TS Inter 1st Year Botany Question Paper March 2018

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TS Inter 1st Year Botany Question Paper March 2018

Time: 3 Hours
Maximum Marks: 60

General Instructions:
Note : Read the following instructions carefully.

  1. Answer all questions of Section ‘A’. Answer any six questions out of eight in Section ‘B’ and answer any two questions out of three in Section – ‘C’.
  2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very short answer type”. Each question carries two marks. Every answer may be limited to five lines. Answer all the questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short answer type”. Each question carries four marks. Every answer may be limited to 20 lines.
  4. In Section ‘C’, questions from Sr. Nos. 19 to 21 are of “Long answer type”. Each question carries eight marks. Every answer may be limited to 60 lines.
  5. Draw labelled diagrams wherever necessary for questions in Section ‘B’ and C.

Section – A (10 × 2 = 20)

Note : Answer all questions. Each answer may be limited to 5 lines.

Question 1.
Who is popularly known as father of Botany ? What was the book written by him ?
Answer:
’Theophrastus” is regarded as the Father of Botany. He wrote a book “de Historia plantarum”.

Question 2.
What does ICBN stands for ?
Answer:
ICBN stands for International code for Botanical Nomenclature.

Question 3.
What does the term phycobiont and mycobiont signify ?
Answer:
The algal component in a lichen is called phycobiont and the fungal component in a lichen is called mycobiont.

TS Inter 1st Year Botany Question Paper March 2018

Question 4.
Define placentation. Which placentation is seen in Dianthus?
Answer:
The arrangement of ovules within the ovary is known as placentation. In Dianthus free central placentation is seen.

Question 5.
Define venation. How do Dicots differ from Monocots with respect to venation ?
Answer:
The arrangement of veins and vein lets in the lamina is called venation. In Dicots Reticulate venation is seen {Net like arrangement of veins} where as in monocots – parallel venation is seen {Parallel arrangement of veins}.

Question 6.
Define Alpha taxonomy and omega taxonomy.
Answer:
Classification based on morphological characters is called alpha taxonomy. Classification based on information from other branches i.e., from Embryology, cytology, palynology, phytochemistry and serology along with morphological characters is called omega taxonomy.

Question 7.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells take to become 32 cells ?
Answer:
80 minutes.

Question 8.
Give one example for each of aminoacids, sugars, nucleotides and fatty acids.
Answer:
Amino acids : Glycine, Alanine .
Sugars : Cellulose, {Glucose, Ribose}
Nucleotides : Adenylic acid Fatty acids : Lecithin, Glycerol

Question 9.
In which phase of the cell cycle DNA synthesis occur ?
Answer:
Synthesis phase of Inter phase.

Question 10.
Define communities. Who classified communities into hydrophytes, mesophytes and xerophytes ?
Answer:
An assemblage of all the populations belonging to different species occur in an area is called community. Eugene Warming classified communities into three types.

TS Inter 1st Year Botany Question Paper March 2018

Section – B (6 × 4 = 24)

Note : Answer any six questions. Each answer may be limited to 20 lines.

Question 11.
Give a brief account of Dinoflagellates.
Answer:

  1. Dinoflagellates are mostly marine and photosynthetic. They appear yellow green, brown, blue or red depending on the pigments in their cells.
  2. The cell wall has stiff cellulose plates on the outer surface.
  3. They have two flagella and produce spinning movements. So these protists are cailed “whirling ships”.
  4. The nucleus has condensed chromosomes which are without histones. This is called mesokaryon.
  5. Some dinoflagellates like Noctiluca show bioluminescene.
  6. Red dinoflagellates like Gonyaulax undergo rapid multiplication and make the sea appear red (Red tides in Medeterraniansea).
  7. Toxins produced by them may kill fishes.

Question 12.
What is meant by Homosporous and heterosporous pteridophytes ? Give two examples.
Answer:
The condition where only one type of spores are produced is called Homosporous pteridophytes.
Ex: Psilotum, Lycopodium

The condition where different types of spores are pro-duced is called Heterosporous pteridophytes.
Ex: Selagenella, Salvinia.

Question 13.
Discuss the various types of pollen tube entry into ovary with the help of diagrams.
Answer:
Pollen tube enters into the ovule from ovary by three ways.
They are
a) Porogamy : Pollen tube enters into the ovule through Micropyle. Ex : Ottelia, Hibiscus.
b) Chalazogamy : “Pollen tube enters into the ovule through Chalaza”. Ex: Casuarina.
c) Mesogamy : “Pollen tube enters into the ovule through the integuments”. Ex : Cucurbita.
TS Inter 1st Year Botany Question Paper March 2018 1

Question 14.
Describe the non-essential floral parts of plants belonging to Fabaceae.
Answer:
In Fabaceae, the non-essential floral parts are Calyx and Corolla.

Calyx : Sepals 5, gamosepalous, valvate aestivation and odd sepal is anterior.

Corolla : Polypetalous, papilionaceous type consisting of large posterior petal called standard petal, two lateral petals called wings and the two anterior fused petals are called keel petals which enclose essential organs. They show descendingly imbricate aestivation.

TS Inter 1st Year Botany Question Paper March 2018

Question 15.
Differentiate between Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER).
Answer:

Rough Endoplasmic Reticulum Smooth Endoplasmic Reticulum
1) ER studded with ribosomes is called Rough ER. 1) ER without ribosomes is called smooth ER.
2) These are mainly composed of cisternae. 2) These are mainly composed of tubules.
3) They are associated with nuclear membrane. 3) They are associated with plasma membrane.
4) They mainly involved in Proteinsynthesis. 4) They mainly involved in Lipid synthesis.

Question 16.
Give a brief account of the types of chromosomes based on the position of Centromere.
Answer:
TS Inter 1st Year Botany Question Paper March 2018 2
Basing on the position of the centromere, four types of chromosomes are recognised.
They are

  1. Metacentric : Centromere is present in between the two ‘ arms of the chromosome. It is V shaped and consists of two equal arms.
  2. Sub-Metacentric : Centromere is present slightly away from the midpoint of a chromosome. It is ‘L’ shaped and consists of two unequal arms.
  3. Acrocentric : Centromere is present at the sub terminal position of the chromosome. It appears ‘J’ shaped and consists of one long arm and one short arm.
  4. Telocentric : Centromere is present at the terminal posi-tion of the chromosome. It appears ‘I’ shaped and con-sists of one arm.

Question 17.
What is Periderm ? How does Periderm formation take place in the dicot stem ?
Answer:
Phellogen, Phellem and Phelloderm together called periderm. In Dicots, stem continues to increase in girth due to activity of vascular cambium. As a result, the other cortical cells and epidermal layers ruptures and need to be replaced to provide new pulective layers. Hence another meinstematic layer develops in the cortical region called cork cambium or phellogen. It is made up of narrow, thin walled cells which cuts of cells in both sides. The outer cells differentiates into work or phellem, while the Inner cells differentiates into secondary cortex or phellodym. The work in imperious to water due to subain deposition in the cell Nail. The secondary cortical cells are parenchymatous, Phellogen, Phellem and phellodlum are collectively known as puiderm.

Question 18.
Give a brief account on the classification of Xerophytes.
Answer:
Xerophytes are classified into three types.
They are :

  1. Ephemerals: These are annuals which complete their life cycle within a short period. Ex : Tribulus.
  2. Succulents : They absorb large quantities of water during rainy season and store it in different plant parts in the form of mucilage. As a result the plant parts become fleshy or succulent. The store water is used during dry periods. Ex: Stem succulents : Opuntia. Root succulents : Asparagus. Leaf succulents : Aloe.
  3. Non succulents: These are perennial plants can withstand prolonged period of drought. Ex : Casuarina.

TS Inter 1st Year Botany Question Paper March 2018

Section – C (2 × 8 = 16)

Note : Answer any two questions. Each answer may be limited to 60 lines.

Question 19.
Define Root modification. Explain how root is modified to perform different functions ?
Answer:
A change in the normal structure of root to carry out new functions according to environment is called root modification.

Root Modifications : In some plants, roots change their shape and structure to perform functions other than absorption and conduction of water and minerals called root modifications.

They are of different types.

  1. In Carrot, turnip (Tap roots), Sweet potato (Adventitious roots), Asparagus (Fibrous roots) become swollen due to storage of food called storage roots.
  2. In Banyan tree, roots arise from the branches grow into the soil, become pillar like and give additional support called prop roots or pillar roots.
  3. In maize, sugarcane, roots arise from the lower nodes of the stem, give additional support called stilt roots.
  4. In Mangroves like Rhizophora and Avicennia, many roots come out of the ground and grow vertically upwards, help in respiration called Pneumatophores.
  5. In Epiphytes like Vanda, special adventitious roots arise help in absorption of moisture from atmosphere called Velamen roots.
  6. In partial parasites like viscum and strga, some Haustorial roots enter into xylem of the host plant to get water and minerals. In complete parasitic like cuscuta and Rafflesia, the haustorial roots enter into xylem and phloem of the host plant and obtain water and minerals and food materials called Parasitic roots.
  7. In the members of Fabaceae, the roots are inhabited by Rhizobium bacteria which helps in N2 fixation called Nodular roots.
  8. In some plants like Taeniophyllum, the roots are chloro phyllous and perform photosynthesis so called photo synthetic roots.

TS Inter 1st Year Botany Question Paper March 2018 3
TS Inter 1st Year Botany Question Paper March 2018 4
TS Inter 1st Year Botany Question Paper March 2018 5

Question 20.
Draw the diagram of a Microsporangium and label its wall layers. Write briefly about the wall layers.
Answer:
TS Inter 1st Year Botany Question Paper March 2018 6
A typical angiospermics anther is bilobed with each lobe having two theca. The anther is a four sided structure consisting of four microsporangia located at the comers, two in each lobe.

In a transverse section, a typical microsporangium is circular in out line and is surrounded by four wall layers, the (a) epidermis
(b) endothecium
(c) wall layers
(d) tapetum.

a) Epidermis : The epidermis is one celled thick, the cells present between the pollen sacs are th thin walled and their region is called as stomium which is useful for the dehiscence of pollen sacs.

b) Endothecium : It is present below the epidermis and expands radically with fibrous thickenings, at maturity these cells loose water and contract and help in. the dehiscence of pollen sacs.

c) Wall layers : Beneath the Endothecium, there are thin walled cells, arranged in one to five layers, which also help in dehiscence of Anther.

d) Tapetum : The innermost wall layer is Tapetum, the cells are large, with thin cell walls, abundant cytoplasm and have more than one nuclei. Tapetum is a nutritive tissue which nourishes the developing pollen grains.

TS Inter 1st Year Botany Question Paper March 2018

Question 21.
Describe the internal structure of a Dicot root.
Answer:
A thin transverse section of dicot root shows three parts namely
i) Epidermis
ii) Cortex and
iii) Stele.

i) Epidermis: It is the outer most layers made of thin walled cells. Some cells protrude in the form of unicellular root hairs. So called Epiblema. It protects the inner parts. Root hairs help in absorption of water from the soil.

ii) Cortex : It consists of several layers of thin walled parenchyma cells with inter cellular spaces. The Innermost layer of cortex is called Endodermis. It comprises a single layer of barrel shaped cells without intercellular spaces. The tangential as well as the radial walls of Endoderm cells show suberin thickenings called casparian strips. Some cells opposite to protoxylem lack these strips called passage cells. They help in the movement of water and dissolved salts from cortex into xylem.

iii) Stele : It is the central part, consists of 4 layers.
a) Pericycle : It is single layered, made of thin walled parenchyma cells, present next to endodermis. It produces lateral roots and become vascular cambium during secondary growth.

b) Vascular Bundle : Xylem and phloem constitutes vascular Bundle. They are arr-anged on different radius (in alternate manner) so called Radial vascular Bun-dle, Xylem is exarch, where protoxylem is towards periphery and metaxylem is towards the center. Xylem is diarch to tetrarch condition. Xylem helps in conduction of water and minerals and phloem helps in conduction of food materials.
TS Inter 1st Year Botany Question Paper March 2018 7
c) Medulla : It is absent or small, made of parenchyma cells. When present, it helps in the storage of food and water.

d) Conjunctive tissue : The parenchyma present between xylem and phloem is called conjunctive tissue which also involves in secondary growth.

AP Inter 1st Year Zoology Question Paper May 2017

Varied difficulty levels in AP Inter 1st Year Zoology Model Papers and AP Inter 1st Year Zoology Question Paper May 2017 cater to students with diverse academic strengths and challenges.

AP Inter 1st Year Zoology Question Paper May 2017

Time: 3 Hours
Max. Marks: 60

Note: Read the following instructions carefully.

  1. Answer All the questions of Section A. Answer any SIX questions in Section B and answer any TWO questions in Section C.
  2. In Section A, questions from SI. Nos. 1 to 10 are of Very short answer type. Each question carries two marks. Every answer may be limited to 5 lines. Answer all questions at one place in the same order.
  3. In Section B, questions from Si. Nos, 11 to 18 are of Short Answer Type. Each question canles four marks. Every answer may be limited to 20 lines.
  4. In Section C, questions from Si. Nos. 19 to 21 are of Long answer type. Each question carries eight marks. Every answer may be limited to 60 lines.
  5. Draw labelled diagrams wherever necessary in Sections B and C.

Section – A
10 x 2 = 20

Note: Answer All the questions in 5 lines each.

Question 1.
What is meant by tautonymy? Give two examples.
Answer:
The practice of naming the animals or organisms, in which the generic name and species name are the same, is called Tautonymy. Ex: Axis axis – spotted dear. Naja naja – The Indian Cobra.

Question 2.
What is Cephalization? How is it useful to its possessors?
Answer:
Cephalisation: The concentration of nerve (Brain) and sensory cells at the anterior end of body is called as cephalisation. As a result of cephalisation, these animals can sense the new environment and move efficiently than the other animals in seeking food, locating matter and in avoiding from predators.

Question 3.
Distinguish between a tendon and a ligament.
Answer:
Tendons are the collagen fiber tissue of dence regular connective tissue which attach the skeletal muscles to bones. Ligaments are also the collagen fibers tissue of dence regular connective tissue which attach bones to other bones.

AP Inter 1st Year Zoology Question Paper May 2017

Question 4.
What is haematocrit value?
Answer:
The percentage of total volume occupied by RBCs in blood is called haematocrit value.

Question 5.
What is Aristotle’s Lantern? Give one example of an animal possessing it?
Answer:
In the mouth of sea Urchin a complex five-jawed masticatory apparatus called Aristotle’s Lantern. Ex: Echinus.

Question 6.
What are pneumatic bones? How do they help birds?
Answer:
Main bones in birds are extensions of air sacs without bone marrow are called pneumatic bones. These are helpful in flying to birds.

Question 7.
Distinguish between lobopodium and filopodium. Give an example to each of them.
Answer:
Lobopodium: The blunt and finger-like tubular pseudopodia containing both ectoplasm and endoplasm is called lobopodium. Ex: Amoeba proteus.
Filopodium : The slender filamentous pseudopodia with pointed tips, composed of only ectoplasm are called Filopodium. Ex: Euglypha.

Question 8.
What are dynein armes? What is their significance?
Answer:
‘A’ tubule of each peripheral doublet bears paired arms along its length called dynein arms made up of protein dynein. The dynein arms of ‘A’ tubule face the tubule ‘B’ of the adjacent doublet.

Question 9.
In which way does tobacco affect the respiration? Name the alkaloid found in tobacco.
Answer:
Tobacco increases the carbon monoxide (CO) level and reduces the oxygen level in the blood. The alkaloid found in the tobacco is “Nicotine”.

AP Inter 1st Year Zoology Question Paper May 2017

Question 10.
Define commensalism. Give one example.
Answer:
This is the interaction in which one species benefits and other is neither harmed nor benefited. Ex: Barnacles growing on the back of a whale benefit while the whale derives no noticeable benefit.

Section – B
6 x 4 = 24

Note: Answer any SIX questions in 20 lines.

Question 11.
Explain “Rivet Popper’’ hypothesis.
Answer:
What if we lose a few species? Will it affect man’s life? Paul Ehrlich experiments Rivet popper, hypothesis, taking an aeroplane as an ecosystem, explains how removal of one by one ‘rivets’ (species of an ecosystem) of various parts can slowly damage the plane (ecosystem)-shows how important a ‘species’ is in the overall functioning of an ecosystem. Removing a rivet from a seat or some other relatively minor important parts may not damage the plane, but removal of a rivet from a part supporting the wing can result in a crash. Likewise, removal of a ‘critical species’ may affect the entire community and thus the entire ecosystem.

Question 12.
Explain the Haversian system.
Answer:
The compact bone consists of several structural units called Ostens or Haversion systems arranged around and parallel to the bone marrow cavities.
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 1
Haversian system:
consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein and a lymphatic vessel. Haversian canal is surrounded by concentric lamellae. Small fluid-filled spaces called lacunae provided with minute canaliculi lie in between the lamellae. Canaliculi connect the lacunae with one another and with Haversian canal. Each lacuna encloses one osteocyte (inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another; with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

Question 13.
What are the chief characters of the Crustaceans?
Answer:

  1. This includes prawns, crabs, lobsters, crayfishes etc.
  2. Mostly marine, a few are fresh water and some are adapted to terrestrial life.
  3. In most species, head and thorax fuse to form cephalo – thorax.
  4. Cephalic appendages are five pairs – first antennae (antennules) second antennae, mandibles, first maxilla and second maxillae.
  5. Thoracic and abdominal appendages are typically biramous.
  6. Respiration is by gills.
  7. Excretory organs are green glands or antenna! glands.
  8. Sense organs include statocysts, compound eyes and antennae.
  9. Gonopores are paired.
  10. Development is direct or indirect involving several larval stages. Basic larva is nauplius. Ex: Palaernon (Prawn), Cancer (Crab).

AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 2

AP Inter 1st Year Zoology Question Paper May 2017

Question 14.
Compare and contrast cartilaginous and bony fishes.
Answer:

Cartilaginous fishes Bony fishes
1. These are marine farms. 1. These are live in all kinds of aquatic habitats.
2. Endoskeleton made by cartilaginous. 2. Endoskeleton made by bone.
3. Body covered by placoid scales. 3. Body covered by cosmoid, ganoid, cycloid or ctenoid scales.
4. Caudal fin is heterocercal. 4. Caudal fin is homocercal.
5. Operculum absent. 5. Operculum present.
6. Air bladder absent. 6. Air bladder present.
7. Gills are lamelliform and are five to seven in each four in each side. 7. Gills are filamentous and are side.
8. These are ureotelic.
Ex: Scoliodori, Pristic, Torpedo
8. These are mostly ammono telic.
Ex: Catla, Labeo, Exocetus, Hippocarnpus.

Question 15.
Describe the process of transverse binary fission in paramecium.
Answer:
Transverse binary fission is performed by Paramecium. Binary fission is the most common method of a sexual reproduction in protozoans.

During favourable conditions, Paramecium stops feeding after attaining its maximum growth. At first, the micronucleus divides by mitosis and the macronucleus divides into two daughter nuclei by arnitosis. The oral groove disappears. After karyokinesis, a transverse constriction appears in the middle of the body, which deepens and divides the parent cell into two daughter individuals, the anterior proter and the posterior opisthe. The proper receives the anterior contractile vacuole, cytopharynx and cytosome from its parent individual.

It develops posterior contractile vacuole and a new oral groove. The opisthe receives the çosterior contractile vacuole of its parent individual. It develops posterior contractile vacuole and a new oral groove. The opisthe receives the posterior contractile vacuole of its parent. It develops a new anterior contractile vacuole, cytopharynx, cytostome and a new oral groove.

Binary fission is completed in almost two hours, in favourable conditions and Paramecium can produce four generations of daughter individuals by binary fission in a day.
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 3
The transverse binary fission is also called homothetogenic fission because the plane of fission is at right angles to the longitudinal axis of the body. As it occurs at right angles to the kineties, it is also called perinatal fission.

Question 16.
What is the need for parasites to develop special adaptations developed by the parasites?
Answer:
Parasites have to evolve mechanisms to counteract and neutralize the host’s defence in order to be successful with in the host. For this purpose, the parasites have developed many special adaptations such as the loss of unnecessary sensory organs, formation of organs for adhesion, high reproductive capacity; etc.

Parasitic adaptations: Parasites have evolved special adaptations to meet the requirements and lead a successful life in the hosts.

  • In order to live in the host some parasites have developed structures like hooks, suckers, rostellum etc., for anchoring. e.g: Taenia solium.
  • Some intestinal parasites have developed protective cuticles to withstand the action of the digestive enzymes of the host. e.g : Ascaris lumbricoides.
  • Some intestinal parasites produce antienzyme to neutralize the effect of host’s digestive enzymes. e.g: Taenia solium.
  • Some parasites live as obligatory anaerobes as the availability of oxygen is very rare for them, e.g.: Entarnoebu histological, Taenia solium etc.
  • Some intestinal parasites live as facultative anaerobes. i.e., if oxygen is not available, they live anaerobically and if oxygen is available, they respire aerobically. e.g: Ascaris lurnbricoides.

Question 17.
Draw a neat labelled diagram of the mouthparts of Cock roach.
Answer:
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 4

AP Inter 1st Year Zoology Question Paper May 2017 with Solutions

Question 18.
Describe ‘Green house effect.
Answer:
The term Greenhouse effect has been derived from a phenomenon that occurs in a greenhouse. Greenhouse is a small glass house and is used for growing plants especially during winter. In a greenhouse, the glass panel allows the passage of light into it, but does not allow heat to escape (as it is reflected back). Therefore, the greenhouse warms up, very much like inside a car that has been parked in the sun for a few hours.

The greenhouse effect is a naturally occurring phenomenon that is responsible for heating of the Earth’s surface and atmosphere. It would be surprising to know that without greenhouse effect the average temperature of the Earth’s surface would have been a chilly 18°C rather than the present average of 15°C.

When sunlight reaches the outermost layer of the atmosphere, clouds and gases reflect about one-fourth of the incoming solar radiation and absorb some of it. Almost half of the incoming solar radiation falls on the Earth’s surface and heats it up. While a small proportion is reflected back.

Section – C
2 x 8 = 16

Note: Answer any two questions in 60 lines each:

Question 19.
Describe the structure and life cycle of Ascaris lumbricoides with the help of a neat and labelled diagram.
Answer:
Ascaris lumbricoids is commonly called the common roundworm. It lives in the small intestine of man, more frequently in children. It is cosmopolitan in distribution. Mode of infection is through contaminated food and water. Infective stage is the embryonated egg with the 2 stage rhabditi form larva.

Structure: Sexes are separate and the sexual dimorphism is distinct. In both males and females. The body is elongated and cylindrical. Mouth is present at the extreme anterior end and is surrounded by three chitinous lips close to the mouth mid-ventrally there is a small aperture called excretory pore.

Male: It has a curved posterior end which is considered the tail. The posterior end possesses a cloacal aperture and a pair of equal-sized chitinoùs pineal spicules or pineal setae which serve to transfer the sperms during copulation.

Female: It has a straight posterior end the tail. The female genital pore or vulva is present mid-ventrally at about one third the length from mouth. The anus is present a little in front of the tail end.

Life history: Copulation takes place in the small intestine of man. After copulation, the female releases approximately two lakh eggs per day. Each egg is surrounded by a protein coat with rippled surface. Hence the eggs of Ascaris are described as mammillated eggs. The protein coat is followed by a chitinous shell and a lipid layer internally. These eggs come out along with faecal matter. In the moist soil, development takes place inside the egg so that the 1 stage rhabditi form larva is produced. It undergoes the 1 moulting and becomes the 2nd” stage rhabditi form larva which is considered the stage infective to man. They reach the alimentary canal of man through contaminated food and water.
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 5
In the small intestine, the shell gets dissolved so that the 2-stage larva is released. Now it undergoes extra intestinal migration. First it reaches the liver through the hepatic portal vein. From there it reaches the heart through the post caval vein. It goes to the lungs through the pulmonary arteries. In the alveoli of lungs, it undergoes the 2nd moulting to produce the 3rd’ stage larva.
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 6
It undergoes the 3 moulting so that the 4th stage larva is produced in the alveoli only. It leaves the alveoli and reaches the small intestine again through the bronchi, trachea, larynx, glottis, pharynx, oesophagus and stomach. In the small intestine. It undergoes the 4th and final moulting to become a young one which attains sexual maturity within 8 to 10 weeks.
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 7

Question 20.
Describe the blood circulatory system of Periplaneta in detail and draw a neat and labelled diagram of it.
Answer:
The circulatory system helps in the transportation of digested food, hormones etc., from one part to another in the body. Periplaneta has an open type of circulatory system as the blood or haemolymph, flows freely within the body cavity or haemocoel. Blood vessels are poorly developed and open into spaces. Visceral organs located in the haemocoel are bathed in the blood. The three main parts associated with the blood circulatory system of Periplaneta are the haemocoel, heart and blood.

Haemocoel: The haemocoel of cockroaches is divided into three sinuses by two muscular, horizontal membranes called dorsal diaphragm or pericardial septum and ventral diaphragm. Both the diaphragms have pores. There is a series of paired triangular muscles called alary muscles. Every segment has one pair of these muscles situated on the lateral sides of the body. These are attached to the pericardial septum by their broad bases and to the terga by their broad bases and to the terga by their pointed ends or apices.

The three sinuses of the haemocoel are known as pericardial haemocoel or the dorsal sinus, the perivisceral haemocoel or the middle sinus and sternal haemocoel or ventral sinus or perineural sinus. The middle sinus is very large as it contains most of the viscera. The dorsal and ventral sinusecs are small as they have only heart and nerve cord, respectively.

Heart: The heart lies in the pericardial haemocoel or dorsal sinus. It is a long muscular, contractile tube found a long the middorsal line, beneath the terga of the thorax and abdomen.
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 8
It consists of 13 chambers. Every chamber opens into the other present in front of it. Three of the thirteen chambers are situated in the thorax and ten in the abdomen. Its posterior end is closed while the anterior end is continued forward as the anterior aorta.

At the posterior side of each chamber, except the last, there is a pair of small apertures called ‘Ostia’ one on each side. Ostia have valves which allow the blood to pass only into the heart from the dorsal sinus.

Blood: The blood of Periplaneta is colourless and is called haemolymph. It consists of a fluid called plasma and free blood corpuscles or haemocytes, which are phagocytic. The phagocytic, the phagocytes are large in size and can ‘ingest foreign particles such as bacteria. There is no respiratory pigment in the blood and so it plays no major role in respiration.

AP Inter 1st Year Zoology Question Paper May 2017

Question 21.
Describe different types of food chains that exist in an ecosystem.
Answer:
The food energy passes from one trophic level to another trophic level mostly from the lower to higher trophic levels. When the path of food energy is ‘linear’ the components resemble the links of a chain and it is called ‘food chain. Generally, a food chain ends with decomposers. The three major types of food chains in an ecosystem are Grazing Food Chain, Parasitic Food Chain and Detritus Food Chain.
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 9
i) Grazing food chain: It is also known as predatory food chain, it begins with the green plants (producers) and the second third and fourth trophic levels are occupied by the herbivores, primary carnivores and secondary carnivores respectively. In some food chains these is yet another trophic level the climax carnivores.

The number of trophic levels in food chains varies from 3 to 5 generally. Some examples from grazing food chain (GFC) are given below.
AP Inter 1st Year Zoology Question Paper May 2017 with Solutions 10
ii) Parasitic food chain: Some authors included the Parasitic Food Chains as a part of the GFC. As in the case of GFCs, it also begins with the producers, the plants (directly or indirectly). However, the food energy passes from large organisms to small organisms in the parasitic chains. For instance, a tree which occupies the si trophic level provides shelter and food for many birds. These birds host many ectoparasites and endoparasites. Thus, unlike in the predator food chain, the path of the flow of energy includes fewer, large-sized organisms in the lower trophic levels and numerous, small-sized organisms in the successive higher trophic levels.

iii) Detritus Food Chain: The detritus food chain (DFC) begins with dead organic matter (such as leaf litter, bodies of dead organisms). It is made up of decomposers which are heterotrophic organisms, mainly the fungi and bacteria. They meet their energy and nutrient requirements by degrading dead organic matter or detritus. These are also known as saprotrophs (sapro: to decompose).

Decomposers: Secrete digestive enzymes that break down dead and waste materials (such as faeces) into simple absorbable substances. Some examples of detritus food chains are:

  1. Detritus (formed from leaf litter)-Earthworms – Frogs – Snakes.
  2. Dead animals – Flies and maggots – Frogs – Snakes.

In an aquatic ecosystem, GFC is the major conduit’ for the energy flow. As against this in a terrestrial ecosystem, a much larger fraction of energy flows through the detritus food chain than through the GFC. The Detritus food chain may be connected with the grazing food chain at some levels. Some of the organisms of DFC may form the prey of the GFC animals. For example, in the detritus food chain given above, the earthworms of the DFC may become the food of the birds of the GFC. It is to be understood that food chains are not ‘isolated always.

TS Inter 1st Year Botany Question Paper May 2017

Successful navigation through TS Inter 1st Year Botany Model Papers and TS Inter 1st Year Botany Question Paper May 2017 builds students’ confidence in their exam-taking abilities.

TS Inter 1st Year Botany Question Paper May 2017

Time: 3 Hours
Maximum Marks: 60

General Instructions:
Note : Read the following instructions carefully.

  1. Answer all questions of Section ‘A’. Answer any six questions out of eight in Section ‘B’ and answer any two questions out of three in Section ‘C’.
  2. In Section A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Erach question carries two marks. Every answer may be limited to 5 lines. Answer all the questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 20 lines.
  4. In Section ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 60 lines.
  5. Draw labelled diagrams whereever necessary for questions in Sections ‘B’ and ‘C’.

Section – A (10 × 2 = 20)

Note : Answer all questions. Each answer may be limited to 5 lines.

Question 1.
Who discovered the cell ? What was the book written by him ?
Answer:
The cell was discovered by “Robert Hooke”. He wrote a Book “Micrographia”.

Question 2.
What are the first organisms to colonize rocks ? Give the generic name of the moss which provides peat
Answer:
Moss plants, sphagnum.

Question 3.
Name two diseases caused by Mycoplasma.
Answer:
Witches broom in plants, pleuropneumonia in Cattle.and Mycoplasma urethritis in Human beings.

TS Inter 1st Year Botany Question Paper May 2017

Question 4.
Needle like phylloclades are found in which plant ?
Answer:
Casuarina.

Question 5.
Differentiate actinomorphic from zygomorphic flower.
Answer:

Actinomorphic Zygomorphic
1. The flower can be cut into two equal halves in any plane is called actinomorphic flower.
Ex : Datura
1. The flower can be cut into two equal halves in one longitudinal plane only is called zygomorphic flower. Ex : Bean

Question 6.
What is geocarpy ? Name the plant which exhibits this phenomenon.
Answer:
Development of fruits inside the soil is called Geocarpy. It is between in Arachis hypogea (Ground nut).

Question 7.
What is referred to as Satellite chromosome ?
Answer:
A few chromosomes have non staining secondary constriction at a constant location which gives the appearance of a small fragment called satellite. The chromosome with satellite is called satellite chromosome [SAT Chromosome].

Question 8.
Select an appropriate chemical bond among ester bond glycosidic bond, peptide bond and hydrogen bond. Write against each of the following :
a) Polysaccharide
b) Protein
c) Fat
d) Water
Answer:
a) Polysaccharide : Glycosidic bond
b) Protein : Peptide bond .
c) Fat : Ester bond
d) Water : Hydrogen bond

Question 9.
If a tissue has at a given time 1024 cells, how many cycles of mitosis had the original parental single cell undergone ?
Answer:
10 mitotic divisions.

Question 10.
Define population and community ?
Answer:
A group of similar Individuals belonging to the same specieg sound in an area is called population. An assemblage of all the populations occuring in an area is called community.

TS Inter 1st Year Botany Question Paper May 2017

Section – B (6 × 4 = 24)

Note : Answer any six questions. Each answer may be limited to 20 lines.

Question 11.
Explain binomial nomenclature.
Answer:
Naming the plants with two words is called Binomial nomen clature. It was first inroduced by Gaspad Bauhin (1623) but the credit goes to Linnaeus (1753).

In Binomial nomenclature, each and every plant has a name, with two words, they should be in Latin. The first word in Genus and the second word is Species. The generic name will be in noun form and always starts with capital letter. The specific name will be in adjective form and starts with small letter. The Botanical name Should be printed in Italics and should be underlined. Ex: Solanum – tuberosum – Potato plant in which Solanum is the generic name and tuberosum is the species name.

Question 12.
What are the characteristic features of Euglenoids ?
Answer:

  1. Most of the Euglenoids are fresh water organisms found in stagnant water.
  2. They have a protein rich layer called pellicle which makes body flexible.
  3. They have two flagella, a short and a long one.
  4. The anterior part of the cell bears an in vagination consisting of cytostome, cytopharynx and reservoir.
  5. Eye spot is present in the reservior.
  6. They behave as Heterotrophs when deprived of sunlight.
  7. The reproduction in by longitudinal binary fission.

Question 13.
List the changes observed in angiosperm flower subsequent to pollination and fertilization.
Answer:

  1. Calyx, corolla, stamens, style and stigma wither away.
  2. Ovary develops into fruit.
  3. Ovules develops into seeds.
  4. The zygote develops into an embryo.
  5. Primary endosperm nucleus develops into endosperm.
  6. The synergid and 2 antipodals digenerate after fertilisation.
  7. Funicle of the ovule develops into stalk of the seed.
  8. Outer integument Of the ovule changes into testa (Outer seed coat).
    Inner integument of the ovule changes into tegmen (inner seed coat).
  9. Micropyle of the ovule chaijges into seed pore.
  10. Hilum changes into scar of the seed.

Question 14.
Give economic importance of plats belonging to Fabaceae.
Answer:
Economic importance :

  1. Pulses like red gram (Cajanus cajari), black gram (Phaseolus mungo), green gram (Phaseolus aureus) , Bengal gram (Cicer arietinum) are rich source of proteins.
  2. Pods of Dolichos, Glycine are used as vegetables.
  3. Seeds of Pisum and Arachis are edible.
  4. Groundnut oil from Arachis hypogaea seeds and soyabean oil from seeds of Glycine max are used in cooking.
  5. The oil cake from Arachis hypogaea is used as fodder.
  6. The oil from the seeds of Derris indica is used in the making of medicines.
  7. Goldsmiths use the seeds of Abrus precatorius for weighing.
  8. Several crops are used in crop rotation due to their nitrogen fixing ability.
  9. Seeds of Trigonella are used as condiment and medicine. The leaves are used as vegetable.
  10. Sesbania and Tephrosia are used as green manure.
  11. Crotalaria and Phaseolus are used as fodder.
  12. Fibre from Crotalaria is used in making ropes.
  13. Indigofera yields blue dye, which is used as a fabric whitener.
  14. Wood from pterocarpus is used for making musical instruments.
  15. Wood from Dalbergia is used for making furniture.

TS Inter 1st Year Botany Question Paper May 2017

Question 15.
What are nucleosomes ? What are they made of ?
Answer:
Under electron microscope, chromatin appears as “beads on string” These seeds are known as nucleosomes. A typical nucleosome consists of 200 base pairs of DNA wrapped around a core of histone octamer having two copies of each of four types of histone proteins. They are H2 A, H2 B, H3 and H4. H1 Histone lies outside the nucleosome core and seals the two turns of DNA by binding at the point when DNA enters and leaves the core. The DNA continues between two nucleosomes is called linker DNA. This chromatin gradually condenses into chromosomes.

Question 16.
Explain prophase – I of meiosis.
Answer:
Meiosis I is longer phase and consists of 5 sub phases namely Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

a) Leptotene : The nucleus increases in size by absorbing water from the cytoplasm. The chromatin material organises into a constant number of chromosomes. The chromosomes are long, slender and show bead like structures called chromomeres. The ends of the chromosomes converge towards one side in the nucleus, where the centrosome lies. This arrangement is called Bouquet stage.

b) Zygotene: The chromosomes become shorter and thicker. They approach each other and form pairs. This homologous pair is called bivalent and the process of pairing is called synapsis. It is accompanied by the formation of Synaptonemal complex. The synapsis occurs at proterminal point or procentric or random means.

c) Pachytene : Bivalent chromosomes now clearly appear as tetrads. This stage is characterised by the presence of recombination modules, the sites of which crossing over occurs between the non-sister chromatids of the homologous chromosomes, Crossing – over is the exchange of genetic material between the two homologous chromosomes. It is also an enzyme mediated process by ‘recombinase’ crossing over leads to recombination of genetic material on the two chromosomes.

d) Diplotene : The homologous chromosomes of a bivalent begin to separate from each other except at the sites of cross overs to dissolution of synaptonemal complex. The ‘X’ shaped structures are called chaismata. The chromatids undergo condensation, contraction and thickening.

e) Diakinesis: It is marked by terminalisation of chaismata. In this phase, the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. By the end of this phase, the nuclear membrane breaks down the nucleolus disappears.

Question 17.
What are the differences between lenticels and stomata ?
Answer:

Lenticels Stomata
1. Lens shaped openings in the cork of woody stems. 1. Openings present in the epidermis of the leaf.
2. They contain closely arranged parenchymatous cells. 2. Stomata is surrounded by specialised Guard cells.
3. They permit the exchange of gases between the outer atmosphere and the inner tissues of the woody organs. 3. They perform transpiration and exchange of gases.
4. They are always open. 4. Opening and closing mecha-nism is present.

Question 18.
What are hydrophytes ? Briefly discuss the different kinds of hydrophytes with examples.
Answer:
Hydrophytes are the plants which grow in water or in very wet places. According to their relation to water and air hydrophytes are classified into five categories.
They are

  1. Free floating Hydrophytes : They float freely on the surface of the water and have no contact with soil.
    E.g. : Pistia, Eichhornia, Wolffia, Salviriia, Azalia.
  2. Rooted hydrophytes with floating leaves : They are attached to the muddy soil by roots. Their leaves have long petioles which keep them floating on the surface of water. E.g. : Nymphaea, Nelumbo and Victoria regia
  3. Submerged suspended hydrophytes: They are completely submerged and suspended in water, but not rooted in the mud and have no contact with air. E.g. : Hydrilla, Ceratophyllum and Utricularia, Najas.
  4. Submerged rooted hydrophytes : These plants are completely submerged in water. They are attached to the muddy soil by roots. E.g. : Vallisneria, Potamogeton etc.
  5. Amphibious plants: These live partly in water and partly in air. E.g. : Sagittaria, Ranunculus, Limnophila.
    Some amphibious plants grow around water bodies, . with water touching them. They will survive in dry periods also. E.g.: Typha, Cyperus etc.

TS Inter 1st Year Botany Question Paper May 2017

Section – C (2 × 8 = 16)

Note : Answer any two questions. Each answer may be limited to 60 lines.

Question 19.
Define root modification. Explain how root is modified to perform different functions. Give arty four modifications with diagrams and examples.
Answer:
A change in the normal structure of root to carry out new functions according to environment is called root modification.

Root Modifications: In some plants, roots change their shape and structure to perform functions other than absorption and conduction of water and Minerals called root modifications.

They are of different types.

  1. In Carrot, turnip (Tap roots), Sweat potato (Adventitious roots), Asparagus (Fibrous roots) become swollen due to storage of food called storage roots.
  2. In Banyan tree, Roots arise from the branches grow into the soil, become pillar like and give additional support called prop roots or pillar roots.
  3. In Maize, sugarcane, roots arise from the lower nodes of the stem, give additional support called stilt roots.
  4. In Mangroves like Rhizophora and Avicennia, Many roots come out of the ground and grow vertically upwards, help in respiration called Pneumatophores.
  5. In Epiphytes like Vanda, special adventitious roots arise, help in absorption of moisture from atmosphere called Velamen roots.
  6. In partial parasites like viscum and strga, some Haustorial roots enter into xylem of the host plant to get water and Minerals. In complete parasitic like cuscuta and Rafflesia, the haustorial roots enter into xylem and phloem of the host plant and obtain water and Minerals and food materials called Parasitic roots.
  7. In the members of Fabaceae, the roots are inhabited by Rhizobium bacteria which helps in N2 fixation called Nodular roots.
  8. In some plants like Taeniophyllum, the roots are chloro- phyllous and perform photosynthesis so called photo- synthetic roots.

TS Inter 1st Year Botany Question Paper May 2017 1
TS Inter 1st Year Botany Question Paper May 2017 2
TS Inter 1st Year Botany Question Paper May 2017 3

Question 20.
With a neat labelled diagram, describe the parts of a matur angiosperm embryo sac. Mention the role of synergids.
Answer:
Mature angiosperm embryosac shows three parts.
They are
1) Egg apparatus
2) Antipodals
3) Central cell
TS Inter 1st Year Botany Question Paper May 2017 4
1) Egg Apparatus: Three cells present towards the micropyle of the embryosac together called egg apparatus. Of which, the central, largest one is called egg and two lateral cells are called synergids. Synergids show finger like projections towards the micropyle called filliform apparatus.

2) Antipodals : Three cells present towards the chalazal end of the ovule are called antipodals. They are also referred to as vegetative cells of the embryosac and disintegrates before or after fertilisation.

3) Central cell : It is the largest cell of the embryosac. It is formed by the fusion of two polar nuclei. It is also called secondary nucleus. It shows central vacuole and 2 haploid polarnuclei.

Role of synergies :

  1. The filliform apparatus of the synergids absorbs foo materials from the Nu cells and supplies to embryosac.
  2. It also secretes some chemicals which direct the growth of the pollen tube towards embryosac.

TS Inter 1st Year Botany Question Paper May 2017

Question 21.
Describe the internal structure of a monocot root. Explain with a labelled diagram.
Answer:
The internal structure of Monocot root shows 3 zones.
They are :
1) Epidermis
2) Cortex and
3) Stele.

1) Epidermis : It is the outermost layer formed by thin walled, rectangular cells, which are compactly arranged without intercellular spaces. Cuticle and stomata are absent. Some epidermal cells (tricho blasts) produce tubular extensions called root hairs. They absorb capillary water from the soil. The epidermis of root is also known as rhizodermis of epiblema or piliferous layer.

2) Cortex : It is a wide and extensive tissue present between the epidermis and stele. It is a bigger than the stele. It can be divided into three sub-zones.
They are :
a) Exodermis
b) General cortex and
c) Endodermis.

a) Exodermis: It is the outer part of the cortex and composed of one to two rows of thick walled, dead, suberised calls. In mature roots, when the outer epidermis is removed, the exodermis acts as a protective layer. It helps in preventing the exit of water from the root tissues.

b) General Cortex : It is formed below the exodermis layer. It is composed of several rows of thin walled living cells that are arranged loosely showing intercellular spaces. The cells of cortex help in the storage of food materials and lateral conduction of water from the epidermis to the stele.

c) Endodermis : The innermost layer of cortex and is composed of single layer of barrel shaped cells that are arranged compactly without intercellular spaces. The radical and transverse walls are wrapped by ligno-suberised bands called casparian bands. Some cells situated opposite to the protoxylem cells are thin walled and without, casparian bands. These are known as passage cells which help in the entry of water from the cortex into the stele.

3) Stele : The central conducting cylinder. It is very prominent and bigger in size. The stele shows Pericycle, Vascular bundles and Medulla.

i) Pericycle : The layer of cells found beneath the endermis is known as pericycle. The cells are thin walled, parenchymatous, rectangular and compact without intercellular spaces The cells are meristematic and divide actively producing late roots. In old and mature roots, the pericycle is sclesclerenchymat and gives mechanical strength.

ii) Vascular bundles : Bundles of xylem and phloem a found separately on different radii, one alternating with the other at the peripheral boundary of the stele. These are known as radical or separate vascular bundles. The xylem is exarch and polyar More than six xylem bundles.

The ground tissue formed between the xylem and phloem stands is known as ‘conjunctive tissue’. It is usually parenchymatous. It helps in storage of food materials and provides mechanical strength.

iii) Medulla or Pith : The wide central part of the stele called ‘medulla or pith’. It is made up of thin walled parenchymatous which primary helps in the storage of food. In some monocot root the medulla is composed of thick walled lignified dead cells a helps in giving mechanical strength.
TS Inter 1st Year Botany Question Paper May 2017 5

TS Inter 1st Year Botany Question Paper March 2017

Successful navigation through TS Inter 1st Year Botany Model Papers and TS Inter 1st Year Botany Question Paper March 2017 builds students’ confidence in their exam-taking abilities.

TS Inter 1st Year Botany Question Paper March 2017

Time: 3 Hours
Maximum Marks: 60

Section – A (10 × 2 = 20)

Answer All questions

Question 1.
Give the scientific name of Mango. Identify the generic name and specific epithet.
Answer:
Mangifera Indica. Generic name is ‘Mangifera’ and Specific epithet is ‘indica’.

Question 2.
Why is Mendel considered as the Father of Genetics ?
Answer:
Mendel conducted Hybridization experiments on Pea plants and also proposed Laws of inheritance. So Mendel is considered as Father of Genetics.

Question 3.
What is palaeobotany ? What is its use ?
Answer:
It deals with the study of Fossil plants. It helps in understanding the course of evolution in plants.

Question 4.
What is the morphology of cup like structure in Cyathium ? In which family it is found ?
Answer:
In Cyathium, Cup like structure is formed from involucre of Bracts. It is seen in “Euphorbiaceae” Family.

TS Inter 1st Year Botany Question Paper March 2017

Question 5.
Define venation. How do dicots differ from monocots with respect to venation ?
Answer:
The arrangement of Veins and Veinlets in the lamina of the leaf is called venation. In Dicots, Reticulate venation is seen (Net like arrangement of Veins) where as in Monocots Parallel venation is seen (parallel arrangement of veins).

Question 6.
What is geocarpy ? Name the plant which exhibits this phenomenon.
Answer:
The development of fruit inside the soil is called Geocarpy. It is seen in Arachis hypogea.

Question 7.
What does ‘s’ refer in a 70s and 80s ribosome ?
Answer:
In both 70s and 80s ribosomes, ‘s’ stands for the sedimentation coefficient expressed in Svedberg units.

Question 8.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them ?
Answer:
300 pollen mother cells.

Question 9.
Define population and community.
Answer:
A group of similar individuals belonging to same species found in an area is called population. An assemblage of all the populations belonging to different species occurring in an area is called community.

Question 10.
What is the difference between a nucleoside and nucleotide ?
Answer:
Nitrogen base and a sugar molecule is called nucleoside. Nitrogen base, as sugar molecule and phosphate group is called nucleotide.

TS Inter 1st Year Botany Question Paper March 2017

Section – B (6 × 4 = 24)

Answer any Six Questions.

Question 11.
Give the salient features and importance of chrysopytes.
Answer:
a) Chrysophytes includes Diatoms and Golden algae.
b) They are found in fresh water as well as in marine water.
c) Most of them are photosynthetic.
d) In Diatoms, the cell walls form two thin overlapping shells, epitheca over hypotheca which fit together as soap box.
e) The walls are embedded with silica and thus the walls are indestructible.
f) Diatoms leave large amount of cell wall deposits in their habitat, this accumulation is referred to as diatomaceous earth or ‘Kiesulghar’.
g) They reproduce asexually by binary fission and sexually by gametes.

Importance:
a) The diatomaceous soil is used in polishing, filtration of oils and syrups.
b) Diatoms are the chief producers in the ocean.

Question 12.
Differentiate between red algae and brown algae.
Answer:

red algae Brown algae
1) They belong to class Rhodophyceae. 1) They belong to class Phaeophyceae.
2) Most of them are marine and some are fresh water forms. 2) They live in fresh water, brackish and salt water.
3) Cell wall is made up of cellulose, pectin and polysulphate esters. 3) Cell wall is made up of Cellulose and algin.
4) The thallus is multicellular. 4) The thallus range from simple branched filamentous forms to profusely branched forms.
5) Flagella are absent. 5) Flagella are 2, unequal and lateral.
6) The major pigments are Chlorophyll-a, d, and r-phycoerythrin. 6) The major pigments are Chlorophyll-a, c, carotenoids and fucoxanthin.
7) Food materials are stored in the form of Floridian starch. 7) Food materials are stored in the form of mannitol and laminarin.
8) Asexual reproduction occurs by non motile spores. 8) Asexual reproduction occurs by biflagellate zoospores.
9) Sexual reproduction is by non-motile gametes. 9) Sexual reproduction is by motile gametes.
10) E.g : Polysiphonia, Porphyra. 10) E.g: Ectocarpus, laminaria.

Question 13.
Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n) :
a) Ovary
b) Anther
c) Egg
d) Pollen
e) Male gamete
f) Zygote
g) Antipodal
h) Megaspore mother cell
Answer:
a) Ovary : Diploid (2n)
b) Anther : Diploid (2n)
c) Egg : Haploid (n)
d) Pollen : Haploid (n)
e) Male gamete : Haploid (n)
f) Zygote : Diploid (2n)
g) Antipodal : Haploid (n)
h) Megaspore mother cell : Diploid (2n)

Question 14.
Write a brief account on the class of Dicotyledonae of Bentham and Hooker’s classification.
Answer:
In Bentham and Hooker’s system of classification, the class Dicotyledonae was divided into three sub-classes namely Polypetalae, Gamopetalae and Monochlamydae. Polypetalae, subclass is again divided into three series namely Thalamiflorae (6 orders), Disciflorae (4 orders) and Calyciflorae (5 orders). Gamopetalae, sub class is again divided into three series namely Inferae (3 orders), Heteromerae (3 orders) and Bicarpellatae (4 orders). Monochlamydae, sub class was divided into 8 series.

TS Inter 1st Year Botany Question Paper March 2017

Question 15.
Describe the structure and functions of power houses of cell.
Answer:
Mitochondria is referred to as power houses of the cell. It is a rod shaped cell organelle having a diameter of 0.2 to 1.0 millimicrons and a length of 1.0 to 4 millimicrons. They are the sites of Aerobic respiration and produce energy in the form of ATP hence they are called power houses of the cell.
TS Inter 1st Year Botany Question Paper March 2017 1
Each mitochondrion is a double membrane bound cell organelle which leaves a perimitochondrial space in between them. The outer membrane is smooth and the inner membrane forms a number of infoldings towards inside called cristae. Inner to inner membrane, there is a fluid filled space called matrix. It possesses a single circular DNA molecule, few RNA molecules, 70s ribosomes and the components required for the synthesis of proteins. The cristae possess stalked particles named F1 particles or oxysomes. These are the sites of Aerobic respiration and produce energy in the form of ATP.

Question 16.
Though redundantly described as a resting phase, interphase does not involve rest. Comment.
Answer:
The interphase also called phase of non apparent division though called the resting phase. It is the time during which the cell is preparing for division by undergoing cell growth and DNA replication. The interphase is divided into three further phases. They are a.G1 phase, S-phase and G2 phase.

a. G1 phase : It corresponds to the interval between mitosis and initiation of DNA replication. In this, the cell is metabolically active and grows continuously.
b. S-phase: DNA replication takes place. The amount of DNA per cell doubles.
c. G2 phase : Proteinsynthesis continues. Cell organelles increases in number.

Question 17.
What are the differences between Lenticels and Stomata ?
Answer:

Lenticels Stomata
a) Lens shaped openings in the cork of woody trees are called lenticels. a) Pores present in the epidermis of young stems and leaves.
b) They show closely arranged parenchymatous cells. Guard cells are absent b) Stomata is covered by a bean shaped guard cells.
c) Lenticels do not involve in photosynthesis. c) Guard cells contain chloroplasts and involve in photosynthesis.
d) They allow gaseous exchange through the compactly arranged cork cells of the bark. d) Stomata regulate the process of transpiration and gaseous exchange.

Question 18.
Give in detail the anatomical adaptations shown by Xerophytes.
Answer:
a) Epidermis is covered with thick cuticle to reduce the rate of transpiration.
b) Epidermal cells may have silica crystals.
c) Epidermis may be multilayered as in leaves of Nerium.
d) Stomata are generally confined to lower epidermis of leaves called hypostomatous. They are present in pits called sunken stomata.
e) Mechanical tissues are very well developed.
f) Vascular tissues are very well developed.

TS Inter 1st Year Botany Question Paper March 2017

Section – C (2 × 8 = 16)

Answer any two Questions.

Question 19.
Define root modification. Explain (with diagrams) how root is modified to perform different functions.
Answer:
A change in the normal structure of root to carry out new functions according to environment is called root modification.

Root modifications: In some plants, roots change their shape and structure to perform functions other than absorption and conduction of water and Minerals called root modifications.
They are of different types.

  1. In Carrot, turnip (Tap roots), Sweat potato (Adventi-tious roots), Asparagus (Fibrous roots) become swollen due to storage of food called storage roots.
  2. In Banyan tree, Roots arise from the branches grow into the soil, become pillar like and give additional support called prop roots or pillar roots.
  3. In Maize, sugarcane, roots arise from the lower nodes of the stem, give additional support called stilt roots.
  4. In Mangroves like Rhizophora and Avicennia, Many roots come out of the ground and grow vertically upwards, help in respiration called Pneumatophores.
  5. In Epiphytes like Vanda, special adventitious roots arise, help in absorption of moisture from almosphere called Velamen roots.
  6. In partial parasites like viscum and strga, some Haustorial roots enter into xylem of the host plant to get water and Minerals. In complete parasitic like cuscuta and Rafflesia, the haustorial roots enter into xylem and phloem of the host plant and obtain water and Minerals and food materials called Parasitic roots.
  7. In the members of Fabaceae, the roots are inhabited by Rhizobium bacteria which helps in N2 fixation called Nodular roots.
  8. In some plants like Taeniophyllum, the roots are chloro – phyllous and perform photosynthesis so called photosynthetic roots.

TS Inter 1st Year Botany Question Paper March 2017 2

TS Inter 1st Year Botany Question Paper March 2017 3

Question 20.
Write a brief account on agents of pollination.
Answer:
Plants use two abiotic and one biotic agents to achieve pollination.
Abiotic agents : Wind and water are the abiotic agents.
A) Anemophily : Pollination by wind is called anemophily. Wind pollinated plants show well exposed stamens, produce light pollen grains and feathery stigma to trap pollen.

B) Hydrophily : Pollination by water called hydrophily. In Vallisneria-pollination occurs on the surface of water called Epihydro- phily. In Zoostera-pollination occurs under water called Hypohydrophily.

Biotic agents : Animals are the biotic agents help in pollination. Pollination by animals is called Zoophily. It is various types.
a) Entomophily: Pollination by insects is called Entomophily. Insect pollinated plants are large, colorful, fragrant and rich in nectar.
b) Omithophily: Pollination by birds is called Omithophily.
c) Chiropterophily: Pollination by bats is called Chiropterophily.
d) Therophily: Pollination by Squirrels is called Therophily.
e) Ophiophily: Pollination by snakes is called Ophiophily.

TS Inter 1st Year Botany Question Paper March 2017

Question 21.
Describe the internal structure of dorsiventral leaf with the help of a diagram.
Answer:
Transverse section of a dorsiventral leaf (dicot leaf) shows 3 important parts.
They are :
1. Epidermis
2. Mesophyll and
3. Vascular bundles.

1) Epidermis : Epidermis is present on the both the upper surface (upper epidermis) (adaxial) and the lower surface (abaxial) of the leaf (Lower Epidermis). The epidermis is made up of one row of barrel shaped Achlorophyllous cells, which are arranged compactly without intercellular spaces. The cells are filled with vacuolated and nucleated protoplast. On outerside of the epidermis a waxy layer called Cuticle is present. Stomata are present, more on the lower surface than on the upper surface. Each stoma is surrounded by two kidney shaped guard cells. They are chlorophyllous and regulate the opening and closing of stomata. Epidermis shows multicellular uniseriate hairs. The cells of leaf hairs are filled with water. They protect the inner tissues by absorbing the heat and prevents evaporation of water from the leaf surface. The stomata help in the gaseous exchange and also promote transpiration.

2) Mesophyll : The ground tissue that extends between the upper and lower epidermal layers is called the mesophyll. It is composed of thin walled parenchyma with chloroplasts. It is chiefly concerned with the synthesis of carbohydrates. In dicot leaves mesophyll is differentiated into two parts namely,
i) Palisade parenchyma and
ii) Spongy parenchymea.

i) Palisade parenchyma: Part of the mesophyll found beneath the upper epidermis is called ‘palisade tissue’. It shows elongated, columnar cells arranged in 1 – 3 vertical rows. Narrow intercellular spaces are present between the cells. In these cells, large numbers of chloroplasts are found nearer to the cell wall. Palisade tissue is primarily concerned with the manufacture of carbohydrates in the presence of sunlight.

ii) Spongy parenchyma: Part of the mesophyll found towards the lower epidermis is called spongy tissue. It shows 3 – 5 rows of irregular shaped cells that are arranged loosely with large intercellular spaces. Some intercellular spaces present in the vicinity of the stomata are very large, forming air chambers (air cavities). In these cells, number of chloroplasts is less. That is why the upper surface of leaf is dark green and the lower surface is light green in colour. Spongy tissue has a primary role in gaseous exchange, apart from the synthesis of food materials.

3) Vascular bundles : Vascular bundles are extended in the mesophyll in the form of veins. They help in supplying water, mineral salts and food materials all over the leaf surface. Veins also provide mechanical strength to the leaf.

The vascular bundles are conjoint, collateral and closed. The xylem is present on the upper side and phloem is on the lower side. Cambium is absent between them. Xylem shows vessles, tracheids, parenchyma and fibres. Phloem shows sieve tubes companion cells and phloem parenchyma.
TS Inter 1st Year Botany Question Paper March 2017 4
Each vascular bundle is surrounded by a layer of specialised – mesophyll cells that are arranged closely and compactly without intercellular spaces. This layer is called bundle sheath or border parenchyma. The bundle sheath cells divide and grow towards the upper and lower epidermal layers. These are calledbundle sheath extensions. They help in the conduction of food materials form the mesophyll to the vascular bundles.

TS Inter 1st Year Botany Question Paper May 2016

Successful navigation through TS Inter 1st Year Botany Model Papers and TS Inter 1st Year Botany Question Paper May 2016 builds students’ confidence in their exam-taking abilities.

TS Inter 1st Year Botany Question Paper May 2016

Time: 3 Hours
Maximum Marks: 60

General Instructions:
Note : Read the following instructions carefully.

  1. Answer all the questions of Section ‘A’. Answer any six questions out of eight in Section ‘B’ and answer any two questions out of three in Section – ‘C’.
  2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to five lines. Answer all the questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 20 lines.
  4. In Section ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 60 lines.
  5. Draw labelled diagrams wherever necessary for questions in Section ‘B’ and C.

Section – A (10 × 2 = 20)

Note : Answer all questions. Each answer may be limited to 5 lines.

Question 1.
What does ICBN stand for ?
Answer:
ICBN stands for ‘International code for Botanical Nomenclature’.

Question 2.
What does the term phycobiont and mycobiont signify ?
Answer:
The algal component of lichen is called phycobiont. The fungal component of lichen is called mycobiont.

Question 3.
Why is Mendel considered as the “Father of Genetics” ?
Answer:
Mendel conducted Hybridization experiments in pea plants and also introduced the laws of inheritance in 1866. Hence, he is considered as the “Father of genetics”.

TS Inter 1st Year Botany Question Paper May 2016

Question 4.
Differentiate actinomorphic from Zygomorphic flower.
Answer:

Actinomorphic flower Zygomorphic flower
“A flower that can be cut into two equal halves in any vertical plane”.
Ex: Datura
“A flower that can be cut into two equal halves in one vertical plane”.
Ex: Bean

Question 5.
What are aggregate fruits ? Give an example.
Answer:
The fruits which develop from multicarpellary, apocarpous ovary called aggregate fruits. Each carpel develops into fruitlet and all these fruitlets aggregate together into a single fruit. Ex : Anona, Naravalia.

Question 6.
Write the floral formula of Solanum plant.
Answer:
Br or Ebr, Ebrl, O, K(5) C(5) A5 G(2).

Question 7.
What does “S” refer in a 70S and 80S ribosome ?
Answer:
In both 70S and 80S ribosomes, “S” refer to Sedimentation coefficient expressed in Svedberg units.

Question 8.
Give one example for each of amino acids, sugars, nucleotides and fatty acids.
Answer:
Amino acids : Glycine, alanine
Sugars : Glucose, ribose.
Nucleotides : Adenylic acid
Fatty acids : Glycerol, Lecithin

Question 9.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them ?
Answer:
300 Pollen mother cells.

Question 10.
Define communities. Who classified plant communities into hydrophytes, mesophytes and xerophytes ?
Answer:
An assemblage of all the populations belonging to different species occurring in an area is called community. Eugen Warming classified plant communities into hydrophytes, Mesophytes and xerophytes.

TS Inter 1st Year Botany Question Paper May 2016

Section – B (6 × 4 = 24)

Note : Answer any six questions. Each answer may be limited to 20 lines.

Question 11.
Write the role of Fungi in our daily life.
Answer:
Uses :

  1. Yeast is used to make bread and beer.
  2. Some fungi like Pencillium are the source of antibiotics.
  3. Some fungi like Agaricus are edible mushrooms.

Disuses :

  1. Some fungi cause rot of orange fruits.
  2. Fungi spoils bread
  3. Albugo causes white rust diesease on Brassica leaves
  4. Puccinia causes rust in wheat

Question 12.
Differentiate between red algae and brown algae.
Answer:

Red algae Brown algae
1. They belong to class Rhodophyceae. 1. They belong to class Phaeophyceae.
2. Most of them are marine and some are fresh water forms. 2. They live in fresh water, brackish and salt water.
3. Cell wall is made up of cellulose, pectin and polysulphate esters. 3. Cell wall is made up of Cellulose and algin.
4. The thallus is multicellular. 4. The thallus range from simple branched filamentous forms to profusely branched forms.
5. Flagella are absent. 5. Flagella are 2, unequal and lateral.
6. The major pigments are Chlorophyll-a,d, and r-phycoerythrin. 6. The major pigments are Chlorophyll-a, c, carotenoids and fucoxanthin.
7. Food materials are stored the form of Floridian starch. 7. Food materials are stored in the form of mannitol and laminarin.
8. Asexual reproduction occurs by non motile spores. 8. Asexual reproduction occurs by biflagellate zoospores.
9. Sexual reproduction is by non-motile gametes. 9. Sexual reproduction is by motile gametes.
10. E.g : Polysiphonia, Porphyra. 10. E.g : Ectocarpus, laminaria.

Question 13.
Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction ?
Answer:

Asexual Sexual
1. Single parent is involved 1. Two parents are involved
2. Offsprings are genetically identical to each other and to their parent. 2. Offsprings are not identical to parents.
3. No fertilization occurs 3. Fertilization occurs
4. No gametes are involved 4. Gametes are involved
5. No mixing of hereditary material. 5. Mixing of hereditary material.

In multicellular or colonial forms of algae, moulds and mushrooms, the body may split or break into smaller fragments. Each fragment develops into a mature individual and this process is called fragmentation. In liverworts, specialized vegetative structures are present for reproduction called Gemmae.

In flowering plants, the units of vegetative propagation such as runners, stolons, suckers, offsets, rhizomes, corms stem tubers, bulbs and bulbils are capable of producing new offsprings. These units are called vegetative propagules. In all these plants, no involvement of sex organs takes place so vegetative reproduction is also called asexual reproduction.

TS Inter 1st Year Botany Question Paper May 2016

Question 14.
Give economic importance of plants belonging Fabaceae.
Answer:

  1. Pulses like red gram (Cajanus), black gram (Phaseolus) green gram (Phaseolus aureus), Bengal gram (Cicer arietinum) are rich source of proteins.
  2. Pods of Dolichos and Glycine are used as vegetables.
  3. Seeds of Pisum and Arachis are edible.
  4. Ground nut oil from Arachis seeds and soyabean oil from Glycine max are used in cooking.
  5. The oil cake from Arachis is used as fodder.
  6. The oil from the seeds of Derris indica is used in making medicines.
  7. The seeds of Abrus precatorius are used as Goldsmith’s weight.
  8. Seeds of Trigonella are used as condiment and medicine. The leaves are used as vegetable.
  9. Sesbania and Tephrosia are used as green manure.
  10. Crotalaria and Phaseolus are used as fodder.
  11. Fibre from Crotalaria is used in making ropes.
  12. Indigofera yields blue dye, which is used in colouring clothes.
  13. Wood from pterocarpus is used in making musical instruments.

Question 15.
Give a brief account of the types of chromosomes based on the position of centromere.
Answer:
Basing on the position of the centromere, four types of chromosomes are recognized.
TS Inter 1st Year Botany Question Paper May 2016 1
They are

  1. Metacentric : Centromere is present in between the two arms of the chromosome. It is ‘V’ shaped and consists of two equal arms.
  2. Sub-Metacentric : Centromere is present slightly away from the midpoint of a chromosome. It is ‘L’ shaped and consists of two unequal arms.
  3. Acrocentric : Centromere is present at the sub terminal position of the chromosome. It appears ‘J’ shaped and consists of one long arm and one short arm.
  4. Telocentric : Centromere is present at the terminal position of the chromosome. It appears ‘I’ shaped and consists of one arm.

Question 16.
Though redundantly described as a resting phase does not really involve rest. Comment.
Answer:
The interphase also called phase of nop apparent division though called the resting phase. It is the time during which the cell is preparing for division by undergoing cell growth and DNA replication. The interphase is divided into three further phases. They are a. G1 phase, S-phase and G2 phase.

a. G1 phase : It corresponds to the interval between mitosis and initiation of DNA replication. In this, the cell is metabolically active and grows continuously.
b. S-phase : DNA replication takes place. The amount of DNA per cell doubles.
c. G2 phase: Proteinsynthesis continues. Cell organelles increases in number.

Question 17.
State the location and functions of different types of meristems.
Answer:
Based on the position, meristems are classified into three types.
They are

  1. Apical meristems : The meristems that are present at the tip of the stem and at the tip of the root are called apical meristems. They help in linear growth of the plant body.
  2. Intercalary meristems : The meristems that are present in between mature tissues are known as intercalary meristems. They contribute to the formation of the primary plant body and also involves in internodal length.
  3. Lateral meristems : The meristems that occur in the mature regions of roots and shoots peripherally called lateral meristems. They help in increase in thickness of the plant organs, e.g: vascular cambium and Cork cambium.

TS Inter 1st Year Botany Question Paper May 2016

Question 18.
What measures do you suggest to protect the pollinators ?
Answer:

  1. Creating own pollinator-friendly garden using a wide variety of native flowering plants.
  2. Reducing the use of pesticides in and around our home.
  3. Encouraging local clubs or school groups to build butterfly gardens, bee boards and bee boxes.
  4. Supporting agriculture enterprises with pollinator – friendly practices to minimize pesticide use.
  5. Encouraging Government agencies t take into account the full economic benefits of wild pollinators when formulating – policies for agriculture and other land uses.
  6. To develop techniques for cultivating native pollinator species for cropn pollination.

Section – C (2 × 8 = 16)

Note : Answer any two questions. Each answer may be limited to 60 lines.

Question 19.
Explain how stem is modified variously to perform different functions (write any 6 modifications)
Answer:
Stems are modified in several ways to perform different functions.
They are :
1) Tendrils : Slender, spirally coiled structures which may develop either from auxiliary bud (cucumber) or from terminal bud (grapes) are called tendrils. They help in climbing.

2) Thorns : Buds are modified into woody, straight pointed thorns which protect plants from grazing animals.
Ex : Citrus, Bougain villaea.

3) Phylloclade : In some plants of acid zones, leaves are modified into scales or spines to reduce the rate of transpiration. In such plants, stems are modified into flattened, green structure which carryout photosynthesis. Such stems are called phylloclades.
Ex : In euphorbia stem is cylindrical, in casuarina needle like, and in opuntia – flattened, fleshy green.

4) Bulbils : In some plants, the vegetative buds or floral buds store food materials. At maturity, may detach from the parent plants, develop.adventitious roots, grow as new plants thus help in vegetative reproduction. Ex : Diascoria.
TS Inter 1st Year Botany Question Paper May 2016 2
5) Underground stems : In some plants, stem grows into soil, store food materials, show perennation, to resist unfavourable conditions and also help in vegetative reproduction. Such stems are called underground stems.
Ex : Rhizome – Ginger, com – colacacia.

6) Sub aerial stems : In some plants, some part of the stem is underground and some part is aerial. Such stems are called sub aerial stems. In such plants, slender, lateral branches arises from the base of the main axis, grow vertically, arches downwards, produce adventitious roots when touches the ground. When they separates from the parent plant, they develop into new plants they help in vegetative reproduction.
Ex : Stolons – Nerium, Jasmine
Suckers – Chrysanthemum, Mertha.

Question 20.
With a neat labelled diagram describe the parts of a mature embryo sac.
Answer:
Mature angiosperm embryosac shows three parts. They are
1) Egg apparatus
2) Antipodals
3) Central cell
TS Inter 1st Year Botany Question Paper May 2016 3
1) Egg Apparatus: Three cells present towards the micropyle of the embryosac together called egg apparatus. Of which, the central, largest one is called egg and two lateral cells are called synergids. Synergids show finger like projections towards the micropyle called filliform apparatus.

2) Antipodals : Three cells present towards the chalazal end of the ovule are called antipodals. They are also referred to as vegetative cells of the embryosac and disintegrates before or after fertilisation.

3) Central cell : It is the largest cell of the embryosac. It is formed by the fusion of two polar nuclei. It is also called secondary nucleus. It shows central vacuole and 2 haploid polarnuclei.

Role of synergies :

  1. The filliform apparatus of the synergids absorbs foo materials from the Nu cells and supplies to embryosac.
  2. It also secretes some chemicals which direct the growth of the pollen tube towards embryosac.

TS Inter 1st Year Botany Question Paper May 2016

Question 21.
Describe the internal structure of a Dicot root with a labelled diagram.
Answer:
A thin transverse section of dicot root shows three parts namely
(i) Epidermis
(ii) Cortex and
(iii) Stele.

(i) Epidermis – It is the outer most layers made of thin walled cells. Some cells protrude in the form of unicellular root hairs. So called Epiblema. It protects the inner parts. Root hairs help in absorption of water from the soil.

(ii) Cortex – It consists of several layers of thin walled parenchyma cells with inter cellular spaces. The Innermost layer of cortex is called Endodermis. It comprises a single layer of barrel shaped cells without Intercellular spaces. The tangential as well as the radial walls of Endoderm cells show suberin thickenings called casparian strips. Some cells opposite to protoxylem lack these strips called passage cells. They help in the movement of water and dissolved salts from cortex into xylem.

(iii) Stele : It is the central part, consists of 4 layers.
TS Inter 1st Year Botany Question Paper May 2016 4
(a) Pericycle : It is single layered, made of thin walled parenchyma cells, present next to endodermis. It produces lateral roots and become vascular cambium during secondary growth.

(b) Vascular Bundle : Xylem and phloem constitutes vascular Bundle. They are arranged on different radius (in alternate manner) so called Radial vascular Bundle, Xylem is exarch, where protoxylem is towards periphery and metaxylem is towards the center. Xylem is diarch to tetrarch condition. Xylem helps in conduction of water and minerals and phloem helps in conduction of food materials.

(c) Medulla It is absent or small, made of parenchyma cells. When present, it helps in the storage of food and water.

(D) Conjunctive tissue: The parenchyma present between xylem and phloem is called conjunctive tissue which also involves in secondary growth.

AP Inter 2nd Year Commerce Question Paper March 2023 with Solutions

Thoroughly analyzing AP Inter 2nd Year Commerce Model Papers and AP Inter 2nd Year Commerce Question Paper March 2023 helps students identify their strengths and weaknesses.

AP Inter 2nd Year Commerce Question Paper March 2023 with Solutions

Time: 3 Hours
Maximum Marks: 100

Part – I (50 Marks)
Section – A
(2 × 10 = 20)

Note: Answer ANY TWO of the following questions in not exceeding 40 lines each.

Question 1.
Define Banking. Explain the functions of Banking.

Question 2.
Explain the objectives and functions of SEBI.

AP Inter 2nd Year Commerce Question Paper March 2023 with Solutions

Question 3.
Explain the redressal mechanism available to consumers under the Consumer Protection Act, of 1986.

Section – B
(4 × 5 = 20)

Note: Answer ANY FOUR of the following questions in not exceeding 20 lines each.

Question 4.
Explain any five functions of entrepreneurs.

Question 5.
Explain the types of entrepreneurs.

Question 6.
What is international trade? Explain various types of international trade.

Question 7.
What is SEZ? Explain their objectives.

Question 8.
Explain the term insurance. Explain the functions of insurance.

Question 9.
The distinction between primary and secondary markets.

Section – C
(5 × 2 = 10)

Note: Answer ANY FIVE of the following questions in not exceeding 5 lines each.

Question 10.
Define entrepreneurship.

Question 11.
Write any one of the characteristics of an entrepreneur.

Question 12.
Wholesaler.

Question 13.
Double Insurance.

Question 14.
National Highway.

AP Inter 2nd Year Commerce Question Paper March 2023 with Solutions

Question 15.
NIFTY.

Question 16.
Commercial Papers.

Question 17.
What is meant by Consumer Protection?

Part – II (50 Marks)
Section – D
(1 × 20 = 20)

Note: Answer the following question.

Question 18.
A and B share profits in the proportions of 3/5 and 2/5. Their Balance Sheet on 31-12-2021 was as follows:
AP Inter 2nd Year Commerce Question Paper March 2023 with Solutions Q18
On that date, C was admitted into the Partnership on the following terms:
(a) That C pays Rs. 10,000 as his capital and Rs. 5,000 as goodwill for his 1/6th share in profits.
(b) That stock and fixtures be reduced by 10% and 5%. Provision for doubtful debts is created on Sundry debtors and Bills receivables.
(c) That the value of land and buildings be appreciated by 20%.
Prepare Revaluation A/c, Partners Capital Accounts, and New balance sheet of the firm.

Section – E
(1 × 10 = 10)

Note: Answer ANY ONE of the following questions.

Question 19.
Krishna of Mumbai and Gopal of Chennai are in the consignment business. Gopal sent goods to Krishna for Rs. 10,000. Gopal paid freight Rs. 500 and insurance Rs. 1,500. Krishna met sales expenses of Rs. 900, Krishna sold the entire stock for Rs. 20,000 and he is entitled to a commission of 5% on sales. Prepare necessary accounts in the books of Gopal.

Question 20.
From the following Receipts and Payments A/c of Nethajee Sports Club. Prepare Income and Expenditure A/c for the year ended on 31-03-2021.
AP Inter 2nd Year Commerce Question Paper March 2023 with Solutions Q20
Additional Information:
(a) Subscriptions include Rs. 1,000 received for the last year.
(b) Rent includes Rs. 600 paid for the last year.

Section – F
(2 × 5 = 10)

Note: Answer ANY TWO of the following questions.

Question 21.
Sandhya sold goods for Rs. 14,000 to Rajeswari on 1st March 2021 and drew upon her a bill of exchange payable after 2 months. Rajeswari accepted the bill and handed over the same to Sandhya. Sandhya immediately discounted the bill with her bank @ 12% p.a. on the due date Rajeswari met her acceptance. Pass the necessary Journal entries in the books of Sandhya.

Question 22.
On 1st January 2019, Suma purchased Furniture for, Rs. 80,000. Depreciation is to be provided annually at 10% under the straight-line method. On 31st December 2021, Furniture was sold for Rs. 40,000. Show the Furniture account assuming that the books are closed on 31st December every year.

Question 23.
Explain the categories 6f share capital.

Question 24.
Find out the profit from the following data:
Capital at the beginning of the year Rs. 40,000
Capital at the end of the year Rs. 45,000
Drawings during the year Rs. 5,000
Capital introduced during the year Rs. 2,500.

Section – G
(5 × 2 = 10)

Note: Answer ANY FIVE of the following questions.

Question 25.
Explain the parties involved in a bill of exchange.

Question 26.
Write any two causes for depreciation.

Question 27.
Delcredere Commission.

AP Inter 2nd Year Commerce Question Paper March 2023 with Solutions

Question 28.
Revenue expenditure.

Question 29.
Preference share.

Question 30.
Partnership deed.

Question 31.
Write any two advantages of a Computerised accounting system.

Question 32.
From the following data, find the profit or loss by a trader.
Capital at the beginning of the year Rs 7,500.
Capital at the end of the year Rs. 10,000.

AP Inter 2nd Year Commerce Question Paper April 2022 with Solutions

Thoroughly analyzing AP Inter 2nd Year Commerce Model Papers and AP Inter 2nd Year Commerce Question Paper April 2022 helps students identify their strengths and weaknesses.

AP Inter 2nd Year Commerce Question Paper April 2022 with Solutions

Time: 3 Hours
Maximum Marks: 100

Part – I (50 Marks)
Section – A
(2 × 10 = 20)

Note: Answer any two of the following questions in not exceeding 40 lines each.

Question 1.
What is transport? What are the benefits of transport?

Question 2.
Define stock exchanges and explain their functions.

Question 3.
Describe the rights and responsibilities of a consumer as per CPA 1986.

Section – B
(4 × 5 = 20)

Note: Answer any four of the following questions in not exceeding 20 lines each:

Question 4.
Explain any five functions of an entrepreneur.

Question 5.
Explain the role of entrepreneurship in economic development.

AP Inter 2nd Year Commerce Question Paper April 2022 with Solutions

Question 6.
What is International Trade? Write various types of International trade.

Question 7.
Write the objectives of SEZ.

Question 8.
Write the facets of electronic banking.

Question 9.
The distinction between primary and secondary markets.

Section – C
(5 × 2 = 10)

Note: Answer any five of the following questions in not exceeding 5 lines each:

Question 10.
Write any two types of entrepreneurs.

Question 11.
Explain any one characteristic of an entrepreneur.

Question 12.
Who is a Retailer?

Question 13.
Double Insurance.

Question 14.
Recurring deposit.

Question 15.
SENSEX.

Question 16.
Capital Market.

AP Inter 2nd Year Commerce Question Paper April 2022 with Solutions

Question 17.
District Forums.

Part – II (50 Marks)
Section – D
(1 × 2 = 20)

Note: Answer the following question:

Question 18.
Kumar and Suresh are partners sharing profit and losses in the ratio of 3 : 2 respectively. Their Balance sheet as of March 31, 2015, was as under:
AP Inter 2nd Year Commerce Question Paper April 2022 with Solutions Q18
On that date, they admit Deepak into partnership for 1/3 share in future profit on the following terms:
1. Furniture and stock are to be depreciated by 10%.
2. Building is appreciated by Rs. 20,000
3. A 5% provision is to be created on Debtors for doubtful debts.
4. Deepak is to bring in Rs. 50,000 as his capital and Rs. 30,000 as goodwill.
Make necessary Ledger Account and Balance Sheet of the new firm.

Section – E
(1 × 10 = 10)

Note: Answer any one of the following questions.

Question 19.
Vijaya of Vijayawada consigned goods worth Rs. 20,000 to his agent Bhavani of Bangalore on consignment. Vijaya spent Rs. 1,000. on transport, and Rs. 500. on insurance. Bhavani sent Rs. 5,000 as advance. After two months Vijaya received the account sales as follows:
1. Half of the goods were sold for Rs. 24,000.
2. Selling expenses were Rs. 1,200.
3. 10% commission on sales.
Give ledger accounts in the books of Vijaya.

Question 20.
Amaravathi Town Club provided Receipts and Payments A/c for the year ended 31 March 2013. Prepare Income and Expenditure A/c.
AP Inter 2nd Year Commerce Question Paper April 2022 with Solutions Q20

Section – F
(2 × 5 = 10)

Note: Answer any two of the following questions.

Question 21.
Sireesha Traders purchased a secondhand machine for Rs. 72,000 on 1st January 2011 and spent Rs. 8,000, on repairs and installed the same. Depreciation is written off at 10% p.a. on the straight line method. On 30th June, 2013 the machine was sold for Rs. 50,000. Prepare machinery accounts assuming that the accounts are closed on 31st December every year.

Question 22.
Satyam sold goods to Sivam worth Rs. 9,000 on 1st June 2013 and drew a bill for 2 months for the same amount. Sivam accepted the bill and returned it to Satyam. Satyam endorsed the bill to his creditor Sundaram on 1st July 2013. The bill was honored on the due date. Pass necessary journal entries in the books of Satyam.

Question 23.
Explain the categories of share capital.

Question 24.
Mr. X keeps books in the single-entry system. Find the profit from the following particulars:
Capital on 31-03-2014 – Rs. 80,000
Capital on 01 – 04 – 2013 – Rs. 70,000
Additional capital as of – Rs, 4,000
Drawings made during the year – Rs. 3,000

Section – G
(5 × 2 = 10)

Note: Answer any five of the following questions in not exceeding 5 lines each:

Question 25.
Days of Grace.

Question 26.
Write any two causes of depreciation.

Question 27.
Del credere commission.

Question 28.
Legacy.

Question 29.
X and Y are partners sharing profit and losses in the 1:2 ratio. They have decided to admit ‘Z’ by giving him 1/4 share in future profits. Calculate the new profit-sharing ratio.

AP Inter 2nd Year Commerce Question Paper April 2022 with Solutions

Question 30.
What is an equity share?

Question 31.
Ready to use.

Question 32.
What is meant by a single entry system?

TS Inter 1st Year Botany Question Paper March 2016

Successful navigation through TS Inter 1st Year Botany Model Papers and TS Inter 1st Year Botany Question Paper March 2016 builds students’ confidence in their exam-taking abilities.

TS Inter 1st Year Botany Question Paper March 2016

Time: 3 Hours
Maximum Marks: 60

General Instructions:
Note : Read the following instructions carefully.

  1. Answer all questions of Section ‘A’. Answer any six questions out of eight in Section ‘B’ and answer any two questions out of three in Section – ‘C’.
  2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to five lines. Answer all the questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 20 lines.
  4. In Section ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 60 lines.
  5. Draw labelled diagrams wherever necessary for questions in Section ‘B’ and C.

Section – A (10 × 2 = 20)

Note : Answer all questions. Each answer may be limited to 5 lines.

Question 1.
Who is popularly known as father of Botany ? What was the book written by him ?
Answer:
“Theophrastus” is regarded as the father of Botany. He wrote a book “de Historia plantarum”.

Question 2.
Define the terms, couplet and lead in taxonomic key.
Answer:
The contrasting characters occur in a pair is called ‘couplet’. Each statement in the key is called lead’.

Question 3.
State two economically important uses of heterotrophic bacteria.
Answer:
Heterotrophic bacteria help in making curd from milk, production of antibiotics and also help in nitrogen fixation in legume roots.

TS Inter 1st Year Botany Question Paper March 2016

Question 4.
Which part of the flower in cashew plant forms the false fruit ?
Answer:
The pedicel develops into false fruit in cashew plant.

Question 5.
Define mericarp. In which plant you find it ?
Answer:
The Schicarpic fruit splits into one seeded bits called mericarps. These are seen in Acacia.

Question 6.
Give the technical description of anthers of Allium cepa.
Answer:
In Allium cepa, the anthers are dithecous, basifixed, introrse and dehisce longitudinally.

Question 7.
Give one example for each of amino acids, sugars, nucleotides and fatty acids.
Answer:
Amino acids : Glycine, alanine
Sugars : Glucose, ribose.
Nucleotides : Adenylic acid.
Fatty acids : Glycerol, Lecithin.

Question 8.
What is middle lamella made of ? What is its functional significance ?
Answer:
Middle lamella is made up of calcium pectate and magnesium pectate. It holds or glues the different neighbouring cells.

Question 9.
Which of the four chromatids of a bivalent at prophase -1 of meiosis can involve in cross over ?
Answer:
Crossing over occurs between non-sister chromatids of the homologous chromosomes.

TS Inter 1st Year Botany Question Paper March 2016

Question 10.
Define heliophytes and sciophytes.
Answer:
Plants grow in direct sunlight are called Heliophytes. Plants grow in in shady places are called Sciophytes.

Section – B (6 × 4 = 24)

Note : Answer any six questions. Each answer may be limited to 20 lines.

Question 11.
Write a note on economic importance of Algae and Bryophytes.
Answer:
Economic importance of Algae :

  1. A half of the carbon dioxide fixation is carried out by Algae through photosynthesis and increases the level of oxygen in the environment.
  2. Many sps. Of Porphyra, Laminaria and Sargassum are used as food.
  3. Some marine Brown algae and red algae produce large amounts of hydrocolloids like Algin and Carrageen.
  4. Agar – Agar, a commercial product obtained from Gelidium and Gracilaria is used to grow microbes and also in the preparation of ice-creams and gellies.
  5. Iodine is extracted from Laminaria.
  6. Chlorella and Spirulina are used as food supplements even by space travelers.

Economic importance of Bryophytes :

  1. Some mosses provide food for herbaceous mammals, birds and other animals.
  2. Species of Sphagnum. A moss provide peat used as fuel and because of its capacity to hold water as packing material for trans-shipment of living material.
  3. Mosses along with lichens are the first organisms to colonise rocks.
  4. Mosses form dense mats on the soil, thus prevents soil erosion.
  5. They play a significant role in plant succession.

Question 12.
Give a brief account of dinoflagellates.
Answer:

  1. They are mostly marine and photosynthetic. They appear yellow, green, blue or red depending on the pigments in their cells.
  2. The cell wall has stiff cellulose plates on the outer surface.
  3. They have two flagella and produce spinning movements. So these protists are called ‘Whirling whips’.
  4. The nucleus has condensed chromosomes which are without histones.
  5. Red dinoflagellates like Gonyaulax undergo rapid multiplication and make the sea appear red. (Red tides of Mediterranean Sea).
  6. Toxins produced by them may kill fishes.

Question 13.
Write a brief account on gametogenesis with examples.
Answer:
The process of formation of male and female gametes is called Gametogenesis. In some Algae, the two gametes are similar in appearance. They are called Homogametes (isogametes).e.g: Cladophora. In majority of sexually reproducing organisms, the two gametes formed are morphologically different. They are called Heterogametes. The male gamete is called antherozoid – and the female gamete is called the egg. e.g: Funaria and Pteris.

TS Inter 1st Year Botany Question Paper March 2016

Question 14.
Describe the nonessential floral parts of plants belonging to Fabaceae.
Answer:
In Fabaceae, the nonessential floral parts are Calyx and Corolla.

Calyx: Sepals 5, gamosepalous, valvate aestivation and odd sepal is anterior.

Corolla : polypetalous, papilionaceous type consisting of large posterior petal called standard petal, two lateral petals called wings and the two anterior fused petals are called keel petals which enclose essential organs. They show descendingly imbricate aestivation.

Question 15.
Give a brief account of the types of chromosomes based on the position of centromere.
Answer:
Basing on the position of the centromere, four types of chromosomes are recognized.
They are

  1. Metacentric : Centromere is present in between the two arms of the chromosome. It is ‘V’ shaped and consists of two equal arms.
  2. Sub-Metacentric : Centromere is present slightly away from the midpoint of a chromosome. It is ‘L’ shaped and consists of two unequal arms.
  3. Acrocentric : Centromere is present at the sub terminal position of the chromosome. It appears ‘J’ shaped and consists of one long arm and one short arm.
  4. Telocentric : Centromere is present at the terminal position of the chromosome. It appears T shaped and consists of the one arm.

TS Inter 1st Year Botany Question Paper March 2016 1

Question 16.
Describe the cell organelle which contains chlorophyll pigments.
Answer:
TS Inter 1st Year Botany Question Paper March 2016 2
The cell organelle, certains the chlorophyll pigments is chlorplast.

They are found in pesophyll cells of leaves adn are oval or spherical or discoid or in ribbon shape. They are 5 – 10mm in length and 2 – 4 mm in width.

Each chloroplast is double membrane bound cell organelle which encloses clouless amorphous stroma.

A number of organised, flattened membranous sacs are present in stroma, which are aranged like in stacks (pile of coins) called grana Thylacoids. These are connected by flat membranous tubules called stroma lamellae. The strome contains enzymes necessary for the synthesis of proteins, small DNA molecules and 70’s’ type of ribosomes. The Thylakids consists of pigments necessary for light reaction (ie) chlorophyll ‘a’, ‘b’, caroteres and xanthophylls.

Question 17.
State the location and function of the different types of meristems.
Answer:
TS Inter 1st Year Botany Question Paper March 2016 3
Based on the position, meristems are classified into three types.
They are

  1. Apical meristems : The meristems that are present at the tip of the stem and at the tip of the . root are called apical meristems. They help in linear growth of the plant body.
  2. Intercalary meristems : The meristems that are present in between mature tissues are known as intercalary meristems. They contribute to the formation of the primary plant body and also involves in internodal length.
  3. Lateral meristems : The meristems that occur in the mature regions of roots and shoots peripherally called lateral meristems. They help in increase in thickness of the plant organs.
    e.g : vascular cambium and cork cambium.

TS Inter 1st Year Botany Question Paper March 2016

Question 18.
Write a brief account on the classification of xerophytes.
Answer:
Xerophytes are classified into three types based on the
morphology, physiologyand life cycle pattern.
They are :
1. Ephemerals : The plants which complete their life cycle within 6 – 8 weeks are called ephemerals. They are also called drought evaders or drought escapers. E.g : Tribulus.

2. Succulents : These plants absorb large amounts of water during rainy season and store it in different parts in the form of mucilage and become fleshy or succulent. The stored water is sparingly utilized during dry periods. These plants are also called drought avoiding plants.
E.g : a. Stem succulents : Opuntia,
b. Leaf succulents : Bryophyllum, Agave.
c. Root succulents : Asparagus.

3. Non-succulents : They are also called true xerophytes. These are perennial plants which can withstand prolonged drought periods, e.g: Casuarina, Nerium.

Section – C (2 × 8 = 16)

Note : Answer any two questions. Each answer may be limited to 60 lines.

Question 19.
Explain how stem is modified variously to perform different functions (write any 6 modifications)
Answer:
Stems are modified in several ways to perform different functions.
They are :
1) Tendrils : Slender, spirally coiled structures which may develop either from auxiliary bud (cucumber) or from terminal bud (grapes) are called tendrils. They help in climbing.

2) Thorns : Buds are modified into woody, straight pointed thorns which protect plants from grazing animals.
Ex : Citrus, Bougain villaea.

3) Phylloclade : In some plants of acid zones, leaves are modified into scales or spines to reduce the rate of transpiration. In such plants, stems are modified into flattened, green structure which carryout photosynthesis. Such stems are called phylloclades.
Ex : In euphorbia stem is cylindrical, in casuarina needle like, and in opuntia – flattened, fleshy green.

4) Bulbils : In some plants, the vegetative buds or floral buds store food materials. At maturity, may detach from the parent plants, develop.adventitious roots, grow as new plants thus help in vegetative reproduction. Ex : Diascoria.
TS Inter 1st Year Botany Question Paper March 2016 4
5) Underground stems : In some plants, stem grows into soil, store food materials, show perennation, to resist unfavourable conditions and also help in vegetative reproduction. Such stems are called underground stems.
Ex : Rhizome – Ginger, com – colacacia.

6) Sub aerial stems : In some plants, some part of the stem is underground and some part is aerial. Such stems are called sub aerial stems. In such plants, slender, lateral branches arises from the base of the main axis, grow vertically, arches downwards, produce adventitious roots when touches the ground. When they separates from the parent plant, they develop into new plants they help in vegetative reproduction.
Ex : Stolons – Nerium, Jasmine
Suckers – Chrysanthemum, Mertha.

Question 20.
With a neat labelled diagram describe the parts of a mature embryo sac. Mention the role of synergids.
Answer:
Mature angiosperm embryosac shows three parts.
They are
1) Egg apparatus
2) Antipodals
3) Central cell
TS Inter 1st Year Botany Question Paper March 2016 5
1) Egg Apparatus: Three cells present towards the micropyle of the embryosac together called egg apparatus. Of which, the central, largest one is called egg and two lateral cells are called synergids. Synergids show finger like projections towards the micropyle called filliform apparatus.

2) Antipodals : Three cells present towards the chalazal end of the ovule are called antipodals. They are also referred to as vegetative cells of the embryosac and disintegrates before or after fertilisation.

3) Central cell : It is the largest cell of the embryosac. It is formed by the fusion of two polar nuclei. It is also called secondary nucleus. It shows central vacuole and 2 haploid polarnuclei.

Role of synergies :

  1. The filliform apparatus of the synergids absorbs foo materials from the Nu cells and supplies to embryosac.
  2. It also secretes some chemicals which direct the growth of the pollen tube towards embryosac.

TS Inter 1st Year Botany Question Paper March 2016

Question 21.
Describe the internal structure of a Dicot root.
Answer:
A thin transverse section of dicot root shows three parts namely
(i) Epidermis
(ii) Cortex and
(iii) Stele.
(i) Epidermis – It is the outer most layers made of thin walled cells. Some cells protrude in the form of unicellular root hairs. So called Epiblema. It protects the inner parts. Root hairs help in absorption of water from the soil.

(ii) Cortex – It consists of several layers of thin walled parenchyma cells with inter cellular spaces. The Innermost layer of cortex is called Endodermis. It comprises a single layer of barrel shaped cells without Intercellular spaces. The tangential as well as the radial walls of Endoderm cells show suberin thickenings called casparian strips. Some cells opposite to protoxylem lack these strips called passage cells. They help in the movement of water and dissolved salts from cortex into xylem.

(iii) Stele : It is the central part, consists of 4 layers.
TS Inter 1st Year Botany Question Paper March 2016 6
(a) Pericycle : It is single layered, made of thin walled parenchyma cells, present next to endodermis. It produces lateral roots and become vascular cambium during secondary growth.

(b) Vascular Bundle : Xylem and phloem constitutes vascular Bundle. They are arranged on different radius (in alternate manner) so called Radial vascular Bundle, Xylem is exarch, where protoxylem is towards periphery and metaxylem is towards the center. Xylem is diarch to tetrarch condition. Xylem helps in conduction of water and minerals and phloem helps in conduction of food materials.

(c) Medulla : It is absent or small, made of parenchyma cells. When present, it helps in the storage of food and water.

(D) Conjunctive tissue : The parenchyma present between xylem and phloem is called conjunctive tissue which also involves in secondary growth.

AP Inter 1st Year Zoology Question Paper March 2019 with Solutions

Varied difficulty levels in AP Inter 1st Year Zoology Model Papers and AP Inter 1st Year Zoology Question Paper March 2019 cater to students with diverse academic strengths and challenges.

AP Inter 1st Year Zoology Question Paper March 2019 with Solutions

Time: 3 Hours
Max. Marks: 60

Note: Read the following instructions carefully.

  1. Answer all the questions of Section A. Answer any six questions in Section B and answer any two questions in Section C.
  2. In Section A, questions from Sr. No. 1 to 10 are of Very Short Answer Type. Each question carries two marks. Every answer may be limited to 5 lines. Answer all these questions at one place in the same order.
  3. In Section B, questions from Sr. Nos. 11 to 18 are of Short Answer Type. Each question carries four marks. Every answer may be limited to 20 lines.
  4. In Section C, questions from Sr. Nos. 19 to 21 are of Long Answer Type. Each question carries eight marks. Every answer may be limited to 60 lines.
  5. Draw labelled diagrams wherever necessary in Sections B and C.

Section – A
10 x 2 = 20

Note: Answer All questions In 5 lines each.

Question 1.
Define the term histology. What is it otherwise called?
Answer:
Histology is the study of the microscopic structure of different tissues. It is also called as Micro anatomy.

Question 2.
Radial symmetry is an advantage to the sessile or slow-moving organisms. Justify this statement.
Answer:
Animals showing radial symmetry live in water and they can respond equally to stimuli that arrive from all directions. Thus, radial symmetry is an advantage to sessile or slow-moving animals.

Question 3.
Distinguish between a tendon and a ligament.
Answer:
Tendons are the collagen fibrous tissue of dense regular connective tissue which attaches the skeletal muscles to bones. ligaments are also the collagen fibers tissue of dense regular connective tissue which attach bones to other bones.

Question 4.
What is the hematocrit value?
Answer:
The percentage of the total volume occupied by RBCs in blood is called hematocrit value.

AP Inter 1st Year Zoology Question Paper March 2019 with Solutions

Question 5.
Distinguish between amphids and phasmids.
Answer:
Amphids: These are the cuticular depressions present on the lips surrounding the mouth in the nematodes such as Aphasmidia animals and serve as Chemoreceptors.
Phasmids: These are the well-developed sensory organs and they occur in some nematodes such as phasmidia animals.

Question 6.
What are pneumatic bones? How do they help birds?
Answer:
Main bones in birds are extensions of air sacs without bone marrow are called pneumatic bones. These are helpful in flying to birds.

Question 7.
List any two differences between a flagellum and a cilium.
Answer:

Flagellum Cillum
1. Flagellum helps in locomotion only. 1. Cilium helps in locomotion, feeding and acts as sensory structures.
2. Flagellum produce undular movement. 2. Cilium produce pendular movement.
3. Flagellum is about 150 µ in length. 3. Cilium small in size 5-10 µ in length.

Question 8.
Why do we refer to the offspring, formed by asexual method of reproduction, a clone?
Answer:
As a result of asexual method the offsprings are not only identical to one another but also exact copies of their parent. The term ‘clone’ is used to describe such morphologically and genetically similar individuals.

Question 9.
The eggs of Ascaris are called Mammillated eggs. Justify.
Answer:
Each egg of Ascaris is surrounded by a protein coat with rippled surface. Hence the eggs of Ascaris are called “mammilated eggs’.

Question 10.
What is cyclomorphosis? Explain its importance in Daphnia.
Answer:
The cyclic seasonal morphological variations among certain organisms is called “Cyclomorphosis”. In the case of Daphnia, it is an adaptation to “stabilize the movement’ in water and can resist the water currents better” to stay in the water rich in food materials.

Section -B
6 x 4 = 24

Note: Answer any six questions in 20 lines each:

Question 11.
Explain in ‘brief Biodiversity Hot Spots.
Answer:
Biodiversity hot spots: A Biodiversity hot spot is a Biogeo graphic Region that is both a significant reservoir of biodiversity and is threatened with destruction. The concept of biodiversity originated by Norman Myers. There are about 34 biodiversity hot spots in the world. As these regions are threatened by destruction habitat loss is accelerated.
e.g.:

  • Western Ghats and Srilanka;
  • Indo Burma;
  • Hima layas in India.

Ecologically unique and biodiversity-rich regions are legally protected as in

  • Biosphere Reserves-14,
  • National Parks- 90,
  • Sanctuaries-448.

Biosphere Reserves: An area which is set aside, minimally disturbed for the conservation of the resources of the biosphere is Biosphere reserve’. Latest biosphere reserve (17th biosphere reserve in India) is Seshachalam Hills.

National Parks: A National Park is a natural habitat strictly reserved for protection of natural life. National Parks, across the country, offer a fascinating diversity of terrain, flora, and fauna. Some important National Parks in India are – Jim Corbett National Park (the first National Park in India located in Uttarakhand), Kaziranga National Park (Assam), Kasu Brahmananda Reddy National Park, Mahavir HarinaVanaSthali National Park (AP). Keoladeo Ghana National Park (Rajasthan), etc.

Sanctuaries: Specific endangered faunal species are well protected in wildlife sanctuaries which permits eco-tourism (as long as animal life is undisturbed). Some important Sanctuaries in India (AP) include-Koringa Sanctuary, Eturnagaram Sanctuary, Papikon Dalu Sanctuary.

AP Inter 1st Year Zoology Question Paper March 2019 with Solutions

Question 12.
Describe the structure of a multipolar neuron.
Answer:
Multipolar neurons have one axon and two or more dendrites. Most neurons in our body are multipolar neurons.
A neuron usually consists of a cell body with one to many dendrites and a single axon.
AP Inter 1st Year Zoology Question Paper March 2019 with Solutions 1
Neurons: Neurons are the functional units of nervous tissue. These are electrically excitable cells which receive, initiate, and conduct /transmit impulses. When a neuron is stimulated, an electric disturbance (action potential) is generated which swiftly travels along its plasma membrane. A neuron usually consists of a ‘cell body” with one to many dendrites and a single axon.

Cell body: It is also called perikaryon, cyton or soma. It contains abundant granular cytoplasm and a large spherical nucleus. The cytoplasm has Nissi bodies (they represent RER, the sites of protein synthesis), neurofibrils, and lipofuscin granules.

Dendrites: Several short, branched processes which arise form the cyton are called dendrites. They also contain Nissi bodies and neurofibrils. They conduct nerve impulses towards the cell body (afferent processes).

Axon: An axon is a single, long, cylindrical process that originates from a region of the cyton called axon hillock. Plasmalemma of an axon is called axolemma, and the cytoplasm is called axoplasm, which contains neurofibrils. However, Nissi bodies are absent. An axon may give rise to collateral branches. Distally it branches into many fine filaments called telo dendrites, (axon terminals), which end in bulb-like structures called synaptic knobs or terminal boutons. Synaptic knobs possess ‘synaptic vesicles’ containing chemicals called neurotransmitters.

Question 13.
What are the salient features of the echinoids?
Answer:

  1. It includes sea urchins, heart urchins, sand dollars, etc. The body is ovoid or discoidal and covered by movable spines.
  2. Arms are absent, tube feet are arranged in five bands, and bear suckers.
  3. Ossicles of the body unite to form a rigid test or corona or case.
  4. Pedicellaria are ‘three jawed’.
  5. Anus and Madreporte are aboral in position.
  6. Ambulacral grooves are closed.
  7. A complex five-jawed masticatory apparatus called Aristotle’s lantern is present just inside the mouth. It is absent in heart urchins.
  8. Life history includes larval form called echinopluteus.
  9. Specialized gills called peristomial gills as present in sea urchins. Eg: Salmacis (Sea urchin), Echino Cardium (Heart urchin), Clypeastoer (Cake Urchin).

Question 14.
List out eight characteristics that help distinguish a fish from the other vertebrates.
Answer:
General characters:

  1. Fishes are completely aquatic poikilothermic (cold-blooded) animals.
  2. Body of a fish is usually streamlined and differentiated into head, trunk, and tail
  3. The exoskeleton consists of mesodermal scales or bony plates. A few are scaleless.
  4. The endoskeleton may be cartilaginous or bony. Skull is monocondylic. Vertebrae are amphicoelous. Centrum is concave at both anterior and posterior faces.
  5. Locomotion is assisted by unpaired (median and caudal) fins along with paired (pectoral and pelvic) fins.
  6. Mouth is ventral or terminal. Teeth are usually acrodont, homodont, and polyphyodont.
  7. Exchange of respiratory gases is performed by the gills. Heart is ‘two-chambered.
  8. Kidneys are mesonephric. Fishes aremty ammonoltelic and some are ureotelic. (cartilaginous fishes).
  9. Cranial nerves are 10 pairs, Meninx Primitiva is the only meninx enveloping the central nervous system.
  10. Internal ear consists of three semicircular canals. The lateral-line sensory system (to detect movement and vibration in the surrounding water) is well-developed. Eyes are without eyelids and each eye ball is protected by a nictitating membrane.
  11. Sexes are separate. Fertilization is internal or external. Development may be direct or indirect.

AP Inter 1st Year Zoology Question Paper March 2019 with Solutions 2

Question 15.
Give an account of pseudopodia.
Answer:
Locomotion in protozoans is performed by cellular extensions such as pseudopodia are found in rhizopods organisms. The pseudopo dia are temporary extensions of cytoplasm that develop in the direction of the movement. These temporary structures are useful to move on the substratum as our legs do, hence the name ‘pseudopodia. There are four kinds of pseudopodia in protozoans.

  1. Lobopodia: blunt + finger-like Pseudopodia. Ex : Amoeba, Entamoeba.
  2. Filopodia: fiber like pseudopodia contain ectoplasm. Ex : Euglypha
  3. Reticulopodia : net like pseudopodia. Ex: Elphidium.
  4. Axopodia or heliopodla: Sun ray-like pseudopodia. Ex : Actinophrys.

AP Inter 1st Year Zoology Question Paper March 2019 with Solutions 3

AP Inter 1st Year Zoology Question Paper March 2019 with Solutions

Question 16.
What are the adverse effects of tobacco?
Answer:
Effect: Smoking increases the carbon monoxide (CO) level and reduces the oxygen level in the blood. Nicotine stimulates the adrenal gland to release adrenaline and nor-adrenaline into the blood. These hormones raise the blood pressure and increase the heart rate. Smoking is associated with bronchitis, emphysema, coronary heart disease, gastric ulcers and increases the incidence of cancers of throat, lungs, urinary bladder, etc. Smoking also paves the way to hard drugs. Yet smoking is very prevalent in society, both among young and old. Tobacco chewing is associated with an increased risk of cancer of the oral cavity.

Question 17.
Draw a neat and labelled diagram of the salivary apparatus of cockroaches.
Answer:
AP Inter 1st Year Zoology Question Paper March 2019 with Solutions 4

Question 18.
What are the deleterious effects of depletion of ozone in the stratosphere?
Answer:
The depletion of ozone is particularly marked over the Antarctic region. This has resulted in the formation of a large area thinned ozone layer commonly called as the ozone hole.

UV radiation with wavelengths shorter than that of UV-B, are almost completely absorbed by Earth’s atmosphere, provided that the ozone layer is intact. But UV-B damages DNA and may induce mutations. It causes ageing of skin, damage to skin cells, and various types of skin cancers. In human eye, cornea absorbs UV-B radiation and a high dose of UV-B causes inflammation of cornea called snow-blindness, cataracts, etc., such exposure may permanently damage the cornea.

Section – C
2 X 8 = 16

Note: Answer any TWO questions in 60 lines each.

Question 19.
Describe the life cycle of Plasmodium vivax in mosquito.
Answer:
Life cycle of plasmodium in mosquito (The mosquito phase) Ross cycle:
When a female Anopheles mosquito bites and sucks the blood of a malaria patient the gametocytes along with the other stages of the erythrocytic cycle reach the crop of mosquitoes. Here all the stages are digested except the gametocytes. Further part of the life cycle consists of

  • Gametogony
  • Fertilization
  • Formation of ookinete & oocysts
  • Sporogony.

i) Gametogony: The formation of male and female gametes from the gametocytes is called gametogony. it occurs in the lumen of the crop of mosquitoes.

Formation of male gametes: During this process, the nucleus of the microgametocyte divides into eight daughter nuclei called pronuclei which reach the periphery. The cytoplasm is pushed out in the form of eight flagella-like processes. into each flagellum-like process, one pronucleus enters and forms a micro gamete or male gamete. These male gametes show lashing movements like flagella and get separated from the cytoplasm of microgametocyte. This process is called exflagellation.

ii) Formation of female gamete: The female gámetocyte undergoes a few changes and transforms into a female gamete. This process is called maturation. The nucleus of the female gamete moves towards the periphery and the cytoplasm at that point forms a projection. This projected region is called the fertilization cone.
AP Inter 1st Year Zoology Question Paper March 2019 with Solutions 5
Fertilization: The fusion of male and female gametes is called fertilization. It also occurs in the lumen of the crop of the mosquito. When an actively moving male gamete comes into contact with the fertilization cone of the female gamete, it enters it, the pronuclei and cytoplasm of these two gametes fuse with each other, resulting in the formation of a synkaryon. Since the two gametes are dissimilar in size this process is known as anisogomy. The female gamete that bears the synkaryon is called the zygote which is round and non-motile.

iii) Formation of ookinete and oocysts: The zygote remains inactive for some time and then transforms into a long, slender, motile, vermiform ookinete or vermicule within 18 to 24 hours. It pierces the wall of the crop and settles beneath the basement membrane. It becomes round and secretes a cyst around its body. This encysted ookinete is now called an oocyst. About 50 to 500 oocysts are formed on the wall of the crop and appear in the form of small nodules.

iv) Sporogony: The formation of sporozoites in the oocysts is called sporogony. According to Bano, the nucleus of the oocyst first undergoes reduction division followed by repeated mitotic divisions resulting in the formation of about 1,000 daughter nuclei. Each bit of nucleus is surrounded by a little bit of the cytoplasm and transforms into a sickle-shaped sporozoite. Oocyst with such sporozoites is called sporocyst.

When this sporocyst raptures, the sporozoites are liberated into the hemocoel of the mosquito. From there, they travel into the salivary glands and are ready for infection. The life cycle of plasmodium in mosquitoes completes in about 10 to 24 days.

Question 20.
Describe the respiratory system of cockroaches with the help of neat and labelled diagrams.
Answer:
Due to the absence of respiratory pigment, the blood of cock roach is colourless and is cannot carry oxygen to different tissues. Therefore a tracheal system is developed to carry the air directly to the tissues. The respiratory system of cockroaches consists of stigmata, tracheae, and tracheoles.

Stigmata or spiracles: The tracheal system communicates with the exterior by ten pairs of openings called stigmata or spiracles. The first two pairs of spiracles are present in the thoracic segments, one pair in mesothorax and one pair in the metathorax. The remaining eight pairs abdominal segments. Spiracles are located in the pleura of their respective segments. The respiratory system in insects is classified on the basis of number and nature of spiracles.

The spiracles of cockroaches are polytheistic (as they are more than 3 pairs) and holocaustic (as all of them are functional). All spiracles are valvular and each of them is surrounded by a chitinous ring called peritreme. All spiracles bear small hair-like structures called trichomes to filter the dust particles.
AP Inter 1st Year Zoology Question Paper March 2019 with Solutions 6
Tracheae: From the atrium of each thoracic spiracle several horizontal tracheae run inside. They join with each other in the thorax to form many tracheal trunks like dorsal cephalic, ventral cephalic trunks and their branches. These branches enter all organs of the head. The thoracic region also contains lateral longitudinal trunks.

The abdominal spiracles lead into atria. From the atrium of each abdominal spiracle three tracheal tube arise. All these tracheal tubes of one side open into three separate longitudinal tracheal trunks. They are lateral dorsal and ventral longitudinal trunks. Lateral longitudinal trunks are the longest tracheal trunks.

The three pairs of longitudinal tracheal trunks of both the sides are interconnected by many commissural tracheae. From all the tracheal trunks several branches are given out, which enter different organs. All tracheal branches entering into an organ end in a special cell called tracheoles cell.

The wall of the tracheae is made of three layers. They are an outer basement membrane, a middle one-cell thick epithelium and an inner layer of cuticle called intima. The intima is produced into spiral thickening called taenidia. The taenidia keep the tracheae always open and prevent it from collapsing.

Tracheoles: The terminal cell of trachea is called tracheoblast or tracheole cell. It has several intracellular tubular extensions called tracheoles. Tracheoles are devoid of intima and taenidia. They are formed of a protein called trachein. Tracheolar fluid is present inside the tracheoles. The level of the bachelor fluid varies with the metabolic activity of the insect.

It is more when the insect is inactive and completely reabsorbed into the tissues when the insect is more active. Tracheoles penetrate the cell and are intimately associated with mitochondria (to supply oxygen to them).

AP Inter 1st Year Zoology Question Paper March 2019 with Solutions

Question 21.
Describe different types of food chains that exist in an ecosystem.
Answer:
The food energy passes from one trophic level to another trophic level mostly from the lower to higher trophic levels. When the path of food energy is linear’ the components resemble the links of a chain and it is called ‘the food chain. Generally, a food chain ends with decomposers. The three major types of food chains in an ecosystem are Grazing Food Chain, Parasitic Food Chain, and Detritus Food Chain.
AP Inter 1st Year Zoology Question Paper March 2019 with Solutions 7
i) Grazing food chain: It is also known as predatory food chain, it begins with the green plants (producers), and the second, third, and fourth trophic levels are occupied by the herbivores, primary carnivores and secondary carnivores respectively. In some food chains, these is yet another trophic level – the climax carnivores. The number of trophic levels in food chains varies form 3 to 5 generally. Some examples from grazing food chain (GFC) are given below:
AP Inter 1st Year Zoology Question Paper March 2019 with Solutions 8
ii) Parasitic food chain: Some authors included the Parasitic Food Chains as a part of the GFC. As in the case of GFCs, it also begins with the producers the plants (directly or indirectly). However, the food energy passes from Large organisms to small organisms in the parasitic chains. For instance, a tree which occupies the 1st trophic level provides shelter and food for many birds.

These birds host many ectoparasites and endo parasites. Thus, unlike in the predator food chain, the path of the flow of energy includes fewer, large-sized organisms in the lower trophic levels and numerous, small-sized organisms in the successive higher trophic levels.

iii) Detritus Food Chain: The detritus food chain (DFC) begins with dead organic matter (such as leaf litter, bodies of dead organisms). It is made up of decomposers which are heterotrophic organisms mainly the ‘fungi and bacteria’. They meet their energy and nutrient requirements by degrading dead organic matter or detritus. These are also known as (sapro: to decompose).

DecomposerS: Secrete digestive enzymes that break down dead and waste materials (such as faeces) into simple absorbable sub stances. Some examples of detritus food chains are :

  1. Detritus (formed from leaf litter) – Earthworms – Frogs – Snakes
  2. Dead animals – Flies and maggots – Frogs – Snakes.

In an aquatic ecosystem, GFC is the major ‘conduit for the energy flow. As against this in a terrestrial ecosystem, a much larger fraction of energy flows through the detritus food chain than through the GFC.

Detritus food chain may be connected with the grazing food chain at some levels. Some of the organisms of DFC may form the prey of the GFC animals. For example, in the detritus food chain given above, the earthworms of the DFC may become the food of the birds of the GFC. It is to be understood that food chains are not ‘isolated’ always.

AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions

Thoroughly analyzing AP Inter 2nd Year Commerce Model Papers and AP Inter 2nd Year Commerce Question Paper May 2019 helps students identify their strengths and weaknesses.

AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions

Time: 3 Hours
Maximum Marks: 100

Part – I (50 Marks)
Section – A
(2 × 10 = 20)

Note:

  • Answer any two of the following questions in 40 lines each.
  • Each question carries 10 marks.

Question 1.
Define Insurance. Discuss the principles of insurance.
Answer:
Insurance means protection against the risk of loss. It compensates against any loss or damage due to the happening of an event. It is a contract between the two parties by which one of them undertakes to indemnify the other person against a loss that may arise due to some events.

Principles of Insurance
1. Insurable Interest:
A person cannot enter into a contract of insurance unless he has an insurable interest in the subject matter of insurance. It is an essential feature of insurance. Without this insurable interest, the contract of insurance will be treated as a wager or gambling contract. A person has an insurable interest in his own life or the life of his wife and a creditor has insurable interest in the debtor.

2. Utmost Good Faith:
Insurance is based on the principle of utmost good faith. It means both the parties of the contract must disclose all the facts relating to the subject matter of insurance. If the insured does not disclose all material facts, the contract between them is void. A person who had suffered from T.B. in the past had not disclosed it in the proposal form. Later on, the insurer comes to know of this fact. He may declare the contract as void.

3. Indemnity:
This is the chief principle of insurance. Indemnity means security against the risk of loss. Under this principle, the insured gets only the loss suffered by the insurer but not profits out of the contract of insurance. The principle of indemnity applies to contracts of fire and marine insurance only, but not to life insurance contracts.

4. Contribution:
Sometimes, goods are insured by more than one company. It is double insurance. The insured can get compensation only for the total loss from all insurance companies put together, but not the total loss from each company. The insurance companies will pay the compensation on a pro-rata basis.

5. Subrogation:
According to this principle, the insurer after compensating the loss of the insured, the right of ownership of the damaged goods is shifted from the insured to the insurance company.
Ex: Mr. X owns a scooter worth ₹ 36,000 and it was insured with an insurance company for full value. Later it was met with an accident and damaged beyond repairs. The insurance company paid the full value as compensation. Then all the rights on the scooter will be passed on to the insurance company.

6. Causa Proxima:
According to this principle, if the nearest and direct factor causes the loss, then only the insurer will have to bear the loss.
Ex: Biscuits in a ship are insured and are destroyed because of the seawater entered through a hole made by the mouse in the bottom of the ship and water entered into the ship. The nearest and direct cause is seawater. Hence, the insurer will have to bear the loss.

7. Mitigation of Loss:
It is the duty of the insured to take necessary steps to minimize the loss that happened due to some event. He should not act carelessly and negligently at the time of loss to the insured property.

AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions

Question 2.
What is SEBI? Explain the objectives and functions of SEBI.
Answer:
Introduction:
The Securities and Exchange Board of India (SEBI) was established by the Government of India in April 1988 to promote healthy growth of the securities market and for investor protection. It was to function under the administrative control of the Ministry of Finance of the Government of India. It got statutory status in January 1992.

Objectives of SEBI:

  • To regulate stock exchanges and the securities industry to promote their orderly functioning.
  • To protect the rights and interests of investors and to guide and educate them.
  • To prevent trading malpractices in the securities industry.
  • To regulate and develop a code of conduct and fair practices by intermediaries like brokers, merchant bankers, etc. to make them competitive and professional.

Functions of SEBI:
SEBI assigned regulatory, development, and protective functions to control and regulate the capital market.

(A) Regulatory Functions:

  • Registration of brokers, sub-brokers, and other players in the market.
  • Registration of collective investment schemes and mutual funds.
  • Regulation of stock brokers, portfolio exchanges, merchant bankers in stock exchanges, and any other securities market.
  • Regulation of takeover bids by companies.
  • Calling for information by undertaking inspections, conducting inquiries, and audits of stock exchanges and intermediaries.
  • Levying fee or other charges for carrying out the purpose of the act.
  • Performing and exercising such powers under the Securities Contracts Regulation Act 1956, as may be delegated by the Government of India.

(B) Development Functions:

  • Training for intermediaries of the securities market.
  • Conducting research and publishing information useful to all market participants.
  • Undertaking measures to develop the capital markets by adopting a flexible approach.

(C) Protective Functions:

  • Prohibition of fraudulent and unfair trade practices like making misleading statements, manipulations, price rigging, etc.
  • Controlling insider trading and imposing penalties for such practices.

Question 3.
Explain the redressal mechanism available to consumers under the Consumer Protection Act, of 1986.
Answer:
Consumer Protection refers to the measures adopted the protect consumers from unethical malpractices by businesses and to provide them with speedy redressal of their grievances. The judicial mechanism set up under the Consumer Protection Act 1986, consists of consumer courts (forums) at the district, state, and national levels. These are known as district forums, the State Consumer Disputes Redressal Commission (State Commission), and the National Consumer Disputes Redressal Commission (National Commission).

1. District Forum:
District forum is established by the State Government in each district.

  • Composition: The district forum consists of a Chairman and two other members, one of them shall be a woman candidate. They are headed by the person of the rank of a District Judge.
  • Jurisdiction: A written complaint can be filed before the District Consumer Forum, where the value of goods or services and the compensation claimed does not exceed ₹ 20 lakhs.
  • Appeal: If a consumer is not satisfied by the decision of the District Forum, he can challenge the same before the State Commission, within 30 days of the order.

2. State Commission:
State commission is established by the state governments in their respective states.

  • Composition: It consists of a president and two members, one of them shall be a woman. It is headed by a person of the level of High Court Judge.
  • Jurisdiction: A written complaint can be filed before the State Commission where the value of goods or services and the compensation claimed exceeds ₹ 20 lakhs but does not exceed ₹ 1 crore.
  • Appeal: In case the aggrieved party is not satisfied with the order of the state commission he can appeal to the National Commission within 30 days of passing of the order.

3. National Commission:
It was constituted in 1988 by the central government. It is the highest authority to settle consumer disputes at the national level.

  • Composition: It consists of a president and not less than four members, one of them shall be a woman. It is headed by a sitting or retired Judge of the Supreme Court.
  • Jurisdiction: All the complaints about those goods and services and compensation value of more than ₹ 1 crore can be filed directly before the National Commission.
  • Appeal: An appeal can be filed against the order of the National Commission to the Supreme Court within 30 days from the date of the order passed.

Section – B
(4 × 5 = 20)

Note:

  • Answer any four of the following questions in not exceeding 20 lines each.
  • Each question carries 5 marks.

Question 4.
Explain the characteristics of entrepreneurs.
Answer:
Entrepreneurs tend to have specific characteristics that distinguish them from other people. The following are some characteristics that every successful entrepreneur must possess.

1. Innovation:
Innovation is an important characteristic of an entrepreneur in modern business. Innovation may take the form of the introduction of a new method or introducing improvements in the existing method. Innovation helps in increasing production and reducing the cost of production.

2. Risk-taking:
The entrepreneur has to pay for all factors of production in advance. There is a chance that he may be rewarded with a handsome profit or he may suffer a heavy loss. Therefore, the risk-bearing is the final responsibility of an entrepreneur.

3. Organisation of Production:
He makes arrangements for land, labor, capital, raw materials, etc, required for setting up a production process. He assesses the viability of having different production processes and selects one that is most suitable.

4. Decision Making:
Every activity of the business requires decision-making. An entrepreneur has to make decisions about the establishment of the business its management and the coordination of various resources.

5. Leadership:
An entrepreneur has to be a leader because he is such a person who organizes, directs, commands, and controls the functioning of the organization. His personality will influence the working of his subordinates because he is taken as a role model. He motivates them to achieve goals quickly and efficiently.

AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions

Question 5.
Explain the relationship between entrepreneur and entrepreneurship.
Answer:
Entrepreneur is the person, entrepreneurship is the process and enterprise is the creation of the person and output of the process. Though the term entrepreneur is often used interchangeably with entrepreneurship, they are conceptually different. The relationship between the two is just like the two sides of the same coin. The following points highlight the relationship between entrepreneurs and entrepreneurship.

  • An entrepreneur is a person: Entrepreneurship is a process.
  • Entrepreneur is organizer: Entrepreneurship is the organization.
  • Entrepreneur is an innovator: Entrepreneurship is innovation.
  • An entrepreneur is a risk bearer: Entrepreneurship is risk-bearing.
  • Entrepreneur is motivator: Entrepreneurship is motivation.
  • Entrepreneur is the creator: Entrepreneurship is the creation.
  • Entrepreneur is visualizer: Entrepreneurship is vision.
  • Entrepreneur is the leader: Entrepreneurship is leadership.
  • Entrepreneur is imitator: Entrepreneurship is imitation.

Question 6.
Distinguish between Home Trade and Foreign Trade.
Answer:
The following are the differences between Home trade and Foreign trade.

Home Trade Foreign Trade
1. Trade Trade carries within the country. Trade carried with other countries.
2. Currency It does not involve any exchange of currency. It involves the exchange of currencies.
3. Restrictions It is not subjected to any restrictions. It is subject to many restrictions.
4. Risk Transport costs and risks are less. Transport costs and risks are higher.
5. Nature It consists of sales, transfer, or exchange of goods within the country. It involves the import and export of goods.
6. Transport of Goods The movement of goods depends on the internal transport system. e.g.: Roads and railways. The movement of goods takes place usually by sea wherever possible.
7. Specialisation It helps to derive benefits of specialization within the country. It helps all trading countries to derive the benefits of specialization.

Question 7.
Explain the advantages of SEZs.
Answer:
Advantages:
The following major benefits can be attributed to Special Economic Zones (SEZ).

  • Employment generation: Special Economic Zones are considered to be highly effective tools for job creation.
  • Economic development: India can be made a transformed economy if special economic zones are implemented properly because special economic zones are engines for economic development.
  • Growth of labor-intensive manufacturing industry: The establishment of special economic zones may lead to faster growth of labor-intensive manufacturing and service industries in the country.
  • Balanced regional development: Special economic zones are beautifully crafted initiatives for achieving balanced regional development.
  • Capacity building: Special economic zones are necessary for stronger capacity building.
  • Export promotion: Special economic zones induce dynamism in the export performance of a country by eliminating
    • Distortions resulting from tariffs and other trade barriers.
    • The corporate tax system and excessive bureaucracy.

Question 8.
Write the advantages of E-Banking.
Answer:
E-banking brings certain advantages:

  • It reduces costs: The cost of banking transactions is considerably reduced. It increases the profitability of the banks.
  • Prompt in Services: There is a high degree of personalization and fast and flexible execution. Thus E-Banking prompt service and there is greater customer satisfaction.
  • Anywhere and any time banking: It is 24-hour in a day and 7 days in a weekly banking service. Bank accounts can be accessed from anywhere. So the customer can obtain information on his account and conduct transactions from his home or office.
  • Cashless banking: Handling of cash is not necessary in E-Banking.
  • Global coverage: It provides global network coverage of bank services. NRIs can monitor their bank account in Indian banks, from abroad.
  • Central database: The database of each branch is centralized. Customers can deposit, withdraw, or remit money from any branch of their bank.

Question 9.
What is a capital market? What is its importance?
Answer:
Capital Market:
A capital market is a market for long-term funds for more than one year. The main instruments traded in the capital market are equity shares, preference shares, debentures, bonds, etc. Development banks, stock exchanges, and investment. Companies play a major role in the capital market.

Importance of Capital Market:

  • The link between Savers and Investors: The capital market plays an important role in mobilizing the savings and diverting them into productive investment.
  • Encouragement of Savings: With the development of the capital market, financial institutions provide a vast range of instruments that encourage people to save them.
  • Encouragement of Investments: Various assets like Shares, bonds, etc., encourage savers to lend to the government or to invest in industry. Thus the capital market facilitates lending to the businessmen and the government.
  • Stability in Price: The capital market tends to stabilize the values of stocks and securities. In this process of stabilization, the capital market provides capital to borrowers at a lower rate of interest.
  • Promotes Economic Growth: Economic growth is possible in any country with the proper allocation of resources among the industries. The capital market not only reflects the general conditions of the economy but also smoothens and accelerates the process of economic growth.

Section – C
(5 × 2 = 10)

Note:

  • Answer any Five of the following questions in not exceeding 5 lines each.
  • Each question carries 2 marks.

Question 10.
Define entrepreneurship.
Answer:
According to P.F. Drucker, “Entrepreneurship is neither a science nor an art. It has a knowledge base. Knowledge in entrepreneurship is a means to an end. Indeed, the ends largely define what contributes knowledge in practice”.

Question 11.
Explain any two functions of an entrepreneur.
Answer:
1. Formation of New Producing Organisation:
It is the function of an entrepreneur, to arrange land, labor, capital, raw materials, etc., required for setting up a production process. According to J.B.Say, the function of a producer/entrepreneur is to rationally combine the forces of production into a new producing organization.

2. Decision Making:
An entrepreneur as a decision maker takes various decisions regarding the following:

  • Ascertaining the objective of the enterprise
  • Sources of finance
  • Product mix
  • Pricing policies
  • Promotion strategies etc.

Question 12.
Define wholesaler.
Answer:
Wholesale trade means buying and selling goods in large quantities or bulk. The trader who engages in wholesale trade is called a wholesaler. He buys goods in bulk from producers and sells them in small lots to retailers or industrial users.

AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions

Question 13.
Bonded warehouse.
Answer:
Bonded warehouses are licensed by the government to accept imported goods before payment of tax and customs duty. Importers are not permitted to remove goods from the docks or the airport till customs duty is paid.

Question 14.
National Highway.
Answer:
National highways are meant for intertransport. These roads also connect state capitals, major cities, etc. The National Highway Authority of India (NHAI) has the responsibility for the development, maintenance, and operation of the national highways.

Question 15.
Commercial Papers.
Answer:
Commercial paper is a short-term money market instrument, which is issued by companies to raise short-term funds at a lower rate of interest than market rates. It usually has a maturity period of 15 days to one year.

Question 16.
NIFTY.
Answer:
NIFTY is an index of the NSE, which consists of 50 companies from 24 different sectors listed on the NSE. The base year for the index is 1995-96 with the base value as 1000.

Question 17.
Give the meaning of Consumer.
Answer:
Under the Consumer Protection Act 1986, The word consumer has been defined separately for goods and services. For goods, a consumer buys any goods for consideration and any user of such goods other than the person who buys it, provided such use is made with the approval of the buyer. For services, a consumer has any service or services for consideration and any beneficiary of such services provided the service is availed with the approval of the person who had hired the service for consideration.

Part – II (50 Marks)
Section – D
(1 × 20 = 20)

Note: Answer the following question.

Question 18.
Krishna and Radha are partners in a business sharing profits and losses equally. Their Balance Sheet on 31-03-2018 stood as under:
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q18
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q18.1
They decided to admit Sathya into the firm on 01-04-2018 on the following terms and conditions:
(a) Sathya has to pay ₹ 2,50,000 for 1/4 share in future profits.
(b) Sathya has to pay ₹ 60,000 for goodwill.
(c) Plant and machinery to be depreciated by 10%.
(d) Buildings to be appreciated by 20%.
(e) 5% reserve for doubtful debts to be created on debtors.
Prepare necessary accounts in the books of the firm after the admission of Sathya with the new Balance sheet.
Answer:
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q18.2
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q18.3
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q18.4

Section – E
(1 × 10 = 10)

Note: Answer any one of the following questions.

Question 19.
Vanaja of Vijayawada consigned goods worth ₹ 10,000 to his agent Kamala of Kakinada on consignment. Vanaja spent ₹ 500 on transport, and ₹ 250 on insurance, and Kamala sent ₹ 2,500 as advance. After two months. Vanaja received the account sales as follows:
(a) Half of the goods were sold for ₹ 12,000.
(b) Selling expenses were ₹ 600.
(c) 5% commission on sales.
Give ledger accounts in the books of Vanaja.
Answer:
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q19
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q19.1
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q19.2

AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions

Question 20.
From the following Receipts and Payments Account of Vishakha Sports Club for the year ending 31 March 2018, prepare the Income and Expenditure Account.
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q20.1
Additional information:
(a) Outstanding salaries ₹ 300.
(b) Opening value of sports equipment ₹ 500. Closing value ₹ 250.
(c) Interest accrued on investments ₹ 100.
(d) Subscription receivable for the year 2018, ₹ 1,500.
(e) Capitalize entrance fees.
Answer:
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q20

Section – F
(2 × 5 = 10)

Note: Answer any Two of the following questions.

Question 21.
Explain the difference between a bill of exchange and a cheque.
Answer:
Bill of Exchange:
According to sec. 5 of the Negotiable Instruments Act, 1881, “A Bill of Exchange is an instrument in writing containing an unconditional order signed by the maker, directing a certain person to pay a certain sum of money only to or to the order of a certain person or to the bearer of the instrument”.

Cheque:
According to the Negotiable Instruments Act, 1881, Sec. 6. “A cheque is a Bill of Exchange drawn on a specified banker and payable on demand”.

The differences between a Bill of Exchange and a Cheque:

Basis of Difference Bill of Exchange Cheque
1. Acceptance Bill of Exchange requires acceptance to become a valuable instrument. Cheques do not require any acceptance.
2. Stamp Duty It requires the necessary stamp as per the act. It does not require any stamp.
3. Crossing It will not have any crossing on the instrument. It may have crossed.
4. Due to Date The bill proceeds will be payable on the due date of the instrument. The cheque amount should be paid for payment immediately as and when it is presented to the bank for payment.

Question 22.
Vinod & Co. purchased plant and machinery for ₹ 35,000 on 1st January 2015 and spent ₹ 5,000 for installation expenses. Depreciation is to be provided at 10% on the Reducing Balance Method. Books are closed on 31st December every year. Prepare Plant and Machinery Accounts for the first three years.
Answer:
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q22

Question 23.
Ram Ltd. issued 10,000 shares of ₹ 100 each for the subscription payable at ₹ 20 per share on application. ₹ 40 per share on the allotment and the balance ₹ 40 on the first and final call. All the amounts were duly received. Make Journal entries in the books of the company.
Answer:
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q23
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q23.1

Question 24.
Capital at the beginning of the year i.e. 01-04-2017 – ₹ 2,50,000
Capital at the end of the year i.e., 31-03-2018 – ₹ 2,25,000
Capital brought in by the proprietor during the year – ₹ 12,000
Withdrawals by the proprietor during the year – ₹ 50,000
The above information to prepare the statement of profit or loss.
Answer:
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q24
AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions Q24.1

Section – G
(5 × 2 = 10)

Note: Answer any Five of the following questions.

Question 25.
What is the due date of a bill?
Answer:
The date on which the bill falls due is called the “Due date” or “Date of Maturity”. The due date is calculated by adding 3 Grace days to the nominal expiry of the billing period.

Question 26.
What is Obsolescence?
Answer:
Obsolescence implies an existing asset becoming outdated due to technological changes or improvements in production methods. In other words, obsolescence means a reduction in the value of fixed assets due to new inventions, new improvements, and changes in customers’ tastes and preferences.

Question 27.
What do you mean by consignment?
Answer:
The word consignment is derived from the French word ‘Consignor’ which means “to hand over”, “to transmit” or “to send”. Consignment means one person sending goods to another person to sell them. The person who sends goods to another is called a “consignor” and the person who receives goods is called a “consignee”. In consignment consignee sold goods on behalf of the consignor on a commission basis.

Question 28.
What is Legacy?
Answer:
An Amount received by nonprofit organizations as per the will of a deceased person is called a “Legacy”. It is treated as capital income and should be shown on the liabilities side of the Balance sheet.

Question 29.
Goodwill.
Answer:
Goodwill is an intangible asset. Goodwill is the value of the reputation of a firm concerning the profits expected in the future over and above the normal profits.

Question 30.
What is authorized capital?
Answer:
Authorized capital is the amount of share capital that a company is authorized to issue to the public by the memorandum of association. It is also called Nominal or Registered capital.

AP Inter 2nd Year Commerce Question Paper May 2019 with Solutions

Question 31.
Write any two advantages of the computerized accounting system.
Answer:

  1. Speed: Computerized accounting systems process accounting data faster than manual efforts. This is because computers require less time than human beings to perform a task.
  2. Reliability: Computerized accounting systems are more reliable than manual accounting systems, because, computer systems perform repetitive operations without tiredness and boredom.

Question 32.
Write any two advantages of incomplete records.
Answer:

  • A single-entry system is a simple method of recording transactions.
  • It is less expensive when compared to double-entry system bookkeeping.
  • It is suitable for small business concerns.
  • Ascertainment of profit or loss is very easy.

TS Inter 1st Year Botany Question Paper May 2015

Successful navigation through TS Inter 1st Year Botany Model Papers and TS Inter 1st Year Botany Question Paper May 2015 builds students’ confidence in their exam-taking abilities.

TS Inter 1st Year Botany Question Paper May 2015

Time: 3 Hours
Maximum Marks: 60

General Instructions:
Note : Read the following instructions carefully.

  1. Answer all questions of Section ‘A’. Answer any six questions out of eight questions in Section ‘B’ and answer any two questions out of three in Section – ‘C’.
  2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to five lines. Answer all the questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 20 lines.
  4. In Section – C, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every ansvyer may be limited to 60 lines.
  5. Draw labelled diagrams wherever necessary for questions in Sections ‘B’ and C.

Section – A (10 × 2 = 20)

Note : Answer all questions. Each answer may be limited to 5 lines.

Question 1.
Give the scientific name of Mango. Identify the generic name and specific epithet.
Answer:
Magnifera indica. Generic name is Magnifera. Specific epithet is indica.

Question 2.
Name two diseases caused by Mycoplasmas.
Answer:
a) Witches broom disease in Plants.
b) Pleuropneumonia in Cattle.
c) Mycoplasmal Urethritis in Human beings.

Question 3.
Explain howtheterm Botany has emerged.
Answer:
In ancient Greek language, ‘BOUSKEIN’ means cattle feed. The term Bouskein gave rise to ‘Botane’ from the which the term botany has emerged.

TS Inter 1st Year Botany Question Paper May 2015

Question 4.
Differentiate fibrous roots from adventitious roots.
Answer:

Firbous roots Adventitious roots
Large number of roots arise from the base of the stem are called fibrous roots. Roots arise from any other part of the plant other than the radicle are called adventitious roots.

Question 5.
Define venation. How do dicots differ from monocOts with respect to venation.
Answer:
‘The arrangement of veins and veinlets in the lamina is called venation”.
In dicots, veins and veinlets are arranged in the form of a net so called Reticulate venation.
In monocots, veins and veinlets are arranged in the form of a parallel manner so called Parallel venation.

Question 6.
What is Omega Taxonomy ?
Answer:
Taxonomy based on Information from other Branches i.e., Embryology, Cytology, Palynology, Phytochemistry, Serology etc.
along with morphological characters is called omega taxonomy.

Question 7.
What is referred to as satellite chromosome ?
Answer:
A few chromosomes have non-straining secondary constrictions at a constant location which gives the appearence of a small fragment called satellite. The chromosome with satellite is called Satellite chromosome (SAT chromosome).

Question 8.
Glycine and Alanine are different with respect to one substituent on the α carbon. What are the other common substituent groups ?
Answer:
Hydrogen, Carboxyl group, amino group and a variable group designated as R group.

Question 9.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them ?
Answer:
\(\frac{1200}{4}\) = 300

Question 10.
Hydrophytes show reduced xylem. Why ?
Answer:
All submerged organs in Hydrophytes are capable of absorbing water so xylem is reduced.

TS Inter 1st Year Botany Question Paper May 2015

Section – B (6 × 4 = 24)

Note : Answer any six questions. Each answer may be limited to 20 lines.

Question 11.
What is heterospory ? Briefly comment on its significance. Give two examples.
Answer:
Production of different types of spores is called heterospory.

Significance :

  1. Microspores formed from Microspore mother cells are small with 0.015 – 0.05 μm. Megaspores formed from megaspore mother cells are big and are with 1.5 μp.
  2. Microspores develop into male gametophytes and megaspores develop into female gametophytes which lead to unisexuality.
  3. The female gametophytes are retained on the parent sporophyte for variable periods and obtain food from the sporophyte.
  4. This dependency of the gametophyte reduces the risk from adverse external conditions. So it gives better to the new embryo than independent gametophyte.
  5. The seed plants form the dominant flora of the present day vegetation. The season is due to seed habit which originated from Heterospory.
  6. The gametophytic tissue is reduced reduction in the number megaspores and male gametes show a tendency towards seed habit. Eg : Selagenella Salvinia.

Question 12.
Give the salient features and importance of Chrysophytes.
Answer:

  1. Chrysophytes includes diatoms and golden algae.
  2. They are found in fresh water as well as in Marine water. Most of them are photosynthetic.
  3. In Diatoms, the cell wall forms two thin overlapping shells (Epitheca over Hypotheca) which fit together as soap box.
  4. The walls are embedded with silica and thus they are Industructable.
  5. Diatoms leave large amounts of cell wall deposits in their habitat, this accumulation is referred as “diatomaceous earth.
  6. They reproduce asexually by Binary fission and sexually by Gametes.

Importance :
1) The diatomaceous soil is used in polishing, filleration of oils and syrups.

Question 13.
Discuss the various types of pollen tube entry into ovary with the help of diagrams.
Answer:
Pollen tube enters into the ovule from ovary by three ways.
They are
a) Porogamy : Pollen tube enters into the ovule through Micropyle.
Ex : Ottelia, Hibiscus.

b) Chalazogamy : “Pollen tube enters into the ovule through Chalaza”.
Ex : Casuarina.

c) Mesogamy : “Pollen tube enters into the ovule through the integuments”.
Ex : Cucurbita.
TS Inter 1st Year Botany Question Paper May 2015 1

TS Inter 1st Year Botany Question Paper May 2015

Question 14.
Describe the essential organs of solanaceae with labelled diagrams.
Answer:
Androecium and Gynoecium are the essential organs.

Androecium : 5 Stamens, Epipetalous, Alternate to petals. Anthers are large, dithecous, basifixed, Introrse and dehisce longitudinally (Datura) or Porously (solanum).

Gynoecium : Bicarpellary, syncarpous, superior ovary with ovules on swollen axile placentation. Ovary is oblique in position due to the tilting of posterior carpel to the right and anterior carpel to the left at an angle of 45°. The style is terminal and stigma is capitat.
TS Inter 1st Year Botany Question Paper May 2015 2

Question 15.
Name two cell organelles that are double membrane bound. Draw labelled diagrams of both.
Answer:
TS Inter 1st Year Botany Question Paper May 2015 3

Question 16.
Explain Prophase – I of meiosis.
Answer:
Meiosis I is longer phase and consists of 5 sub phases namely Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

a) Leptotene : The nucleus increases in size by absorbing water from the cytoplasm. The chromatin material organises into a constant number of chromosomes. The chromosomes are long, slender and show bead like structures called chromomeres. The ends of the chromosomes converge towards one side in the nucleus, where the centrosome lies. This arrangement is called Bouquet stage.

b) Zygotene : The chromosomes become shorter and thicker. They approach each other and form pairs. This homologous pair is called bivalent and the process of pairing is called synapsis. It is accompanied by the formation of Synaptonemal complex. The synapsis occurs at proterminal point or procentric or random means.

c) Pachytene : Bivalent chromosomes now clearly appear as tetrads. This stage is characterised by the presence of recombination modules, the sites of which crossing over occurs between the nonsister chromatids of the homologous chromosomes. Crossing – over is the exchange of genetic material between the two homologous chromosomes. It is also an enzyme mediated process by ‘recombinase’ crossing over leads to recombination of genetic material on the two chromosomes.

d) Diplotene: The homologous chromosomes of a bivalent begin to separate from each other except at the sites of cross overs to dissolution of synaptonemal complex. The ‘X’ shaped structures are called chaismata. The chromatids undergo condensation, contraction and thickening.

e) Diakinesis : It is marked by terminalisation of chaismata. In this phase, the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. By the end of this phase, the nuclear membrane breaks down the nucleolus disappears.

TS Inter 1st Year Botany Question Paper May 2015

Question 17.
What is the difference between lenticels and stomata ?
Answer:

Lenticels Stomata
1. Lens shaped openings in the cork of woody stems. 1. Openings present in the epidermis of the leaf.
2. They contain closely arranged parenchymatous cells. 2. Stomata is surrounded by specialised Guard cells.
3. They permit the exchange of gases between the outer atmosphere and the inner tissues of the woody organs. 3. They perform transpiration and exchange of gases.
4. They are always open. 4. Opening and closing mechanism is present.

Question 18.
Give in detail the anatomical adaptations shown by xerophytes.
Answer:

  1. Epidermis is covered with thick cuticle to reduce the rate of transpiration.
  2. Epidermal cells may have silica crystals.
  3. Epidermis may be multilayered.
  4. Stomata are generally confined to lower epidermis of leaves and present in pits (Sunken) in some plants.
  5. Mechanical tissues are well developed.
  6. Vascular tissues are relatively well developed.

Section – C (2 × 8 = 16)

Note : Answer any two questions. Each answer may be limited to 60 lines.

Question 19.
Describe the post fertilization changes in a flower.
Answer:

  1. The sepals, petals, stamens, style and stigma wither away.
  2. Ovary accumulates food materials and develops into a fruit.
  3. Ovules develop into seeds.
  4. Zygote develops into embryo.
  5. Primary endosperm nucleus undergoes several mitotic divisions to form endosperm (a nutritive tissue to embryo).
  6. Synergid and antipodals disintegrates.
  7. The outer integument changes into outer seed coat (Testa) and the inner integument changes into inner seed coat (Tegmen).
  8. Funicle develops into seed stalk.
  9. Micropyle develops into seed pore.
  10. As the seed matures, some amount of endosperm is left out called perisperm (Pepper).

Question 20.
Define root modification. Explain with labelled diagrams, how root is modified to perform different functions.
Answer:
A change in the normal structure of root to carry out new functions according to environment is called root modification.

Root modifications : In some plants, roots change their shape and structure to perform functions other than absorption and conduction of water and Minerals called root modifications.
They are of different types.

  1. In Carrot, turnip (Tap roots), Sweat potato (Adventitious roots), Asparagus (Fibrous roots) become swollen due to storage of food called storage roots.
  2. In Banyan tree. Roots arise from the branches grow into the soil, become pillar like and give additional support called prop roots or pillar roots.
  3. In Maize, sugarcane, roots arise from the lower nodes of the stem, give additional support called stilt roots.
  4. In Mangroves like Rhizophora and Avicennia, Many roots come out of the ground and grow vertically upwards, help in respiration called Pneumatophores.
  5. In Epiphytes like Vanda, special adventitious roots arise, help in absorption of moisture from atmosphere called Velamen roots.
  6. In partial parasites like viscum and strga, some Haustorial roots enter into xylem of the host plant to get water and Minerals. In complete parasitic like cuscuta and Rafflesia, the haustorial roots enter into xylem and phloem of the host plant and obtain water and Minerals and food materials called Parasitic roots.
  7. In the members of Fabaceae, the roots are inhabited by Rhizobium bacteria which helps in N2 fixation called Nodular roots.
  8. In some plants like Taeniophyllum, the roots are chlorophyllous and perform photosynthesis so called photosynthetic roots.

TS Inter 1st Year Botany Question Paper May 2015 4
TS Inter 1st Year Botany Question Paper May 2015 5

TS Inter 1st Year Botany Question Paper May 2015

Quetsion 21.
Describe the internal structure of a monocot root which the help of a labelled diagram.
Answer:
The internal structure of Monocot root shows 3 zones.
They are :
1) Epidermis
2) Cortex and
3) Stele,

1) Epidermis : It is the outermost layer formed by thin walled, rectangular cells, which are compactly arranged without intercellular spaces. Cuticle and stomata are absent. Some epidermal cells (tricho blasts) produce tubular extensions called root hairs. They absorb capillary water from the soil. The epidermis of root is also known as rhizodermis or epibiema or piliferous layer.

2) Cortex : It is a wide and extensive tissue present between the epidermis and stele. It is bigger than the stele. It can be divided into three sub-zones.
They are :
a) Exodermis
b) General cortex and
c) Endodermis.

a) Exodermis : It is the outer part of the cortex and composed of one to two rows of thick walled, dead, suberised calls. In mature roots, when the outer epidermis is removed, the exodermis acts as a protective layer. It helps in preventing the exit of water from the root tissues.

b) General Cortex : It is formed below the exodermis layer. It is composed of several rows of thin walled living cells that are arranged loosely showing intercellular spaces. The cells of cortex help in the storage of food materials and lateral conduction of water from the epidermis to the stele.

c) Endodermis : The innermost layer of cortex and is composed of single layer of barrel shaped cells that are arranged compactly without intercellular spaces. The radical and transverse walls are wrapped by ligno-suberised bands called casparian bands. Some cells situated opposite to the protoxylem cells are thin walled and without casparian bands. These are known as passage cells which help in the entry of water from the cortex into the stele.

3) Stele : The central conducting cylinder. It is very prominent and bigger in size. The stele shows Pericycle, Vascular bundles and Medulla.

i) Pericycle : The layer of cells found beneath the endodermis is known as pericycle. The cells are thin walled, parenchymatous, rectangular and compact without intercellular spaces. The cells are meristematic and divide actively producing lateral roots. In old and mature roots, the pericycle is sclerenchymatous and gives mechanical strength.

ii) Vascular bundles : Bundles of xylem and phloem are found separately on different radii, one alternating with the other, at the peripheral boundary of the stele. These are known as radial’ or separate vascular bundles. The xylem is exarch and polyarch. More than six xylem bundles.

The ground tissue formed between the xylem and phloem stands is known as ‘conjunctive tissue’. It is usually parenchymatous. It helps in storage of food materials and provides mechanical strength.

iii) Medulla or Pith : The wide central part of the stele is called ‘medulla or pith’. It is made up of thin walled parenchyma which primary helps in the storage of food. In some monocot roots, the medulla is composed of thick walled lignified dead cells and helps in giving mechanical strength.
TS Inter 1st Year Botany Question Paper May 2015 6

TS Inter 1st Year Botany Question Paper March 2015

Successful navigation through TS Inter 1st Year Botany Model Papers and TS Inter 1st Year Botany Question Paper March 2015 builds students’ confidence in their exam-taking abilities.

TS Inter 1st Year Botany Question Paper March 2015

Time: 3 Hours
Maximum Marks: 60

General Instructions:
Note : Read the following instructions carefully.

  1. Answer all questions of Section ‘A’. Answer any six questions out of eight questions in Section ‘B’ and answer any two questions out of three in Section – ‘C’.
  2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to five lines. Answer all the questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 20 lines.
  4. In Section-C, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 60 lines.
  5. Draw labelled diagrams wherever necessary for questions in Section ‘B’ and C.

Section – A (10 × 2 = 20)

Note : Answer all questions. Each answer may be limited to 5 lines.

Question 1.
Define population and Community.
Answer:
“A group of similar individuals belonging to the same species found in an area’1 is called population. An assemblage of all the population occuring in an area” is called community.

Question 2.
What is Geocarpy ? Name the plant which exhibits this pheno-menon.
Answer:
Development of fruits inside the soil is called Geocarpy. It is seen in Arachis hypogea (Ground nut).

Question 3.
What is Paleobotany ? What is its use ?
Answer:
Study of fossil plants is called Palaeobotany. It helps in under – standing the course of evolution in plants.

TS Inter 1st Year Botany Question Paper March 2015

Question 4.
Give the main criteria used for classification by Whittaker.
Answer:
The main criteria for classification proposed by Whittaker include cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships.

Question 5.
What is Flora ?
Answer:
The actual account of Habitat, distribution and systematic listing of plants of a given area is called flora.

Question 6.
Mention a single membrane bound organelle which is rich in hydrolytic enzymes.
Answer:
Lysosome.

Question 7.
Name the stage of meiosis in which actual reduction in chromosome number occurs.
Answer:
Starts in anaphase I and is complete by telophase I.

Question 8.
Starch, Cellulose, Glycogen, Chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
a) Cotton fibre
b) Exoskeleton of Cockroach,
c) Liver
d) Peeled Potato.
Answer:
a) Cotton fibre : Cellulose
b) Exoskeleton of cockroach : Chitin
c) Liver : Glycogen
d) Peeled potato : Starch

TS Inter 1st Year Botany Question Paper March 2015

Question 9.
Differentiate between Racemose and Cymose Inflorescence.
Answer:

Racemose Inflorescence Cymose inflorescence
1) Peduncle grows indefinitely. 1) Peduncle grows definitely.
2) Flowers are arranged in acropetal manner on the peduncle. 2) Flowers are arranged in Basipetal manner on the peduncle.
3) Flowers open in a centripetal manner. 3) Flowers open in a centrifugal manner.

Question 10.
Define Placentation. What type of placentation is found in Dianthus ?
Answer:
The arrangement of ovules with in the ovary is known as placentation. In dianthus. Free central placentation is found.

Section – B (6 × 4 = 24)

Note : Answer any six questions. Each answer may be limited to 20 lines.

Question 11.
Describe the essential organs of Solanaceae.
Answer:
Androecium and Gynoecium are the essential organs.
Androecium : Stamens 5, epipetalous, alternating with petals anthers are dithecous, basifixed, Introrse and dehisce longitudinally ‘Datura) or porous (Solanum).

Gynoecium : Bicarpellary, syncarpous, bilocular, superior ovary with many ovules on swollen axile placentation. Style is terminal and stigma is capitate. Carpels are arranged obliquly at 45°. (Posterior carpel to the right and anterior carpel to the left).

Question 12.
What is Heterospory ? Briefly comment on its significance. Give two examples.
Answer:
Production of different types of spores (2 types) is called Heterospory.

Significance : The smaller microspore forms male gametophyte which intern produce male gametes. The larger megaspore forms female gametophyte which produce female gamete, the Egg. The Male gamete fuses with the egg forms Diploid zygote.

This retention and germination of the megaspore with in the megasporangium ensures proper development of zygote. The zygote develops into the future sporophyte. The evolution of the seed habit is released to the retention of the Megaspore. Heterospory is considerd as an important step in evolution as it is a precursor to the seed habit.
Ex : Selagenella, Salnnia.

Question 13.
Give a brief account of Dinoflagellates.
Answer:

  1. Dinoflagellates are mostly marine and photosynthetic. They appear yellow green, brown, blue or red depending on the pigments in their cells.
  2. The cell wall has stiff cellulose plates on the outer surface.
  3. They have two flagella and produce spinning movements. So these prOtists are called “whirling whips”.
  4. The nucleus has condensed chromosomes which are without histones. This is called mesokaryon.
  5. Some dinoflagellates like Noctiluca show bioluminescene.
  6. Red dinoflagellates like Gonyaulax undergo rapid multiplication and make the sea appear red (Red tides in Medeterranian sea).
  7. Toxins produced by them may kill fishes.

Question 14.
Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction ?
Answer:

Asexual Reproduction Sexual Reproduction
1. Single parent is involved. 1. Two parents take part in reproduction.
2. Offsprings are genetically identical to each other and to their parent. 2. Variations occurs in offspring,
3. No internal fertilization or external fertilization. 3. Fertilisation takes place.
4. No gamete formation. 4. Gametes are formed.
5. No mixing of hereditary material. 5. Mixing of hereditary material.

In multicellular or colonial forms of algae, moulds and mushrooms, the body may split/break or get separated into smaller fragments. Each fragment thus formed develops into a mature individual. This process is called fragmentation.

Some plants have specialized structures for reproduction called gemmae (in liverworts).

In flowering plants, the units of vegetative propagation such as runners, stolons, suckers, offset, rhizomes, corm, stem tuber, bulb, bulbil, reproductive leaves are capable of giving rise to new offsprings. These structures are called vegetative propagules. In all these plants, no involvement of sex organs takes place, so vegetative reproduction is also a sexual reproduction.

TS Inter 1st Year Botany Question Paper March 2015

Quetsion 15.
What are Hydrophytes ? Briefly discuss the different kinds of hydrophytes with examples.
Answer:
The plants which grow in water or in very wet places. According to the way in which they develop in water, they are further subdivided into the following five categories.

i) Free floating hydrophytes : They float freely on water surface and have no contact with soil.
Eg : Pistia, Eichhornia, Wolffia, Salvinia.

ii) Rooted hydrophytes with floating leaves : Their roots are fixed in mud, but leaves have long petioles which keep them floating on the water surface.
Eg : Nymphaea, Nelumbo and Victoria regia.

iii) Submerged suspended hydrophytes : These plants are in contact with only water, being completely submerged and not rooted in the mud.
Eg : Hydrilla, Ceratophyllum and Utricularia.

iv) Submerged rooted hydrophytes: These plants are completely submerged in water and attached to the substratum by their root system.
Eg : Vallisneria, Potamogeton etc.

v) Amphibious plants : These live partly in water and partly in air.
Eg : Sagittaria, Rannunculus, Llmnophila, Typha etc.

Question 16.
State the location and function of different types of meristems.
Answer:
Based on the position, meristems are classified into three types.
A) Apical Meristems: The meristems that are present at the tip of the root and at the tip of the stem or branches are called Apical Meristems. They help in the linear growth of the plant body.

B) Intercalary Meristems : The Meristems that are present in between mature tissues are known as Intercalary Meristems. They occur in grasses. They also contribute to the formation
of the primary plant body and involve in internodal growth.

C) Lateral Meristems : The meristems that occur in the mature regions of roots and shoots of many plants are called lateral meristems. They help in increase in thickness of the plant organs (Root, Stem). Ex : Vascular cambium, cork cambium.
TS Inter 1st Year Botany Question Paper March 2015 1

Question 17.
Describe the structure of Nucleus.
Answer:
It is a spherical ball like structure which controles and coordinates various life activities of a cell and is considered to be the most important cell organella or cell brain. It was discovered by Robert Brown. It shows four parts.

They are :
a) Nuclear membrane : The Nucleus is covered by 2 layered envelope which are seperated by perinuclear space. It is lipo-proteinous in nature, acts as a bridge between cytoplasm and Nucleoplasm. It is porous at some places called Nuclear pores.

b) Nucleoplasm : The homogenous, semi solid substance inside the Nuclear membrane is called Nucleoplasm or Karyoplasm. It is composed of Glyco Proteins, Ribonucleo Proteins Hydrolysing enzymes, DNA & RNA polymerases, chromatin material and nucleolus.

c) Chromatin Material: The deeply stained network like material is called chromatin material. It is associated with DNA & Histone proteins. It transforms into thick rod shaped structures called chromosomes during cell division.

d) Nucleus : It is a spherical body found in nucleoplasm. It was discovered by Fontana. It is the seat for the production of Ribosomes.

Functions :

  1. It controls and co-ordinates the function of all cell organelles so called cell brain.
  2. It involves in heridity.
  3. It plays an important role in reproduction in unicellular organisms.

TS Inter 1st Year Botany Question Paper March 2015 2

TS Inter 1st Year Botany Question Paper March 2015

Question 18.
Nucleic acid exhibits secondary structure. Justify with example.
Answer:
Nucleic acids exhibits a wide variety of secondary structure. For example, one of the secondary structures exhibited by DNA is Watson- Crick model. According to this, DNA exists as a double helix. The two stands of polynucleotides are antiparallel i.e., run in the opposite direction. The back bone is formed by the sugar – phosphate, sugar chain. The nitrogen bases are projected more or less perpendicular to this backbone but face inside. A combines with T by two hydrogen bonds where G combines with C by three hydrogen bonds. One full turn of the helical strand would involve ten base pairs. The length of one coil is 34 A and the distance between base pairs is 3.4 A. It is also called B-DNA.
TS Inter 1st Year Botany Question Paper March 2015 3

Section – C (2 × 8 = 16)

Note : Answer any two questions. Each answer may be limited to 60 lines.

Quetsion 19.
Define Root modification. Explain how root is modified to perform different functions (write any 6 modifications).
Answer:
A change in the normal structure of root to carryout new functions according to environment is called root modification.

Root modifications :

  • Storage Roots : In Carrot, turnip (Tap roots). Sweat potato (Adventitious roots), Asparagus (Fibrous roots) become swollen due to storage of food called storage roots.
  • Prop Roots : In Banyan tree, roots arise from the branches grow into the soil, become pillar like and give additional support called prop roots or Pillar roots.
  • Stilt Roots : In Maize, sugarcane, roots arise from the lower nodes of the stem, give additional support called stilt roots.
  • Respiratory Roots : In Mangroves like Rhizophora and Avicennia, Many roots come out of the ground and grow vertically upwards, help in respiration called Pneumato – phores.
  • Epiphytic Roots: In epiphytes like Vanda, special adventitious roots arise, help in absorption of moisture from atmosphere called Velamen roots.
  • Parasitic Roots : In partial parasites like viscum and Striga, some haustorial roots enter into xylem of the host plant to get water and minerals. In complete parasities like cuscuta and Rafflesia, the haustorial roots enter into xylem and phloem of the host plant and obtain water, minerals and food materials called parasitic roots.
  • Nodular Roots : In the members of Fabaceae, the roots are inhabited by Rhizobium bacteria which helps in N? fixation called nodular roots.
  • Photo synthetic Roots : In some plants like Taeniophyllum, the roots are chlorophyllous and perform photosynthesis so called photosynthetic roots.

TS Inter 1st Year Botany Question Paper March 2015 4

Question 20.
Describe the process of fertilization in Angiosperms with labelled diagrams.
Answer:
The fusion of male and female gametes is called fertilization. The process of fertilization in angiosperms is described under the following five steps.

A) Entry of the pollen tube into the ovule : The pollen tube enters into the ovule in 3 ways.

  1. Pologamy : Pollen tube enters into the ovule through microphyle. Ex. : Ottelia, Hibiscus.
  2. Chalazogamy : Pollen tube enters into the ovule through chalaza. Ex. : Casuarina.
  3. Mesogamy : Pollen tube enters into the ovule through integu-ments or funiculus. Ex. : Cucurbita.

TS Inter 1st Year Botany Question Paper March 2015 5
B) Entry of pollen tube into the embryosac : The pollen tube enters into the embryosac only through the microphylar region either by destroying one of the synergids or in between egg cell and synergid. The entry of pollen tube is directed by filiform apparatus.

C) Dicharge of male gametes or sperms : After entry of pollen tube into the embryosac, the male gametes are liberated by one of the following ways.
a) Tip of the pollen tube may burst out.
b) Degeneration of the tip of the pollen tube.
c) Formation of an apical pore at the tip of the pollen tube. The pollen tube finally releases the intact male gametes and vegetative nucleus.

D) Gametic Fusion: One of the two nuclei, one nucleuses sperm nucleus [first sperm] fuses with egg cell and forms a dipliod zygote (2n). This fusion is called syngamy or fertilization. It was first discovered by Strasburger (1884).

E) Triple fusion and Double fertilization : The second nucleus fuses with the secondary nucleus of the embryosac and forms Primary endosperm nucleus [PEN]. This fusion is called as Triple fusion. It was first observed by Nawaschin in Lilium and Fertillaria. In angiosperms the two male gametes release into the embryo sac take part in two fertilizations.

The first sperm nucleus combines with egg cell to form zygote and the second sperm nucleus combines with secondary nucleus to form primary endosperm nucleus. With the occurrence of two fertilizations, this phenomenon is, i.e., Double fertilization includes syngamy as well as triple fusion. It results in the formation of fertile endospermic seeds.

TS Inter 1st Year Botany Question Paper March 2015

Question 21.
Describe the T.S. of Dicot stem with a neat labelled diagram.
Answer:
The structure of young dicot stem can be clearly understood by observing the transverse section of stem of Helianthus annus (sunflower). It shows three major zones, namely epidermis, cortex and stele.

1. Epidermis : It is the outer most layer of rectangular or tabular cells arranged compactly without any intercellular spaces. On outer surface of epidermis, a waxy layer called Cuticle is found. The cuticle is chemically composed of a substance cutin. The cell walls of epidermis also show the presence of cutin. Stomata are present in the epidermis. Multicellular trichomes develop on the epidermis. The cuticle and the trichomes check the evaporation of water and protect the stem from high temperature.

The epidermal layer gives protection to the inner tissues and also prevents the evaporation of water from the plant body. Through stomata, the epidermis allows the exchange of gases and promotes transpiration. Trichomes prevent entry of pathogens.

2. Cortex : The part extending between the epidermis and the stele is known as cortex. The cortex is smaller than the stele. It shows three sub-zones, namely
i) Hypodermis
ii) General Cortex and
iii) Endodermis.

i) Hypodermis: This is the outermost part of cortex and composed of 3-6 rows of collenchymatous cells. It is found beneath the epidermis and helps in providing tensile strength (elasticity) to the stem. The cells are arranged compactly without intercellular spaces and show excessively thickened corners. The cells are filled with active vacuolated cytoplasm possessing chloroplasts. Thus, the hypodermis also helps in the assimilation of food materials. It also gives mechanical strength.

ii) General Cortex : It is found beneath the hypodermal layer and is made up of 5 – 10 rows of thin walled, living parenchyma cells with or without intercellular spaces. The outer layers of cells contain chloroplasts and in the inner layers, leucoplasts are found. The general cortex is primarily concerned with the assimilation and storage of food materials.

iii) Endodermis : The inner most layer of cortex is called endo- dermis. The cells are barrel shaped, compactly arranged without intercellular spaces. The endodermis cells contain vacuolated protoplasts and show starch grains. So it is also known as ‘Starch Sheath’.

3) Stele : The central conducting cylinder is called the ‘Stele’. It occupies a major part of the stem. It is composed of 4 parts.
i) Pericyde
ii) Vascular bundles
iii) Pith or Medulla and
iv) Medullary rays.

i) Pericyde : It is present in the form of a discontinuous ring and is made up of 3 – 5 rows of the thick walled, dead, lignified cells which gives mechanical strength to the stele. It appears as semilunar patches of sclerenchyma above the vascular bundles with intervening masses of parenchyma.

ii) Vascular bundles : About 15 – 20 vascular bundles are arranged in the form of a ring. This arrangement is called Eustele. Each vascular bundle is wedge or top shaped. In the vascular bundels, xylem and phloem are arranged on the same radius. A meristematic layer of cells called cambium is present in between the xylem and phloem. So they are called conjoint, collateral and open vascular bundles. Xylem is at the lower side and phloem at the upper side of the vascular bundle.

Xylem consists of vessels and xylem parenchyma. There may be few tracheids and xylem fibres. The metaxylem is towards the pericyde and protoxylem towards the pith. This is called endarch xylem. Phloem consists of sieve tubes companion cells and phloem parenchyma. Xylem and phloem are vascular tissues which conduct water, mineral salts and organic solutes respectively.

iii) Medulla (or) pith : It is the central part of the stele and filled with thin walled parenchymatous cells, showing intercellular spaces. It is well – developed, extensive and occupies a large part of the stele. The chief function of the medulla is to store food materials.

iv) Medullary rays : They are found in between the vascular bundles. They show many horizontal rows of thin walled radially elongated living cells which help in lateral conduction of food materials.
TS Inter 1st Year Botany Question Paper March 2015 6