Lab Manual Class 10 Biology

Homology and Analogy of Plants and Animals Experiment Class 10 Practical Science NCERT

CBSE Class 10 Biology Lab Manual Homology and Analogy of Plants and Animals

Board	CBSE
Textbook	NCERT Books
Class	Class 10
Subject	Biology
Experiment Name	Homology and Analogy of Plants and Animals
Category	Class 10 Science Lab Manual

The experiment to determine Homology and Analogy of Plants and Animals are part of Science Lab Manual for Class 10 CBSE Experiments is designed to help students bridge the gap between theoretical concepts and practical applications through hands-on experiments in Biology.

Science Lab Manual Class 10 CBSE Homology and Analogy of Plants and Animals Experiment

Determine Homology and Analogy of Plants and Animals Class 10 Practical

Introduction

Evolution: It may be defined as the formation of wide varieties of organisms which have been evolved from pre-existing organisms through their gradual changes since the beginning of life.

CBSE Class 10 Science Solutions

The fossil records, vestigial organs, homologous and analogous organs provide evidence for common ancestry.
The embryological studies of the early developmental stages in all the vertebrates show some common similarities like notochord, gill slits, embryonic tail, etc.
Vestigial organs are those organs which were functional earlier but now are rudimentary. These are important in tracing the evolutionary relationship with other vertebrates. For example, nictitating membrane and appendix in human beings.
Divergent evolution and convergent or homology and analogy helps us to trace evolutionary relationship with other organisms.
Convergent evolution: Type of evolution when two unrelated species undergo several changes and adaptations to become more similar, this is called convergent evolution. Usually, these two species live in similar climates and environments in different parts of the world that favour the same adaptations.
Analogy: Analogy, or analogous structures, is actually the one that indicates there is no common ancestor between two organisms. The anatomical structures may look similar and may perform the same functions but they are actually a product of convergent evolution.
Analogy in animals:
1. The bats, flying insects and birds use their wings to fly, but bats are actually mammals and not related to birds or flying insects.
2. The fins of a shark and a dolphin are favourable adaptations for animals that need to swim and move in the water. But sharks are classified within the fish family while dolphins are mammals.
Analogy in plants: Potato and sweet potato perform same function, i.e., storage of food but one is modified stem while the other is modified root.
Divergent evolution: Type of evolution where closely related species become less similar in structure and function due to the adaptations they acquire during the natural selection process. Migration to new climates, competition for niches with other species, and even micro-evolutionary changes like DNA mutations can contribute to divergent evolution.
Homology: The study of organisms with similar anatomical structures is called homology. The homologous organs did evolve from a recent common ancestor. Organisms with homologous structures are more closely related to each other in the evolution tree.
Homology in animals: The forelimbs of humans and wings of bat have similar structure though they perform different functions.
Homology in plants: The prickly spines on a cactus and the leaves on an oak tree look very dissimilar, but they are actually homologous structures. They have very different functions. In cactus spines are primarily for protection and to prevent water loss in its hot and dry environment, the oak tree does not have these adaptations.

CBSE Class 10 Science Lab Manual Experiment – 5

Aim
To study homology and analogy with the help of models/charts/specimens of either animals or plants.
Theory

Homologous organs: The organs which perform different functions in different species but have similar basic structure and similar embryonic origin are called homologous organs. E.g., limbs of human being, frog, bird and
lizard.
Homology: Similar in characteristics resulting from shared ancestry.
Homologous features arise from adaptive behaviour, to adapt to different environmental conditions and modes of life.
Homology in Plants: (Leaves)
Analogous Organs: The organs which are quiet different in fundamental structure and embryonic origin but perform same function and may superficially look alike are called analogous organs. For e.g., wings of bird, bat, insects are used for flying but the internal structure is different.
Analogy: The organisms showing analogy do not share common ancestors.
Analogous feature arise when two unrelated species adapt themselves to similar climate and environmental condition.
Analogy in Plants: Thoms and spines are modified organs seen in plants are analogous structures. Thom is modification of stem and spine is modification of leaf.
Tendrils in plant show similar function but they are different in origin.

Materials Required

Preserved specimens
Animals: Limbs of frog, lizard and bat. Wings of insect, bat and bird. .
Plants: Pitcher plant, venus fly trap and cactus Plants with tendrils:
Pea plant – Leaf tendril
Grape plant – Stem tendril
Smilax – Stipular tendril

A. TO STUDY HOMOLOGY
I. IN ANIMALS

Procedure

Observe carefully the preserved specimens of limbs of frog, limbs of lizard and limbs of bat.
Draw diagrams and record your observations.

Observations

The limbs of frog, lizard and birds are similar in structure.
Each limb has humerus, ulna, radius, carpal and five sets of digits.

Conclusion
The similarity in structure but difference in function proves that all these homologous organs are evolved from common ancestor.

II. IN PLANTS

Procedure

Observe carefully the given specimens of pitcher plant, venus fly trap and cactus plant.
Record your observations to study the homologous organs and draw diagrams.

Observations

The leaves are modified for different functions, but the structure is similar.
In pitcher plant, the leaves are modified into pitchers to trap insects.
In venus fly trap plant the leaves are modified into jaws to trap insects.
In cactus plant, the leaves are modified into spines to reduce water loss through transpiration.

Conclusion
The modification of leaves in different plants showing similar origin but different functions shows the homology in plants.

B. TO STUDY ANALOGY
I. IN ANIMALS

Procedure

Observe carefully the preserved specimens of wings of insect, bat and bird.
Draw diagrams and record your observations.

Observations

The function of wings in all the three specimens is same but the structure is different.
The wings of insect has no limbs.
The wings of bat has limbs with five digits whereas the wings of bird has only three digits.

Conclusion
The wings of birds, insects and bats has common use, i.e., flying but the structure is different. These organs are called analogous organs.

II. IN PLANTS

Procedure

Observe the specimens/samples of plants showing leaf tendril, stem tendril and stipular tendril.
Record your observations with the help of diagram.

Observations

The function of all the three types of tendrils is same.
The structure of each tendril and its origin is different.

Conclusion
The types of tendril in plants show the analogy in plants and all these tendrils are analogous organs seen in plants.

CBSE Class 10 Science Lab Manual Viva Voce

Question 1:
What does the similarity in organs due to homology prove?
Аnswer:
Common ancestors.

Question 2:
Name an analogous organ of spine.
Аnswer:
Thom

Question 3:
What is the modification of leaves in venus fly trap?
Аnswer:
In venus fly trap, the leaves are modified into jaws.

Question 4:
What is common in limbs of frog and lizard?
Аnswer:
Their structure is same but function is different.

Question 5:
Out of potato, sweet potato, radish and carrot; make pairs of homologous organs and analogous organs.[Delhi 2012]
Аnswer:
Radish and carrot both are homologous organs as these are modified roots.
Homologous structures are similar in origin but perform different functions. Carrot and radish are underground roots. So, they represent the correct homologous structures. But potato is a stem and sweet potato is an underground modified root, so they represent the analogous organs.

CBSE Class 10 Science Lab Manual Practical Based Questions

Question 1:
What are homologous organs?
Аnswer:
Homologous organs are those organs which are similar in structure but different in function.

Question 2:
Give two examples of homologous organs.
Аnswer:
Limbs of frog, lizard and bat, and leaves of cactus, pitcher plant and venus fly trap.

Question 3:
What are analogous organs?
Аnswer:
Analogous organs have same function but different structures.

Question 4:
Give two examples of analogous organ.
Аnswer:
The examples of analogous organs are wings of birds and wings of insects and tendrils in plants like leaf tendril, stem tendril and stipular tendril.

Question 5:
Name one plant that has stem tendril.
Аnswer:
Grape plant.

Question 6:
Name one plant that has leaf tendril.
Аnswer:
Pea plant.

Question 7:
What does the analogy in animals or plants show?
Аnswer:
Analogy shows that the organisms (animals/plants) showing analogy never had common ancestors.

Question 8:
Give an example of analogous organs in marine animals.
Аnswer:
Fins of fish and flipper of whale are analogous organs.

Question 9:
Give an example of analogous organs in flying creatures.
Аnswer:
Wings of bird and wings of insect are analogous organs.

Question 10:
Name two homologous organs in plants.
Аnswer:
The storage organs, i.e., potato and ginger (store food for plants) are homologous.

CBSE Class 10 Science Lab Manual Multiple Choice Questions (MCQs)

Questions based on Procedural and Manipulative Skills
Question 1:
The forelimbs of human and wings of bat are an example of:
(a) vestigial organs
(b) analogous structures
(c) homologous structures
(d) none of these

Question 2:
Wings of insects and birds are an example of
(a) vestigial organs
(b) homologous organs
(c) analogous organs
(d) none of these

Question 3:
The similarity of bone structure in the forelimbs of many vertebrates is an example of
(a) diversity
(b) homology
(c) analogy
(d) variations

Question 4:
Similar structures that evolved independently are called
(a) homology
(b) analogy
(c) both (a) and (b)
(d) none of these.

Question 5:
Pitcher plant and venus fly trap plant represent the examples of:
(a) analogous organs
(b) homologous organs
(c) vestigial organs
(d) none of these

Question 6:
Potato and sweet potato are the examples of:
(a) analogous organs
(b) vestigial organs
(c) homologous organs
(d) none of these

Question 7:
Potato and ginger are the examples of:
(a) analogous organs
(b) homologous organs
(c) vestigial organs
(d) both (a) and (b)

Question 8:
Potato and sweet potato are respectively the modified form of:
(a) root and stem
(b) stem and leaf
(c) stem and root
(d) stem and flower

Question 9:
Thoms of citrus and tendrils of cucurbits are the examples of:
(a) analogous organs
(b) homologous organs
(c) vestigial organs
(d) both (a) and (c)

Questions based on Observational Skills
Question 10:

The above figures demonstrate
(a) analogy
(b) homology
(c) adaptation
(d) vestigial organs

Question 11:

The figures given above show
(a) modification in leaves
(b) plants with similar origin
(c) homology in plants
(d) all of the above

Question 12:

The thorn and the tendril is the modified organ of:
(a) stem
(b) bud
(c) leaf
(d) thorn

Question 13:

Both potato and sweet potato store food in plants, they are the modified parts of:
(a) potato – stem
sweet potato – stem
(b) potato – root
sweet potato – root
(c) potato – stem
sweet potato – root
(d) potato – root
sweet potato – stem

Questions based on Reporting and Interpretation Skills
Question 14:
During the early stage of development, the embryo of reptiles, birds and mammals look very similar. This suggests that reptiles, birds and mammals
(a) have a common functions
(b) live in same environment
(c) have shown evolution
(d) have common vestigial organs

Question 15:
The correct labelling is

(a) (1) carpal (2) humerus (3) digits
(b) (1) humerus (2) carpal (3) digits
(c) (1) digits (2) carpal (3) humerus
(d) (1) humerus (2) digits (3) carpal.

Question 16:

The students were shown the specimens of limbs of lizard and bat. The correct conclusion is
(a) they have similar structures but different functions.
(b) they have originated from common ancestors.
(c) these are homologous organs.
(d) all of the above.

Question 17:

The above specimens were shown to students for recording the observations and conclusion. The correct answer is

Observations	Conclusion
(a) Wings of animals	common structure, common function
(b) Wings of variety of animals	different structures but same function
(c) Analogous organs	structures with common ancestor
(d) Analogous organs	organs with same structure and function

Question 18:
In pea plant, the tendrils are the modification of:
(a) stem
(b) leaf
(c) root
(d) flower.

Question 19:
In grape plant, the tendrils are the modification of:
(a) stem
(b) leaf
(c) root
(d) flower

Question 20:
The correct examples of homologous organs are:
(a) limbs of bat, frog and lizard
(b) sweet potato, carrot and raddish
(c) leaves of pitcher, venus fly trap and cactus
(d) all of these

Question 21:
One of the examples of two analogous organs can be the wings of parrot and [Delhi 2011]
(a) flipper of whale
(b) foreleg of horse
(c) front leg of frog
(d) wings of housefly

Question 22:
Study the different conclusions drawn by students of a class on the basis of observations of preserved available specimens of plants and animals. [Delhi 2013]
I. Potato and sweet potato are analogous organs in plants.
II. Wings of insects and wings of birds are homologous organs in animals.
III. Wings of insects and wings of bats are analogous organs in animals.
IV. Thoms of citrus and tendrils of cucurbita are analogous organs in plants.
The correct conclusions are:
(a) I and II
(b) II and IV
(c) I and III ,
(d) III and IV

Question 23:
Study the different conclusions drawn by students on the basis of their observations of fresh available specimens of plants and animals: [Outside Delhi 2013]
I. Potato and sweet potato are homologous organs.
II. Wings of insects and wings of bird are analogous organs.
III. Wings of insects and wings of bats are homologous organs.
IV. Thoms of citrus and tendrils of cucurbita are homologous organs.
The correct conclusions are:
(a) I and II
(b) II and IV
(c) I and III
(d) III and IV

Question 24:
You have a basket of vegetables which contains carrot, potato, tomato, ginger, radish, sweet potato. Select two vegetables to represent the correct homologous structures. [Outside Delhi 2013]
(a) Potato and sweet potato
(b) Carrot and radish
(c) Potato and carrot
(d) Carrot and tomato

Question 25:
Study the following statements: [Oittside Delhi 2014]
I. Wings of birds and wings of bats are homologous organs.
II. Wings of birds and wings of insects are modified forelimbs.
III. Wings of birds and wings of insects are analogous organs.
IV. Wings of birds and forelimbs of horse are homologous organs.
The correct statements are:
(a) I and II
(b) II and III
(c) III and IV
(d) I and IV

Question 26:
Which one of the following pairs of vegetables is an example of homologous structures? [Delhi 2014]
(a) Potato and sweet potato
(b) Carrot and radish
(c) Carrot and tomato
(d) Tomato and radish

Аnswers:

CBSE Class 1o Science Lab Manual Scoring Key With Explanation

(c) These organs are similar in structure but different in function.
(c) These organs have same function but different structures.
(b) Same structures.
(a) Same structures.
(b) Same structures of leaves but different functions.
(a) Both have different structures (stem and root) but same function (stores food).
(b) Both are stems (same structure) but different functions.
(c) Potato is stem and sweet potato is root (modified in both cases).
(b) Similar structures but different functions.
(a) Different structures (tendrils of stem and leaf) but same function.
(d) These are leaves of cactus and venus fly trap.
(a) In both, stem is modified.
(c) Potato is stem and sweet potato is root.
(c) The embryonic stage is same for all.
(b) Option (b) has correct order of labels.
(d) These organs have same structures with one origin.
(b) The designs of the three wings, their structure and components are very different. They look similar because they have a common use for flying.
(b) Pea plant has leaf tendril.
(a) Grape tendril is from stem.
(d) All are correct.
(d) Analogous means same function, wings have same function.
(c) Analogous organs.
(b) Homologous and analogous organs.
(b) Both are roots. Homologous structures are similar in origin but perform different functions. Carrot and radish are underground roots. So, they represent the correct homologous structures.
(c) Examples of homologous and analogous organs.
(b) Both are roots.

More Resources CBSE Class 10 Lab Manual Practical Skills:

NCERT Class 10 Science Lab Manual

Stomata Experiment Class 10 Practical Science NCERT

CBSE Class 10 Biology Lab Manual Stomata

Board	CBSE
Textbook	NCERT Books
Class	Class 10
Subject	Biology
Experiment Name	Stomata
Category	Class 10 Science Lab Manual

The experiment to determine Stomata are part of Science Lab Manual for Class 10 CBSE Experiments is designed to help students bridge the gap between theoretical concepts and practical applications through hands-on experiments in Biology.

Science Lab Manual Class 10 CBSE Stomata Experiment

Determine Stomata Class 10 Practical

Biology Practicals For Class 10 CBSE

Stomata Experiment Class 10 Introduction

All over the plant body epidermis is present. The epidermis is made up of single layer of cells. Epidermis on the aerial parts of the plants often secrete a waxy, water-resistant layer on their outer surface. This prevents water loss, mechanical injury and invasion by parasitic bacteria or fungi. The epidermal cells form a central layer without intercellular spaces. Most of these cells are flat with thicker outer and side walls and thin inner wall. In this epidermic layer of leaves we can see small pores called stomata.
Stomata are small holes or openings present on the leaf surface in epidermis. The lower side of the leaf has more stomata (singular stoma).

NCERT Class 10 Science Solutions

Guard cells

Stomata has a small pore which is guarded by the guard cells. The guard cells control the opening and the closing of the stomata.
The guard cells are bean-shaped in dicots and dumb-bell shaped in monocots.
The inner wall of the guard cell is thick whereas the outer wall is thin.
Plants need gases like oxygen for respiration and carbon dioxide for photosynthesis. It is through this small opening called stomata the gases diffuse inside the leaf and are taken in.

Opening and Closing of stomata

During photosynthesis the stomatal pore opens whenever there is demand of carbon dioxide and the opening of these stomata also causes the loss of water by the leaves through the process called transpiration.
The pressure built in the guard cells due to the water in it is called turgor pressure. .
The guard cells opening and closing mechanism depends on the turgidity they get due to water.
The guard cells absorb water and become turgid which results in opening of stomata during the day.
At night the guard cells are flaccid and so the stomata get closed. But in some desert plants the opening of stomata happens only at night to prevent the loss of water.

Stomata distribution

Stomata are spread out over leaves so that the waste gases produced by the leaf can diffuse away quickly, this stops the build-up of excreted products, which would slow gas exchange.
More stomatal pores are present on the dorsal side of leaf than the ventral side.
In desert plants, stomata are more sunken type compared to tropical plants.
Distribution of stomata in the plants differ from species to species present in the same region.
In cactus plants, leaves are reduced to spines so stomata are present on the stem.
In aquatic plants, stomata are either absent or non-functional stomata are absent in roots.
Fewer stomata on the upper surface prevent excessive loss of water due to transpiration as this surface is directly exposed to sunlight.

NCERT Class 10 Science Lab Manual Experiment – 1

Aim
To prepare a temporary mount of a leaf peel to show stomata.
Theory

Plants need oxygen for respiration and carbon dioxide for photosynthesis. The exchange of gases in plants occurs through the surface of stems, roots and leaves.
On leaves there are plenty of small tiny pores called stomata.
On the dorsal side of leaf more stomatal pores are present than the ventral surface of leaf.
Through these pores, plants can also lose water by the process called transpiration.
To avoid excess loss of water, the stomata pores closes and when gases are required, these pores open.
This opening and closing of pores is monitored by guard cells.
The guard cells swell when water flows into them, causing the stomata pore to open. When the guard cells shrink the stomata pores close.
The guard cells contain chloroplast and nucleus in it. They are bean-shaped in dicots and dumb-bell shaped in monocots.

Materials Required
Freshly plucked leaf of Rheo or Tradescantia, petri dish, slide, coverslip, needle, forceps, brash, dropper, watch glass, filter paper, glycerine, safranin solution and microscope.
Procedure

Take a freshly plucked leaf (Rheo or Tradescantia).
Stretch the leaf with its dorsal (lower) part facing upwards.
Break the leaf by applying suitable pressure so that the epidermis projects from the leaf.
Cut the epidermis and put it in a petri dish.
Take a watch glass, add few drops of water and a drop of stain in it.
Transfer the small piece of epidermis from petri dish into the watch glass with the help of brash.
Allow the peel to remain in the stain for 2-3 minutes, so that it can take up the stain.
With the help of brush transfer the stained peel into a petri dish with water to remove the extra stain.
Now take a clean slide and place it on a filter paper. In the centre of the slide put a drop of glycerine and transfer the stained peel from petri dish on the slide.
Gently hold the coverslip with the needle and place it on the peel. Avoid air bubbles formation.
Use the filter paper to clean the excess stain, water or glycerine that comes out from the coverslip sides.
Ensure that the slide is clean and place it under the microscope. First view it under low power (10X) and then under high power(45X).
Record your observations.

Observations

In an epidermal peel we see single layer of cells.
In between the epidermal layer small spots are seen.
When focused under powerful microscope the stomata pores are clearly seen.
Each stomata pore has two kidney-shaped cells called guard cells.
Each guard cell has one nucleus and many chloroplasts.

Conclusion
Epidermal layer of leaf peel has many stomata pores. Each stomatal pore has two kidney shaped guard cells, in dicots plants. Each guard cell has one nucleus and many chloroplasts.
Precautions

While removing the epidermal peel, ensure that you pluck the thinner scrap of leaf.
Do not overstain the peel.
Avoid air-bubbles formation while placing the coverslip.
The peel should not be folded.
The slide should be clean and dry before placing it under microscope.

NCERT Class 10 Science Lab Manual Viva Voce

Question 1:
What controls the opening and closing of stomata?
Аnswer:
The guard cells controls the opening and closing of stomata.

Question 2:
When does the pore close?
Аnswer:
When guard cells shrink the stomatal pore closes.

Question 3:
What influences the opening of stomata?
Аnswer:
The sunlight and the swelling of guard cells.

Question 4:
Name the layer of tissue in which stomatas are seen.
Аnswer:
It is the epidermal tissue in which stomatas are present.

Question 5:
Name the stain used for preparing slide for stomata.
Аnswer:
Safranin solution

NCERT Class 10 Science Lab Manual Practical Based Questions

Question 1:
Where do you find more stomatas?
Аnswer:
Stomatas are present on leaves and stem, but more stomatas are present on the lower side of a leaf.

Question 2:
What is the importance of stomata for a leaf?
Аnswer:
Stomata helps in exchange of gases like oxygen and carbon dioxide required by plants. It also removes extra water from leaves by transpiration.

Question 3:
How does guard cells help in opening of stomata?
Аnswer:
The guard cells are bean-shaped cells which on swelling opens the pore.

Question 4:
Desert plants cannot afford to lose water through stomata. How do they exchange gases?
Аnswer:
In desert plants the stomata are sunken, the exchange of gases takes place at night. In some desert plants leaves are adapted to thorns.

Question 5:
What things will you observe if you focus the stomata slide under high power objective of a microscope?
Аnswer:
Under high power objective we can see the stomata surrounded by guard cells. Each guard cell have nucleus and chloroplast.

NCERT Class 10 Science Lab Manual Questions

Question 1:
What is the function of guard cells in stomata?
Аnswer:
Guard cells control the opening and closing of pore in stomata.

Question 2:
Why is the number of stomata greater on the lower surface of a leaf?
Аnswer:
To maintain the amount of water in the cells of leaf and control the rate of transpiration the lower surface of the leaf has more stomata than the upper surface of the leaf.

Question 3:
Why are stomata absent in roots?
Аnswer:
The role of stomata is to exchange gases during photosynthesis and respiration and helps in transpiration. The roots helps in the absorption of water and hence stomata is not required on the roots.

Question 4:
What is the shape of guard cells in stoma of grass leaf?
Аnswer:
Grass leaf has dumb-bell shaped stomata.

Question 5:
Do guard cells have rigid or elastic walls? Justify your answer.
Аnswer:
Guard cells have elastic walls. It is the elastic walls of the guard cells which changes its shape to open or close the stomata. When the guard cells are turgid the pore opens due to the concave shape formation of the guard cells.

NCERT Class 10 Science Lab Manual Multiple Choice Questions (MCQs)

Questions based on Procedural and Manipulative Skills
Question 1:
The stomata openings are seen in the single-layered tissues of
(a) epidermal tissue
(b) sclerenchyma tissues
(c) parenchyma tissue
(d) collenchyma tissues.

Question 2:
The stain used to prepare stomatal slide is
(a) methylene blue
(b) safranin
(c) acetocarmine
(d) cotton blue.

Question 3:
While preparing the stomatal slide, glycerine is used to
(a) prevent air bubbles
(b) avoid drying of peel
(c) to stain the peel
(d) make it visible

Question 4:
To prepare a temporary mount of a leaf peel one should take:
(a) a leaf plucked a day in advance
(b) a freshly plucked leaf.
(c) a dried and preserved leaf
(d) a leaf plucked a few hours before the mounting.

Question 5:
The opening and closing of stomata is controlled by
(a) transpiration
(b) turgidity of cells
(c) intensity of light
(d) all of these.

Question 6:
To remove the extra stain, water or glycerine collected on the slide, the correct method is
(a) wash the sides of the slide
(b) clean it with handkerchief
(c) use cotton swabs to collect it
(d) use filter/blotting paper to clean it.

Question 7:
A student was asked to prepare a temporary mount of stomata from a leaf. He should select:
(a) tip of the leaf
(b) midrib of the leaf
(c) epidermis of stem
(d) lower epidermis.

Question 8:
To prepare a temporary mount of a leaf peel for observing stomata the chemicals used for staining and mounting respectively are
(a) safranin and iodine
(b) safranin and glycerine
(c) iodine and safranin
(d) glycerine and iodine.

Question 9:
Which one of the following is the correct method of obtaining a leaf peel for observing stomata?
(a) Crack the leaf and pluck the peel.
(b) Take a leaf and with blade make thin sections.
(c) Use needle and forcep to pluck a thin layer of leaf.
(d) Break the leaf and remove the projections of epidermal layer.

Question 10:
The steps involved in making a slide of epidermal peel of leaf are as follows:
I Put out a thin peel from the lower surface of the leaf.
II Place a drop of glycerine on the slide.
III Stain the peel in safranin.
IV Place the stained peel on the glycerine.
V Remove the extra stain by washing with water. VI Place the coverslip over the peel.
Which one is the correct sequence of steps to be followed?
(a) I, II, III, IV, V, VI
(b) I, III, V, II, IV, VI
(c) I, III, IV, II, V, VI
(d) I, II, IV, III, V, VI.

Question 11:
A student focussed the epidermal peel of leaf under the low power of a microscope but could not see the parts. He should
(a) use the coarse adjustment knob again to focus the slide.
(b) use the fine adjustment knob to increase magnification.
(c) focus under high power using coarse adjustment knob.
(d) focus under high power using fine adjustment knob.

Question 12:
Given below are the steps in the preparation of a temporary mount of a stained leaf peel.
(i) Cover the material with the cover slip.
(ii) Transfer the stained peel to the clean glass slide and add a drop of glycerine.
(iii) Remove the peel from the lower surface of the leaf,
(iv) Drop it in the water in a petri dish and add a drop of safranin stain.
The correct sequence of steps is:
(a) (iii), (iv), (ii), (i)
(b) (i), (if), (iii), (iv)
(c) (ii), (iii), (iv), (i)
(d) (iii), (iv), (i), (ii).

Questions based on Observational Skills
Question 13:
In the figure given below of stomata the correct label for A and B are

(a) A – nucleus, B – guard cells
(b) A – guard cells, B – nucleus
(c) A – chloroplast, B – nucleus
(d) A- chloroplast, B – guard cell. .

Question 14:
The temporary mount of stomata was seen under the microscope which appeared pink in colour. The stain used was
(a) iodine
(b) acetocarmine
(c) phenolphthalein
(d) safranin.

Question 15:
A well stained leaf peel mount, when observed under the high power of a microscope, shows nuclei in
(a) epidermal cells
(b) guard cells
(c) both guard cells and epidermal cells
(d) none of these.

Question 16:
A well stained leaf peel preparation, when focused under high power of microscope, would show
(a) epidermal cells, stomata, guard cells each with one nucleus and many chloroplasts.
(b) epidermal cells, stomata, guard cells with many nuclei and one chloroplast.
(c) stomata and guard cells without nuclei or chloroplasts.
(d) stomata but no guard cells or epidermal cells.

Question 17:
In the diagram of the stomatal pore given below the marking corresponding to the chloroplast is:

(a) A
(b) B
(c) C
(d) D

Question 18:
In the sketch of the stomatal apparatus which of the following is missing:
(a) cell membrane of the cells
(b) cell wall of the cells
(c) nuclei of guard cells
(d) chloroplast in guard cells.

Questions based on Reporting and Interpretation Skills
Question 19:
On using the high power objective of a microscope an epidermal leaf shows:
(a) stomata surrounding many guard cells
(b) stomata surrounded by a pair of guard cells
(c) stomata surrounded by several epidermal cells
(d) stomata surrounded by several guard cells.

Question 20:
Four students A, B, C and D, make the following records given below, for the parts marked X and Y in this diagram.

Student	X	Y
A	Stoma	Nucleus
B	Nucleus	Stoma
C	Epidermal cell	Stoma
D	Cell wall	Epidermal cell

The correct record, is that of student
(a) A
(b) B
(c) C
(d) D

Question 21:
The labelling for the slide of leaf peel showing stoma by the four students who made the diagram and tabulated the labels, is as follows:

Student	I	II	III	IV
A	Stoma	Nucleus	Epidermal cell	Cell wall
B	Nucleus	Stoma	Epidermal cell	Cell wall
C	Epidermal cell	Stoma	Nucleus	Cell wall
D	Cell Wall	Epidermal cell	Nucleus	Stoma

The student who made the correct labelling is:
(a) Student A
(b) Student B
(c) Student C
(d) Student D.

Question 22:
In the following sketch of the stomatal apparatus, the parts I, II, III and IV were labelled differently by four students

The correct labelling, out of the following is:
(a) (I) guard cell, (II) stoma, (III) starch granules, (IV) nucleus
(b) (I) chloroplast, (II) nucleus, (III) stoma, (IV) guard cell
(c) (I) guard cell, (II) starch, (III) nucleus, (IV) stoma
(d) (I) cytoplasm, (II) chloroplast, (III) stoma, (IV) nucleus.

Question 23:
The correct labellings for the given diagram are respectively

(a) stoma, guard cell, nucleus
(b) stoma, chloroplast, epidermal cell
(c) stomatal pore, nucleus, guard cell
(d) stoma, guard cell, epidermal cell.

Question 24:
A student draws the following sketch of the stomatal apparatus:

The chloroplast and nucleus is denoted by
(a) 5 and 3 respectively
(b) 4 and 3 respectively
(c) 3 and 5 respectively
(d) 4 and 5 respectively.

Question 25:
The correct sequence, out of the following options, for focusing a slide of epidermal peel of a leaf under a microscope to show the stomatal pore is:
I. Observe under low power.
II. Adjust mirror to get maximum light.
III. Place the slide on the stage.
IV. Focus under high power.
(a) II, III, I, IV
(b) I, II, III, IV
(c) III, II, IV, I
(d) III, II, I, IV

Аnswers:

NCERT Class 10 Science Lab Manual Scoring Key With Explanation

(a) Stomata is present in epidermal layer.
(b) Safranin is used for plants slide mount.
(b) Glycerine has the property to retain water.
(b) A fresh leaf gives better mount.
(d) All the factors are responsible for the opening and closing of stomata.
(d) Filter paper helps in absorbing the extra liquid near coverslip on the slide.
(d) On lower epidermis of leaf, more stomatas are present.
(b) Safranin stains and glycerine prevents drying of the mount.
(d) A single layer of epidermis is best obtained by this method.
(b) It is the correct procedure for making a temporary slide.
(a) The right focus adjustment can make the slide rightly focussed.
(a) It is the correct procedure for making a temporary slide.
(d) Guard cells are bean shaped and makes the pore and chloroplast is present in the cell.
(d) Safranin stain is pink in colour.
(c) Both the cells have nucleus in them.
(a) Under high power, the cells and its components are seen clearly.
(d) Chloroplast is present in the cell and are many in number.
(c) Guard cells have prominent nucleus.
(b) Under high power, the guard cells are distinctly seen.
(a) Pore is stoma and the nucleus is present in the cell.
(b) Nucleus is in the cell, the pore is made by guard cells and the layer is epidermis.
(b) Nucleus is in the cell, the pore is made by guard cells and the layer is epidermis.
(c) Pore is surrounded by guard cells and nucleus is in the cell.
(b) Chloroplasts are many in number and nucleus is big and one prominent structure in cell.
(d) A slide of epidermal peel of a leaf under a microscope to show the stomatal pore is to place the slide on the stage then adjust mirror to get maximum light first observe the side under low power and then under high power.

More Resources CBSE Class 10 Lab Manual Practical Skills:

NCERT Class 10 Science Lab Manual

CO2 is Released During Respiration Experiment Class 10 Practical Science NCERT

CBSE Class 10 Biology Lab Manual CO2(Carbon Dioxide) is Released During Respiration Experiment

Board	CBSE
Textbook	NCERT Books
Class	Class 10
Subject	Biology
Experiment Name	CO2 is Released During Respiration
Category	Class 10 Science Lab Manual

The experiment to determine CO2 is Released During Respiration are part of Science Lab Manual for Class 10 CBSE Experiments is designed to help students bridge the gap between theoretical concepts and practical applications through hands-on experiments in Biology.

Science Lab Manual Class 10 CBSE CO2 is Released During Respiration Experiment

Determine CO2 is Released During Respiration Class 10 Practical

Respiration Experiment Class 10 Introduction

Respiration is the biochemical process during which organic food, mainly glucose that is present in the cell, breaks down into simpler substances and liberates carbon dioxide and energy. There are two types of respiration- aerobic and anaerobic respiration.

NCERT Class 10 Science Solutions

Aerobic Respiration: This type of respiration requires oxygen, so it is called aerobic respiration. During aerobic respiration, complete oxidation of carbohydrates takes place. Glucose is broken down by oxygen to release energy, while carbon dioxide and water are the by-products of the reaction. The released energy is used to make a special energy molecule called Adenosine triphosphate (ATP). Aerobic respiration occurs in plants as well as animals and takes place in the mitochondria.
The word equation for aerobic respiration is:
Glucose + Oxygen → Carbon dioxide + Water + Energy
The chemical equation is:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + 38 ATP (energy)
Anaerobic Respiration: Sometimes there is not enough oxygen available for animals and plants to respire, but they still need energy to survive, so they carry out respiration in the absence of oxygen/less oxygen to produce the energy they require. As the respiration takes place in the absence of oxygen, incomplete oxidation of food occurs and much less energy is released. This is called anaerobic respiration and the process occurs in the cytoplasm.
The word equation for anaerobic respiration in plants is:
Glucose → Ethanol + Carbon dioxide + Energy
The chemical equation is:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ + 2 ATP (energy)
The word equation for anaerobic respiration in animals is:
Glucose → Lactic acid + Energy
The chemical equation is:
C₆H₁₂O₆→ 2C₃H₆O₃ + 2 ATP (energy)

Factors Affecting Respiration
Here are some of the few factors that affect the rate of respiration.

Temperature: At a very high temperature, the rate of respiration decreases with time and at very low temperature,
the respiration rate is insignificant. Optimum temperature for respiration is 20 – 30°C.
Carbon dioxide concentration: Increase in CO₂ concentration and absence of O₂ adversely affects the rate of aerobic respiration.
Light: Light controls respiration by raising the temperature of an organism.
Water: The respiratory rate increases with the increase in water content of the respiring organism.

NCERT Class 10 Science Lab Manual Experiment 3

Aim
To show experimentally that carbon dioxide is given out during respiration.
Theory

All living things show respiration.
It is a chemical process that occurs inside the cell, hence called cellular respiration.
It involves the breaking down of food to release energy and carbon dioxide.
Its reaction is the reverse of photosynthesis.
There are two types of respiration in animals: Aerobic and anaerobic respiration.
Aerobic respiration needs oxygen and anaerobic respiration occurs in the absence of oxygen.
There are three pathways of respiration as shown below:
The energy released in cellular respiration is immediately used to synthesise a molecule called ATP.
Plants also release CO₂ during respiration.
The exchange of gases during respiration takes place through small pores on the leaf called stomata.
Carbon dioxide can be tested by lime water test.
A freshly prepared lime water is Ca(OH)₂ When CO₂ is allowed to pass through it an insoluble compound called CaCO₃ is formed which makes the lime water milky.

(A) Test for release of CO₂ during respiration in animals.
Materials Required
Two test tubes, a cork with two holes, two glass tubes, syringe, lime water.
Procedure

Take some freshly prepared lime water in two test tubes.
Fit cork with two holes in test tubes A and B.
Fix two glass tubes in this cork of test tube A as shown in the figure.
Exhale air into the tube and record your observations.
In another test tube B, which has lime water, pass air through syringe and record your observations.

Observation

In test tube A, the lime water turns milky sooner than in test tube B.

Conclusion

The exhaled air contains lot of CO₂ which turns lime water milky.
This proves that during respiration we exhale CO₂ gas.

Precautions

The glass tube should be dipped in the lime water.
The lime water should be freshly prepared.

(B) To test release of C02 by plants during respiration.
Materials Required
A conical flask, small test tube, cork, thread, germinating seeds, a bent tube, a beaker, water and freshly prepared lime water.
Procedure

Take two conical flasks, add germinating seeds with little water sprinkled over it.
Fix the mouth of conical flasks with cork in which a bent tube is fixed.
Suspend a small test tube containing KOH solution in it with the help of a thread in conical flask A.
Allow the mouth of the bent tube to be immersed in water in set-up A and in lime water in set-up B as shown below.
Record your observations after few hours.

Observations

In set-up A, the water level in the bent tube dipped in beaker increases after few hours.
This is because the oxygen present in the conical flask is taken up by germinating seeds and CO₂ released due to respiration is absorbed by KOH present in small tube. Hence, the air pressure in the flask reduces and water level rises.
In set-up B, the freshly prepared lime water turns milky. This is due to excess CO₂ released into the test tube during respiration of germinating seeds.

Conclusion
This shows that CO₂ is given out during respiration.
Precautions

Lime water should be freshly prepared.
KOH solution should be freshly prepared.
Germinating seeds should have lot of moisture in them.

Respiration Practical Class 10 Viva Voce

Question 1:
Name the parts of the plant through which exchange of gases takes place.
Аnswer:
Stomata and lenticles.

Question 2:
Is respiration a physical or chemical process?
Аnswer:
Respiration is a chemical process. It takes place in cells.

Question 3:
Name the part of cell in which final breaking down of food takes place and energy is released
Аnswer:
Mitochondria

Question 4:
Name two organisms that respire in absence of oxygen.
Аnswer:
Bacteria and yeast.

Question 5:
What are the end products of anaerobic respiration?
Аnswer:
It is ethanol, carbon dioxide and energy.

Question 6:
Name the energy currency for most of the cellular processes.
Аnswer:
ATP, i.e., Adenosine tri phosphate.

NCERT Class 10 Science Lab Manual Practical Based Questions

Question 1:
State two types of respiration.
Аnswer:
Aerobic (in presence of oxygen) and anaerobic respiration (in absence of oxygen).

Question 2:
What are the end products of respiration that takes place in lack of oxygen?
Аnswer:
In lack of oxygen, lactic acid and energy is released.

Question 3:
On passing CO₂ gas through freshly prepared lime water why does it turn milky?
Аnswer:
The freshly prepared lime water is Ca(OH)₂ and when CO₂ is passed through it, the compound formed is CaCO₃ which forms insoluble precipitate that is white in colour.

Question 4:
If you insert a thermometer in a sealed beaker containing germinating seeds, the temperature of thermometer increases. Why?
Аnswer:
The germinating seeds respire and release CO₂ gas alongwith heat energy. It is due to this heat energy the temperature of thermometer increases.

NCERT Class 10 Science Lab Manual Questions

Question 1:
What is the role of KOH in this experiment?
Аnswer:
KOH is used to absorb the carbon dioxide released during respiration of germinating seeds which creates a vacuum in the flask.

Question 2:
When we say that plants and animals respire, where exactly is the process occurring?
Аnswer:
Respiration is a chemical process and occurs in mitochondria.

Question 3:
Why do we use germinating seeds in this experiment?
Аnswer:
The rate of respiration in germinating seeds is faster and the results are visible in less time.

NCERT Class 10 Science Lab Manual Multiple Choice Questions (MCQs)

Questions based on Procedural and Manipulative Skills
Question 1:
When respiration occurs in presence of oxygen, the end products are
(a) CO₂ + H₂O
(b) C₂H₅OH + energy
(c) C₂H₅OH + CO₂ + energy
(d) CO₂ + H₂O + energy.

Question 2:
The end products of anaerobic respiration is
(a) C₂H₅OH + energy
(b) C₂H₅OH + CO₂
(c) C₂H₅OH + CO₂ + energy
(d) C₂H₅OH + H₂O + CO₂

Question 3:
Cellular respiration occurs in
(a) cytoplasm
(b) mitochondria
(c) nucleus
(d) cytoplasm and mitochondria

Question 4:
Plants exchange gases through
(a) stomata
(b) lenticles
(c) both (a) and (b)
(d) none of these

Question 5:
To test the release of CO₂ gas during respiration, the chemical used is
(a) lime
(b) lime water
(c) calcium carbonate
(d) marble

Question 6:
Fermentation is the process used by yeast/bacteria. The end product is released and used in the following except for one.
(a) idli dough
(b) bread dough
(c) pakoda
(d) dosa dough.

Question 7:
To study the release of CO₂ by plants we should use
(a) a potted plant
(b) germinating seeds
(c) dry seeds
(d) boiled seeds.

Question 8:
If a thermometer is kept in a conical flask containing germinating seeds, the temperature of thermometer will
(a) increase
(b) decrease
(c) remain the same
(d) first increases then decrease.

Question 9:
At levels of atmospheric O₂ below 1%, the amount of CO₂ released is relatively high. This is probably because
(a) The Krebs cycle is very active.
(b) O₂ is being converted to H₂O.
(c) Alcohol fermentation is occuring.
(d) photosynthesis cannot function at night.

Question 10:
In the experiment demonstrating respiration in germinating seeds, KOH is used to
(a) absorb carbon dioxide produced by the seeds
(b) absorb oxygen present in the flask
(c) absorb water vapour released by the seeds
(d) liberate oxygen to be used by the seeds.

Question 11:
Why is some KOH placed in a small test tube in the flask with germinating seeds in the experiment to demonstrate the occurrence of respiration in germinating seeds?’
(a) To provide oxygen required by the seeds for respiration.
(b) To absorb carbon dioxide and create partial vacuum in the flask.
(c) To absorb water from the seeds to make them dry.
(d) To make the air present in the flask alkaline.

Question 12:
In the experiment to show that CO, is given out during respiration, the student uses:
(a) Lime water
(b) Alcohol
(c) KOH solution
(d) Iodine solution.

Question 13:
Seeds that are taken in the flask during respiration experiment must be
(a) just wet
(b) boiled
(c) dry
(d) germinating

Questions based on Observational Skills
Question 14:
A student sets up the apparatus for the experiment to show that CO, is released during respiration by germinating seeds. After 2 hours, he would observe.
(a) KOH turning milky
(b) Water level rising in the bent tube in the beaker.
(c) Water level decreasing in the bent tube in the beaker.
(d) Water turning turbid in the beaker.

Question 15:
In the following set-up which shows that “carbon dioxide is given out during respiration” the KOH kept in the flask

(a) makes the air in the flask alkaline
(b) creates partial vacuum in the flask
(c) absorbs moisture present in the flask.
(d) provides oxygen for respiration to the germinating seeds.

Question 16:
What is the use of KOH solution in this experiment?

(a) Absorb CO₂ released by germinating seeds
(b) Absorb O₂ released by germinating seeds
(c) Absorb moisture released by the seeds.
(d) None of these.

Question 17:
In the following experimental set-up to show that CO₂ is evolved during respiration, substances A and B are:

(a) A is water and B is lime water
(b) A is lime water and B is water
(c) A is CaSO₄ crystals and B is water
(d) A is KOH solution and B is water

Question 18:
After performing the experiment to show that germinating seeds give out carbon dioxide during respiration students drew the following diagram.

The correct labelled diagram is (a) A (b) B (c) C (d) D

Question 19:
The following experimental set-ups were kept in the laboratory to show CO₂ is given out during respiration

After two hours, students observed that water rises in the delivery tube
(a) only in set-up (A)
(b) only in set-up (B)
(c) in both (A) and (B)
(d) neither in (A) and nor in (B)

Questions based on Reporting and Interpretation Skills
Question 20:
The function of KOH in the experimental set-up to show that CO₂ is released during respiration is
(a) to enhance respiration
(b) to release oxygen for respiration
(c) to absorb CO₂ released by germinating seeds.
(d) to remove water vapour from the flask.

Question 21:
Respiration occurs only in living cells like germinating seeds because
(a) living cells need ATP
(b) living cells have glucose
(c) living cells have O₂
(d) all of these

Question 22:
An experimental set-up is given below to demonstrate that CO₂ is given out during respiration.

Four students made the following observations marked I, II, III, and IV.
I. Level of water remained the same in both the beaker and the delivery tube.
II. Level of water increased in delivery tube.
III. Level of water gets reduced in both the beaker and the delivery tube.
IV. Water ascends into the delivery tube and again descends.
Which one of the above is correct observation?
(a) I
(b) II
(c) III
(d) IV

Question 23:
The following experiment was set up to show that a gas is given out during respiration. But there was no rise in the level of water. This was because

(a) germinating seeds have not been kept under water in the flask
(b) water is kept in the beaker instead of lime water
(c) the cork on the flask is made of rubber
(d) no substance is kept in the flask to absorb the gas given out by the seeds.

Аnswers:

NCERT Class 10 Science Lab Manual Scoring Key With Explanation

(d) In presence of air, the end products are the same with energy.
(c) In absence of air, the respiration products are ethanol, CO₂ and less energy.
(d) Respiration in cells occur in cytoplasm and mitochondria.
(c) Both stomata on leaf and lenticels on stem/root helps in exchange of gases.
(b) Lime water test helps in identifying the CO₂ gas, by turning milky.
(c) Yeast grows in all except pakora.
(b) Germinating seeds respire more to release CO₂.
(a) The release of CO₂ is co-related with the release of heat energy.
(c) When O₂ is absent (or very low), anaerobic respiration (alcohol fermentation) is initiated. Alcohol fermentation releases CO₂. Photosynthesis, which would consume CO₂ to produce glucose, is obviously not occurring.
(a) KOH absorbs CO₂ from the air.
(b) To study the rise of water level in the bent tube, CO₂ needs to be absorbed.
(a) Lime water turns milky with CO₂.
(d) Only germinating seeds will respire more.
(b) The air in the tube is used up for respiration and hence the air pressure inside the tube is reduced.
(b) KOH absorbs the CO₂ released and the O₂ is used up by plant and hence, reduces the pressure in the bent tube.
(a) KOH combines with CO₂.
(d) KOH is kept near germinating seeds and the bent tube is dipped in water.
(d) KOH tube is kept in a flask with germinating seeds and bent tube is dipped in water.
(a) Cotton plug allows the air to pass through it.
(c) KOH reacts with CO₂.
(d) The respiration reaction occurs only in living cells where it is supplied with all the raw materials.
(b) The water level in the bent tube will rise because the pressure inside the tube is reduced.
(d) The gases exchanged remains in the tube and in the flask and hence the pressure will not change.

More Resources CBSE Class 10 Lab Manual Practical Skills:

NCERT Class 10 Science Lab Manual

Light is Necessary for Photosynthesis Experiment Class 10 Practical Science NCERT

CBSE Class 10 Biology Lab Manual Light is Necessary for Photosynthesis

Board	CBSE
Textbook	NCERT Books
Class	Class 10
Subject	Biology
Experiment Name	Light is Necessary for Photosynthesis
Category	Class 10 Science Lab Manual

The experiment to determine Light is Necessary for Photosynthesis are part of Science Lab Manual for Class 10 CBSE Experiments is designed to help students bridge the gap between theoretical concepts and practical applications through hands-on experiments in Biology.

Science Lab Manual Class 10 CBSE Light is Necessary for Photosynthesis Experiment

Determine Light is Necessary for Photosynthesis Class 10 Practical

Light Is Necessary For Photosynthesis Experiment Class 10 Introduction

Photosynthesis is a process by which plants prepare food. During this reaction, carbon dioxide and water are converted into glucose by chlorophyll in the presence of light energy.

NCERT Class 10 Science Solutions

Photosynthesis takes place in leaf cells. These contain chloroplasts, which are tiny objects containing chlorophyll.
Plants absorb water through their roots, and carbon dioxide through their leaves. They makes food in the form of glucose. Glucose is further converted into starch. Oxygen is the by-product released during photosynthesis.
The following events occur during the process of photosynthesis:
(a) Light is absorbed by the chlorophyll pigments present in the leaf.
(b) Light energy is converted into chemical energy and splits water into hydrogen and oxygen molecules.
(c) Carbon dioxide is reduced to carbohydrates.
The above step may not occur simultaneously in some plants. For example, desert plant take up CO₂ at night and prepare an intermediate which is acted upon by the chlorophyll during the day.
Destarching: When a plant is kept in darkness for about forty eight hours all the carbohydrates made and stored by the plants are utilised for providing energy to the plants. This process is called destarching.
Boiling of leaf in alcohol: The leaf to be tested for presence of starch is boiled in alcohol to remove the pigment chlorophyll otherwise it will interfere with the starch test. Alcohol is not heated directly over the flame because it will evaporate quickly without being much in contact with the leaves. It is therefore important to heat the test tube containing alcohol and leaves in water bath.
Starch test: On adding iodine solution to the starch, it forms starch-iodine complex which is blue-black in colour. When partially covered leaf is treated with iodine solution then the covered portion do not turn blue-black on adding iodine solution whereas the uncovered portion which received the light turns blue-black due to synthesis of starch in them.

NCERT Class 10 Science Lab Manual Experiment 2

Aim
To show experimentally that light is necessary for photosynthesis.
Theory

Plants prepare their food by the process called photosynthesis. To make food, plants need CO₂ water, chlorophyll and light/sunlight. In absence of any of these plants cannot prepare their food.
Plants can prepare their food in blue light.
The rate of photosynthesis depends on all three factors i.e.—light, temperature, availability of components, i.e.,— CO₂ and H₂0.
If the intensity of light increases the rate of photosynthesis also increases.
When light falls on plants they show light reaction. In this light reaction the water in leaves undergo photolysis
i.e.,—the water splits to form oxygen and hydrogen due to photons of light. The oxygen gas is released out in the atmosphere but hydrogen is kept by the plant. It is this hydrogen that combines with CO₂ to form carbohydrate (reduction reaction). Hence; photosynthesis is an oxidation-reduction reaction.
Photosynthesis reaction:

Materials Required
A healthy potted plant, beaker, a pair of forceps, tripod stand, wire gauze, bunsen burner, black paper, paper clips, iodine solution, alcohol, water bath etc.
Procedure I

Take a healthy potted plant and keep it in a dark room for 48 hours so that all the starch gets used up.
Now cover one leaf of a plant with a black paper using paper clip.
Keep this plant in sunlight for about six hours.
Pluck two leaves from the plant, one that is covered and the other one that is uncovered.
Dip the leaves in boiling water for a few minutes.
Now immerse the leaves in a beaker containing alcohol.
Carefully place this beaker in water bath and heat it till the alcohol begins to boil.
Observe the colour of the leaves and solution.
Wash the leaves with lot of fresh water.
Now dip the leaves in iodine solution for a few minutes.
Now observe the colour of leaves and compare them.

Observations

When leaves are boiled in alcohol, the alcohol solution becomes green and the leaves become colourless.
When iodine solution is added on the leaves
(a) a leaf covered with black paper showed no colour changes with iodine solution.
(b) another leaf which was not covered with black paper when dipped in dilute iodine solution, the colour of leaf changed to blue-black.

Inference

During photosynthesis plants prepare starch.
When the leaf is covered, it is not allowed to take sunlight and hence, no starch was prepared in the leaf.
Iodine solution becomes blue-black in presence of starch. On adding iodine solution to covered leaf no colour change was observed. It indicates that no starch was made by this leaf.
Whereas the uncovered leaf got sunlight for 6 hours and when iodine solution was added to it, the colour changed to blue-black.
This proves that sunlight is required for photosynthesis.

Procedure II

Select a potted plant, keep it in dark room for 48 hours.
Select a healthy leaf and clip a portion of it with dark colour paper using clips.
Keep this plant in sunlight for 6 hours.
Then do the same steps (4-11) as in procedure 1 on previous page.

Important Note

When you cover a portion of leaf with dark paper the results are not clearly visible. There is a possibility of translocation of food from uncovered leaf to a covered part of leaf.
The above experiment can be done by covering a portion of leaf with black paper.

Precautions

Select a small healthy, herbaceous potted plant.
Do not destarch the plant for more than 48 hours.
Choose a leaf and clip it carefully so that it does not break or crack from the stem.
Alcohol is highly inflammable, be careful while boiling leaf in alcohol using water bath.
Wash the alcohol from the leaves and then do the iodine test.
Satisfactory results will not be obtained if the plant is not completely de-starched.

Biology Practical Class 10 Viva Voce

Question 1:
If plant is kept in a room with lights on, can it perform photosynthesis?
Аnswer:
Yes, plants can prepare their food in any light i.e.-tube light, bulb-light, sunlight.

Question 2:
In which colour of light, plants can prepare their food best.
Аnswer:
Plants can prepare their food best in blue light.

Question 3:
In which part of the plant the photolysis reaction initiates?
Аnswer:
In chloroplast.

Question 4:
What is the original colour of iodine solution?
Аnswer:
Brown.

NCERT Class 10 Science Lab Manual Practical Based Questions

Question 1:
What are the raw materials required by plants for photosynthesis?
Аnswer:
Plants need light, carbon dioxide, water and chlorophyll pigments in leaf for photosynthesis.

Question 2:
What is the use of light in photosynthesis?
Аnswer:
In the presence of light, photolysis reaction takes place in a plant.
In photolysis reaction, the water decomposes to form hydrogen and oxygen gas.

Question 3:
In a plant if there are no leaves for a short period of time, i.e.,-Autumn, how do such plants survive?
Аnswer:
Plants store the food in roots, stems and this stored food is used/transported by plant to all the parts wherever the food is required.

Question 4:
Why do we boil the leaves in alcohol solution?
Аnswer:
Alcohol helps in the removal of colour pigment chlorophyll from the leaf which may otherwise interfere in the starch test, as starch test is the colour change test.

Question 5:
Why do we need to decolourise the leaf?
Аnswer:
To see the iodine colour change with starch.

Question 6:
Why should we always use water bath while boiling the leaf in alcohol solution?
Аnswer:
The alcohol is highly inflammable solution. It will catch fire if we heat the beaker containing alcohol directly on the flames of burner. To prevent such accidents, it is always advised to use water bath.

NCERT Class 10 Science Lab Manual Questions

Question 1:
What is meant by de-starching? Why do plants get de-starched when kept in continuous darkness for about forty eight hours?
Аnswer:
The removal of all the starch present in plant is called de-starching. When plants are kept in continuous darkness for about forty eight hours, all the starch present in them is used up for various biological processes and no new starch is formed due to absence of light.

Question 2:
Will you get the same result if you perform the experiment without de-starching the plant? Give reason.
Аnswer:
The variable we are testing here is light that affects photosynthesis and the product formed is starch. If the starch is already present in the plant, the experiment will not give the fair test.

Question 3:
Why do we warm the leaves in alcohol?
Аnswer:
Alcohol helps in the removal of colour pigment chlorophyll from the leaf which may otherwise interfere in the starch test, as starch test is the colour change test.

Question 4:
Arrange the following steps in correct sequence:
(i) de-starching the plant;
(ii) treatment with iodine;
(iii) attaching black paper strips to the leaf;
(iv) keeping the set-up in sunlight;
Аnswer:
The correct order is: (i), (iii), (iv), (ii).

Question 5:
Why do we keep the experimental plant in bright sunlight?
Аnswer:
After destarching, we keep the plant in the bright sunlight so that it undergoes photosynthesis to produce starch.

Question 6:
Can this experiment be performed with a de-starched leaf detached from the plant? Give reasons.
Аnswer:
The leaf cannot undergo photosynthesis if it is detached from the plant and the experiment will not give fair test.

NCERT Class 10 Science Lab Manual Multiple Choice Questions (MCQs)

Questions based on Procedural and Manipulative Skills
Question 1:
The first step in photosynthesis is
(a) convert light energy into chemical energy
(b) reduction of C0₂ gas to carbohydrates
(c) photolysis of water
(d) absorption of light energy by chlorophyll.

Question 2:
In desert plants, the first step of photosynthesis can be
(a) absorption of sunlight
(b) photolysis of water
(c) intake of C0₂ at night
(d) convert light energy into chemical energy.

Question 3:
In an experiment to show that sunlight is necessary for photosynthesis, the leaf is boiled in alcohol for a few minutes using a water bath. It is essential because:
(a) alcohol is highly volatile
(b) the steam from water bath heats the leaf rapidly
(c) steam from water bath dissolves the chlorophyll
(d) alcohol is flammable.

Question 4:
Before testing the leaf for starch at the end of the experiment, “light is necessary for photosynthesis”, the experimental leaf, should be boiled in
(a) water
(b) alcohol
(c) KOH solution
(d) hydrochloric acid.

Question 5:
For the experiment that, “light is necessary for photosynthesis” the potted plant is first kept in darkness for a day. This is to:
(a) deactivate the chloroplast
(b) destarch leaves
(c) activate chloroplasts
(d) prepare leaves for photosynthesis.

Question 6:
The steps, necessary for setting up the experiment “To demonstrate that light is necessary for photosynthesis” are not given here in proper sequence. The correct order is:
I. keep the potted plant in sunlight for 3 to 4 hours
II. keep the potted plant in darkness for about 48 hours
III. cover a leaf of the plant with a strip of black paper
IV. pluck the leaf and test it for starch.
(a) I, III, IV, II
(b) I, IV, III, II
(c) II, IV, III, I
(d) II, III, I, IV.

Question 7:
Before carrying out the test for the presence of starch in a leaf-on exposure to sunlight, the leaf is put into alcohol contained in a beaker and boiled over a water bath. This step is carried out to
(a) extract starch
(b) dissolve chlorophyll
(c) allow water to move into the leaf
(d) make membrane of leaf cells more permeable.

Question 8:
A student performed the starch test on a leaf, some steps involved are shown below:
The correct sequence of steps should be:

(a) (iv), (iii), (ii), (i)
(b) (i), (ii), (iii), (iv)
(c) (ii), (iii), (iv), (i)
(d) (i), (iii), (iv), (ii).

Question 9:
What is the right procedure to remove chlorophyll from a destarched leaf?
(a) Boil the destarched leaf in lime water.
(b) Boil the destarched leaf in alcohol.
(c) Boil the destarched leaf in water only.
(d) Boil the destarched leaf in a mixture of alcohol and water.

Questions based on Observational Skills
Question 10:
A potted plant is kept in different coloured lights. The best rate of photosynthesis is seen in
(a) green light
(b) blue light
(c) white light
(d) yellow light.

Question 11:
When leaf is boiled with ethanol and treated with iodine solution, its colour changes into:
(a) pink
(b) blue
(c) blue-black
(d) black.

Question 12:
The best result of the experiment that light is necessary for photosynthesis would be yielded by using leaves from a plant kept for over twenty four hours:
(a) in a pitch dark room
(b) in a dark room with table lamp switched on
(c) outside in the garden
(d) outside in the garden covered by glass case.

Question 13:
The figure that correctly depicts the removal of chlorophyll is:

Question 14:
A leaf from a destarched plant is covered with black paper strip as shown in figure I. The starch test is done on the leaf after 8 hours of exposure to light.

The result will be as shown in diagram:
(a) A
(b) B
(c) C
(d) D

Questions based on Reporting and Interpretation Skills
Question 15:
The best rate of photosynthesis is in blue light because:
(a) the leaves absorb maximum blue light
(b) the leaves reflect maximum blue light
(c) the blue light has maximum wavelength
(d) the blue light stimulates the chloroplast maximum.

Question 16:
A potted plant was kept in dark room for 3 days and then 4 leaves were covered with different coloured papers as shown below.

The leaves were tested for starch, the leaf which showed no starch in it is
(a) the leaves covered with red, green and blue strips
(b) the leaves covered with green and black strips
(c) the leaf covered with green strip
(d) the leaf covered with any of the above strips.

Question 17:
On completion of the experiment to demonstrate that
“light is necessary for photosynthesis”, four students reported the inference as follows. Identify the correct inference.
(a) Part of the leaf covered with strip can only undergo photosynthesis
(b) Uncovered parts of the leaf cannot synthesises starch.
(c) Photosynthesis takes place only in the presence of sunlight
(d) Light is necessary for photosynthesis of starch in green plants.

Question 18:
Given below is a sketch of a leaf partially covered with black paper and which is to be used in the experiment to show that light is compulsory for the process of photosynthesis. At the end of the experiment, which one of the leaf parts labelled I, II and III will become blue-black when dipped in iodine solution?

(a) I only
(b) II only
(c) III only
(d) I and III only.

Question 19:
In an experiment of photosynthesis, a student fixed a strip of black paper on the dorsal surface of leaf in the morning, in the evening he tested the leaf for starch. The result was:
(a) The dorsal surface of the leaf was white but the ventral surface turned blue.
(b) Both the surfaces of the leaf were white.
(c) The entire leaf turned blue-black.
(d) The entire leaf remained white.

Question 20:
A part of a destarched leaf of a potted plant was covered with black paper strips on both sides and the plant was kept in sunlight for 8 hours. The leaf was then tested with iodine after boiling it in a alcohol. Only the uncovered part of the leaf turned blue-black. The inference is that
(a) C0₂ is necessary for photosynthesis
(b) light is necessary for photosynthesis
(c) chlorophyll is necessary for photosynthesis
(d) water is necessary for photosynthesis.

Аnswers:

NCERT Class 1o Science Lab Manual Scoring Key With Explanation

(d) The light energy is absorbed to initiate the reaction.
(c) To avoid the loss of water during day times, the stomata will not open.
(d) Water bath prevents the direct heating of alcohol which is highly inflammable and can catch fire.
(b) Alcohol helps in removing the chlorophyll.
(b) Destarching helps in removing the starch from the leaf.
(d) It is the right procedure for the experiment.
(b) As starch test is colour change test, hence coloured pigment chlorophyll needs to be removed.
(d) It is the correct procedure.
(b) Alcohol helps in dissolving the chlorophyll without affecting the other cells.
(b) Blue light is better absorbed among the other given lights (red is the best).
(c) Starch test is iodine solution changes blue-black with starch.
(a) This helps in de-starching the leaves.
(d) Leaf should be in alcohol and this container should be heated in water.
(b) The covered part of leaf will not have starch and will not show colour change.
(d) The blue light is absorbed the best by chloroplast.
(d) Once the leaf is covered by paper, it is deprived of light and no photosynthesis occurs in it.
(d) Green plants can prepare starch by photosynthesis.
(d) I and III parts of leaf gets light and hence starch is made but II part gets no light.
(c) Covering the dorsal side of leaf will not deprive it from light.
(b) Light is must for photosynthesis as it initiates the reaction in cells.

More Resources CBSE Class 10 Lab Manual Practical Skills:

NCERT Class 10 Science Lab Manual

Dicot Seed Experiment Class 10 Practical Science NCERT

CBSE Class 10 Biology Lab Manual Dicot Seed

Board	CBSE
Textbook	NCERT Books
Class	Class 10
Subject	Biology
Experiment Name	Stomata
Category	Class 10 Science Lab Manual

The experiment to determine Dicot Seed are part of Science Lab Manual for Class 10 CBSE Experiments is designed to help students bridge the gap between theoretical concepts and practical applications through hands-on experiments in Biology.

Science Lab Manual Class 10 CBSE Dicot Seed Experiment

Determine Dicot Seed Class 10 Practical

Dicot Seed Experiment Class 10 Introduction

Seed: A seed is formed by the fertilized ovule and pollen egg. It contains, reserve food and protective coat. When the seed is sown in soil/ kept in soaked wet cotton a new plant appears from the embryo.

CBSE Class 10 Biology Solutions

The seed consists of an outer coat called testa. Enclosed in it is the stored food and an embryo plant. The food may be stored in the cotyledon or around the embryo on the endosperm.
Monocot seed: The seed having only one cotyledon is called monocot seed. For e.g., maize and wheat seed.
Dicot seed: The seed having two cotyledons is called dicot seed. For e.g., gram and bean seed.
On the basis of the presence or absence of endosperm, both in dicot and monocot seed, they are classified as endospermic or non-endospermic seeds.
Endospermic seed:
(a) Seeds are with endosperm.
(b) Food is stored in the endosperm.
(c) Seed’s may be of monocot or dicot.
E.g. Castor, maize
(d) Cotlyledons are thin like paper.
Non-endospermic seed:
(a) Seeds are without endosperm.
(b) Food is stored in the cotyledon.
(c) Seeds are of dicot type.
E.g. Pea, bean
(d) Cotyledons are thick and fleshy.
Structure of dicot seed (bean seed): The seed coats have characteristic colours. When the seeds are soaked in water, they swell considerably and the seed coats become soft. In this condition the seed coats are easily removed. The cotyledons or seed leaves, can be easily seen.
Structure of monocot seed (Maize seed): Each grain is made up of following^parts:
1. Seed coat: It is the outer brownish layer of the grain. In this, seed and fruit wall are fused together.
2. Endosperm: It comprises the major part of grain and is filled with reserve food. It is composed of two regions:
(a) Outer single layer is made up of proteins.
(b) Inner starchy endosperm. It is separated from embryo by a layer called epithelium.
3. Embryo: It contains a single lateral cotyledon and embryo axis with plumule and radicle are at its two ends.

CBSE Class 10 Biology Lab Manual Experiment 6

Aim
To identify the different parts of an embryo of a dicot seed (pea, gram or red kidney bean).
Theory

Seed: Seed is a small embryonic plant present in a safe coating of seed coat, it stores food.
Seed formation: The male gamete of plant, i.e., pollen grains and female gamete of a plant, i.e., ovules fuse together to form seed. The seed formation takes place due to fertilization, and it is the product of reproduction in plants. The embryo of seed is formed from the zygote.
Food in seed: The food is stored in the cotyledons of embryo in some plants and in the endosperm, a special tissue outside the embryo in other plants.
Three basic parts of a seed:
1. An embryo
2. Nutrient for embryo
3. Seed coat.
Embryo: The embryo of seed is an immature plant from which a new plant can grow.
The radicle that comes out of the embryo is the embryonic root. The plumule is the embryonic shoot.
Cotyledons: It is the seed leaf present in seed. If the embryo has one seed leaf it is monocotyledon and if it has two seed leaves it is dicotyledon.
Epicotyl: The part of the embryonic stem above the point of attachment of the cotyledon is the epicotyl.
Hypocotyl: The area between the radicle and the place of origin of cotyledons is termed as hypocotyl.
Nutrients for the Embryo: Seed stores nutrients for the growth of an embryo during germination. The nutrients/ stored food is in the form of oil, fat and protein.
Seed Coat: The seed coat protects the embryo from mechanical injury and from drying out. It can be a paper thin as in case of peanut or may be very thick e.g. coconut.

Materials Required
Water soaked seeds of pea, gram or red kidney beans, petridish, forcep, needle, brush and simple microscope and slide.
Procedure

Take 8-10 soaked seeds of pea/gram/red kidney beans, place them on wet cotton in petridish overnight. The seed coat becomes soft which helps in the opening of the seeds.
With the help of forcep, slowly remove the seed coat and study different parts of seed embryo.
Now, slowly remove the embryo axis with needle and place it on the slide.
Observe these three parts of the seed obtained, record your observations and draw diagrams.

Observations

The seed has a small pore called micropyle.
It is a dicot seed, i.e., the seed has two cotyledons.
The embryo axis shows radicle and plumule, (as shown in the figure), the radicle is future root and the plumule is future shoot.
The food is stored in cotyledons.

Conclusion
The different parts of an embryo of a dicot seed were identified as plumule (future shoot), radicle (future root), seed coat (outer covering) and cotyledons (food store)
Precautions

The best quality seeds should be used for study.
Soak the seeds overnight to make the seed coat soft.
Observe the parts under simple microscope/lens and record your observations.
Remove the seed coat very gently.

Biology Practicals For Class 10 CBSE Viva Voce

Question 1:
What is cotyledon?
Аnswer:
Cotyledon is the seed leaf.

Question 2:
What is epicotyl?
Аnswer:
The part of the embryonic stem above the point of attachment of the cotyledon is called epicotyl.

Question 3:
What is hypocotyl?
Аnswer:
The area between the radicle and the place of origin of cotyledon is termed as hypocotyl.

Question 4:
Name two dicot seeds.
Аnswer:
Pea, gram.

Question 5:
Differentiate between monocot and dicot seeds.
Аnswer:
Monocot seeds are made up of one cotyledon. Dicot seeds have two cotyledons.

CBSE Class 10 Biology Practical Based Questions

Question 1:
Name two types of seed.
Аnswer:
Moncot and dicot.

Question 2:
Name two monocot seeds.
Аnswer:
Maize and wheat.

Question 3:
Name three parts of seed.
Аnswer:
The seed has seed coat, micropyle and cotyledon.

Question 4:
What is the function of endosperm in the seed?
Аnswer:
Endosperm stores food/nutrients in the seed.

Question 5:
Name three different forms of food storage in endosperm/seed.
Аnswer:
Food in seed is stored in the form of oil, fat and proteins.

Question 6:
What is present on embryo axis?
Аnswer:
The embryo axis has plumule and radicle.

Question 7:
What do plumule and radicle grow into?
Аnswer:
Plumule grows into shoot and radicle grows into root.

Question 8:
When we open a dicotyledon seed, then its embryo shows two parts. Name these two parts and write their functions. [Outside Delhi 2011]
Аnswer:
Two parts of the embryo are: Plumule and Radicle.
Plumule after germination develops into shoot of the new baby plant.
Radicle after germination develops into the root of the new baby plant.

Question 9:
What are cotyledons? How are the number of cotyledons different in gram seed and the maize (com) seed?
[Outside Delhi 2012]
Аnswer:
Cotyledons are special structures in the seeds of the plants. Cotyledons store food for the embryo of the plant.
In gram seeds, there are two cotyledons and in maize (corn) seeds, there is only one cotyledon.

CBSE Class 1o Biology Lab Manual Multiple Choice Questions (MCQs)

Questions based on Procedural and Manipulative Skills
Question 1:
Food is stored in the seed in the form of
(a) oil
(b) protein
(c) fats
(d) all of these

Question 2:
The condition needed by dormancy is
(a) exposure to heat
(b) Exposure to moisture
(c) exposure to cold
(d) Sowing in soil

Question 3:
The leaves of the seed is called
(a) epicotyl
(b) hypocotyl
(c) micropyle
(d) cotyledon

Question 4:
The small pore, through which water enters in few
(a) cotyledon
(b) plumule
(c) radicle
(d) micropyle

Question 5:
The future root present on embryonic axis in the seed is called
(a) radicle
(b) plumule
(c) epicotyl
(d) cotyledon

Question 6:
The future shoot present on embryonic axis in the seed is called
(a) radicle
(b) plumule
(c) epicotyl
(d) cotyledon

Question 7:
The baby plant inside the seed is called
(a) cotyledon
(b) ovule
(c) pollen
(d) embryo

Question 8:
When the pollen unites with ovule to form seed it is called
(a) germination
(b) pollination
(c) fertilization
(d) reproduction

Question 9:
Which of the following flower parts develop into a fruit after pollination?
(a) Ovule
(b) Ovary
(c) Stamen
(d) Stigma

Question 10:
What is the name given to the food storage structure that surrounds the embryo in a dicot seed?
(a) Cotyledon
(b) Plumule
(c) Radicle
(d) Ovary

Questions based on Observational Skills
Question 11:
It is the thin layer of seed called

(a) seed coat
(b) embryo
(c) micropyle
(d) endosperm

Question 12:
The labels(1)and(2) in the alongside figure ovule

(a) (1) radicle (2) endosperm (3) cotyledon
(b) (1) plumule (2) endosperm (3) radicle
(c) (1) plumule (2) cotyledon (3) radicle
(d) (1) plumule (2) seed coat (3) radicle.

Questions based on Reporting and Interpretation Skills
Question 13:
A seed develops from
(a) pollen grain
(b) ovule
(c) ovary
(d) fertilized ovule

Question 14:
The label (I) in pea seed is

(a) cotyledon
(b) micropyle
(c) seed coat
(d) plumule

Question 15:
The correct labels (1), (2) and (3) made by a student while studying different parts of gram seed are

(a) (1) radicle (2) endosperm (3) cotyledon
(b) (1) plumule (2) endosperm (3) radicle
(c) (1) plumule (2) cotyledon (3) radicle
(d) (1) plumule (2) seed coat (3) radicle.

Question 16:
In the figure, the parts marked A, B and C are sequentially: [Delhi 2013]

(a) Plumule, Radicle and Cotyledon
(b) Radicle, Plumule and Cotyledon
(c) Plumule, Cotyledon and Radicle
(d) Radicle, Cotyledon and Plumule

Question 17:
In the following diagram showing the structure of embryo of a dicot seed, what are the parts marked I, II and III sequentially? [Outside Delhi 2014]

(a) Plumule, Cotyledon, Radicle
(b) Plumule, Radicle, Cotyledon
(c) Cotyledon, Plumule, Radicle
(d) Radicle, Plumule, Cotyledon

Аnswers:

CBSE Class 10 Biology Lab Manual Scoring Key With Explanation

(d) Stored forms of food.
(b) Moisture helps in softening of seed coat and further initiates the outgrowth from seed.
(d) Cotyledons are called seed leaves.
(d) Micropyle is the pore in seeds.
(a) Radicle becomes root.
(b) Plumule becomes shoot.
(d) Embryo is the first stage of development.
(c) Fertilization is the fusion of male and female gametes.
(b) Ovary becomes fruit and fertilized ovule becomes a seed.
(a) It is present in seed as leaves structure.
(a) Seed coat protects the seed.
(c) Radicle is longer and grows towards gravity.
(d) Ovules are fertilized by pollens to form zygote and then changed into a seed.
(b) A tiny hole in the testa is called the micropyle.
(c) The leaves are cotyledon, root grows down towards gravity.
(a) Option (a) has correct labels.
(a) Correct parts are labelled in option (a).

More Resources CBSE Class 10 Lab Manual Practical Skills:

NCERT Class 10 Science Lab Manual

MCQ Questions	NCERT Solutions
CBSE Sample Papers
NCERT Exemplar Solutions	LCM and GCF Calculator
TS Grewal Accountancy Class 12 Solutions
TS Grewal Accountancy Class 11 Solutions

Lab Manual Class 10 Biology

Science Lab Manual Class 10 CBSE Homology and Analogy of Plants and Animals Experiment

Determine Homology and Analogy of Plants and Animals Class 10 Practical

CBSE Class 10 Science Lab Manual Experiment – 5

CBSE Class 10 Science Lab Manual Viva Voce

CBSE Class 10 Science Lab Manual Practical Based Questions

CBSE Class 10 Science Lab Manual Multiple Choice Questions (MCQs)

CBSE Class 1o Science Lab Manual Scoring Key With Explanation

Science Lab Manual Class 10 CBSE Stomata Experiment

Determine Stomata Class 10 Practical

NCERT Class 10 Science Lab Manual Experiment – 1

NCERT Class 10 Science Lab Manual Viva Voce

NCERT Class 10 Science Lab Manual Practical Based Questions

NCERT Class 10 Science Lab Manual Questions

NCERT Class 10 Science Lab Manual Multiple Choice Questions (MCQs)

NCERT Class 10 Science Lab Manual Scoring Key With Explanation

Science Lab Manual Class 10 CBSE CO2 is Released During Respiration Experiment

Determine CO2 is Released During Respiration Class 10 Practical

NCERT Class 10 Science Lab Manual Experiment 3

Respiration Practical Class 10 Viva Voce

NCERT Class 10 Science Lab Manual Practical Based Questions

NCERT Class 10 Science Lab Manual Questions

NCERT Class 10 Science Lab Manual Multiple Choice Questions (MCQs)

NCERT Class 10 Science Lab Manual Scoring Key With Explanation

Science Lab Manual Class 10 CBSE Light is Necessary for Photosynthesis Experiment

Determine Light is Necessary for Photosynthesis Class 10 Practical

NCERT Class 10 Science Lab Manual Experiment 2

Biology Practical Class 10 Viva Voce

NCERT Class 10 Science Lab Manual Practical Based Questions

NCERT Class 10 Science Lab Manual Questions

NCERT Class 10 Science Lab Manual Multiple Choice Questions (MCQs)

NCERT Class 1o Science Lab Manual Scoring Key With Explanation

Science Lab Manual Class 10 CBSE Dicot Seed Experiment

Determine Dicot Seed Class 10 Practical

CBSE Class 10 Biology Lab Manual Experiment 6

Biology Practicals For Class 10 CBSE Viva Voce

CBSE Class 10 Biology Practical Based Questions

CBSE Class 1o Biology Lab Manual Multiple Choice Questions (MCQs)

CBSE Class 10 Biology Lab Manual Scoring Key With Explanation

Footer

Maths NCERT Solutions

SCIENCE NCERT SOLUTIONS