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TN State Board 11th Biology Previous Year Question Paper June 2019 English Medium

Instructions:

  1. The question paper comprises of four parts Questions for Botany and Zoology are asked separately.
  2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
  3. All questions of Part I, II, III, and IV are to be attempted separately.
  4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and Writing the option code and the Corresponding answer.
  5. Question numbers 9 to  4 in Part II are two-marks questions. This also is answered in about one or list sentences.
  6. Question numbers 15 to 19 in Part III are three-marks questions These are to be answered in about three to five short sentences.
  7. Question numbers 20 and 21 in Part  IV are five-marks questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Answer all the questions. Choose the correct answer.  [8 × 1 = 8]

Choose the most appropriate answer from the given four alternative and write the option code and the corresponding answer.

Question 1.
The book of “The structure and reproduction of the Algae” was published by …………………….
(a) F.E. Fritch
(b) F.E. Round
(c) R.E. Lee
(d) M.S. Randhawa
Answer:
(a) F.E. Fritch

Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium

Question 2.
More than one locule ovaries are called ………………………..
(a) Plurilocular
(b) Unilocular
(c) Bilocular
(d) Trilocular
Answer:
(a) Plurilocular

Question 3.
The two subunits of ribosomes remain united at critical ion level of ………………….
(a) Magnesium
(b) Calcium
(c) Sodium
(d) Ferrous
Answer:
(a) Magnesium

Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium

Question 4.
The structure of Glycine ammo acid is …………………………
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 1
Answer:
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 2

Question 5.
Polyderm is found in the roots and underground stems of family is ………………………
(a) Magnesium
(b) Fabaceae
(c) Solanaceae
(d) Lilliaceae
Answer:
(a) Magnesium

Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium

Question 6.
In a fully turgid cell: …………………
(a) DPD = 10 atm; OP = 5 atm; TP = 10 atm.
(b) DPD = 0 atm; OP = 10 atm; TP = 10 atm.
(c) DPD = 0 atm; OP = 5 atm; TP = 10 atm.
(d) DPD = 20 atm; OP = 20 atm; TP = 10 atm.
Answer:
(d) DPD = 20 atm; OP = 20 atm; TP = 10 atm.

Question 7.
Who proposed the “Theory of photosynthesis in guard cells” ?
(a) Von Mohl
(b) Sayre
(c) Levit
(d) Steward
Answer:
(a) Von Mohl

Question 8.
Which prevents flow of electrons from NADH + H+ /FADH2 to CoQ?
(a) Cyanide
(b) Oligomycin
(a) Von Mohl
(d) 2,4 DNP
Answer:
(a) Von Mohl

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
We are using perfume to our body. Write the name of the plant and their product.
Answer:
Name of the perfume yielding plant – Santalum album. Product – Sandal oil.

Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium

Question 10.
Write the floral formulafof Phyllanthus amarus.
Answer:
Floral formula: Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 3

Question 11.
Draw the diagram of conjoint, bicollateral and open of vascular system.
Answer:
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 4

Question 12.
How does the trees increase their girth ?
Answer:
In plants, the increase in the thickness (girth) of stem and root is by means of secondary growth or longitudinal growth.

Question 13.
Define the Theory of Electro Osmosis.
Answer:
Electro-Osmotic theory: The theory of electro osmosis was proposed by Fenson (1957) and Spanner (1958). According to this, an electricpotential across the sieve plate causes the movement of water along with solutes. This theory fails to explain several problems concerning translocation.

Question 14.
Respiratory Quotient is zero in succulent plants. Why?
Answer:
In some succulent plants like Opuntia, Bryophyllum carbohydrates are partially oxidised to organic acid, particularly malic acid without corresponding release of CO2 but O2 is consumed hence the RQ value will be zero.

Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium

Part – III

Answer any 3 questions. Question No. 19 is compulsory. [3 × 3 = 9]

Question 15.
Do you agree that virus is living organism? If you say yes, justify your answer.
Answer:
Yes. Viruses have the characters of living organisms because

  • They possess nucleic acid and protein
  • Ability to multiply inside a. living cell
  • Ability to inject and cause disease
  • Capable of mutation
  • Show irritability
  • Hosts specific.

Question 16.
You are looking like your father or mother in externally. Why? Explain the reason.
Answer:
Externally, we resemble our father or mother upto a certain degree, because the characters of our parents are inherited to us through their genes present in the chromosome of egg (mother) and sperm (father).

Question 17.
Write any three significance of mitosis.
Answer:

  1. Genetic stability – daughter cells are genetically identical to parent cells.
  2. Growth – as multicellular organisms grow, the number of cells making up their tissue increases. The new cells must be identical to the existing ones.
  3. Repair of tissues – damaged cells must be replaced by identical new cells by mitosis.

Question 18.
Which is determined you parental character? Describe about the structure.
Answer:
The parental characters are determined by the chromosome.
The four important features of the chromosome are:

  • The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called centromere.
  • The number of chromosomes per species is fixed: Example: Mouse has 40 chromosomes, Onion has 16 and Human have 46.
  • Chromosomes occur in pairs: Chromosomes of a cell occur in pairs, called homologous pairs. Each of a pair come originally from each parent. Example, human has 46 chromosomes, 23 coming originally from each parent in the process of sexual reproduction.

Question 19.
Write the role of nitrogenase enzyme in nitrogen fixation.
Answer:
Fixation of atmospheric nitrogen: Nitrogen fixation process requires Nitrogenase enzyme . complex, Minerals (Mo, Fe and S), anaerobic condition, ATP, electron and glucose 6 phosphate as H+ donor. Nitrogenase enzyme is active only in anaerobic condition. To create this anaerobic condition a pigment known as leghaemoglobin is synthesized in the nodules which acts as oxygen scavenger and removes the oxygen. Nitrogen fixing bacteria in root nodules appears pinkish due to the presence of this leghaemoglobin pigment.
Overall equation: N2 + 8e + 8H+ + 16 ATP → 2NH3+ + H2 + 16 ADP + 16 Pi

Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
Explain the Aerial modification of stern with an example.
Answer:
Aerial modification of stem:
1. Creepers: These are plants growing closer (horizontally) to the ground and produces roots at each node. e.g., Cynodon dactylon, Oxalis, Centella

2. Trailers (Stragglers): It is a weak stem that spreads over the surface of the ground without rooting at nodes. They are divided into 3 types,

  • Prostrate (Procumbent): A stem that grows flat on the ground, e.g., Evolvulus alsinoides, Indigofera prostrata.
  • Decumbent: A stem that grows flat but becomes erect during reproductive stage. e.g., Portulaca, Tridax, Lindenbergia
  • Diffuse: A trailing stem with spreading branches, e.g., Boerhaavia diffusa, Merremia
    tridentata

3. Climbers: These plants have long weak stem and produce special organs for attachment , for climbing over a support. Climbing helps to display the leaves towards sunlight and to
position the flower for effective pollination.

i. Root climbers: Plants climbing with the help of adventitious roots (arise from nodes) as in species of Piper betel, Piper nigrum, Hedera helix, Pothos, Hoy a.

ii. Stem climbers (twiners): These climbers lack specialised structure for climbing and the stem itself coils around the support, e.g., Ipomoea, Convolvulus, Dolichos, Clitoria, Quisqualis.
Stem climbers may coil around the support clockwise or anti-clockwise. Clockwise coiling climbers are called dextrose. Example: Dioscorea alata. Anti-clockwise coiling climbers are called sinistrose. e.g., Dioscorea bulbifera.

iii. Hook climbers: These plants produce specialized hook like structures which are the modification of various organs of the plant. In Artabotrys inflorescence axis is modified into hook. In calamus (curved hook) leaf tip is modified into hook. In Bignonia unguis- cati the leaflets are modified into curved hook. In Hugonia the axillary buds modified into hook.

iv. Thorn climbers: Climbing or reclining on the support with the help of thorns as in Bougainvillea and Carissa.

v. Lianas (woody stem climber): Woody perennial climbers found in tropical forests are lianas. They twine themselves around tall trees to get light, e.g., Hiptage benghalensis, Bauhinia vahlii, Entada pursaetha.

vi. Tendril climbers: Tendrils are thread-like coiling structures which help the plants in climbing. Tendrils may be modifications of Stem – as in Passiflora, Vitis and Cissus quadrangularis; Inflorescence axis – Antigonon; Leaf – Lathyrus, Leaflets – Pisum sativum; Petiole – Clematis; Leaftip – Gloriosa; Stipules – Smilax. In pitcher plant (Nepenthes) the midrib of the leaf often coils around a support like a tendril and holds the pitcher in a vertical position.

[OR]

The plants also having the hormones like human being. What are they? Explain the physiological effect of any one.
Answer:
The plants also having the hormones like human being. They are called Phylohormones.
Example: Giberillin, Auxin, Cytokinin, ethylene, Abscissic acid.
Physiological effect of of Cytokinins:

  • Cytokinin promotes cell division in the presence of auxin (IAA).
  • Induces cell enlargement associated with IAA and gibberellins.
  • Cytokinin can break the dormancy of certain light-sensitive seeds like tobacco and induces seed germination.
  • Cytokinin promotes the growth of lateral bud in the presence of apical bud.
  • Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.
  • Cytokinin (i) increases rate protein synthesis (ii) induces the formation of inter-fascicular cambium, (iii) Overcomes apical dominance  (iv) induces formation of new leaves, chloroplast and lateral shoots.
  • Plants accumulate solutes very actively with the help of cytokinins.

Question 21.
Explain the Calvin Cycle (Flow chart only).
Answer:
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 5

[OR]

Draw the outline of Bentham & Hooker classification.
Answer:
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 6

Bio-Zoology[Maximum Marks: 35]

Part – I

Answer all the questions. [8 x 1 = 8]

Choose the most appropriate answer from the given four alternative and write the option code and the corresponding answer.

Question 1.
Every unit of classification regardless of its rank is ………………
(a) Taxon
(b) Variety
(c) Species
(d) Strain
Answer:
(a) Taxon

Question 2.
The zoological name of National Bird is ……………..
(a) Pavo Cristatus
(b) Zoothera Salimalii
(c) Columba livia
(d) Chalcophaps indiva
Answer:
(a) Pavo Cristatus

Question 3.
Which one of the following is not true to simple epithelial tissue?
(a) It is composed of a single layer of cells.
(b) Squamous epithelium is a type of simple epithelium which has flattened cells.
(c) Simple epithelium covers dry surface of the skin.
(d) Pseudo-stratified is a type of simple epithelium which lines the epididymis.
Answer:
(d) Pseudo-stratified is a type of simple epithelium which lines the epididymis.

Question 4.
Which plasma protein is involved in the coagulation of blood ?
(a) Globulin
(b) Fibrinogen
(c) Albumin
(d) Serum amylase
Answer:
(b) Fibrinogen

Question 5.
Match Column-I with Column-II and choose the correct option.
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 7
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 8
Answer:
(c) p – (iv), Q – (iii), R – (ii),  S – (i)

Question 6.
Which one of the following diseases is caused due to our own body antibodies act against our own body substance?
(a) Tetany
(b) Osteoporosis
(c) Myasthenia gravis
(d) Gout
Answer:
(c) Myasthenia gravis

Question 7.
Silk is obtained from ………………
(a) Laccifer lacca
(b) Nosema bombycis
(c) Attacus ricini
(d) Attacus mvlitta
Answer:
(d) Attacus mvlitta

Question 8.
The good egg layers and delicious breed is ………………..
(a) Chittagong
(b) Leghorn
(c) Plymouth
(d) Aseel
Answer:
(a) Chittagong

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Differentiate between closed type and open type of circulatory system.
Answer:
Closed circulatory system :

  1. The circulation in which lood is present inside the blood vessels is called closed circulatory system
  2. It is found in higher organisms. e.g. annelids. cephalochordates and vertebrates.

Open circulatory system

  1. The circulation in which blood remains filled in tissue spaces due to the absence of blood vessels is called open circulatory system
  2. It is found in lower organisms. e.g. arthropods, molluscsand echinoderms

Question 10.
How does gall stones are formed?
Answer:
Any alteration in the composition of the bile can cause the formation of stones in the gall bladder. The stones are mostly formed of crystallized cholesterol in the bile. The gall stone causes obstruction in the cystic duct, hepatic duct and also hepato-pancreatic duct causing pain, jaundice and pancreatitis.

Question 11.
At Tuticorin region, the pearl oyster diving people have some changes in their blood. State the reasons for it.
Answer:
When a person descends deep into the sea, the pressure in the surrounding water increases which causes the lungs to decrease in volume. This decrease in volume increases the partial pressure of the gases within the lungs. This effect can be beneficial, because it tends to drive additional oxygen into the circulation, but this benefit also has a risk, the increased pressure can also drive nitrogen gas into the circulation.

This increase in blood nitrogen content can lead to a condition called nitrogen narcosis. When the diver ascends to the surface too quickly a condition called ‘bends’ or decompression sickness occurs and nitrogen comes out of solution while stilUin the blood forming bubbles. Small bubbles in the blood are not harmful, but large bubbles can lodge in small capillaries, blocking blood flow or can press on nerve endings.

Decompression sickness is associated with pain in joints and muscles and neurological problems including stroke. The risk of nitrogen narcosis and bends is common in scuba divers.

Question 12.
Depolarization means reversal of polarity. Mention the iconic charge of both sides of axolemma.
Answer:
Depolarization – Reversal of polarity
When a nerve fibre is stimulated, sodium voltage-gate opens and makes the axolemma permeable to Na+ ions; meanwhile the potassium voltage gate closes. As a result, the rate of flow of Na+ ions into the axoplasm exceeds the rate of flow of K+ ions to the outside fluid [ECF], Therefore, the axolemma becomes positively charged inside and negatively charged outside. This reversal of electrical charge is called Depolarization.

Question 13.
Expand the word SMGA and write its uses.
Answer:
Skeletal Muscle Glycogen Analysis (SMGA) – Used to measure an Athlete’s sporting performance by taking muscle biopsies. It is a standard method to measure muscle glycogen. Muscle glycogen provides the main source of energy during anaerobic exercise. Furthermore, total glycogen stores within the body also contribute significantly to energy metabolism in endurance-type events lasting longer in duration. A single glycogen molecule may contain 5000 glucose units compared to that of 5000 individual glucose molecules.

Question 14.
In your class various height of students over there. Why so? Write the reason.
Answer:
Students in the class are of different heights because of the secretion of growth hormone. Those students who are of maximum height secrete more growth hormone and those students , with low secretion of growth hormone are dwarf or reduced in growth, because growth hormone promotes the growth of all tissue and metabolic processes of body.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Hemichordates posses the characters of echinoderms and chordates. Write any 3 characters of echinoderms as well as chordates seen in hemichordates.
Answer:
Three character of echinoderms seen in hemichordates.

  1. Hemichordates are marine habitats.
  2. Circulatory system is open type.
  3. External fertilization.

Three character of chordates seen in hemichordates.

  1. Sexes are separate.
  2. Triploblastic animals.
  3. Coelomate animals.

Question 16.
Name the visual units of compound eyes of cockroach.
Answer:
Ommatidium (simple eye) is the visual unit of compound eye of cockroach. Each eye of cockroach has 2000 simple eys or ommatidia.

Question 17.
Give an examples for stenohaline and euryhaline.
Answer:
Example of stenohaline animal is Gold fish.
Example for euryhaline animals is Salmon.

Question 18.
Brain is the chief organ of our body. It controls endocrine function by a special structure. What is the name of it ? How does it control ?
Answer:
Brain controls the endocrine system of our body by a special structure called hypothalamus. Hypothalamus is a small cone shaped structure that projects downward from the brain ending into the pituitary stalk. It interlinks both the nervous system and endocrine system.

Though pituitary gland is known as master endocrine glands that controls the other endocrine glands, but it is, in turn controlled by the hypothalamus. Hypothalamus contains groups of neurosecretary cells. It produces neurotransmitters which regulate the secretions of the pituitary. The hormones produced by the hypothalamus acts either as a releasing hormone or as a inhibitory hormone.

Question 19.
Write the different types of silkworm in our country.
Answer:
Different types of silkworm in our country are,

  1. Bombyx mori
  2. Antheraea assamensis
  3. Antheraea mylitta
  4. Attacus ricini

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
Write a short note on buccal cavity of Frog.
Answer:
The alimentary canal consists of the buccal cavity, pharynx, oesophagus, duodenum, ileum and the rectum which leads to the cloaca and opens outside by the cloacal aperture. The wide mouth opens into the buccal cavity. On the floor of the buccal cavity lies a large muscular sticky tongue. The tongue is attached in front and free behind. The free edge is forked. When the frog sights an insect it flicks out its tongue and the insect gets glued to the sticky tongue. The tongue is immediately withdrawn and the mouth closes.

A row of small and pointed maxillary teeth is found on the inner region of the upper jaw. In addition vomerine teeth are also present as two groups, one on each side of the internal nostrils. The lower jaw is devoid of teeth. The mouth opens into the buccal cavity that leads to the oesophagus through the pharynx.
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 9

[OR]

Explain the refractive errors of eye.
Answer:
Refractive errors of eye:
Myopia (near sightedness): The affected person can see the nearby objects but not the distant objects. This condition may result due to an elongated eyeball or thickened lens; so that the image of distant object is formed in front of the yellow spot. This error can be corrected using concave lens that diverge the entering light rays and focuses it on the retina.

Hypermetropia (long sightedness): the affected person can see only the distant objects clearly but not the objects nearby. This condition results due to a shortened eyeball and thin lens; so the image of closest object is converged behind the retina. This defect can be overcome by using convex lens that converge the entering light rays on the retina.

Presbyopia: Due to aging, the lens loses elasticity and the power of accommodation. Convex lenses are used to correct this defect.
Astigmatism is due to the rough (irregular) curvature of cornea or lens. Cylindrical glasses are used to correct this error.
Cataract: Due to the changes in nature of protein, the lens becomes opaque. It can be corrected by surgical procedures.
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 10

Question 21.
Kidney functioning is regulated by hormonal feedback control mechanism. Explain this mechanism.
Answer:
Antidiuretic hormone or Vasopressin, juxtaglomerular apparatus and atrial natriuretic factor regulate the kidney function.

Antidiuretic hormone or Vasopressin: When there is excessive loss of fluid from the body or when there is an increase in the blood pressure, the osmoreceptors of the hypothalamus stimulates the neurohypophysis to secrete the antidiuretic hormone or vasopressin. It facilitates reabsorption of water by increasing the number of aquaporins on the cell surface membrane of the distal convoluted tubule and collecting duct.

When the water loss from the body is less or when you drink excess amount of juice, osmoreceptors stop secreting ADH and the aquaporins of the collecting ducts move into the cytoplasm. Hence dilute urine is produced to maintain the blood volume.

Renin angiotensin: Juxtaglomerular apparatus (JGA) is a specialized tissue in the afferent arteriole of the nephron. It consists of macula densa and granular cells. The macula dcnsa cells sense distal tubular flow and affect afferent arteriole diameter. The granular cells secrete an enzyme called renin. A fall in glomerular blood flow, glomerular blood pressure and glomerular filtration rate, can activate JG cells to release renin. This converts angiotensinogen into angiotensin-I. Angiotensin converting enzyme converts angiotensin-I to angiotensin-II.

Angiotensin-II stimulates Na+ reabsorption in the proximal convoluted tubule by vasoconstriction of the blood vessels and increases the glomerular blood pressure. Angiotensin-II stimulate adrenal cortex to secrete aldosterone that causes reabsorption of Na+, K+ excretion and absorption of water from the distal convoluted tubule and collecting duct. This increases the glomerular blood pressure and glomerular filtration rate. This complex mechanism is generally known as Renin-Angiotensin-Aldosterone System (RAAS).
Tamil Nadu 11th Biology Previous Year Question Paper June 2019 English Medium - 11

Atrial natriuretic factor: Excessive stretch of cardiac atrial cells cause an increase in blood flow to the atria of the heart and release Atrial Natriuretic Peptide or Factor (ANF). It travels to the kidney where it increases Na+ excretion and increases the blood flow to the glomerulus, acting as vasodilator on the afferent glomerular arterioles and as a vasoconstrictor on efferent arterioles.

It decreases aldosterone release from the adrenal cortex and decreases release of renin, thereby decreasing angiotensin-II. ANF acts antagonistically to the renin-angiotensin system, aldosterone and vasopressin.

[OR]

Explain, how the combination of aquaculture and hydroponics techniques can be successfully executed in our homes.
Aquaponics is a technique which is a combination of aquaculture (growing fish) and hydroponics (growing plants in non-soil media and nutrient-laden water). Aquaponics may also prevent toxic water runoff. It also maintains ecosystem balance by recycling the waste and excretory products produced by the fish. In India, aquaponics was started in 2013. Some primary methods of aquaponic gardening that are in use nowadays are as follows:

1. Deep water culture is otherwise known as raft based method. In this method a raft floats in water. Plants are kept in the holes of raft and the roots float in water. This method is applicable for larger commercial scale system. By this method fast growing plants are cultivated.

2. Media based method involves growing plants in inert planting media like clay pellets or shales. This method is applicable for home and hobby scale system. Larger number of fruiting plants, leafy green plants, herbs and other varieties of plants can be cultivated.

3. Nutrient Film technique involves the passage of nutrient rich water through a narrow trough or PVC pipe. Plants are kept in the holes of the pipe to allow the roots to be in free contact with in the water stream.

4. Aqua vertica is otherwise known as vertical aquaponics. Plants are stacked on the top of each other in tower systems. Water flows in through the top of the tower. This method is suitable for growing leafy greens, strawberries and other crops that do not need supporting solid substratum to grow.

Advantages of Aquaponic gardening Water conservation: No need of water discharge and recharge as the water is maintained by recycling process.

Soil: Bottom soil may be loaded with freshwater. Microbes in water can convert the waste materials into usable forms like ammonia into nitrates which are used by the plants. Thus the soil fertility is maintained.

Pesticides: In this system use of pesticides is avoided and hence it is ecofriendly.
Weeds: Since the plants are cultured in confined conditions, growth of weeds is completely absent. The utilization of nutrient by plants is high in this method.

Artificial food for fishes: In this system plant waste and decays are utilized by fishes as food. So, the need for the use of supplementary feed can be minimized.

Fertilizer usage: Artificial or chemical fertilizers is not required for this system since the plants in the aquaponics utilize the nutrients from the fish wastes dissolved in water. Cultivable fishes like tilapia, trout, koi, gold fish, bass etc., are cultured in aquaponics. Common cultivable plants like tomato, pepper, lettuce, cucumber, and rose are co-cultivated in this method.

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