11th | Aster Classes

Balbharati solutions, for, Biology, 11th, Standard, Maharashtra, State Board, Chapter 8, Plant Tissues, and Anatomy, Exercise, [Pages 95 – 96],

Exercise | Q 1. (A) | Page 95

Choose the correct option.

Location or position of meristematic regions is divided into ______ types.

one

two

three

none of the above

SOLUTION

three


Cambium is also called ______.

apical meristem

intercalary meristem

lateral meristem

none of the above

SOLUTION

lateral meristem


Collenchyma is a type of ______ tissue.

living

dead

living and dead

none of the above

SOLUTION

living


______ is a complex permanent tissue.

Parenchyma

Sclerenchyma

Chlorenchyma

Xylem

SOLUTION

Xylem


Mesophyll tissue is present in ______.

root

stem

leaf

flower

SOLUTION

leaf


Exercise | Q 2. (A) | Page 95

Answer the following question.

A fresh section was taken by a student but he was very disappointed because there were only a few green and most colourless cells. Teacher provided a pink colour solution. The section was immersed in this solution and when observed it was much clearer. What is magic?

SOLUTION

The pink coloured solution given by teacher must be a safranin stain. Safranin is used to stain plant tissues, especially lignified tissues such as cell wall and xylem.


While observing a section, many scattered vascular bundles could be seen. Teacher said, in spite of this large number the stem cannot grow in girth. Why?

SOLUTION

1. Students must have observed monocot stems.

2. It is because, monocot stem shows scattered vascular bundles.

3. In monocot stem, vascular bundles are closed i.e. without cambium.

4. Thus, secondary growth does not occur which is required for an increase in girth. Hence, in spite of having large number of scattered vascular bundles, monocot stems do not grow in girth.


A section of the stem had vascular bundles, where one tissue was wrapped around the other. How will you technically describe it?

SOLUTION

Concentric vascular bundle:

1. When one vascular tissue is completely encircling the other, it is called a concentric vascular bundle.

2. When phloem is encircled by xylem, it is called a leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called a hadrocentric vascular bundle.

3. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called an amphivasal vascular bundle.


There were two cut logs of wood lying in the campus. One had growth rings and other didn’t. Teacher said it is due to differences in their pattern of growth which is dependent on the season. How?

SOLUTION

1. It is possible that one of the cut logs was of a tropical tree, whereas the other was of a temperate tree. Since tropical trees grow in a similar manner all year, growth rings are not apparent. Another explanation for this could be that the log which had growth rings must be of an old tree which has experience many seasons, whereas the log without growth rings must be of younger tree, that has not been subjected to seasonal changes and hence not developed prominent growth rings.

2. Growth rings are formed due to cambial activity during favourable and non-favourable climatic conditions.

3. During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, a lumen is narrow and walls are thick with abundant fibres, high density.

4. Spring wood and autumn wood that appears as alternate light and dark concentric rings, constitute an annual ring or growth ring.

5. These growth rings can be used to estimate the age of the tree. These are found more in older trees as compare to a younger tree.


While on the trip to Kashmir, Pintoo observed that cut portions of large trees show distinct rings, which he never found in Maharashtra. Why is so?

SOLUTION

1. Cut portions of large tress show distinct rings which are annual rings formed due to activity of cambium during favourable and non-favourable climatic conditions.

2. Kashmir falls under the temperate region where the climatic conditions are not uniform through the year. In the spring season, conditions are favourable due to which cambium is active, whereas, in the autumn season, conditions are unfavourable due to which cambium is less active. This leads to the formation of spring wood and autumn wood that appears as alternate light and dark concentric rings, constitute an annual ring or growth ring.

3. Maharashtra falls under the tropical region where climatic conditions are favourable throughout the year. In tropical areas, the continuous growth of secondary xylem occurs. Thus, trees growing in tropical regions show less or no annual rings as compared to trees in the temperate region.


A student was observing a slide with no label under a microscope. The section had some vascular bundles scattered in the ground tissue. It is a section of a monocot stem! He exclaimed. No! it is a section of fern rachis, said the teacher. Teacher told to observe vascular bundle again. Student agreed, Why?

SOLUTION

1. In fern rachis, the number of vascular bundles is less as compared to the number of vascular bundles in monocot stem. In monocot stem, vascular bundles are numerous.

2. In fern rachis, xylem consists of only tracheids whereas in monocot stem, xylem consists of vessels (protoxylem and metaxylem) as well as tracheids. Monocot stem shows the presence of lysigenous cavity just below protoxylem.

3. In fern rachis, phloem consists of only sieve cells whereas in monocot stem, phloem consists of sieve tubes and companion cells.

Thus, a student must have observed these differences in the given section and agreed to teacher’s statement that the given section is of fern rachis and not of monocot stem.


Student found a wooden stopper in lab. He was told by an old lab attendant that it is there for many years. He kept thinking how it did not rot?

SOLUTION

1. The wooden stopper or cork is obtained from the phellem (cork) part of a bark.

2. Phellem (cork) is impervious in nature and does not allow entry of water due to suberized walls.

3. Due to this it does not rot and remains as it is for many years.


Student while observing a slide of leaf section observed many stomata on the upper surface. He thought he has placed slide upside down. Teacher confirmed it is rightly placed. Explain.

SOLUTION

1. In a dicot leaf, stomata are generally absent on the upper epidermis but are present on the lower epidermis. Thus, the student must have thought that he has placed slide upside down.

2. According to teacher, the section was placed rightly, thus the given section must be of monocot leaf.

3. It is because in monocot leaf stomata are present on both upper and lower epidermis.


Exercise | Q 3. (A) | Page 95

Write a short note on

Structure of stomata.

SOLUTION

1. Small gateways in the epidermal cells are called stomata.

2. Stoma is controlled or guarded by specially modified cells called guard cells.

3. These guard cells may be kidney-shaped (dicot) or dumbbell-shaped (monocot), collectively called as stomata.

4. Guard cells have chloroplasts to carry out photosynthesis.

5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables the exchange of gases and water vapour.

6. Stomata are further covered by subsidiary cells.

7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.


secondary growth.

SOLUTION

1. Dicotyledonous plants and gymnosperms exhibit increase in girth of root and stem.

2. In dicot stem, secondary growth begins with the formation of a continuous cambium ring.

3. The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.

4. The cells of medullary rays adjoining these intrafascicular cambium strips become meristematic (regain the capacity to divide) and form the interfascicular cambium.

5. Thus, a complete and continuous ring of vascular cambium is formed.

6. The cambium ring cuts off new cells, towards both inner and outer sides.

7. The cells that are cut-off towards pith (inner side) mature into secondary xylem and cells that are cut-off towards the periphery mature into secondary phloem.

8. Generally, amount of secondary xylem is more than the secondary phloem.


peculiarity of a sclerenchyma cell wall.

SOLUTION

1. Cell wall of sclerenchyma is evenly thickened due to uniform deposition of lignin.

2. Cell wall of sclereids is extremely thick and strongly lignified.


Differentiate between Vascular bundle of monocot and Vascular bundle of dicot.

SOLUTION


Differentiate between Xylem and Phloem functioning

SOLUTION


Distinguish Between Internal or anatomical difference between monocots and dicots.

SOLUTION


Exercise | Q 5. (A) | Page 96

Draw a neat labelled diagram.

T. S. of Dicot leaf

SOLUTION

T. S. of Monocot root

SOLUTION

T. S. of dicot stem

SOLUTION


Exercise | Q 6 | Page 96

Write the information related to the diagram given below.

SOLUTION

1. Small gateways in the epidermal cells are called as stomata.

2. Stoma is controlled or guarded by specially modified cells called guard cells.

3. These guard cells may be kidney-shaped (dicot) or dumbbell-shaped (monocot), collectively called as stomata.

4. Guard cells have chloroplasts to carry out photosynthesis.

5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.

6. Stomata are further covered by subsidiary cells.

7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.


Write the information related to the diagram given below.

SOLUTION


Exercise | Q 7. (a) | Page 96

Identify the following diagram, label it and prepare a chart of characteristics.

1. Epiblema:

It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.

2. Cortex:

It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.

3. Exodermis:

After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:

The innermost layer of cortex is called Endodermis. The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele:

It consists of pericycle, vascular bundles and pith.

a. Pericycle:

Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.

b. Vascular bundle:

Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc. Connective tissue: A parenchymatous tissue is present in between xylem and phloem.

c. Pith:

The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.

6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.


Identify the following diagram, label it and prepare a chart of characteristics.

SOLUTION

1. It is a type of simple permanent tissue.

2. Cells in this tissue are thin walled, isodiametric, round, oval to polygonal or elongated in shape.

3. Cell wall is composed of cellulose.

4. Cells are living with prominent nucleus and cytoplasm with large vacuole.

5. Parenchyma has distinct intercellular spaces. Sometimes, cells may show compact arrangement.

6. The cytoplasm of adjacent cells is interconnected through plasmodesmata and thus forms a continuous tissue.

7. This is less specialized permanent tissue.

8. Occurrence:

These cells are distributed in all the parts of a plant body viz. epidermis, cortex, pericycle, pith, mesophyll cells, endosperm, xylem and phloem.

9. Functions:

These cells store food, water, help in gaseous exchange, increase buoyancy, perform photosynthesis and different functions in plant body.

10. Dedifferentiation in parenchyma cells develops vascular cambium and cork cambium at the time of secondary growth.


Identify the following diagram, label it and prepare a chart of characteristics.

SOLUTION

Structure of stomata:

1. Small gateways in the epidermal cells are called as stomata.

2. Stoma is controlled or guarded by specially modified cells called guard cells.

3. These guard cells may be kidney shaped (dicot) or dumbbell shaped (monocot), collectively called as stomata.

4. Guard cells have chloroplasts to carry out photosynthesis.

5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.

6. Stomata are further covered by subsidiary cells.

7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.


Identify the following diagram, label it and prepare a chart of characteristics.

SOLUTION

Structure of dorsiventral leaf:

The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:

Upper epidermis:

It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.

Mesophyll:

Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.

1. Palisade parenchyma:

Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.

2. Spongy parenchyma:

Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.

Vascular system:

It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.

Lower epidermis:

It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.


Exercise | Q 8 | Page 96

Distinguish between dicot and monocot leaf on the basis of following characters.

SOLUTION


Chapter 7: Cell Division

Balbharati solutions, for, Biology, 11th, Standard, fyjc, Maharashtra, State Board, Chapter 7, Cell Division, Exercise, [Pages 83 – 84],

Exercise | Q 1. (A) | Page 83

Choose the correct option.

The connecting link between Meiosis-I and Meiosis-II is ______.

interphase – I

interphase – II

interkinesis

anaphase – I

SOLUTION

interkinesis


Synapsis is pairing of ______.

any two chromosomes

non-homologous chromosomes

sister chromatids

homologous chromosomes

SOLUTION

homologous chromosomes


Spindle apparatus is formed during which stage of mitosis?

Prophase

Metaphase

Anaphase

Telophase

SOLUTION

Metaphase


Chromosome number of a cell is almost doubled up during ______.

G 1 – phase

S – phase

G 2 – phase

G 0 – phase

SOLUTION

S – phase


How many meiotic divisions are necessary for the formation of 80 sperms?

80

40

20

10

SOLUTION

20


How many chromatides are present in anaphase-I of meiosis-I of a diploid cell having 20 chromosomes?

4

6

20

40

SOLUTION

40


In which of the following phase of mitosis chromosomes are arranged at the equatorial plane?

Prophase

Metaphase

Anaphase

Telophase

SOLUTION

Metaphase


Find incorrect statement.

Condensation of chromatin material occurs in prophase.

Daughter chromatids are formed in anaphase

Daughter nuclei are formed at metaphase.

Nuclear membrane reappears in telophase

SOLUTION

Daughter nuclei are formed at metaphase


Histone proteins are synthesized during ______.

G 1 phase

S – phase

G 2 phase

Interphase

SOLUTION

S – phase


Exercise | Q 2. (A) | Page 83

Answer the following question.

While observing a slide, student observed many cells with nuclei. But some of the nuclei were bigger as compared to others but their nuclear membrane was not so clear. Teacher inferred it as one of the phase in the cell division. Which phase may be inferred by the teacher?

SOLUTION

Prophase


Students prepared a slide of onion root tip. There were many cells seen under a microscope. There was a cell seen under a microscope. There was a cell with two groups of chromosomes at opposite ends of the cell. This cell is in which phase of mitosis?

SOLUTION

Anaphase


Students were shown some slides of cancerous cells. Teacher made a comment as if there would have been control at one of its cell cycle phase, there wouldn’t have been a condition like this. Which phase the teacher was referring to?

SOLUTION

The phase teacher was referring would be G 1 phase.


Some Mendelian crossing experimental results were shown to the students. Teacher informed that there are two genes located on the same chromosome. He enquired if they will be ever separated from each other?

SOLUTION

1. Genes are located on chromosomes at specific distances and positions.

2. The greater this distance, the greater the chance that a crossover can occur between the genes and the greater the chances of recombination.

3. The chances of recombination are less between the genes that are placed closed to each other on the chromosome.

4. Therefore, due to recombination, the two genes located on the same chromosome have the possibility of separating from each other.


Students were observing a film on Paramoecium. It underwent a process of reproduction. Teacher said it is due to cell division. But students objected and said that there was no disappearance of the nuclear membrane and no spindle formation, how can it be cell division? Can you clarify?

SOLUTION

1. Paramoecium is a unicellular organism. The division in Paramoecium occurs by amitosis.

2. It is the simplest mode of cell division.

3. In amitosis, nucleus elongates and a constriction appears. This constriction deepens and divides the nucleus into two daughter nuclei followed by the division of cytoplasm.


Exercise | Q 2. (F) | Page 84

Answer the following question 

Is the meiosis responsible for evolution? Justify your answer.

SOLUTION

1. Meiosis ensures that organisms produced by sexual reproduction contain the correct number of chromosomes.

2. Meiosis exhibits genetic variation by the process of recombination.

3. Variations increase further after the union of gametes during fertilization creating offspring with unique characteristics. Thus, it creates diversity of life and is responsible for evolution.


Why mitosis and meiosis – II are called as homotypic division?

SOLUTION

1. In mitosis, the chromosome number and genetic material of daughter cells remain the same as that of the parent cell.

2. In meiosis – II, two haploid cells formed during the first meiotic division divide further into four haploid cells.

This division is identical to mitosis.

The daughter cells formed in the second meiotic division are similar to their parent cells with respect to the chromosome number formed in meiosis – I.

Hence mitosis and meiosis – II are called homotypic division.


Explain the significance of mitosis.

SOLUTION

(i) It helps to maintain linear heredity of an organism by keeping the chromosome number constant in daughter cells.

(ii) It helps in the development of an organism from zygotic stage to adult stage.

(iii) It is the means of repair and regeneration of cells.

(iv) Asexual reproduction is accomplished only through mitosis.

(v) Details of mitosis are similar in all organisms which emphasizes the unity of life.

SOLUTION 2

1. As mitosis is equational division, the chromosome number is maintained constant.

2. It ensures equal distribution of the nuclear and the cytoplasmic content between the daughter cells, both quantitatively and qualitatively. Therefore, the process of mitosis also maintains the nucleo-cytoplasmic ratio.

3. The DNA is also equally distributed.

4. It helps in the growth and development of organisms.

5. Old and worn-out cells are replaced through mitosis.

6. It helps in the asexual reproduction of organisms and vegetative propagation in plants.


Answer the following question.

Enlist the different stages of prophase – I.

SOLUTION

It is the most complicated and longest phase of the meiotic division. It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

1. Leptotene:

The volume of the nucleus increases. The chromosomes become long distinct and coiled. They orient themselves in a specific form known as bouquet stage. This is characterized by the ends of chromosomes converged towards the side of the nucleus where the centrosome lies. The centriole duplicates into two and migrates to opposite poles.

2. Zygotene:

Pairing of non-sister chromatids of homologous chromosomes takes place by the formation of the synaptonemal complex. This pairing is called synapsis. Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

3. Pachytene:

Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids. The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points. These points are called chiasmata (Appear like a cross-X). Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

4. Diplotene:

The chiasma becomes clearly visible in diplotene due to the beginning of repulsion between synapsed homologous chromosomes. This is known as desynapsis. Synaptonemal complex also starts to disappear.

5. Diakinesis:

The chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes. The displacement of chiasmata is termed as terminalization.

The terminal chiasmata exist till the metaphase. The nucleolus and nuclear membrane completely disappear and spindle fibres begin to appear.


Exercise | Q 3. (A) | Page 84

Draw labelled diagram and write an explanation.

With the help of a suitable diagram, describe the cell cycle.

SOLUTION

1. Series of events occurring in the life of a cell is called cell cycle. Interphase and M – phase are the two phases of cell cycle.

2. Interphase: It is the stage between two successive cell divisions. It is the longest phase of a cell cycle during which the cell is highly active and prepares itself for cell division.

The interphase is subdivided into three sub-phases as G 1 -phase, S-phase and G 2 -phase.

i. G 1 – phase (First gap period/First Gap Phase):

It begins immediately after cell division. RNA (mRNA, rRNA and tRNA) synthesis, protein synthesis and synthesis of membranes take place during this phase.

ii. S – phase (Synthesis phase):

In this phase DNA is synthesized (replicated), so that amount of DNA per cell doubles. Synthesis of histone proteins takes place in this phase.

iii. G 2 – phase (Second growth phase/Second Gap Phase):

Metabolic activities essential for cell division occur during this phase.

Various proteins which are necessary for the cell division are also synthesized in this phase.

Apart from this, RNA synthesis also occurs during this phase.

In animal cells, a daughter pair of centrioles appears near the pre-existing pair.


Draw labelled diagrams and write an explanation.

Solution


Distinguish between mitosis and meiosis.

SOLUTION


Mitosis

Prophase:

Metaphase:

Anaphase:

Telophase:

Meiosis:

Prophase I:

Metaphase-I:

Anaphase-I:

Cytokinesis-I:


Draw labelled diagrams and write an explanation.

Draw the diagram of metaphase.

SOLUTION

1. Chromosomes are completely condensed and appear short.

2. Centromere and sister chromatids become very prominent.

3. All the chromosomes are arranged at the equatorial plane of cells. This is called the metaphase plate.

4. The mitotic spindle is fully formed in this phase. e. Centromere of each chromosome divides horizontally into two, each being associated with a chromatid.


Exercise | Q 4 | Page 84

Match the following column A with column B.

SOLUTION


Is a given figure correct? why?

SOLUTION

1. The given figure is incorrect as the spindle fibres are not attached to the centromere of the chromosomes.

2. During metaphase, chromosomes are attached to spindle fibres with the help of centromeres.


Exercise | Q 6 | Page 84

If an onion has 16 chromosomes in its leaf cell, how many chromosomes will be there in its root cell and grain?

SOLUTION

1. The chromosomes in root cell will be 16 as root cell is a diploid cell.

2. The chromosomes in pollen grain will be 8 as pollen grain is a haploid cell.


Exercise | Q 7 | Page 84

Identify the following phase of mitosis and label the ‘A’ given in the diagram.

SOLUTION

The diagram shown is of Metaphase.

A: Chromosomes arranged on the metaphase plate


Identify the following phase of mitosis and label the ‘B’ given in the diagram.

SOLUTION

The diagram shown is of Anaphase.

B: Chromatids moving to opposite poles.


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Balbharati, solutions, for, Biology, 11th, Standard, Maharashtra, State Board, Chapter 6, Biomolecules, Exercise, [Pages 74 – 75

Exercise | Q 6 | Page 75

If double-stranded DNA has 14% C (cytosine) what percent A (adenine), T (thymine) and G (gaunine) would you expect?

SOLUTION

A purine always pairs with pyrimidine.

Adenine pairs with thymine and cytosine pairs with guanine.

Therefore, as per the given data

If cytosine = 14% then guanine = 14%.

According to Chargaff’s rule,

(C+G) = 14 + 14 = 28%

Therefore, (A+T) = 72%

So, A= 36%, T= 36%, G = 14%.

Balbharati, solutions, for, Biology, 11th, Standard, Maharashtra, State Board, Chapter 6, Biomolecules, Exercise, [Pages 74 – 75, Long answer question,

Exercise | Q 5. (A) | Page 75

Long answer question.

What are biomolecules?

SOLUTION

Biomolecules are essential substances produced by our body which are necessary for life.


Explain the biomolecules building blocks of life.

SOLUTION

The building blocks of life are carbohydrates, lipids, proteins and nucleic acids.

Carbohydrates:

1. Carbohydrates are biomolecules made from carbon, hydrogen and oxygen.

2. The general formula of carbohydrates is (CH 2 O) n.

3. They contain hydrogen and oxygen in the same ratio as in water (2:1).

4. Carbohydrates can be broken down to release energy.

5. Based on sugar units, carbohydrates are classified into three types: Monosaccharides, disaccharides and polysaccharides.

Lipids:

1. These are group of substances with greasy consistency with long hydrocarbon chains containing carbon, hydrogen and oxygen.

2. In lipids hydrogen to oxygen ratio is greater than 2:1.

3. Lipid is a broader term used for fatty acids and their derivatives.

4. They are soluble in organic solvents (non-polar solvents).

5. Fatty acids are organic acids which are composed of hydrocarbon chain ending in carboxyl group (COOH).

6. These are divided into: Saturated fatty acids and unsaturated fatty acids.

7. Fatty acids are basic molecules which form different kinds of lipids.

8. Lipids are classified into three types: Simple lipids, Compound lipids, Derived lipids.

Proteins:

1. Proteins are large molecules containing amino acid units ranging from 100 to 3000.

2. They have higher molecular weight.

3. In proteins, amino acids are linked together by peptide bonds which join the carboxyl group of one amino acid residue to the amino group of another residue.

4. A protein molecule consists of one or more polypeptide chains.

5. Proteins contain any or all twenty naturally occurring amino acid types.

6. Proteins have different structures like primary structure, secondary structure, tertiary structure and quaternary structure.

7. Proteins are classified into three types:

Simple proteins: Simple proteins on hydrolysis yield only amino acids. E.g. Histones and albumins.

Conjugated proteins: It consists of a simple protein united with some non-protein substance. E.g. Haemoglobin.

Derived proteins: These proteins are not found in nature as such but are derived from native protein molecules on hydrolysis. E.g. Metaproteins, peptones.

Nucleic Acids:

1. Nucleic acids are macromolecules composed of many small units or monomers called nucleotides.

2. Each nucleotide is formed of three components i.e. pentose sugar, a nitrogen base and a phosphate (phosphoric acid).

3. When sugar combines with a nitrogenous base it forms nucleoside. Nucleotides can be called as nucleoside phosphate.

4. There are two types of nucleic acids, i.e. DNA and RNA. DNA (Deoxyribonucleic acid) is the genetic material of a cell. It is a double-stranded helix. Each strand of helix is made up of deoxyribose nucleotides. RNA (Ribonucleic Acid) is a single-stranded structure having fewer nucleotides as compared to DNA. The strands may be straight or variously folded upon itself. It is made up of nucleotides.


Explain the classes of carbohydrates with examples.

SOLUTION

Based on number of sugar units, carbohydrates are classified into three types namely, monosaccharides, disaccharides and polysaccharides.

Monosaccharides:

1. Monosaccharides are the simplest sugars having crystalline structure, sweet taste and soluble in water.

2. They cannot be further hydrolyzed into smaller molecules.

3. They are the building blocks or monomers of complex carbohydrates.

4. They have the general molecular formula (CH 2 O)n, where n can be 3, 4, 5, 6 and 7. e. They can be classified as triose, tetrose, pentose, etc.

5. Monosaccharides containing the aldehyde (–CHO) group are classified as aldoses e.g. glucose, xylose, and those with a ketone(–C=O) group are classified as ketoses. E.g. ribulose, fructose.

Disaccharides:

1. Disaccharide is formed when two monosaccharide react by condensation reaction releasing a water molecule. This process requires energy.

2. A glycosidic bond forms and holds the two monosaccharide units together.

3. Sucrose, lactose and maltose are examples of disaccharides.

4. Sucrose is a nonreducing sugar since it lacks free aldehyde or ketone group.

5. Lactose and maltose are reducing sugars.

6. Lactose also exists in beta form, which is made from β-galactose and β-glucose.

7. Disaccharides are soluble in water, but they are too big to pass through the cell membrane by diffusion.

Polysaccharides:

1. Monosaccharides can undergo a series of condensation reactions, adding one unit after the other to the chain till a very large molecule (polysaccharide) is formed. This is called polymerization.

2. Polysaccharides are broken down by hydrolysis into monosaccharides.

3. The properties of a polysaccharide molecule depends on its length, branching, folding and coiling.

4. Examples: Starch, glycogen, cellulose.


Describe the types of lipids and mention their biological significance.

SOLUTION

Lipids are classified into three main types:

Simple lipids:

1. These are esters of fatty acids with various alcohols. Fats and waxes are simple lipids.

2. Fats are esters of fatty acids with glycerol (CH 2 OH-CHOH-CH 2 OH).

3. Triglycerides are three molecules of fatty acids and one molecule of glycerol.

4. Unsaturated fats are liquid at room temperature and are called oils. Unsaturated fatty acids are hydrogenated to produce fats e.g. Vanaspati ghee.

Biological significance:

1. Fats are a nutritional source with high calorific value and they act as reserved food materials.

2. In plants, fat is stored in seeds to nourish embryo during germination.

3. In animals, fat is stored in the adipocytes of the adipose tissue.

4. Fats deposited in subcutaneous tissue act as an insulator and minimize loss of body heat.

5. Fats deposited around the internal organs act as cushions to absorb mechanical shocks.

6. Wax is another example of a simple lipid. They are esters of long-chain fatty acids with long-chain alcohols.

7. They are found in the blood, gonads, and sebaceous glands of the skin.

8. Waxes are not as readily hydrolyzed as fats.

9. They are solid at ordinary temperatures.

10. Waxes form water-insoluble coating on hair and skin in animals, waxes form an outer coating on stems, leaves, and fruits.

Compound lipids:

1. These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.

2. They contain a molecule of glycerol, two molecules of fatty acids and a phosphate group or simple sugar.

3. Some phospholipids such as lecithin also have a nitrogenous compound attached to the phosphate group.

4. Phospholipids have both hydrophilic polar groups (phosphate and nitrogenous group) and hydrophobic non-polar groups (hydrocarbon chains of fatty acids).

5. Glycolipids contain glycerol, fatty acids, simple sugars such as galactose. They are also called cerebrosides.

Biological significance:

1. Phospholipids contribute in the formation of the cell membranes.

2. Large amounts of glycolipids are found in the brain white matter and myelin sheath.

Derived Lipids:

1. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.

2. One of the most common sterols is cholesterol.

Biological significance:

1. It is widely distributed in all cells of the animal body, but particularly in nervous tissue.

2. Cholesterol exists either free or as a cholesterol ester.

3. Adrenocorticoids, sex hormones (progesterone, testosterone) and vitamin D are synthesized from cholesterol.

4. Cholesterol is not found in plants.

5. Sterols exist as phytosterols in plants.

6. Yam Plant (Dioscorea) produces a steroid compound called diosgenin. It is used in the manufacture of antifertility pills. i.e. birth control pills.


Explain the chemical nature, structure and role of phospholipids in the biological membrane.

SOLUTION

Chemical nature: Phospholipids are amphiphilic in nature. As they have a hydrophilic head and hydrophobic tail.

Structure: It contains alcohol, two fatty acid chains and a phosphate group.

Role: Phospholipids forms the membranes around the cells and cellular organelles. They form a lipid bilayer membrane. The phospholipids are arranged tail to tail. It serves as a barrier against the movement of any ions or polar compounds into and out of the cell.


Describe classes of proteins with their importance.

SOLUTION

On the basis of structure, proteins are classified into three categories:

Simple proteins:

1. Simple proteins on hydrolysis yield only amino acids.

2. These are soluble in one or more solvents.

3. Simple proteins may be soluble in water.

4. Histones of nucleoproteins are soluble in water.

5. Globular molecules of histones are not coagulated by heat.

6. Albumins are also soluble in water but they get coagulated on heating.

7. Albumins are widely distributed e.g. egg albumin, serum albumin, and legumelin of pulses are albumins.

Importance: They are involved in structural components; they also act as a storage kind of protein. Some are associated with nucleic acids in nucleoproteins of cells.

Conjugated proteins:

1. Conjugated proteins consist of a simple protein united with some non-protein substance.

2. The non-protein group is called the prosthetic group e.g. haemoglobin.

3. Globin is the protein and the iron-containing pigment haem is the prosthetic group.

4. Similarly, nucleoproteins have nucleic acids.

5. Proteins are classified as glycoproteins and mucoproteins.

6. Mucoproteins are carbohydrate-protein complexes e.g. mucin of saliva and heparin of blood.

7. Lipoproteins are lipid-protein complexes e.g. conjugate protein found in the brain, plasma membrane, milk etc.

Importance: They are involved in structural components of cell membranes and organelles. They also act as a transporter. Some conjugated proteins are important in the electron transport chain in respiration.

Derived proteins:

1. These proteins are not found in nature as such.

2. These proteins are derived from native protein molecules on hydrolysis.

3. Metaproteins, peptones are derived proteins.

Importance: They act as a precursor for many molecules which are essential for life.


What are enzymes?

SOLUTION

Enzymes are biological macromolecules which act as a catalyst and accelerates the reaction in the body.


How are enzymes classified? Mention an example of each class.

SOLUTION

Enzymes are classified into six classes:

1. Oxidoreductases: These enzymes catalyze oxidation and reduction reactions by the transfer of hydrogen and/or oxygen.

e.g. alcohol dehydrogenase

\[\ce{Alcohol + NAD^+ ->[Alcohol][dehydroenase] Aldehyde + NADH2}\]

2. Transferases: These enzymes catalyse the transfer of certain groups between two molecules. e.g. glucokinase

\[\ce{Glucose + ATP ->[Glucokinase] Glucose – 6 – Phosphate + ADP}\]

3. Hydrolases: These enzymes catalyze hydrolytic reactions. This class includes amylases, proteases, lipases etc. e.g. Sucrase

\[\ce{Sucrose + water ->[Sucrase] Glucose + Fructose}\]

4. Lyases: These enzymes are involved in elimination reactions resulting in the removal of a group of atoms from the substrate molecule to leave a double bond. It includes aldolases, decarboxylases, and dehydratases, e.g. fumarate hydratase.

\[\ce{Histidine ->[Histidine][deca rbox ylase]Histamine + CO2}\]

5. Isomerases: These enzymes catalyze structural rearrangements within a molecule. Their nomenclature is based on the type of isomerism. Thus, these enzymes are identified as racemases, epimerases, isomerases, mutases, e.g. xylose isomerase.

\[\ce{Glucose – 6 – Phosphate ->[Isomerase] Fructose – 6 – Phosphate}\]

6. Ligases or Synthetases: These are the enzymes which catalyze the covalent linkage of the molecules utilizing the energy obtained from hydrolysis of an energy-rich compound like ATP, GTP e.g. glutathione synthetase, Pyruvate carboxylase.

\[\ce{Pyruvate + CO2 + ATP ->[Pyruvate][ca rbox ylase] Oxaloacetate + ADP + Pi}\]


Explain the properties of an enzyme?

SOLUTION.

Proteinaceous Nature:

All enzymes are basically made up of protein.

Three-Dimensional conformation:

1. All enzymes have specific 3-dimensional conformation.

2. They have one or more active sites to which substrate (reactant) combines.

3. The points of the active site where the substrate joins with the enzyme is called the substrate-binding site.

Catalytic property:

1. Enzymes are like inorganic catalysts and influence the speed of biochemical reactions but themselves remain unchanged.

2. After completion of the reaction and release of the product, they remain active to catalyze again.

3. A small number of enzymes can catalyze the transformation of a very large quantity of the substrate into an end product.

4. For example, sucrase can hydrolyze 100000 times of sucrose as compared with its own weight.

Specificity of action:

1. The ability of an enzyme to catalyze one specific reaction and essentially no other is perhaps its most significant property. Each enzyme acts upon a specific substrate or a specific group of substrates.

2. Enzymes are very sensitive to temperature and pH.

3. Each enzyme exhibits its highest activity at a specific pH i.e. optimum pH.

4. Any increase or decrease in pH causes a decline in enzyme activity e.g. enzyme pepsin (secreted in stomach) shows the highest activity at an optimum pH of 2 (acidic).

5. Trypsin (in the duodenum) is most active at an optimum pH of 9.5 (alkaline).

6. Both these enzymes viz. pepsin and trypsin are protein-digesting enzymes.

Temperature:

1. Enzymes are destroyed at a higher temperature of 60-70°C or below, they are not destroyed but become inactive.

2. This inactive state is temporary and the enzyme can become active at a suitable temperature.

3. Most of the enzymes work at an optimum temperature between 20°C and 35°C.


Describe the models for enzyme actions.

SOLUTION

There are two types of models:

1st

Lock and Key model:

1. The lock and Key model was first postulated in 1894 by Emil Fischer.

2. This model explains the specific action of an enzyme with a single substrate.

3. In this model, the lock is the enzyme and the key is the substrate.

4. The correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).

2nd

Induced Fit model (Flexible Model):

1. Induced Fit model was first proposed in 1959 by Koshland.

2. This model states that the approach of a substrate induces a conformational change in the enzyme.

3. It is the more accepted model to understand the mode of action of an enzyme.

4. The induced-fit model shows that enzymes are rather flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it.

5. It is also the point at which the final form and shape of the enzyme are determined.


Describe the factors affecting enzyme action.

SOLUTION

The factors affecting enzyme activity are as follows:

Concentration of substrate:

1. An increase in the substrate concentration gradually increases the velocity of enzyme activity within the limited range of substrate levels.

2. A rectangular hyperbola is obtained when velocity is plotted against the substrate concentration.

3. Three distinct phases (A, B and C) of the reaction are observed in the graph.

Where V = Measured velocity, V max = Maximum velocity, S = Substrate concentration, K m = Michaelis-Menten constant.

4. K m or the Michaelis-Menten constant is defined as the substrate concentration (expressed in moles/lit) to produce half of the maximum velocity in an enzyme-catalyzed reaction.

5. It indicates that half of the enzyme molecules (i.e. 50%) are bound with the substrate molecules when the substrate concentration equals the K m value.

6. K m value is a constant and a characteristic feature of a given enzyme.

7. It is a representative for measuring the strength of ES complex.

8. A low K m value indicates a strong affinity between enzyme and substrate, whereas a high K m value reflects a weak affinity between them.

9. For majority of enzymes, the K m values are in the range of 10 -5 to 10 -2 moles.

Enzyme Concentration:

1. The rate of an enzymatic reaction is directly proportional to the concentration of the substrate.

2. The rate of reaction is also directly proportional to the square root of the concentration of enzymes.

3. It means that the rate of reaction also increases with the increasing concentration of enzyme and the rate of reaction can also decrease by decreasing the concentration of the enzyme.

Temperature:

1. The temperature at which the enzymes show maximum activity is called Optimum temperature.

2. The rate of a chemical reaction is increased by a rise in temperature but this is true only over a limited range of temperature.

3. Enzymes rapidly denature at a temperature above 40°C.

4. The activity of enzymes is reduced at low temperatures.

5. The enzymatic reaction occurs best at or around 37°C which is the average normal body temperature in homeotherms.

Effect of pH:

1. The pH at which an enzyme catalyzes the reaction at the maximum rate is known as optimum pH.

2. The enzyme cannot perform its function beyond the range of its pH value.

Other substances:

1. The enzyme action is also increased or decreased in the presence of some other substances such as coenzymes, activators, and inhibitors.

2. Most of the enzymes are a combination of a co-enzyme and an apo-enzyme.

3. Activators are the inorganic substances which increase the enzyme activity.

4. The inhibitor is the substance that reduces the enzyme activity.


What are the nucleic acids?

SOLUTION

Nucleic acids are macromolecules composed of many small units or monomers called nucleotides.


Enlist the point of differences among DNA and RNA.

SOLUTION

DNA:

1. It is the genetic material of the majority of the organisms.

2. It is double-stranded.

3. Deoxyribose sugar is present.

4. Nitrogen bases like Adenine, Guanine, Cytosine, Thymine are present.

5. Specific base pairing is observed.

6. Total number of purines is equal to the total number of pyrimidine. Thus, the purine to pyrimidine ratio is 1:1.

7. It is present in the nucleus.

8. It is responsible for determining hereditary characters and for formation of RNA.

RNA:

1. It is a genetic material only of some viruses.

2. It is single-stranded.

3. Ribose sugar is present.

4. Nitrogen bases like Adenine, Guanine, Cytosine, Uracil are present.

5. Nitrogen bases do not form pairs.

6. The amount of purine and pyrimidine may or may not be equal.

7. It is present in the nucleus and cytoplasm.

8. It takes part in protein synthesis.


What are the types of RNA? Mention the role of each class of RNA.

SOLUTION

There are three types of cellular RNAs:

1. messenger RNA (mRNA),

2. ribosomal RNA (rRNA),

3. transfer RNA (tRNA).

Messenger RNA (mRNA):

a. It is a linear polynucleotide.

b. It accounts 3% of cellular RNA.

c. Its molecular weight is several million.

d. mRNA molecule carrying information to form a complete polypeptide chain is called cistron.

e. Size of mRNA is related to the size of the message it contains.

f. Synthesis of mRNA begins at 5’ end of DNA strand and terminates at 3’ end.

Role of messenger RNA:

It carries genetic information from DNA to ribosomes, which are the sites of protein synthesis.

Ribosomal RNA (rRNA):

a. rRNA was discovered by Kurland in 1960.

b. It forms 50-60% part of ribosomes.

c. It accounts 80-90% of the cellular RNA.

d. It is synthesized in nucleus.

e. It gets coiled at various places due to intrachain complementary base pairing.

Role of ribosomal RNA:

It provides a proper binding site for m-RNA during protein synthesis.

Transfer RNA (tRNA):

a. These molecules are much smaller consisting of 70-80 nucleotides.

b. Due to presence of complementary base pairing at various places, it is shaped like clover-leaf.

c. Each tRNA can pick up a particular amino acid.

d. Following four parts can be recognized on tRNA

1. DHU arm (Dihydroxyuracil loop/ amino acid recognition site

2. Amino acid binding site

3. Anticodon loop/codon recognition site

4. Ribosome recognition site.

e. In the anticodon loop of tRNA, three unpaired nucleotides are presently called anticodon which pair with codon present on mRNA.

f. The specific amino acids are attached at the 3′ end in the acceptor stem of clover leaf of tRNA.

Role of transfer RNA:

It helps in the elongation of the polypeptide chain during the process called translation.


How metabolic pool is formed in the cell.

SOLUTION

1. The metabolic pool in the cell is formed due to glycolysis and Krebs cycle.

2. The catabolic chemical reaction of glycolysis and Krebs cycle provides ATP and biomolecules. These biomolecules form the metabolic pool of the cell.

3. These biomolecules can be utilized for the synthesis of many important cellular components.

4. The metabolites can be added or withdrawn from the pool according to the need of the cell.


What is metabolism?

SOLUTION

Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new organic material.


Balbharati, solutions, for, Biology, 11th, Standard, Maharashtra, State Board, Chapter 6, Biomolecules, Exercise, [Pages 74 – 75, Complete the following chart,

Complete the following chart.

SOLUTION


Answer the following with reference to the following figure.

1. Name the type of bond formed between two polypeptides.

2. Which amino acid is involved in the formation of such bond?

3. Amongst I, II, III and IV structural level of protein, which level of structure includes such bond?

SOLUTION

1. Disulfide bond

2. Cysteine

3. Tertiary structure.


Match the following items given in column I and II.

SOLUTION


Balbharati, solutions, for, Biology, 11th, Standard, Maharashtra, State Board, Chapter 6, Biomolecules, Exercise, [Pages 74 – 75, Answer the following question,

Exercise | Q 3. (A) | Page 74

Answer the following question.

What are building blocks of life?

SOLUTION

Life is composed of four main building blocks: Carbohydrates, proteins, lipids, and nucleic acids.


Explain the peptide bond.

SOLUTION

1. The covalent bond that links the two amino acids is called a peptide bond.

2. Peptide bond is formed by condensation reaction.


How many types of polysaccharides you know?.

SOLUTION

There are two types of polysaccharides:

1. Homopolysaccharides: It contains same type of monosaccharides. E.g. Starch, glycogen, cellulose.

2. Heteropolysaccharides: It contains two or more different monosaccharides. E.g. Hyaluronic acid, heparin, hemicellulose.


Enlist the significance of carbohydrates.

SOLUTION

Significances of carbohydrates are as follows:

1. Carbohydrates provide energy for metabolism.

2. Glucose is the main substrate for ATP synthesis.

3. Lactose, a disaccharide present in the milk provides energy to babies.

4. Polysaccharide serves as a structural component of cell membrane, cell wall and reserved food as starch and glycogen.


What is reducing sugar?

SOLUTION

1. A sugar that serves as a reducing agent due to the presence of free aldehyde or ketone group is called a reducing sugar.

2. These sugars reduce Benedict’s reagent (Cu 2+ to Cu + ) since they are capable of transferring hydrogens (electrons) to other compounds, a process called reduction.

3. All monosaccharides are reducing sugars.


What is the basic difference between saturated and unsaturated fatty acid?

SOLUTION


Enlist the examples of simple protein and add their significance.

SOLUTION

Examples of simple proteins are: E.g.: Albumins and histones.

Significance:

1. Albumin:

a. It is the main protein in the blood.

b. It maintains the pressure in the blood vessels.

c. It helps in the transportation of substances like hormone and drugs in the body.

2. Histones:

a. It is the chief protein of chromatin.

b. They are involved in the packaging of DNA into structural units called nucleosomes.


Explain the secondary structure of a protein with examples.

SOLUTION

1. There are two types of the secondary structures of protein: α-helix and β-pleated sheets.

2. The polypeptide chain is arranged in a spiral helix. These spiral helices are of two types: α-helix (right-handed) and β-helix (left-handed).

3. This spiral configuration is held together by hydrogen bonds.

4. The sequence of amino acids in the polypeptide chain determines the location of its bend or fold and the position of formation of hydrogen bonds between different portions of the chain or between different chains. Thus, peptide chains form an α-helix structure.

5. Example of α-helix structure is keratin.

6. In some proteins two or more peptide chains are linked together by intermolecular hydrogen bonds. Such structures are called β-pleated sheets.

7. Example of a β-pleated sheet is silk fibres.

8. Due to the formation of hydrogen bonds peptide chains assume a secondary structure.


Explain the induced fit model for the mode of enzyme action.

SOLUTION

The induced fit model shows that enzymes are flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it. It is also the point at which the final form and shape of the enzyme are determined


What is RNA? Enlist types of RNA.

SOLUTION

1. RNA stands for Ribonucleic Acid. It is a long single-stranded polynucleotide chain that helps in protein synthesis, functions as a messenger and translates messages coded in DNA into protein.

2. There are three types of RNA:

mRNA (messenger RNA), rRNA (ribosomal RNA) and tRNA (transfer RNA).


Describe the concept of metabolic pool.

SOLUTION

1. Metabolic pool is the reservoir of biomolecules in the cell on which enzymes can act to produce useful products as per the need of the cell.

2. The concept of metabolic pool is significant in cell biology because it allows one type of molecule to change into another type E.g. Carbohydrates can be converted to fats and vice-versa.


How do secondary metabolites useful for mankind?

SOLUTION

1. Drugs developed from secondary metabolites have been used to treat infectious diseases, cancer, hypertension, and inflammation.

2. Morphine, the first alkaloid isolated from Papaver somniferum is used as a pain reliver and cough suppressant.

3. Secondary metabolites like alkaloids, nicotine, cocaine and the terpenes, cannabinol are widely used for recreation and stimulation.

4. Flavours of secondary metabolites improve our food preferences.

5. Tannins are added to wines and chocolate for improving astringency.

6. Since most secondary metabolites have antibiotic property, they are also used as food preservatives.

7. Glucosinolates is a secondary metabolite which is naturally present in cabbage imparts a characteristic flavour and aroma because of nitrogen and sulphur–containing chemicals. It also offers protection to these plants from many pests.


Chapter 5: Cell Structure and Organization.

EXERCISE [PAGES 57 – 58]

Balbharati, solutions, for, Biology, 11th, Standard, Maharashtra, State, Board, Chapter 5, Cell Structure and Organization, Exercise, Pages 57, 58,

  1. Choose the correct option 👈 Click here for Solution.

  1. Answer the following question. 👈 Click here for Solution.

  1. Answer the following questions. 👈 Click here for Solution.

  1. Label the diagram and write down the detail of concept in your word. 👈 Click here for Solution.

  1. Complete the flow chart. 👈 Click here

  1. Label the A, B, C, and D in the above diagram and write the functions of organells A and B. 👈 Click here for solution.

  1. Identify cell structure or organelle from its description below. 👈 Click here for solution.
Click above link for solution

  1. Onion cells have no chloroplast. How can we tell they are plants? 👈 Click here for solution

Onion cells have no chloroplast. How can we tell they are plants?

SOLUTION

1. The bulb of an onion is a modified form of leaves.

2. While photosynthesis takes place in the leaves (present above the ground) of an onion containing chloroplast, the little glucose that is produced from this process is converted into starch (starch granules) and stored in the bulb.

3. Starch act as reserved food material in plants.

4. Using an iodine solution, we can test for the presence of starch in onion cells. If starch is present, the iodine changes from brown to blue-black or purple. Hence, we can say that though onion cells have no chloroplast they are considered as plants.


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