is very active in the winter season. This statement is not correct because in temperate regions, the vascular cambium is actually dormant and inactive during the winter season. The cambium becomes very active during the spring season when temperatures rise and growth resumes. During winter, the cambium remains relatively inactive, which is why statement (a) contradicts the actual physiological behavior of cambium in temperate climates. Statements (b), (c), and (d) are all correct descriptions of cambial activity in temperate regions.
Removal of a ring of wood tissue outside the vascular cambium from the tree trunk kills it because a) Water cannot move up b) Food does not travel down and root become starved c) Shoot apex become starved d) Annual rings are not produced
The correct answer is b) Food does not travel down and root become starved. When a ring of wood tissue outside the vascular cambium is removed from the tree trunk, this process is called girdling or ringing. The removal of this tissue destroys the phloem, which is the tissue responsible for the translocation of food (photosynthates) from the leaves downward to the roots and other non-photosynthetic parts. Without this downward movement of food, the roots become starved of nutrients and eventually die, leading to the death of the entire tree. …
1-Mark Questions (MCQ)
When we peel the skin of a potato tuber we remove a) Periderm b) Epidermis c) Cuticle d) Sapwood
a) Periderm
Ch 11Transport in Plants
5-Mark Questions
If the concentration of salt in the soil is too high and the plants may wilt even if the field is thoroughy irrigated. Explain
High salt concentration in the soil creates an osmotic potential that is more negative than that of the plant cells, making it difficult for plants to absorb water even when the soil is thoroughly irrigated. When the concentration of salt in the soil is too high, the osmotic potential of the soil solution becomes very low, meaning the soil solution has a lower water potential than the plant cells. …
How phosphorylase enzyme open the stomata in starch sugar interconversion theory?
The discovery of the enzyme phosphorylase in guard cells by Hanes in 1940 provided strong support for the starch-sugar interconversion theory regarding stomatal movement. This theory explains how the interconversion of starch and sugar, influenced by pH changes, leads to the opening and closing of stomata. During the day, photosynthesis occurs in the guard cells, consuming carbon dioxide and leading to an increase in the pH of the guard cell sap. …
2-Mark Questions
List out the non-photosynthetic parts of a plant that need a supply of sucrose?
The non-photosynthetic parts of a plant that require a continuous supply of sucrose, which is the primary form of sugar transported in plants, include various organs that are actively growing, storing food, or performing metabolic functions without producing their own sugars. These essential sink regions include the roots, which are responsible for absorbing water and nutrients and anchoring the plant; tubers, such as potatoes, which are specialized underground stems that serve as storage organs for carbohydrates; developing fruits, which require a significant amount of energy and building blo …
What is meant by Porin?
Porin is a large transporter protein found in the outer membrane of plastids, mitochondria and bacteria which facilitates the passage of smaller molecules through the membrane. These proteins form channels or pores that allow water and small solutes to cross the membrane while restricting the passage of larger molecules. Porins are essential for maintaining selective permeability and enabling controlled transport of ions and small organic molecules across these organellar membranes.
What are the three types of plasmolysis?
Plasmolysis is the process in which plant cells lose water when placed in a hypertonic solution, causing the protoplast to shrink away from the cell wall. This phenomenon is categorized into three distinct types based on the extent of water loss and protoplast shrinkage. These types are incipient plasmolysis, evident plasmolysis, and final plasmolysis. Incipient plasmolysis is the initial stage where the protoplast just begins to pull away from the cell wall at the corners. Evident plasmolysis occurs when the protoplast has significantly shrunk and detached from most of the cell wall. …
1-Mark Questions (MCQ)
In a fully turgid cell:
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm
Ch 12Mineral Nutrition
5-Mark Questions
Nitrogen is present in the atmosphere in huge amounts but higher plants fail to utilize it. Why?
Higher plants are unable to directly utilize the abundant atmospheric nitrogen because it exists in a gaseous diatomic form (N2), which is highly stable due to a strong triple covalent bond. Plants can only absorb nitrogen in its fixed forms, primarily as nitrate (NO3-) or ammonium (NH4+) ions from the soil. Therefore, atmospheric nitrogen must undergo a process called nitrogen fixation, where it is converted into these usable forms. …
Why is that in certain plants, deficiency symptoms appear first in younger parts of the plants while in others, they do so in mature organs?
Deficiency symptoms appear at different locations in plants depending on the mobility of minerals within the plant body. Minerals are classified into two categories based on their mobility. Actively mobile minerals include nitrogen, phosphorus, potassium, magnesium, chlorine, sodium, zinc, and molybdenum. When these minerals become deficient, the plant translocates them from older leaves to younger leaves because younger leaves are metabolically more active and have higher nutrient demands. …
2-Mark Questions
Identify correct match. 1. Die back disease of citrus -(i) Mo 2. Whip tail disease – (ii) Zn 3. Brown heart of turnip -(iii) Cu 4. Little leaf -(iv) B
The correct answer is (b) 1–(iii) 2–(i) 3–(iv) 4–(ii). Die back disease of citrus is caused by copper (Cu) deficiency, whip tail disease is caused by molybdenum (Mo) deficiency, brown heart of turnip is caused by boron (B) deficiency, and little leaf disease is caused by zinc (Zn) deficiency. These are characteristic deficiency diseases in plants, and each is associated with the lack of a specific micronutrient essential for plant growth and development.
Plant A in a nutrient medium shows whiptail disease plant B in a nutrient medium shows a little leaf disease. Identify mineral deficiency of plant A and B?
Plant A, exhibiting whiptail disease, is deficient in the mineral molybdenum (Mo). Molybdenum is a crucial component of enzymes like nitrogenase and nitrate reductase, which are vital for nitrogen metabolism in plants. Its deficiency leads to characteristic symptoms such as the distortion and necrosis of young leaves, particularly the lamina, giving them a 'whiptail' appearance. Plant B, showing little leaf disease, is deficient in the mineral zinc (Zn). …
Cyanobacteria does not include a) Nostoc b) Anabaena c) Clostridium d) Oscillatoria
The correct answer is (c) Clostridium. Cyanobacteria are photosynthetic prokaryotes, also known as blue-green algae. Nostoc, Anabaena, and Oscillatoria are all well-known genera of cyanobacteria capable of photosynthesis and nitrogen fixation. Clostridium, however, is a genus of anaerobic bacteria that are not photosynthetic and do not belong to the cyanobacteria group. Clostridium species are heterotrophic and are found in soil and other environments, but they are not classified as cyanobacteria.
1-Mark Questions (MCQ)
If a plant is provided with all mineral nutrients but, Mn concentration is increased, what will be the deficiency?
(a) Mn prevent the uptake of Fe, Mg but not Ca
Ch 13Photosynthesis
5-Mark Questions
Two groups (A&B) of bean plants of similar size and same leaf area were placed in identical conditions. Group A was exposed to light of wavelength 400 – 450 nm & Group B to light of wavelength of 500 – 550nm. Compare the photosynthetic rate of the 2 groups giving reasons.
Group A, exposed to light of wavelength 400–450 nm, will show a significantly higher photosynthetic rate compared to Group B. The wavelength range of 400–450 nm corresponds to the blue region of the electromagnetic spectrum, where chlorophyll a exhibits its maximum absorption peak at approximately 450 nm. Since chlorophyll a is the primary photosynthetic pigment in the reaction center, maximum light absorption in this region leads to efficient excitation of electrons and optimal photosynthetic activity. …
A tree is believed to be releasing oxygen during nighttime. Do you believe the truthfulness of this statement? Justify your answer by giving reasons?
This statement is partially true, but the explanation requires clarification. A tree can release oxygen during nighttime, but this occurs only in certain types of plants, specifically CAM (Crassulacean Acid Metabolism) plants, not in typical C3 plants. During the day, CAM plants keep their stomata closed to minimize water loss in arid environments. At night, when temperatures are cooler and water loss is reduced, CAM plants open their stomata and fix carbon dioxide using the enzyme phosphoenolpyruvate carboxylase (PEP carboxylase). …
2-Mark Questions
The correct sequence of flow of electrons in the light reaction is: (a) PS II, plastoquinone, cytochrome, PS I, ferredoxin. (b) PS I, plastoquinone, cytochrome, PS II ferredoxin. (c) PS II, ferredoxin, plastoquinone, cytochrome, PS I. (d) PS I, plastoquinone, cytochrome, PS II, ferredoxin.
The correct answer is (a) PS II, plastoquinone, cytochrome, PS I, ferredoxin. This represents the accurate sequence of electron flow during the light-dependent reactions of photosynthesis. Electrons are first excited in Photosystem II (PS II) by light energy, then pass through plastoquinone and the cytochrome b6f complex, which facilitates the pumping of protons into the thylakoid lumen. The electrons then reach Photosystem I (PS I), where they are re-energized by light, and finally are transferred to ferredoxin, which reduces NADP+ to NADPH. …
Identify true statement regarding light reaction of photosynthesis? (a) Splitting of water molecule is associate with PS I. (b) PS I and PS II involved in the formation of NDPH + H +. (c) The reaction center of PS I is Chlorophyll a with absorption peak at 680 nm. (d) The reaction center of PS II is Chlorophyll a with absorption peak at 700 nm.
(b) PS I and PS II are both involved in the formation of NADPH + H+. Photosystem II (PS II) is responsible for the initial excitation of electrons and the splitting of water molecules (photolysis), releasing electrons, protons, and oxygen. These electrons are then passed through an electron transport chain. Photosystem I (PS I) receives electrons from this chain and further excites them using light energy. The energized electrons from PS I are then transferred to ferredoxin, which ultimately reduces NADP+ to NADPH + H+ with the help of the enzyme NADP+ reductase. …
Choose the rightly matched pair a) Chlorophyll a – Accessory pigments and trap solar energy b) Chlorophyll b – Differs from Chlorophyll a in having CH3 instead of CHO – at 3rd C atom c) Chlorophyll c – Differs from Chlorophyll a by lacking a phytol tail d) Chlorophyll d – It has CHO at 3rd at the 3rd carbon atom at 11 – pyrrole ring
The correct answer is (c) Chlorophyll c – Differs from Chlorophyll a by lacking a phytol tail. Chlorophyll c is structurally distinct from chlorophyll a in that it lacks the phytol tail, which is a long hydrophobic hydrocarbon chain attached to the porphyrin ring in chlorophyll a. This structural difference affects the solubility and localization of chlorophyll c within the thylakoid membrane. …
1-Mark Questions (MCQ)
Which chlorophyll molecule does not have a phytol tail? a) Chl – a b) Chl – b c) Chl – c d) Chl – d
(c) Chl – c
Ch 14Respiration
5-Mark Questions
The respiratory quotient is zero in succulent plants. Why?
In succulent plants such as Opuntia and Bryophyllum, the respiratory quotient (RQ) is zero because these plants undergo a unique type of respiration in which carbohydrates are only partially oxidized to organic acids, particularly malic acid, without the corresponding release of carbon dioxide. During this process, oxygen is consumed as the plants break down carbohydrates, but since CO2 is not released in equivalent amounts, the ratio of CO2 released to O2 consumed becomes zero. …
Explain the reactions taking place in mitochondrial inner membrane.
Electron and hydrogen (proton) transport takes place across four multiprotein complexes (I-IV). They are. 1. Complex-I (NADH dehydrogenase). It contains a flavoprotein (FMN) and associated with non-heme iron Sulphur protein (Fe-S). This complex is responsible for passing electrons and protons from mitochondrial NADFI (Internal) to Ubiquinone (UQ) NADH+H + UQ ⇌ NAD – +UQH 2 2. In plants, an additional NADH dehydrogenase (External) complex is present on the outer surface of inner membrane of mitochondria which can oxidise cytosolic NADH + H +. …
2-Mark Questions
What are enzymes involved in phosphorylation and dephosphorylation reactions in EMP pathway?
In the Embden-Meyerhof-Parnas (EMP) pathway, phosphorylation and dephosphorylation reactions are catalyzed by specific enzymes. The enzymes involved in phosphorylation reactions are hexokinase, which phosphorylates glucose to glucose-6-phosphate, and phosphofructokinase, which catalyzes the phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate. These phosphorylation steps require energy input from ATP. …
The external factors that affect respiration are: (a) temperature, insufficient O 2 and amount of protoplasm (b) temperature, insufficient O 2 and high concentration of CO 2 (c) temperature, high concentration of CO 2 and respiratory substrate (d) temperature, high concentration of CO 2 and amount of protoplasm
The external factors that significantly influence the rate of respiration in organisms are primarily environmental conditions that affect metabolic processes. These include temperature, which directly impacts enzyme activity; insufficient oxygen (O2), as oxygen is the final electron acceptor in aerobic respiration; and a high concentration of carbon dioxide (CO2), which can inhibit certain respiratory enzymes. Temperature plays a critical role because respiratory enzymes have optimal temperature ranges for their activity. …
Which one is wrongly matched Column I Column II A. NADH +H + Three ATP B. Glycolysis Twenty four ATP C. FAD Two ATP D. Cytoplasmic NADH+H + Two ATP
The statement that is wrongly matched in the given options is B. Glycolysis – Twenty four ATP. Let's analyze each option for accuracy regarding ATP production in cellular respiration. NADH + H+ typically yields 3 ATP molecules when its electrons are passed through the electron transport chain in aerobic respiration. FAD, specifically FADH2, yields 2 ATP molecules through the electron transport chain. Cytoplasmic NADH + H+ can yield 2 ATP molecules, particularly when its electrons are transferred into the mitochondria via specific shuttle systems, such as the glycerol phosphate shuttle. …
1-Mark Questions (MCQ)
The number of ATP molecules formed by complete oxidation of one molecule of pyruvic acid is:
(a) 12
Ch 15Plant Growth and Development
5-Mark Questions
If the diameter of the pulley is 6 inches, length of pointer is 10 inches and distance travelled by pointer is 5 inches. Calculate the actual growth in length of plant. a) 3 inches b) 6 inches c) 12 inches d) 30 inches
The correct answer is 1.5 inches. To calculate the actual growth in length of the plant, we use the principle of the arc length subtended by the pointer's movement on the pulley. First, we determine the radius of the pulley. Given the diameter of the pulley is 6 inches, the radius of the pulley is half of that, which is \(6/2 = 3\) inches. The distance traveled by the pointer is 5 inches, and the length of the pointer is 10 inches. …
What are the parameters used to measure growth of plants?
The growth of plants is a complex process that can be quantitatively measured using several parameters, reflecting different aspects of their development. These parameters allow scientists to assess the rate and extent of plant growth accurately. One common method is to measure the increase in length or girth, particularly in organs like roots and stems, which exhibit apical and lateral growth respectively. Another significant parameter is the increase in fresh or dry weight, which provides an indication of the biomass accumulation over time. …
2-Mark Questions
Select the wrong statement from the following: (a) Formative phase of the cells retain the capability of cell division. (b) In elongation phase development of central vacuole takes place. (c) In maturation phase thickening and differentiation takes place. (d) In maturation phase, the cells grow further.
The wrong statement among the given options is (d) In maturation phase, the cells grow further. During the maturation phase, cells primarily undergo thickening and differentiation, developing specialized structures and functions. While they reach their maximum size, significant further growth in terms of increase in size is characteristic of the elongation phase, not the maturation phase. …
Select the correctly matched one A) Humanurine i) Auxin-B B) Corn gram oil ii) GA 3 C) Fungs iii) Abscisic acid II D) Herring fish sperm iv) Kinetin E) Unripcrnaizegrains v) AuxinA F) Young cotton boils vi) Zeatin
The correctly matched option is b) A – v, B – i, C – ii, D – iv, E – vi, F – iii. This pairing correctly associates the source with the plant growth regulator or related substance. Specifically, Auxin A was first isolated from human urine, while Auxin B was isolated from corn germ oil. Gibberellin (GA3) was initially discovered from the fungus Gibberella fujikuroi. Kinetin, a type of cytokinin, was first isolated from herring fish sperm. Zeatin, another cytokinin, was isolated from unripe maize grains. Abscisic acid (ABA) was isolated from young cotton bolls.
The total growth of the plant consists of four phases in the following order. (a) Log phase, lag phase, decelerating phase and maturation phase (b) Log phase, lag phase, maturation phase and decelerating phase (c) Lag phase, log phase, maturation phase and decelerating phase (d) Lag phase, log phase, decelerating phase and maturation phase
The total growth of the plant consists of four distinct phases that occur in a specific sequential order. This progression is typically represented by a sigmoid or S-shaped growth curve. The correct order of these phases is (d) Lag phase, log phase, decelerating phase and maturation phase. The lag phase is the initial period where growth is slow as cells adjust to the new environment and prepare for division. This is followed by the log phase (or exponential phase), characterized by rapid and constant growth, where cells divide and enlarge at their maximum rate. …
1-Mark Questions (MCQ)
In unisexual plants, sex can changed by the application of a) Ethanol b) Cytokinins c) ABA d) Auxin
c) ABA
Frequently asked questions
- Write the distinguishing features of Monera.
- The kingdom Monera encompasses all prokaryotic organisms, which are characterized by the absence of a true nucleus and other membrane-bound organelles. These organisms are typically microscopic and include diverse forms such as Mycoplasma, bacteria, actinomycetes, and cyanobacteria. Monerans exhibit various modes of nutrition; some are autotrophic, performing photosynthesis or chemosynthesis, while others are heterotrophic, acting as decomposers, parasites, or symbionts. …
- Why do farmers plant leguminous crops in crop rotations/mixed cropping?
- Farmers plant leguminous crops in crop rotations and mixed cropping systems because these plants form symbiotic associations with Rhizobium bacteria in their root nodules. Rhizobium is a nitrogen-fixing bacterium that converts atmospheric nitrogen into nitrates and other nitrogen compounds that enrich the soil and increase its fertility. When leguminous crops such as pulses, beans, or clover are grown alternately with non-leguminous crops like paddy or wheat, the soil nitrogen content is naturally replenished without requiring synthetic nitrogen fertilizers. …
- Identify the correctly matched pair a. Actinomycete – a) Late blight b. Mycoplasma – b) lumpy jaw c. Bacteria – c) crown gall d. Fungi – d) sandal spike
- The correctly matched pair is: a. Actinomycete – lumpy jaw, b. Mycoplasma – sandal spike, c. Bacteria – crown gall, d. Fungi – late blight. Lumpy jaw is a disease of cattle caused by Actinomycetes. Sandal spike disease affecting sandalwood trees is caused by Mycoplasma. Crown gall, a plant disease characterized by tumor-like growths, is caused by the bacterium Agrobacterium tumefaciens. Late blight of potato is a fungal disease caused by Phytophthora infestans.
- Differentiate Homoiomerous and Heteromerous lichens.
- Homoiomerous lichens have algal cells evenly and uniformly distributed throughout the entire thallus, mixed with fungal hyphae in a homogeneous manner without any distinct layering or organization. In contrast, heteromerous lichens display a distinct stratified structure with clearly defined layers: an upper cortex of fungal hyphae, a middle algal layer containing photosynthetic algal cells, and a lower cortex of fungal hyphae, with a medulla of loosely arranged hyphae in between. …
These important questions are selected from the Samacheer Kalvi Class 11 Botany textbook book-back exercises to help you revise the most useful questions. Mark weightage (5/2/1) follows the usual exam pattern and may vary by exam — always check your latest syllabus and question pattern. Open each chapter for the complete set of questions and answers.