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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 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 10 Cell Cycle and Cell Division Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Very short answer type questions

Question 1.
During which phase of mitotic cell division, chromosomes gets separated?
Answer:
During anaphase.

Question 2.
Does mitosis occurs before or after the interphase?
Answer:
Yes, mitosis occurs before or after the interphase, as dividing phase (meiosis or mitosis) and interphase are considered only as the major phases of a cell cycle.

Question 3.
Mitosis cell division helps in regeneration process. How?
Answer:
Mitosis helps in regeneration by keeping all the somatic cells of an organism genetically similar, so that they are able to regenerate part or whole of the organism.

Question 4.
Given that average duplication time of E. coli is 20 minutes. How much time will two E. coli cells take to become 32 cells?
Answer:
2 hours (2ⁿ = 2⁵ = 2 × 2 × 2 × 2 × 2 = 32 generations).

Question 5.
If a tissue has 1024 cells at a given time, how many cycles of mitosis had the original parental single cell undergone?
[NCERT Exemplar]
Answer:
10 (2ⁿ, where n =10 generations).

Question 6.
Two key events take place during S-phase in animal cells, i.e., DNA replication and duplication of centriole. In which parts of the cell do these events occur? [NCERT Exemplar]
Answer:
DNA replication in the nucleus. Centriole duplication in the cytoplasm.

Question 7.
At what stage of meiosis, formation of tetrads occurs? Name it.
Answer:
Tetrads are formed during pachytene of prophase-I (meiosis-I).

Question 8.
Meiosis is essential in sexually reproducing organisms. How?
Answer:
Meiosis is essential in sexually reproducing organisms because it keeps the chromosome number constant.

Question 9.
Which cells of our body do not divide?
Answer:
Neuron cells stops dividing soon after the birth of a child.

Question 10.
The diagram shows a bivalent at prophase-I of meiosis. Which of the four chromatids can cross over? [NCERT Exemplar]
Answer:
The non-sister chromatids of homologous pair of chromosome undergo meiosis.

Short answer type questions

Question 1.
Imagine a situation if there was no meiosis. Then what would have happened to the next generation?
Answer:
In the absence of meiosis the next generation would have double the number of chromosomes after fusion of gametes. This would have resulted in the birth of an altogether new species. The maintenance of characters set would have been possible only through asexual reproduction.

Question 2.
Give a description of metaphase I of meiosis.
Answer:
Metaphase I: The bivalent chromosomes align on the equatorial plate. The microtubules from the opposite poles of the spindle attach to the pair of homologous chromosomes.

Question 3.
Describe telophase I of meiosis.
Answer:
Telophase I

• The nuclear membrane and nucleolus reappear, cytokinesis follows and this is called as diad of cells.
• Although in many cases the chromosomes do undergo some dispersion, they do not reach the extremely extended state of the interphase nucleus.
• The stage between the two meiotic divisions is called interkinesis and is generally short lived. Interkinesis is followed by prophase II, a much simpler prophase than prophase I.

Question 4.
What is the process of cell division in prokaryotes?
Answer:
Prokaryotes do not have nucleus. So, there is no elaborate karyokinesis, as seen in eukaryotes. In prokaryotes the replication of DNA starts the process of cell division. Once genetic material is replicated, it is followed by division of cytoplasm. The process is known as binary fission.

Question 5.
How does meiosis facilitate creation of offsprings, with distinct characters?
Answer:
Meiosis happens during gametogenesis and as a result gametes have half the number of chromosomes. During fertilization, when gametes fuse together two different sets of chromosomes make a new set. This results in an offspring, who has distinct characters, compared to parents.

Question 6.
What is the significance of mitosis?
Answer:
Significance of Mitosis

• In multicellular organisms, body growth is by mitotic divisions of the cells.
• Replacement of worn out tissues/cells (e.g., blood cells, skin cells) and repair of the injured tissues is by mitosis.
• In unicellular organisms, mitosis are involved in asexual reproduction
  (multiplication of cells).
• In plants, vegetative propagation involves only mitotic divisions and genetically identical individuals are produced.
• Uncontrolled cell divisions in certain tissues/organ (cancer) result in tumours.

Long answer type questions

Question 1.
Briefly describe the significance of cell division.
Answer:
Cell division is significant in the following ways :

• Cell Multiplication: Cell division is a means of cell multiplication or formation of new cells from pre-existing cells.
• Continuity: It maintains continuity of living matter generation after generation.
• Multicellular Organisms: The body of a multicellular organism is formed of innumerable cells. They are formed by repeated divisions of a single cell or zygote. As the number of cells increases, many of them begin to differentiate, form tissues and organisms.
• Cell Size: Cell division helps in maintenance of a particular cell size which is essential for efficiency and control of cell activities.
• Genetic Similarity: The common type of cell division or mitosis maintains genetic similarity of all the cells in an individual despite being different, i.e., structurally and functionally.

Question 2.
Explain meiosis-II in an animal cell.
Answer:
All these happen in the two haploid nuclei simultaneously.

• Prophase-II: It takes short time. Spindle formation begins and the chromosomes become short. Two chromatids, are joined to a single centromere. Nuclear membrane and nucleolus disintegrate.
• Metaphase-II: At the equator, the chromosomes lie and spindle is formed. The centromere of every chromosomes is joined to the spindle fibre and centromere also divides.
• Anaphase-II: The daughter chromosomes are formed. Chromatids move towards their poles with the spindle fibres.
• Telophase-II: Reaching at the poles, chromosomes form nuclei which are haploid (n) daughter nuclei. Again nuclear membrane is constructed. Nucleolus now becomes clearly visible.
• Cytokinesis: It occurs and four daughter cells are formed which are haploid (n). It may occur once or twice (i.e., in meiosis-I and II) or only after the meiosis-II cell division.

Question 3.
Describe briefly the phases of meiotic division.
Answer:
Meiotic division takes place in germ cells. The number of chromosomes is reduced to half in daughter cells.

Meiotic cell division is divided into two phases, i.e., meiotic-I and II.
In the meiotic-I division, the homologous chromosomes pair to form bivalent. Exchange of genetic material takes place. The chromosomes now separate and get distributed into daughter cells.

Long prophase-I is divided into five sub-stages, i.e., leptotene, zygotene, pachytene, diplotene and diakinesis. During metaphase-I, the bivalents arrange themselves on equatorial plate with their arms on the plate but the centromere is directed towards opposite pole. It is followed by anaphase-I. Now, the homologous chromosomes repel each other, move to the opposite poles with both their chromatids. In this way each pole gets half the chromosomes number of the parent cell.

In telophase-I, the nuclear envelope and nucleolus again appear. The centromere of each chromosome breaks, separating the chromatids, one each to a daughter cell. The meiotic cell division maintains chromosome number of a species.

As a result of meiotic division, the four daughter cells are formed with half the chromosome number (haploid) in each cell.