Important Questions

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 9 Biomolecules Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 9 Biomolecules

Very short answer type questions

Question 1.
Write the name of any one amino acid, sugar, nucleotide and fatty acid. [NCERT Exemplar]
Answer:
Alanine is an amino acid, glucose is a sugar, adenylic acid is a nucleotide and linolenic acid is a fatty acid.

Question 2.
Mention four essential major elements of life.
Answer:
Oxygen, carbon, hydrogen and nitrogen are the four essential elements of life.

Question 3.
Name one element invariably found in proteins but not in all carbohydrates and lipids.
Answer:
Nitrogen is found invariably in proteins, but not in all carbohydrates and lipids.

Question 4.
What is the name given to a polysaccharide composed of two different monomers? Also give example for this.
Answer:
The name given is heteropolysaccharide to the type of polysaccharide, which is composed of different types of monomers, e.g., Pectin.

Question 5.
One of the homopolysaccharide is also known as animal starch. Name it.
Answer:
Glycogen is also known as animal starch.

Question 6.
The macromolecules that forms the hereditary determinants of the living organism. Name it.
Answer:
Nucleic acid.

Question 7.
A nitrogenous base is present in RNA but absent in DNA. Name it. Also give example in which it exists.
Answer:
Uracil (U), is the nitrogenous base present only in RNA, e.g., viruses like hepatitis C.

Question 8.
How many hydrogen bonds are formed between:
(i) Guanine and cytosine
(ii) Adenine and thymine, respectively?
Answer:
(i) Three hydrogen bonds between guanine and cytosine.
(ii) Two hydrogen bonds between adenine and thymine.

Question 9.
Reaction given below is catalysed by oxidoreductase between two substrates A and A, complete the reaction. [NCERT Exemplar] A reduced + A oxidised →
Answer:
A reduced + A’ oxidised → A oxidised + A’ reduced + A’ reduced

Question 10.
Name two physical factors which can affect the enzyme activity?
Answer:
Temperature and pH are the two physical factors that affects activity of an enzyme.

Question 11.
The enzyme that works only in the presence of a co-factor or coenzyme called
Answer:
Apoenzyme works only in the presence of a co-factor or coenzyme.

Question 12.
What do you mean by living state?
Answer:
The living state is a non-equilibrium steady-state to be able to perform work.

Short answer type questions

Question 1.
Give a tabular representation of different constituents of a living cell.
Answer:

Component	% of Total Cellular Mass
Water	70-90%
Proteins	10-15%
Carbohydrates	3%
Lipids	2%
Nucleic Acids	5-7%
Ions	1%

Question 2.
What are polysachharides?
Answer:
Polysaccharides are long chains of sugars. They are threads containing different monosaccharides as building blocks.

In a polysaccharide chain (say glycogen), the right end is called the reducing end and the left end is called the non-reducing end. Starch forms helical secondary structures. In fact, starch can hold 12 molecules in the helical portion.
Examples: Cellulose, chitin

Question 3.
Give a brief description of nucleic acid.
Answer:
For nucleic acids, the building block is a nucleotide. A nucleotide has three chemically distinct components. One is a heterocyclic compound, the second is a monosaccharide and the third is a phosphoric acid or phosphate.

The heterocyclic compounds in nucleic acids are the nitrogenous bases named adenine, guanine, uracil, cytosine, and thymine. Adenine and Guanine are substituted purines while the rest are substituted pyrimidines. The skeletal heterocyclic ring is called as purine and pyrimidine respectively. The sugar found in polynucleotides is either ribose (a monosaccharide pentose) or 2’ deoxyribose. A nucleic acid containing deoxyribose is called deoxyribonucleic acid (DNA) while that which contains ribose is called ribonucleic acid (RNA).

Question 4.
What is the difference between primary and secondary metabolites?
Answer:
Primary metabolites are found in both, animal cells and plant cells. Secondary metabolites are found only in plant cells.
Functions of primary metabolites are known to scientists, while functions of secondary metabolites are not known yet.

Question 5.
Explain the basic structure of an amino acid.
Answer:

Amino acid is an organic compound, which has an amino group and an acidic group, present
as substituents on the same carbon ; i.e., the a-carbon. Because of this amino acids are also called α-amino acids. On four valency positions there are four substituent groups. They are hydrogen, carboxyl group, amino group and a variable group. The variable group is called the ‘R’ group. The nature of R group decides the type of an amino acid.

Question 6.
Describe the classification and nomenclature of enzymes.
Answer:
Classification and Nomenclature of Enzymes: The International Union of Biochemists (IUB) has classified all the enzymes into the following six classes:

(a) Class 1: Oxidoreductases: These enzymes catalyse the oxidation (by adding oxygen or removal of hydrogen or removal of electrons) or reduction (by adding hydrogen or adding electrons to a substrate) of a substance.
S reduced + S’ oxidised → S oxidised + S’ reduced

(b) Class 2: Transferases: These enzymes catalyse the transfer of specific groups from one substrate to another. S – G + S’ → S + S’ – G.

(c) Class 3: Hydrolases: These enzymes catalyse the breakdown of larger molecules into smaller molecules with the addition of water.

(d) Class 4: Lyases: These enzymes catalyse the cleavage of specific covalent bonds and removal of specific group (s), without the use of water.

(e) Class 5: Isomerases: These enzymes catalyse the rearrangement of atoms in a molecule to form isomers.

(f) Class 6: Ligases: These enzymes catalyse covalent bonding (of C-0, C-S, C-N, P-O etc.) between two substrates to form a large molecule, mostly involving utilisation of energy by hydrolysis of ATP.

Long answer type questions

Question 1.
Enumerate the difference between a nucleotide and nucleoside. Give two examples of each with their structure. [NCERT Exemplar]
Answer:
Differences between nucleotide and nucleoside are given below:

Nucleotide	Nucleoside
A nucleotide consists of a nitrogenous base, a sugar (ribose or deoxyribose) and one to three phosphate groups, i. e., sugar + base + phosphate.	A nucleoside consists of a nitrogenous base t covalently attached to a sugar (ribose or deoxyribose), but without the phosphate group, i. e., sugar + base

Question 2.
What is the concept of metabolism? What are the metabolic basis for living?
Answer:
The continuous process of breakdown and synthesis of biomolecules through chemical reactions occurring in the living cells is called metabolism.

• Each of the metabolic reaction results in a transformation of biomolecules.
• Most of these metabolic reactions do not occur in isolation but are always linked with some other reactions.
• In these reactions, the metabolites are converted into another metabolite in a series of linked reactions called metabolic pathways.
• Each metabolite has a definite rate and direction during the flow through a metabolic pathways called the dynamic state.

In living systems, metabolism involves two following types of pathways:
(a) The anabolic pathway is called biosynthetic pathway. It leads to a more complex structure from a simpler structure, e.g., The pathway involving the conversion of acetic acid into cholesterol. These pathways consume energy.

(b) The catabolic pathways lead to simpler structure from a complex structure, e g., The pathway involving conversion of glucose into lactic acid in our skeletal muscles. This pathway lead to the release of energy, e.g., Energy is liberated when glucose is degraded to lactic acid in our skeletal muscles.

Question 3.
Formation of Enzyme-Substrate complex (ES) is the first step in catalysed reactions. Describe the other steps till the formation of product.
Answer:
Mechanism of Enzymatic Action: The catalytic cycle of an enzyme action can he described in the following steps :

• First, the substrate binds to the active site of the enzyme, fitting into the active site.
• The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate. The formation of the ES complex is essential for catalysis.
  E + S → ES → EP → E + P
• The active site of the enzyme, now in close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme product complex is formed.
• The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle one again.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 1 Reproduction in Organisms Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 1 Reproduction in Organisms

Very short answer type questions

Question 1.
What is life span?
Answer:
The period between birth and the natural death of an organism represents its life span.

Question 2.
On what factors does the type of reproduction adopted by an organism depend on? [NCERT Exemplar]
Answer:
The organism’s habitat, physiology and genetic make-up determines the type of reproduction adopted by it.

Question 3.
Name an organism where cell division is itself a mode of reproduction.
Answer:
Protists/Monerans/Amoeba/Paramecium.

Question 4.
Mention two inherent characteristics of Amoeba and Yeast that enable them to reproduce asexually. [NCERT Exemplar]
Answer:

1. They are unicellular organisms.
2. They have a very simple body structure.

Question 5.
What is conidia?
Answer:
The asexual, non-motile spores produced externally/exogenously in some fungi are called conidia, e.g., Penicillium.

Question 6.
Define gemmules.
Answer:
Internal asexual reproductive units or buds in sponges are called gemmules.

Question 7.
Name the vegetative propagules in the following
(a) Agave
(b) Bryophyllum
Answer:
(a) Agave – Bulbil
(b) Bryophyllum – Leaf buds/adventitious buds.

Question 8.
Mention the unique feature with respect to flowering and fruiting in bamboo species.
Answer:
Bamboo flowers once in its life time generally after 50-100 yrs of vegetative growth. It produces large number of fruits and dies.

Question 9.
Is Marchantia monoecious or dioecious? Where are the sex 1 organs borne in this plant? [NCERT Exemplar]
Answer:
Marchantia is dioecious. The male sex organs, antheridia, are borne on j the antheridiophores and female sex organs called archegonia are borne on archegoniophores.

Question 10.
Suggest a possible explanation why the seeds in a pea are arranged in a row, whereas those in tomato are scattered in the juicy pulp. [NCERT Exemplar]
Answer:
The ovary of pea plant is monocarpellary and the ovules are arranged along one margin whereas in tomato the ovary is tricarpellary with axile placentation.

Question 11.
In a developing embryo, analyse the consequences if cell divisions are not followed by cell differentiation.
[NCERT Exemplar]
Answer:
If differentiation does not follow division, embryo will not develop and this will not develop into a new organism.

Question 12.
Name the phenomenon and one bird where the female gamete directly develops into a new organism.
Answer:
The phenomenon is called parthenogenesis. Turkey.

Question 13.
Mention the site where syngamy occurs in amphibians and reptiles, respectively.
Answer:
In amphibians, external fertilisation occurs hence, syngamy occurs in the medium of water. In reptiles, internal fertilisation occurs hence, syngamy occurs within the body of female parent.

Question 14.
Name the group of organisms that produce non-motile gametes. How do they reach the female gamete for fertilisation?
Answer:
Angiosperms produce non-moule gametes. They reach the female gamete with the help of air or water.

Question 15.
The number of taxa exhibiting asexual reproduction is drastically reduced in the higher plants (angiosperms) and higher animals (vertebrates) as compared with lower groups of plants and animals. Analyse the possible reasons for this situation. [NCERT Exemplar]
Answer:
Both angiosperms and vertebrates have a more complex structural organisation. They have evolved very efficient mechanism of sexual reproduction. Since asexual reproduction does not create new genetic pools in the offspring and consequently hampers their adaptability to external conditions, these groups have resorted to reproduction by the sexual method.

Short answer type questions

Question 1.
The cell division involved in gamete formation is not of the same type in different organisms. Justify.
Answer:
The parents may be haploid or diploid but the gametes always have to be haploid, biploid parents undergo meiosis to produce haploid gametes, whereas haploid parents undergo mitosis to produce haploid gametes.

Question 2.
Although sexual reproduction is a long drawn, energy-intensive complex form of reproduction, many groups of organisms in Kingdom Animalia and Plantae prefer this mode of reproduction. Give at least three reasons for this. [NCERT Exemplar]
Answer:

• Sexual reproduction brings about variation in the offspring.
• Since gamete formation is preceded by meiosis, genetic recombination occurring during crossing over (meiosis-I), leads to a great deal of variation in the DNA of gametes.
• The organism has better chance of survival in a changing environment.

Question 3.
Explain the importance of syngamy and meiosis in a sexual life cycle of an organism.
Answer:
Syngamy and meiosis play an important role in sexual life cycle of any diploid organisms, may be a plant or animal syngamy, i.e. fusion of haploid gametes/sex cells (n) (fertilisation) to form diploid egg cell/zygote (2n). Zygote divides repeatedly by mitotic divisons to form an embryo which develop into a new diploid organisms. After attaining sexual maturity, the organisms undergo special type of meiotic divisions – spermatogenesis/ microsporogenesis and megasporogenesis/ oogenesis to form haploid male sex cells and female sex cells. These sex cells (n) again fuse to form diploid zygote.

Question 4.
Why are mosses and liverworts unable to complete their sexual mode of reproduction in dry conditions? Give reasons.
Answer:
For sexual reproduction to take place in mosses and liverworts the motile male gametophytes, antherozoids, have to swim on the water surface to fertilise the immotile female gametophytes, egg. In dry conditions, the antherozoids do not reach the egg and hence fertilisation cannot occur.

Question 5.
Why do algae and fungi shift to sexual mode of reproduction just before the onset of adverse conditions?
Answer:
Algae and fungi shift to sexual mode of reproduction for survival during unfavourable conditions. Fusion of gametes helps to pool their resources for survival. The zygote develops a thick wall that is resistant to dessication and damage which undergoes a period of rest before germination.

Long answer type questions

Question 1.
What is gametogenesis? Describe the different types of gametes and draw labelled diagrams. ,
Answer:
The process of formation of two types of gametes – male and female inside the gametangia is called gametogenesis.

Depending upon the size and motility, the gametes are of following types :
1. Isogametes or Homogamets: The gametes are similar in shape, size, structure arid function. Their fusion is called isogamy e.g., Cladophora. However, when the isogametes are physiologically different and the gametes produced by one parent do not fuse with each other, the gametes belonging to different mating types can only fuse and their fusion is called physiological anisogamy, e.g., Ulothrvc.

2. Anisogametes or Heterogametes: The fusing gametes are different in form, size, structure and behaviour. The larger, non-motile, food-laden gamete is called ovum/egg/oosphere/ macrogamete. The smaller, motile, active gamete is called sperm/male gamete/antherozoid. Such gametes are called anisogametes or heterogametes and their fusion is termed as anisogamy or heterogamy e.g., Cladophora.

Question 2.
Write a note on sexuality in organisms.
Answer:
Sexuality in Organisms: Sexual reproduction in organisms generally involves the fusion of gametes from two different individuals. But this is not always true.

Sexuality in Plants: Plants may have both male and female reproductive structures in the same plant (bisexual) e.g., rose or on different plants (unisexual) e.g., papaya. In several fungi and plants, terms such as homothallic and monoecious are used to denote the bisexual condition and heterothallic and dioecious are the terms used to describe unisexual condition.

In flowering plants, the unisexual male flower is staminate i.e., bearing stamens, while the female is pistillate or bearing pistils. In some flowering plants, both male and female flowers may be present on the same individual (monoecious) or on separate individuals (dioecious). Some examples of monoecious plants are cucurbits and coconuts and of dioecious plants are papaya, mulberry and date palm.

Sexuality in Animals: There are species which possess both the reproductive organs (bisexual). Earthworms, sponge, tapeworm and leech are typical examples of bisexual animals that possess both male and female reproductive organs i.e., they are hermaphrodites. There are large number of animal species which are either male or female and are called as unisexual organisms, e.g., frog, lizard, crow, dog, cat, rabbit, human beings etc.

Punjab State Board Syllabus PSEB 12th Class Biology Important Questions Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 12

PSEB 12th Class Biology Important Questions in Punjabi English Medium

• Chapter 1 Reproduction in Organisms Important Questions
• Chapter 2 Sexual Reproduction in Flowering Plants Important Questions
• Chapter 3 Human Reproduction Important Questions
• Chapter 4 Reproductive Health Important Questions
• Chapter 5 Principles of Inheritance and Variation Important Questions
• Chapter 6 Molecular Basis of Inheritance Important Questions
• Chapter 7 Evolution Important Questions
• Chapter 8 Human Health and Disease Important Questions
• Chapter 9 Strategies for Enhancement in Food Production Important Questions
• Chapter 10 Microbes in Human Welfare Important Questions
• Chapter 11 Biotechnology: Principles and Processes Important Questions
• Chapter 12 Biotechnology and its Applications Important Questions
• Chapter 13 Organisms and Populations Important Questions
• Chapter 14 Ecosystem Important Questions
• Chapter 15 Biodiversity and Conservation Important Questions
• Chapter 16 Environmental Issues Important Questions

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 13 Photosynthesis in Higher Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 13 Photosynthesis in Higher Plants

Very short answer type questions

Question 1.
Photosynthetic pigments are located in which part of the chloroplast? ‘
Answer:
Photosynthetic pigments are located in the lipid part of thylakoid membrane.

Question 2.
Does moonlight support photosynthesis? Find out. [NCERT Exemplar]
Answer:
No, because the intensity of moonlight is several thousand times less than that of direct sunlight, insufficient for the light-dependent phase of photosynthesis.

Question 3.
Oxygen evolved during photosynthesis comes from H₂O or CO₂?
Answer:
Oxygen in photosynthesis evolves from H₂O (i. e., by splitting of water).

Question 4.
Mention conditions under which PS-I functions.
Answer:
Conditions necessary under which PS-I function are as follows :

• When the wavelength of light is higher than 680 nm.
• When NADPH accumulates and CO₂ fixation is retarded.

Question 5.
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic CO₂ requirements? [NCERT Exemplar]
Answer:
Succulents (water-storing) plants such as cacti, fix CO₂ into organic compound using PEP carboxylase at night when the stomata are open.

Question 6.
Which compound is meant for donating hydrogen to carbohydrate in Calvin cycle?
Answer:
NADPH is responsible for donating hydrogen to carbohydrate in Calvin cycle.

Question 7.
Indicate the main steps during Calvin cycle.
Answer:
The main steps during Calvin cycle are Carboxylation, Reduction, Regenaration.

Question 8.
Which of the mechanism of photosynthesis is responsible for trapping of light energy, splitting of water, oxygen release and formation of ATP and NADPH?
Answer:
Light reaction is responsible for all the above-mentioned conditions.

Question 9.
How many molecules of carbon dioxide, ATP and NADPH are required to make one molecule of glucose?
Answer:

1. Carbon dioxide 6 molecules
2. ATP 18 molecules
3. NADPH 12 molecules

Question 10.
What would happen to the rate of photosynthesis in C₃ -plants if CO₂ concentration level almost doubles from its present level in the atmosphere.
Answer:
If the CO₂ concentration in C₃ -plants almost get doubles from its present level in the atmosphere plants will grow much faster and leads to higher productivity due to higher rate of photosynthesis.

Question 11.
Why photosynthesis is an oxidation-reduction process?
Answer:
Photosynthesis is an oxidation-reduction process because water is oxidised to oxygen and carbon is reduced to carbohydrates.

Question 12.
The type of anatomy of leaves possessed by C₄-plant is different from those C₃-plant. Explain.
Answer:
C₄ -plant possess a special anatomy of leaves called Kranz anatomy, which means that the mesophyll tissue is undifferentiated in leaves

Short answer type questions

Question 1.
Give a brief explanation of photosynthesis.
Answer:

• Carbon dioxide is converted into sugars in a process called carbon fixation.
• Carbon fixation is a redox reaction, so photosynthesis needs to supply both a source of energy to drive this process and also the electrons needed to convert carbon dioxide into carbohydrate, which is a reduction reaction.
• In general outline, photosynthesis is the opposite of cellular respiration, where glucose and other compounds are oxidised to produce carbon dioxide, water, and release chemical energy.

Question 2.
Explain some early experiments that led to a gradual development in our understanding of photosynthesis?
Answer:
Early Experiments
(i) Joseph Priestley (1733-1804):
Priestley hypothesised that plants restore to the air whatever breathing animals and burning candles remove.

(ii) Jan Ingenhousz (1730-1799)
He showed that only the green parts of the plants could release oxygen.

(iii) Julius von Sachs

• He showed that the green substance (chlorophyll) is located in special bodies (chloroplasts).
• He provided evidence (1854) for the production of glucose in the green parts of plants and stored in the form of starch.

(iv) Engelmann (1843-1909):

• He split the light using a prism into its component parts and illuminated a green alga, Cladophora placed in a suspension of aerobic bacteria.
• The bacteria were used to detect the sites of oxygen evolution.
• He observed that the bacteria accumulated mainly in the region of blue and red light of the spectrum.
• He first described the action spectrum of photosynthesis; the action spectrum resembles roughly the absorption spectrum of chlorophyll a and b.

(v) Cornelius van Niel (1897-1985):

• He conducted experiments with purple and green sulphur bacteria, he demonstrated that photosynthesis is essentially a light-dependent reaction in which hydrogen from a hydrogen-donor reduces carbon dioxide to carbohydrates.
• He gave the present-day equation of photosynthesis.

• In green plants water (H₂O) is the hydrogen donor and it is oxidised to oxygen.
• Purple and green sulphur bacteria use H₂S as the hydrogen donor and so the oxidation product is sulphur and no oxygen is produced.
• Thus, he inferred that the oxygen evolved by green plants during photosynthesis comes from water (H₂O) and not from carbon dioxide (CO₂).
• This was later proved by using radioactive isotopes of oxygen(H7180).
• The overall correct equation for photosynthesis is as follows :

Question 3.
In chloroplast what are sites for light reactions and dark reactions?
Answer:
There is a clear division of labour in chloroplasts. The membrane system is responsible for Light reactions. Light energy is trapped by the membrane system and synthesis of ATP and NADPH takes place over there. The stroma utilises CO₂ to synthesize sugar; ATP and NADPH from light reaction are also utilized by stroma.

Question 4.
Give a brief account of light reaction.
Answer:
Light reactions or the ‘Photochemical’ phase include following steps :

• light absorption,
• water splitting,
• oxygen release, and
• the formation of high-energy chemical intermediates, ATP and NADPH.

Several complexes are involved in the process. The pigments are organised into two discrete photochemical light harvesting complexes (LHC) within the Photosystem I (PS I) and Photosystem II (PS II). These are named in the sequence of their discovery, and not in the sequence in which they function during the light reaction.

The LHC are made up of hundreds of pigment molecules bound to proteins. Each photosystem has all the pigments (except one molecule of chlorophyll a) forming a light-harvesting system also called antennae. These pigments help to make photosynthesis more efficient by absorbing different wavelengths of light.

The single chlorophyll a molecule forms the reaction centre. The reaction centre is different in both the photosystems. In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm, hence is called P700, while in PS II it has absorption maxima at 680 nm, and is called P680.

Question 5.
Explain the electron transport system in photosynthesis.
Answer:
Electron Transport: The whole scheme of electron transport starting from PS II, uphill to primary electron acceptor, downhill to cytochrome complex and PS I, excitation of PS I, transfer of electrons uphill to another acceptor and finally downhill to NADP+, is called Z-scheme, because of the characteristic shape, Z, formed when all the carriers are placed in a sequence according to their redox potential values.

Question 6.
What do you understand by splitting of water and its importance in photosynthesis?
Answer:
The splitting of water is associated with the PS II; water is split into H⁺, [0] and electrons. This creates oxygen, one of the net products of photosynthesis. The electrons needed to replace those removed from photosystem I are provided by photosystem II.
2H₂O → 4H⁺ +O₂ + 4e^–
2H₂O → 4H⁺ +O₂ + 4e^–.

Question 7.
Write a short note on chemiosmotic hypothesis.
Answer:
Chemiosmotic Hypothesis
ATP synthesis is linked to the development of a proton gradient across the membranes of thylakoids; it results due to the following reasons:

1. Since the splitting of the water molecules or photolysis takes place on the inner side of the thylakoid membrane, the protons -produced accumulate within the lumen of the thylakoids.
2. The primary electron acceptor is located towards the outer side of the membrane and transfers its electrons to the H carrier; so this molecule removes a proton from the stroma while transporting an electron and releases it into the lumen or inner side of the thylakoid membrane.
3. The enzyme NADP reductase is located on the stroma side of the membrane; along with the electrons coming from PS I, protons are also needed to reduce NADP and so these protons are also removed from the stroma.

The gradient is broken down due to the movement of protons across the membrane through transmembrane channel of the Fo of the ATP synthetase; the other portion of ATP synthetase, called Fi undergoes conformational changes with the energy provided by the breakdown of proton gradient and synthesises several molecules of ATP.

Question 8.
What do you understand by biosynthetic phase of photosynthesis?
Answer:
Synthesis of food (sugar) takes place during dark reaction. Since sugar is synthesised in this phase, it is also called as biosynthetic phase.

Long answer type questions

Question 1.
Is it correct to say that photosynthesis occurs only in leaves of a plant? Besides leaves, what are the other parts that may be capable of carrying out photosynthesis? Justify.
Answer:
Although all cells in the green parts of a plant have chloroplasts, most of the energy is captured in the leaves. The cells in the interior tissues of a leaf, called the mesophyll, can contain between 450000 and 800000 chloroplasts for every square millimetre of leaf.

The surface of the leaf is uniformly coated with a water-resistant waxy cuticle that protects the leaf from excessive evaporation of water and decreases the absorption of ultraviolet or blue light to reduce heating. The transparent epidermis layer allows light to pass through to the palisade mesophyll cells, where most of the photosynthesis takes place. The green stems are also capable of performing photosynthesis.

Question 2.
The entire process of photosynthesis consists of a number of reactions. Where in the cell do each of these take place?
(i) Synthesis of ATP and NADPH ……………………………………. .
(ii) Photolysis of water …………………………….. .
(iii) Fixation of CO₂ …………………………… .
(iv) Synthesis of sugar molecule ……………………………….. .
(v) Synthesis of starch ……………………………………… .
Answer:
(i) Synthesis of ATP and NADPH in thylakoids.
(ii) Photolysis of water occurs in inner side of thylakoid membrane.
(iii) Fixation of CO₂ occurs in stroma of chloroplast.
(iv) Synthesis of sugar molecule occurs in chloroplast.
(v) Synthesis of starch occurs in cytoplasm.

Question 3.
Which factors affect the process of photosynthesis? Explain in detail.
Answer:
Factors Affecting Photosynthesis
Photosynthesis is affected by both internal/plant factors and extemal/environmental factors.
The internal or plant factors that affect the rate of photosynthesis include
(i) the number, size, age and orientation of leaves,
(ii) mesophyll cells,
(iii) chloroplasts,
(iv) the amount of chlorophyll and
(v) the internal CO₂ concentration.

Blackman’s Law of Limiting Factors:
When a physiological process is controlled by a number of factors, the rate of the reaction depends on the slowest factor. This means that at a given time, only the factor which is the least (limiting) among all the factors, will determine the rate of the reaction.

(i) Light
Light quality and light intensity influence photosynthesis.
Light of wavelength between 400 nm and 700 nm is effective for photosynthesis and this light is known as photosynthetically active radiation (PAR).

As the intensity of light increases the rate of photosynthesis increases. But at higher light intensities, the rate of photosynthesis does not increase; it may be due to two reasons :
(a) Other factors needed for photosynthesis may be limiting.
(b) Destruction (photooxidation) of chlorophyll.

(ii) Temperature

• The photochemical phase is less affected by temperature.
• But the biosynthetic phase that involves enzyme-catalysed reactions, is more sensitive to temperature.
• The C₄ plants have a higher temperature optimum, while C₃ plants have a lower temperature optimum.

(iii) Carbon dioxide concentration

• In C₃ plants, the rate of photosynthesis increases with increase in CO₂ concentration, and saturation occurs beyond 450 μ/L^-1.
• In C₄ plants, the saturation is reached at a concentration of about 360 μ/L^-1.

(iv) Water
Water influences photosynthesis in two ways :

1. If available water decreases and plants show water stress, the stomata close; hence there will be a decreased supply of carbon dioxide for photosynthesis.
2. The leaves become wilted and the surface area for activities decreases.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 12 Mineral Nutrition Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 12 Mineral Nutrition

Very short answer type questions

Question 1.
Write the criteria for essentiality of an element.
Answer:
Criteria for Essentiality of an Element

• The element must be absolutely necessary for supporting normal growth and reproduction; in the absence of the element, the plants do not complete their life cycle.
• The requirement of the element must be specific and not replaceable by another element, i.e., deficiency of any one element cannot be met by supplying some other element.
• The element must be directly involved in the metabolism of the plant.

Question 2.
Name two imrhobile elements in plants.
Answer:
Calcium and sulphur are two immobile elements.

Question 3.
The deficiency of which element causes the death of stem and root apices?
Answer:
Boron.

Question 4.
The deficiency of this particular element causes the con dition of little leaf or mottle leaf in the plants. Name the element.
Answer:
Deficiency of element zinc.

Question 5.
Deficiency of mineral nutrition is not responsible for etiolation. Yes or No. If no then explain why?
Answer:
No, mineral deficiency is not responsible for etiolation because it is related to absence of light.

Question 6.
Where do the symptoms of deficiency of phosphorus appear first in the plant? And why?
Answer:
The deficiency of phosphorus appears first in older leaves. Because it is a mobile element.

Question 7.
A particular macroelement is obtained by the plants from both mineral and non-mineral sources. Identify the element.
Answer:
Nitrogen is the element which is obtained from both mineral and non-mineral sources.

Question 8.
Yellowish edges appear in leaves deficient in. [NCERT Exemplar]
Answer:
Magnesium.

Question 9.
Name the macronutrient which is a component of all organic compounds but is not obtained from soil. [NCERT Exemplar]
Answer:
Nitrogen.

Question 10.
Ammonification of nitrogen during nitrogen cycle is bring about by a special process. Identify the process.
Answer:
It is done by decomposition.

Question 11.
Organisms like Pseudomonas and Thiobacillus are of great significance in nitrogen cycle. How? [NCERT Exemplar]
Answer:
These organisms carry out denitrification. They help to maintain the constant level of nitrogen in the atmosphere.

Question 12.
A farmer adds Azotobacter culture to soil before sowing maize. Which mineral element is being replenished? [NCERT Exemplar]
Answer:
Nitrogen.

Short answer type questions

Question 1.
What are two macronutrients which are not obtained through soil as mineral nutrition? What is their importance?
Answer:
Oxygen and carbon are not obtained through soil as mineral nutrition. Oxygen is necessary for respiration and carbon dioxide is necessary for photosynthesis.

Question 2.
A fanner adds/supplies Na, Ca, Mg and Fe regularly to his field and yet he observes that the plants show deficiency of Ca, Mg and Fe. Give a valid reason and suggest a way to help the farmer improve the growth of plants.
Answer:
Calcium deficiency can be due to shortage of water so proper irrigation should be done. Iron deficiency can be due to high alkalinity of soil. And excessive use of potassium can result in poor absorption of magnesium. The farmer should take corrective actions.

Question 3.
Which enzyme is found in root nodules of leguminous plants? What role does it play?
Answer:
Nitrogenase is found in root nodules of leguminous plants. The enzyme binds with nitrogen and helps it in getting transported to the plant.

Question 4.
Write short notes on – Reductive animation and Transamination.
Answer:
(i) Reductive Animation: In these processes, ammonia reacts with α-Ketoglutaric acid and forms glutamic acid as indicated in the equation given below :
Glutamate

(ii) Transamination: It involves the transfer of amino group from one amino acid to the keto group of a keto acid. Glutamic acid is the main amino acid from which the transfer of NH₂, the amino group takes place and other amino acids are formed through transamination. The enzyme transaminase catalyses all such reactions. For example,

Long answer type questions

Question 1.
Hydroponics have been shown to be a successful technique for growing of plants, yet most of the crops are still grown on land. Why?
Answer:
The technique of growing plants in a nutrient solution is known as hydroponics.
This method requires purified water and mineral nutrient salts. After a series of experiments, in which the roots of the plants were immersed in nutrient solutions and an element was added/removed or given in varied concentration, a mineral solution suitable for the plant growth was obtained.

By this method, essential elements were identified and their deficiency symptoms discovered. Hydroponics has been successfully employed as a technique for the commercial production of vegetables such as tomato, seedless cucumber and lettuce.

Yet, most of the crops are still grown on land because the nutrient solutions must be adequately aerated to obtain the optimum growth in hydroponis. Moreover, the minerals must be continuously added in the solution. No, such activity is required in the soil.

Question 2.
We find that Rhizobium forms nodules on the roots of leguminous plants. Also Frankia, another microbe forms nitrogen fixing nodules on the roots of non-leguminous plant Alnus.
(i) Can we artificially induce the property of nitrogen fixation in a plant, leguminous or non-leguminous?
(ii) What kind of relationship is observed between mycorrhiza and pine trees?
(iii) Is it necessary for a microbe to be in close association with a plant to provide mineral nutrition? Explain with the help of one example. [NCERT Exemplar]
Answer:
(i) Yes, we can artificially induce the property of nitrogen fixation in a plant, leguminous or non-leguminous by genetic engineering. It involves the introduction of nif genes that cause the synthesis of nitrogenase enzyme by some vector in the plant in which we have to induce symbiosis.

(ii) Symbiotic association.

(iii) Yes, it is necessary for a-microbe to be in close association with a plant to provide mineral nutrition. For example, plants that contribute to nitrogen fixation include the legume family-Fabaceae or Leguminosae, such as clover, soyabeans, alfalfa, lupines and peanuts. They contain symbiotic bacteria called Rhizobia within nodules in their root systems, producing nitrogen compounds that help the plant to grow and compete with other plants.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 11 Transport in Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 11 Transport in Plants

Very short answer type questions

Question 1.
Define translocation.
Answer:
Transport over longer distances proceeds through the vascular system (the xylem and the phloem) and known as translocation.

Question 2.
Osmosis is a special kind of diffusion, in which water diffuses across the cell membrane. The rate and direction of osmosis depends upon both. [NCERT Exemplar]
Answer:
Pressure and concentration gradient.

Question 3.
The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane, because. [NCERT Exemplar]
Answer:
The cell wall is freely permeable to water and other substances in the solution but the plasma membrane is selectively permeable.

Question 4.
Why does rate of transport reach maximum or becomes saturated in facilitated diffusion?
Answer:
The transport rate reach a maximum because all the transport proteins are occupied/saturated.

Question 5.
Imbibition is considered a method of diffusion. Comment.
Answer:
Imbibition is considered as a method of diffusion because the movement of water occurs along the concentration gradient during this process.

Question 6.
Give one basic difference between antiport and uniport.
Answer:
In antiport both the molecules cross the membrane in opposite directions whereas, in uniport molecules moves across a membrane independent of any other passing molecule.

Question 7.
Mention two .factors on which net direction of molecules and rate of osmosis depends.
Answer:
The two factors responsible are:

1. Pressure gradient
2. Concentration gradient.

Question 8.
A flowering plant is planted in an earthen pot and irrigated. Urea is added to make the plant grow faster, but after sometime the plant dies. This may be due to which factor? [NCERT Exemplar]
Answer:
As urea content make the soil hypertonic in nature, therefore, the plant dies due to exosmosis.

Question 9.
The endodermis is impervious to water. Comment.
Answer:
The inner boundary of the cortex, i.e., endodermis is impervious to water because of a band of suberised matrix called the casparian strip.

Question 10.
Identify the vascular tissue responsible for translocation of organic and inorganic substances from leaves to other parts of the plant.
Answer:
Phloem is responsible for this type of translocation.

Question 11.
How do root hairs increase the absorption of water by plants?
Answer:
Root hairs increase the surface area of roots. This helps in making contact with larger volume of water. Thus, the presence of root hairs helps in absorption by plants.

Question 12.
It is seen that the number of stomata are greater on the lower surface of the leaf than the upper surface. Why is it so ?
Answer:
Stomata are present in greater number on the lower surface because if more number of stomata will be present on the upper surface, it would lead to great amount of water loss through transpiration. Thus, to avoid the excessive transpiration, stomata are present in greater number on lower surface of the leaf.

Question 13.
Elucidate the channels of food transport in plants.
Answer:
The channels of food transport are sieve tubes and sieve cells of phloem.

Question 14.
How are companion cells helpful to sieve tubes?
Answer:
The companion cells are connected to the sieve tubes by plasmodesmata and provide them with proteins, ATP and other nutrients.

Short answer type questions

Question 1.
Define facilitated diffusion.
Answer:
Membrane proteins provide sites at which movement of certain molecules takes place. These molecules have hydrophilic moiety and hence it is difficult for them to cross a membrane. They- need assistance of membrane proteins to cross the membrane. This is called facilitated diffusion.

Question 2.
Give a comparison table of different transport mechanisms.
Answer:
Comparison of Different Transport Mechanisms

Question 3.
Photosynthesis needs constant supply of water. But transpiration can hamper this supply. How do plants of desert area manage to get sufficient water in spite of faster transpiration?
Answer:
Desert plants have a different mechanism of photosynthesis and it is called C₄ pathways. The evolution of the C₄ photosynthetic system is probably one of the strategies for maximising the availability of CO₂ while minimising water loss. C₄ plants are twice as efficient as C₃ plants in terms of fixing carbon (making sugar). However, a C 4 plant loses only half as much water as a C₃ plant for the same amount of CO₂ fixed.

Question 4.
What is the significance of transpiration?
Ans. Significance of Transpiration

• Transpiration pull facilitates movement of water from roots.
• Transpiration supplies water for photosynthesis.
• It pulls minerals from soil.
• Helps in cooling of plants.
• Maintains shape of plant cells.

Question 5.
Describe the movement of water in leaves.
Answer:
Evaporation from the leaf sets up a pressure gradient between the outside air and the air spaces of the leaf. The gradient is transmitted into the photosynthetic cells and on the water-filled xylem in the leaf vein. This facilitates movement of water from xylem to the guard cells of stomata.

Long answer type questions

Question 1.
Observe the given figure and answer the following questions:

(i) State the nature of solution marked (1).
(ii) What process has been depicted in figures C to D?
(iii) In which figures turgor pressure will be zero?
(iv) In which figures wall pressure will he positive?
Answer:
(i) The solution marked as (1) will he hypertonic (more concentrated) due to which cell shrinks.
(ii) From C to D, figure is showing the process of deplasmolysis as shrinked cell has again regained its original shape.
(iii) Turgor pressure will be zero in figure B and C because cell is in a flaccid condition.
(iv) Wall pressure will be positive in figure A and D because in these figure cell wall is exerting ‘equal and opposite pressure against the expanding protoplasm.

Question 2.
Minerals are present in the soil in sufficient amount. [NCERT Exemplar]
(i) Do plants need to adjust the types of solute that reach xylem.
(ii) Which molecules help to adjust this?
Answer:
(i) An analysis of the xylem exudates shows that though some of the nitrogen travels as inorganic ions, much of it is carried in the organic form as amino acids and related compounds. Similarly, small amount of P and | S are carried as organic compounds. In addition, small amount of exchange of materials does take place between xylem and phloem.

(ii) Mineral ions are frequently remobilised, particularly from older, sensecing parts. Older dying leaves export much of their mineral content to younger leaves. Similarly, before leaf fall in deciduous plants, minerals are removed to other parts. Elements most readily mobilised are phosphorus, sulphur, nitrogen and potassium. Some elements that are structural components like calcium are not remobilised.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 19 Excretory Products and their Elimination Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 19 Excretory Products and their Elimination

Very short answer type questions

Question 1.
Write the mode of excretion performed by Xenopus.
Answer:
Dual excretion (mainly ammonotelism and partly ureotelism).

Question 2.
In which of the organism antennary glands are found as excretory organ?
Answer:
Crustaceans.

Question 3.
What is the excretory product from the kidney of reptiles? [NCERT Exemplar]
Answer:
Uric acid.

Question 4.
Give the name of vessel of peritubular capillaries that runs parallel to the loop of Henle.
Answer:
Vasa recta.

Question 5.
Give the name of the reservoir of urine in the body.
Answer:
Urinary bladder.

Question 6.
Give the name of cells that are responsible for the formation of filtration slits or slit pores.
Answer:
Podocytes (epithelial cells of Bowman’s capsule).

Question 7.
In which part of excretory system of mammals you can first use the term urine?
Answer:
The filtrate formed by the process of ultrafiltration in the Bowman’s capsule is called glomerular filtrate or primary urine.

Question 8.
What is the ratio of the concentrated filtrate to that of the initial filtrate?
Answer:
The concentrated urine (filtrate) is nearly four times concentrated than the initial filtrate formed.

Question 9.
What will be the effect on the amount of urine released when water is abundant in the body tissues in case of vertebrates?
Answer:
Vertebrates excrete large quantities of dilute urine when water is abundant in the body tissues and vice-versa.

Question 10.
What is the pH of urine? [NCERT Exemplar]
Answer:
It ranges from 4.5-8.2, average pH is 6.0.

Question 11.
What are the two substances responsible for causing the gradient for increasing hyperosmolarity of medullary interstitium?
Answer:
NaCl and urea.

Question 12.
Give the name of the main component that play an important role in the counter-current mechanism.
Answer:
Henle’s loop and vasa recta.

Short answer type questions

Question 1.
Describe the structure of human kidney.
Answer:

• Shape and Size of Kidney: Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 g.
• Structure of Kidney: Towards the centre of the inner concave surface of the kidney is a notch called hilum through which ureter, blood vessels and nerves enter.
• Inner Structure: Inner to the hilum is a broad funnel-shaped space called the renal pelvis with projections called calyces. The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and
• an inner medulla. The medulla is divided into a few conical masses (medullary pyramids) projecting into the calyces. The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini.

Question 2.
What is the function of proximal convoluted tubules in the kidney?
Answer:
Function of Proximal Convoluted Tubule (PCT): PCT is lined by simple cuboidal brush border epithelium which increases the surface area for reabsorption. Nearly all of the essential nutrients, and 70-80 percent of electrolytes and water are reabsorbed by this segment. PCT also helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, ammonia and potassium ions into the filtrate and by absorption of HCO from it.

Question 3.
What is the function of Henle’s loop?
Answer:
Function of Henle’s Loop: Reabsorption in this segment is minimum. However, this region plays a significant role in the maintenance of high osmolarity of medullary interstitial fluid. The descending limb of loop of Henle is permeable to water but almost impermeable to electrolytes. This concentrates the filtrate as it moves down. The ascending limb is impermeable to water but allows transport of electrolytes actively or passively. Therefore, as the concentrated filtrate pass upward, it gets diluted due to the passage of electrolytes to the medullary fluid.

Question 4.
What is the function of distal convoluted tubule?
Answer:
Function of Distal Convoluted Tubule (DCT): Conditional reabsorption of Na⁺ and water takes place in this segment. DCT is also capable of reabsorption of HCO and selective secretion of hydrogen and potassium ions and NH₃ to maintain the pH and sodium-potassium balance in blood.

Question 5.
What is the function of collecting duct?
Answer:
Collecting Duct

• Large amounts of water could be reabsorbed from this region.
• This segment also allows the transport of small amounts of urea into the medullary interstitium, to maintain the osmolarity.
• It also plays a role in maintaining pH and ionic balance of the body fluids by selective section of K⁺ and H⁺ ions.

Question 6.
How does reabsorption take place in the excretory system in human?
Answer:
Reabsorption: A comparison of the volume of the filtrate formed per day (180 litres per day) with that of the urine released (1.5 litres), suggest that nearly 99 per cent of the filtrate has to be reabsorbed by the renal tubules. This process is called reabsorption. The tubular epithelial cells in different segments of nephron perform this either by active or passive mechanisms. For example, substances like glucose, amino acids, Na+, etc., in the filtrate are reabsorbed actively whereas the nitrogenous wastes are absorbed by passive transport. Reabsorption of water also occurs passively in the initial segments of the nephron. During urine formation, the tubular cells secrete substances like H⁺, K⁺ and ammonia into the filtrate. Tubular secretion is also an important step in urine formation as it helps in the maintenance of ionic and acid base balance of body fluids.

Question 7.
Which gland releases ADH? What is the role of ADH in excretion?
Answer:
Hypothalamus releases ADH. ADH facilitates water reabsorption from latter part of tubules. This prevents diuresis. Excess fluid loss, through urine is called diuresis.

Question 8.
Write a short note on-disorders of the excretory system.
Answer:
Disorders of the Excretory System
The disorders related to kidneys are :

• Uremia
• Renal calculi and
• Glomerulonephritis

Hemodialysis is the process of removal of nitrogenous wastes from the blood of a uremia patient.
Kidney transplantation is the ultimate method of correcting urinary failure, in which a functioning kidney from a suitable donor is transplanted.

Hemodialysis

• Blood from the artery of an uremia patient is taken, cooled to 0°C and mixed with an anticoagulant like heparin.
• It is put into the cellophane tubes of the artificial kidney, where cellophane is permeable to micromolecules, but not to macromolecules like plasma protein.
• Outside the cellophane tube is the dialysing fluid, which has the same composition as that of plasma except the nitrogenous molecules like urea, uric acid, creatine, etc.
• Hence, the nitrogenous molecules from within the cellophane tubes flow into the dialysing fluid, following concentration gradient, (dialysis)
• The blood coming out of the artificial kidney is warmed to body temperature, mixed with antiheparin and restored to a vein of the patient.

Long answer type questions

Question 1.
The glomerular filtrate in the loop of Henle gets concentrated in the descending and then gets diluted in the ascending limbs. Explain. [NCERT Exemplar]
Answer:

• The gradient of increasing hyperosmolarity of medullary interstitium is maintained by a counter current mechanism and the proximity between the Henle’s loop and vasa recta.
• This gradient is mainly caused by NaCl and urea. The transport of these substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter current mechanism.
• NaCl is transported by the ascending limb of Henle’s loop, which is exchanged with the descending limb of vasa-recta. NaCl is returned to the medullary interstitium by the ascending part of vasa recta.
• But, contrarily, the water diffuses into the blood of ascending limb of vasa recta and is carried away into the general blood circulation.
• Permeability to urea is found only in the deeper parts of thin ascending limbs of Henle’s loops and collecting ducts. Urea diffuses out of the collective ducts and enters into the thin ascending limb.
• A certain amount of urea recycled in this way is trapped in medullary interstitium by the collecting tubule. This mechanism helps in the maintenance of a concentration gradient in the medullary interstitium.
• Presence of such gradient helps in an easy passage of water from the collecting tubule, resulting in the formation of concentrated urine (filtrate) i.e., nearly four times concentrated than the initial filtrate formed.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 4 Animal Kingdom Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 4 Animal Kingdom

Very short answer type questions

Question 1.
What are coelomates? Give two examples.
Answer:
Animals possessing coelom are called coelomates, e.g., Annelids, molluscs, arthropods, etc.

Question 2.
Body cavity is the cavity present between body wall and gut wall. In some animals the body cavity is not lined by mesoderm. Such animals are called: [NCERT Exemplar]
Answer:
Pseudocoelomates.

Question 3.
In some animal groups, the body is found to be divided into compartments with atleast some organs/organ repeated. Name this characteristic feature. [NCERT Exemplar]
Answer:
The segmentation that simultaneously divides body both externally and internally is called metamerism.

Question 4.
Name the group which lack digestive tract.
Answer:
Porifera.

Question 5.
Name an animal having canal system and spicules. [NCERT Exemplar]
Answer:
Scypha (sycon)/Leucosolenia.

Question 6.
Give an example of animal which exhibit alternation of generation. [NCERT Exemplar]
Answer:
Obelia, shows alternation of asexual and sexual phases.

Question 7.
Name the animal which exhibits the phenomenon of bioluminescence. Mention the phylum to which it belongs.
[NCERT Exemplar]
Answer:
Pleurobrachia, phylum – Ctenophora.

Question 8.
How is mesoglea different in ctenophores that in cnidarians?
Answer:
The mesoglea in ctenophores also contains amoebocytes, elastic fibres and muscle cells.

Question 9.
What is the role of radula in Mollusca? [NCERT Exemplar]
Answer:
It is the rasping organ that help in feeding.

Question 10.
Identify the animal in which adults exhibit radial symmetry and larvae exhibit bilateral symmetry. [NCERT Exemplar]
Answer:
In Echinodermata, the adults are radially symmetrical and larvae are bilaterally symmetrical.

Question 11.
Why are urochordates called tunicates?
Answer:
This is because the soft body of urochordates is surrounded by a thick test or tunic, often transparent or translucent.

Question 12.
In which fishes cartilaginous skeleton is present?
Answer:
Chondrichthyes.

Question 13.
Which amphibians show branchial (gills) respiration?
Answer:
Young ones of most of the amphibians which are aquatic show branchial respiration.

Question 14.
What is the importance of pneumatic bones and air sacs in Aves? [NCERT Exemplar]
Answer:

• Pneumatic bones are light but strong, the feature which helps in flight,
• Air sacs increase the efficiency of respiration and provide buoyancy to the animal.

Short answer type questions

Question 1.
Define tissue level of organization in the Kingdom Animalia. Give some examples of organisms showing tissue level of organization.
Answer:
In tissue level of organization there are specified groups of cells to carry out specific functions. This is a start of division of labour in the Animal Kingdom. Examples: Aurelia and Hydra.

Question 2.
What are diploblastic and triploblastic organisation?
Answer:
Animals in which the cells are arranged in two embryonic layers, an external ectoderm and an internal endoderm, are called diploblastic animals, e.g., coelenterates. An undifferentiated layer, mesoglea, is present in between the ectoderm and the endoderm. Those animals in which the developing embryo has a third germinal layer, mesoderm, in between the ectoderm and endoderm, are called triploblasdc animals (platyhelminthes to chordates).

Question 3.
Write two characters on the basis of which you can say that Coelenterata is more evolved than Porifera.
Answer:
(a) Porifera shows cellular level of organization, while Coelenterata shows tissue level of organisation.
(b) All members of Porifera are sessile, i.e., they are attached to the substratum, while some members of Coelenterata are motile, showing further improvement.

Question 4.
On the basis of which characters you can say that Aschelminthes are more advanced compared to Platyhelminthes?
Answer:
(a) Platyhelminthes are acoelomate,while Aschelminthes are pseudocoelomates. This indicates development of mesoderm.
(b)In Platyhelminthes both sexes are on the same animal, while in Aschelminthes there is segregation of sexes. This shows another point of evolution.

Question 5.
Give a description of the water vascular system in Echinodermata..
Answer:
Water Vascular System in Echinodermata: The water vascular system is a hydraulic system used by echinoderms, such as starfish and sea urchins, for locomotion, food and waste transportation, and respiration. The system is composed of canals connecting numerous tube feet. Echinoderms move by alternately contracting muscles that force water into the tube feet, causing them to extend and push against the ground, then relaxing to allow the feet to retract.

Long answer type questions

Question 1.
Differentiate between polyp and medusa. [NCERT Exemplar]
Answer:
Differences between polyp and medusa are as follows :

Polyp	Medusa
1. It is a fixed zooid.	It is free swimming.
2. It is asexual.	It is sexual.
3. It is cylindrical in outline.	Medusa is umbrella-shaped.
4. Tentacles found at upper end of manubrium.	Tentacles occur along the margin of umbrella.
5. Mouth is circular and terminal over upright manubrium.	Mouth is four sided, lies at the lower end of hanging manubrium.
6. Velum and sense organs are absent.	Medusa have a-circular velum and eight sense organs or statocysts.

Question 2.
Differentiate between Chordates and Non-Chordates.
Answer:
Differences between chordates and non-chordates are given below:

Chordates	Non-Chordates
•» Notocord is present at stages in some stages of development.	Notochord is absent.
•» Central nervous system is dorsal, hollow, single and non-ganglionated.	Central nervous system is ventral, solid, double and ganglionated.
•» Gill slits present on lateral side of pharynx in sum stages of throughout life.	Gill slits are absent.
•» Tail is present in some stages and throughout life.	Tail generally absent.
•» Heart is ventral.	Heart is dorsal.
•» Haemoglobin is present in RBCs.	It is present in plasma.

Question 3.
Which features make mammals as most successful and dominant animals?
Answer:
Features which make mammals as dominant and successful animals are as follows:

• The presence of an insulating and protective hairy exoskeleton
• They are warm blooded so have high rate of metabolism.
• They are viviparous animals and show placentation and intrauterine development which increases the chances for survival of young ones.
• They show high degree of parental care.
• They have more developed hearing efficiency due to the presence of pinna, three ear-ossicles and coiled cochlea in the ear.
• They are able to speak through language.
• They have good power of learning due to the presence of more developed brain.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 3 Plant Kingdom Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 3 Plant Kingdom

Very short answer type questions

Question 1.
Name the types of classification of plants.
Answer:
Artificial, natural and phylogenetic.

Question 2.
Which system indicates evolutionary as well as genetic relationships among organisms?
Answer:
Phylogenetic system of classification.

Question 3.
What is cytotaxonomy?
Answer:
Cytotaxonomy is a method of classification. It is based on cytological structure and their relatedness.

Question 4.
Food is stored as floridean starch in Rhodophyceae. Mannitol is the reserve food material of which group of algae?
[NCERT Exemplar]
Answer:
Phaeophyceae.

Question 5.
Holdfast, stipe and frond constitutes the plant body in case of
[NCERT Exemplar]
(a) Rhodophyceae
(b) Chlorophyceae
(c) Phaeophyceae
(d) All of these
Answer:
The lamina of Phaeophyceae members large sized body with differentiation of holdfast, stipe and lamina.

Question 6.
The plant body in higher plants is well differentiated and well developed. Roots are the organs used for the purpose of absorptions. What is the equivalent of the roots in the less developed lower plants? [NCERT Exemplar]
Answer:
Rhizoids.

Question 7.
Most algal genera show haplontic life cycle. Name an alga which is (i) Haplodiplontic (ii) Diplontic [NCERT Exemplar]
Answer:
Haplodiplontic: Ulva, Dictyota.
Diplontic: Fucus, Cladophora, Glomerata.

Question 8.
How are mosses considered ecologically important?
Answer:
Mosses, along with lichens are the first organism to colonise rocks, hence are ecologically important.

Question 9.
A prothallus is
(a) a structure in pteridophyte formed before the thallus develops
(b) a sporophytic free-living structure formed in pteridophyte
(c) a gametophytic free-living structure formed in pteridophytes
(d) a primitive structure formed after fertilisation in pteridophyte
[NCERT Exemplar]
Answer:
Gametophyte is a free-living small thalloid structure called prothallus. In most ferns, the prothallus is green and autotrophic.

Question 10.
Where are seeds located in gymnosperm?
Answer:
Seeds lie naked or exposed on the surface of megasporophyll.

Question 11.
The embryo sac of an angiosperm is made up of:
(i) 8 cells
(ii) 7 cells and 8 nuclei
(iii) 8 nuclei
(iv) 7 cells and 7 nuclei [NCERT Exemplar]
Answer:
(ii) Embryo sac of angiosperm develops up to 8 nucleate state prior to fertilisation. There is a three celled egg apparatus, three antipodal cells and two polar nuclei.

Question 12.
What is alternation of generations?
Answer:
Alernation of generations is regular switch over from gamete bearing haploid gametophyte to haploid spore producing diploid sporophyte.

Short answer type questions

Question 1.
What are the main differences among Chlorophyceae, Phaeophyceae and Rhodophyceae?
Answer:

Question 2.
What is the general structure of bryophytes?
Answer:
Structure of Bryophytes: It is thallus-like and prostrate or erect, and attached to the substratum by unicellular or multicellular rhizoids. They lack true roots, stem or leaves. They may possess root-like, or stem-like structures.

Question 3.
What is the general structure of pteridophytes?
Answer:

• The main plant body is a sporophyte which is differentiated into true root, stem and leaves. These organs possess well-differentiated vascular tissues.
• The leaves in pteridophyta are small (microphylls) as in Selaginella or large (macrophylls) as in ferns.
• The sporophytes bear sporangia that are subtended by leaf-like appendages called sporophylls. In some cases sporophylls may form distinct compact structures called strobili or cones (Selaginella, Equisetum).

Question 4.
Write short notes on:
(a) Importance of carbon fixation by algae.
(b) Importance of Gymnosperms
(c) Importance of Angiosperms
(d) Medicinal use of algae
Answer:
(a) About 50% of carbon fixation is done by algae. This enables majority of sea organisms to get the required food.
(b) Gymnosperms are mainly used as decorative plants. Certain paints are
prepared from Gymnosperm plants.
(c) Angiosperms are the major providers of food-grains to the mankind.
(d) Spirullina is made by algae and is used as a nutritional supplement.

Long answer type questions

Question 1.
Algae are known to reproduce asexually by a variety of spores under different environmental condition. Name these spores and the conditions under which they are produced. [NCERT Exemplar]
Answer:
Zoospores: Flagellate spores, under favourable conditions.
Aplanospores: Non-flagellate, thin-walled spores under approaching unfavourable conditions.
Hypnospores: Thick-walled, resting spores in unfavourable conditions.
Akinetes: Thin-walled and thick-walled spores formed from whole cells in unfavourable conditions. .
Autospores: Spores which look exactly like parent cell formed under favourable conditions.

Question 2.
Write about habit and habitat of algae.
Answer:
Habit and Habitat of Algae: Algae are chlorophyll-bearing, simple, thalloid, autotrophic and largely aquatic (both fresh water and marine) organisms. They occur in a variety of other habitats : moist stones, soils and wood. Some of them also occur in association with fungi (lichen) and animals (e.g., on sloth bear).

Question 3.
What are the differences betweenpinus and cycas?
Answer:

Pinus	Cycas
1. Roots are micorrhizal.	1. Roots are not micorrhizal.
2. Stems are branched.	2. Stems are unbranched.
3. Male and female strobili are on same tree.	3. They are on different trees.

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics

Very Short Answer Type Questions

Question 1.
Can a system be heated and its temperature remain constant? (NCERT Exemplar)
Answer:
If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.

Question 2.
Air pressure in a car tyre increases during driving. Explain. (NCERT Exemplar)
Answer:

• During driving, temperature of the gas increases while its volume remains constant.
• So, according to Charles’ law, at constant V, p ∝ T.
• Therefore, pressure of gas increases.

Question 3.
Write conditions for an isothermal process.
Answer:
The conditions for an isothermal process are :

• The walls should be diathermic.
• The process should be quasi-static.

Question 4.
Why air quickly leaking out of a balloon becomes cooler?
Answer:
Leaking of air is adiabatic expansion and adiabatic expansion produces cooling.

Question 5.
If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Answer:
Here, heat removed is less than the heat supplied and hence the room become hotter.

Question 6.
Is reversible process is possible in nature?
Answer:
A reversible process is never possible in nature because of dissipative forces and condition for a quasi-static process is not practically possible.

Question 7.
On what factors, the efficiency of a Carnot engine depends?
Answer:
On the temperature of source of heat and the sink.

Question 8.
Which thermodynamic law put restrictions on the complete conversion of heat into work?
Answer:
According to second law of thermodynamics, heat energy cannot converted into work completely.

Short Answer Type Questions

Question 1.
What are the limitations of the first law of thermodynamics?
Answer:
Following are the limitations of the first law of thermodynamics :

• It does not tell us about the direction of flow of heat.
• It fails to explain why heat cannot be spontaneously converted into work.

Question 2.
Two bodies at different temperatures T₁ and T₂ are brought in contact.
Under what condition, they settle to mean temperature? (after they attain equilibrium)
Answer:
Let m₁ and m₂ are masses of bodies with specific heats s₁ and s₂, then if their temperature after they are in thermal equilibrium is T.

Then, if > T₁> T₂ and assuming no heat loss.
Heat lost by hot body = heat gained by cold body
m₁s₁(T₁-T)=m₂s₂(T-T₂)
⇒ \(\frac{m_{1} s_{1} T_{1}+m_{2} s_{2} T_{2}}{m_{1} s_{1}+m_{2} s_{2}}\) = T[equilibrium temperature]
So for, bodies to settle down to mean temperature,
m₁ = m₂ and s₁ = s₂
means bodies have same specific heat and have equal masses.
Then, T = \(\frac{T_{1}+T_{2}}{2}\) [mean temperature]

Question 3.
When ice melts, then change in internal energy, is greater than the heat supplied, why?
Solution:
When ice melts, volume of water formed is less than that of ice. So, surroundings (environment) does work on the system (ice). And by first law,
ΔQ = ΔW+ΔU
⇒ ΔU = ΔQ-ΔW
(ΔW = negative as work is done on the system)
⇒ ΔU>ΔQ

Question 4.
Calculate the work done for adiabatic expansion of a gas.
Solution:
Consider (say µ mole) an ideal gas, which is undergoing an adiabatic expansion.
Let the gas expands by an infinitesimally small volume dV, at pressure p, then the infinitesimally small work done given by
dW = pdV
The net work done from an initial volume V₁ to final volume V₂ is given by
W= ∫v₁v² pdV
For an adiabatic process, pV^γ = constant = K

For an adiabatic process, K = p₁V^γ = p₂V^γ
For an ideal gas, p₁V₁ = μRT₁ and p₂V₂ = μRT₂.
So, we have
W = \(\frac{1}{(1-\gamma)}\left[\mu R T_{2}-\mu R T_{1}\right]=\frac{\mu R}{(\gamma-1)}\left[T_{1}-T_{2}\right]\)

Question 5.
What is a heat engine? What is the best way to increase efficiency of a heat engine? Is it possible to design a thermal engine that has 100% efficiency?
Solution:
A heat engine is a device (or a combination) which converts heat into work.
Its efficiency, η = \(\frac{\text { Work output }}{\text { Heat input }}=1-\frac{T_{2}}{T_{1}}\)
where, T₂ = temperature of sink
T₁ = temperature of source.
From above expression, we can see that for 100% efficiency, T₂ =0
It is impossible to design a thermal engine that has 100% efficiency because it is not possible to have a sink with kelvin temperature.

Question 6.
An ideal engine works between temperatures T₁ and T₂. It derives an ideal refrigerator that works between temperatures T₃ and T₄. Find the ratio Q₃/Q₁ in terms of T₁, T₂, T_3, and T₄.

Solution:
W = work done by engine = Q₁ – Q₂
and W = work done supplied to refrigerator = Q₃ -Q₄
Q₁ – Q₂ =Q₃ – Q₄
Dividing by Q₁, on both sides of the above equation, we get

Question 7.
Under what condition, an ideal Carnot engine has 100% efficiency?
Solution:
Efficiency of a Carnot engine is given by η = \(\left(1-\frac{T_{2}}{T_{1}}\right)\)
where, T₂ = temperature of sink
and T₁ = temperature of sink source
So for η = 1 or 100%, T₂ = 0 K or heat is rejected into a sink at 0 K temperature.

Question 8.
Draw p-V diagram of a Carnot cycle.
Solution:
p-V diagram for Carnot cycle

Long Answer Type Questions

Question 1.
A cycle followed by a machine (made of one mole of a perfect gas in a cylinder with a piston) is shown in figure

A to B: volume constant B to C: adiabatic
C to D: volume constant D to A : adiabatic
V_C = V_D = 2 V_A = 2 V_B
(i) In which part of the cycle, heat is supplied to the machine from outside?
(ii) In which part of the cycle, heat is being given to the surrounding by the machine?
(iii) What is the work done by the machine in one cycle? Write your answer in terms of P_A’ P_B’ V_A.
(iv) What is the efficiency of the machine?
Take γ = \(\frac{5}{3}\) for the gas and C_V =R for one mole.
Solution:
(i) A to B because T_B > T_A, as p ∝ T [ ∴ V = constant]
(ii) C to D because T_C>T_D, as P ∝ T [∴ V=constant]
(iii) W_AB = \(\int_{B}^{C} \) pdV=O and W_CD =0 [∴ V= constant]
Similarly,

(iv) Heat supplied during process A to B
dQ_AB = dU_AB
Q_AB = \(\frac{3}{2} n R\left(T_{B}-T_{A}\right)=\frac{3}{2}\left(p_{B}-p_{A}\right) V_{A}\)
∴ Efficiency = \(=\frac{\text { Net work done }}{\text { Heat supplied }}=\left[1-\left(\frac{1}{2}\right)^{2 / 3}\right]\)

Question 2.
Explain with the suitable example that a reversible process must be carried slowly and a fast process is necessarily irreversible.
Answer:
A reversible process must pass through equilibrium states which are very close to each other so that when process is reversed, it passes back through these equilibrium states. Then, it is again decompressed or it passes through same equilibrium states, system can be restored to its initial state without any change in surroundings. e.g., If a gas is compressed as shown But a reversible process can proceed without reaching equilibrium in intermediate states.