Important Questions

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants

Very short answer type questions

Question 1.
An anther with malfunctioning tapetum often fails to produce viable male gametophytes. Give any one reason.
Answer:
A malfunctioning tapetum does not provide enough nourishment to the developing male gametophytes and thus fail to produce viable male gametophytes.

Question 2.
Complete the following flow chart
Pollen mother cell → Pollen tetrad

[NCERT Exampler]
Answer:
Generative cell.

Question 3.
Gynoecium of a flower may be apocarpous or syncarpous. Explain with the help of an example each.
Answer:
Gynoecium of a flower may be apocarpous means the carpels are free from one another and there is no fusion of any part e.g., Ranunculus, Rose. Gynoecium of a flower is syncarpous, means the carpels are fused by their ovaries. The number of fusing carpels may vary from 2 (Petunia) to ∞ (Hibiscus).

Question 4.
Name the parts of the gynoecium which develop into fruit and seeds. [NCERT Exemplar]
Answer:
Ovary develops into fruit and ovules develop into seeds.

Question 5.
How many haploid cells are present in a mature female gametophyte of a flowering plant? Name them.
Answer:
One dikaryotic polar cell with two haploid nuclei and six haploid cells, viz, 3 antipodal, 2 synergids and 1 egg.

Question 6.
Name the type of pollination in self-incompatible plants. [NCERT Exemplar]
Answer:
Xenogamy.

Question 7.
How do flowers of Vallisneria get pollinated?
Answer:
In Vallisneria, the female flower stalk is coiled to reach the water surface to receive the pollen grains carried by water currents.

Question 8.
What is pollen-pistil interaction and how is it mediated?
Answer:
The pistil accepts the right type (compatible) of pollen and promotes fertilisation and rejects the pollen of other species and incompatible pollen of the same species. It is the result of interaction between the chemicals of the pollen and those of stigma.

Question 9.
State the function of filiform apparatus found in mature embryo sac of an angiosperm.
Answer:
Filiform apparatus plays an important role in guiding the path of pollen tubes into the synergids.

Question 10.
Normally one embryo develops in one seed but when an orange seed is squeezed many embryos of different shapes and sizes are seen. Mention how it has happened?
Answer:
An orange seed has many embryos because of polyembryony.

Question 11.
Name the component cells of the ‘egg apparatus’ in an embryo sac. [NCERT Exemplar]
Answer:
Two synergids and an egg.

Question 12.
Name the common function that cotyledons and nucellus perform. [NCERT Exemplar]
Answer:
Cotyledons and nucellus provide nourishment.

Question 13.
In the embryos of a typical dicot and a grass, which are the true homologous structures? [NCERT Exemplar]
Answer:
Cotyledons and scutellum.

Question 14.
In a case of polyembryony, if an embryo develops from the synergid and *another from the nucellus, which is haploid and which is diploid? [NCERT Exemplar]
Answer:
Synergid embryo is haploid and nucellar embryo is diploid.

Short answer type questions

Question 1.
Name the organic materials the exine and intine of an angiosperm pollen grains are made up of. Explain the role of exine.
Answer:
Exine is made up of sporopollenin and intine is made up of cellulose and pectin.
Due to the sporopollenin, exine can withstand high temperature and strong acids. It is also not affected by enzymes. It is because of this reason that pollen grains are well preserved as fossils.

Question 2.
Differentiate between geitonogamy and xenogamy in plants. Which one between the two will lead to inbreeding depression and why?
Answer:

Geitonogamy	Xenogamy
1. It is transfer of pollen grains from the anther to the stigma of another flower of same plant.	It is transfer of pollen grains from the anther to the stigma of a different plant.
2. The pollen grains are genetically similar to the plant.	The pollen grains are genetically different from the plant.

Geitonogamy will lead to inbreeding depression because the pollen grains are genetically similar, which results in inbreeding. Continued inbreeding will thus reduce fertility and productivity.

Question 3.
Double fertilisation is reported in plants of both, castor and groundnut. However, the mature seeds of groundnut are non-albuminous and castor are albuminous. Explain the post fertilisation events that are responsible for it.
Answer:
The development of endosperm (preceding the embryo) takes place from primary endosperm nucleus (PEN) in both, castor and groundnut. The developing embryo derives nutrition from endosperm.

PEN undergoes repeated division to give free nuclei. Subsequently cell wall is formed and endosperm becomes cellular. At this stage endosperm is retained in castor or is not fully consumed but in groundnut endosperm is consumed by growing embryo.

Question 4.
Differentiate between albuminous and non-albuminous seeds, giving one example of each.
Answer:

• Albuminous seeds have residual endosperm ip them. For example, maize.
• Non-albuminous seeds do not have any residual endosperm. For example, pea.

Question 5.
A non biology person is quite shocked to know that apple is a false fruit, mango is a true fruit and banana is a seedless fruit. As a biology student how would you satisfy this person?
Answer:
In apple only the thalamus (along with ovary) portion contributes to fruit. Therefore, it is a false fruit. Mango develops only from the ovary, therefore it is a true fruit.
Banana develops from ovary but without fertilisation. The method is known as parthenocarpy. Since there is no fertilisation, no seeds are formed.

Question 6.
Why are some seeds referred to as apomictic seeds? Mention one advantage and one disadvantage to a farmer who uses them.
Answer:
Seeds produced without fertilisation are referred to as apomictic.
Advantage: Desired characters are retained in offspring (progeny) as there is no segregation of characters in offspring (progeny). Seed production is assured in absence of pollinators.

Disadvantage: Cannot control accumulation of deleterious genetic mutation. These are usually restricted to narrow ecological niches and lack ability to adapt to changing environment.

Long answer type questions

Question 1.
A flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds.
Answer the following questions giving reasons:
(a) How many ovules are minimally involved?
(b) How many megaspore mother cells are involved?
(c) What is the minimum number of pollen grains that must land on stigma for pollination?
(d) How many male gametes are involved in the above case?
(e) How many microspore mother cells must have undergone reduction division prior to dehiscence of another in the above case?
Answer:
(a) 360 ovules are involved. One ovule after fertilisation forms one seed.
(b) 360 megaspore mother cells are involved. Each megaspore mother cell forms four megaspores out of which only one remains functional.
(c) 360 pollen grains. One pollen grain participates in fertilisation of one ovule.
(d) 720 male gametes are involved. Each pollen grain carries two male gametes (which participate in double fertilisation) (360 × 2 = 720).
(e) 90 microspore mother cells undergo reduction division. Each microspore mother cell meiotically divides to form four pollen grains (360/4 = 90).

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 16 Environmental Issues Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 16 Environmental Issues

Very short answer type questions

Question 1.
Why are lichens regarded as pollution indicators?
Answer:
Lichens are regarded as pollution indicators because they do not grow in areas that are polluted. So their presence indicates no pollution in that area and their absence indicates that the area is polluted.

Question 2.
Why is the use of unleaded petrol recommended for motor vehicles equipped with catalytic converters?
Or
Why are owners of motor vehicles equipped with catalytic converters advised to use unleaded petrol?
Or
Why is it desirable to use unleaded petrol in vehicles fitted with catalytic converters?
Answer:
Vehicles fitted with catalytic converters need to use unleaded petrol because lead inactivates the catalyst in the catalytic converter and increases hydrocarbon emission, thereby harming the environment.

Question 3.
In which year was the Air (Prevention and Control of Air Pollution) Act amended to include noise as air pollutant? [NCERT Exemplar]
Answer:
The Air (Prevention and Control of Air Pollution) Act 1981 was amended in 1987 to include noise as an air pollutant.

Question 4.
Name an industry which can cause air pollution, thermal pollution, and eutrophication. [NCERT Exemplar]
Answer:
Fertilizer factory.

Question 5.
Mention two advantages for preferring CNG over diesel as an automobile fuel.
Answer:
Advantages of CNG :

1. It burns more efficiently than petrol or diesel.
2. It is cheaper than petrol or diesel and cannot be siphoned off by thieves or adulterated.

Question 6.
What is an algal bloom? [NCERT Exemplar]
Answer:
The excessive growth of algae (free-floating) that causes coloration of water bodies is called algal bloom.

Question 7.
Write the name of the organism that is referred to as the ‘Terror of Bengal’.
Or
Why is Eichhomia crassipes nicknamed as* ‘Terror of Bengal’?
Answer:
Eichhornia crassipes is nicknamed as ‘Terror of Bengal’ because it grows very fast in the water body and depletes the dissolved oxygen. Hence, disturbing the ecosystem dynamics of the water body.

Question 8.
What is the raw material for polyblend? [NCERT Exemplar]
Answer:
Polyblends are natural man-made fibres, made by the mixture of two or more polymers, especially plastic waste products.

Question 9.
Mention the effect of UV rays on DNA and proteins in living organisms.
Answer:
The high energy of UV rays breaks the chemical bonds within DNA and protein molecules.

Question 10.
Write the unit used for measuring ozone thickness.
Answer:
Dobson unit.

Question 11.
State the purpose of signing the Montreal Protocol.
Answer:
Montreal Protocol was signed at Montreal, in 1987 to curb the emission of ozone-depleting substances.

Question 12.
What is reforestation? [NCERT Exemplar]
Answer:
Reforestation is the process of restoring a forest that once existed but was removed at some point of time in the past.

Short answer type questions

Question 1.
It is a common practice to plant trees and shrubs near the boundary walls of buildings. What purpose do they serve?
[NCERT Exemplar]
Answer:
The practice of growing and maintaining trees and shrubs near the boundary wall of residential or official buildings is a common practice. This is because it acts as a barrier for sound and check noise pollution. This green belt of trees and shrubs also acts as an effective measure to check primary air pollutants like dust, fly ash, etc.

Question 2.
‘Determination of Biological Oxygen Demand (BOD) can help in suggesting the quality of a water body. Explain.
Answer:
Biochemical Oxygen Demand (BOD) is the amount of dissolved oxygen required for microbial breakdown of biodegradable organic matter. Aerobic organisms use a lot of oxygen and as a result, there is a sharp decline in Dissolved Oxygen (DO) in the water body. This Can cause death of fishes and other aquatic species. The relationship between BOD and sewage can be understood from the graph given below :

Determination of BOD is thus an important parameter in suggesting the quality of a water body. The presence of more organic waste increases the microbial activity thus decreasing the DO. BOD is higher in polluted water and lesser in clean water.

Question 3.
Explain the process of secondary treatment given to the primary effluent up to the point it shows a significant change in the level of biological oxygen demand (BOD) in it.
Answer:
The primary effluent is passed into large aeration tanks where it is constantly agitated. Air is pumped into it mechanically. This allows vigorous growth of useful aerobic microbes into floes. These microbes consume the major part of organic matter in the effluent (this significantly reduces the BOD of the effluent).

Question 4.
Excessive nutrients in a freshwater body cause fish mortality. Give two reasons.
Answer:
Excessive nutrients in the water body result in the excessive and rapid growth of plants and animals, resulting in formation of increased organic matter, death of plants and animals increased the organic matter at the bottom which decomposes, increased BOD deplete the oxygen content, resulting in fish mortality.

Question 5.
Is it true that, if the dissolved oxygen level drops to zero, the water will become septic? Give an example which could lower the dissolved oxygen content of an aquatic body. [NCERT Exemplar]
Answer:
Yes, it is true. In case of zero level of Dissolved Oxygen (DO), the water becomes septic. Organic pollutants like fertilizer in aquatic bodies are responsible for lowering (up to zero) the level of dissolved oxygen.

Question 6.
Explain giving reasons why thermal power plants are not considered eco-friendly?
Answer:
Thermal power plants release particulate and gaseous pollutants in the environment. Inhalation of these pollutants can cause breathing or respiratory symptoms^irritation, inflammation, damage to lungs, and even premature death.

Question 7.
Blend of polyblend and bitumen, when used, helps to increase road life by a factor of three. What is the reason? [NCERT Exemplar]
Answer:
Polyblend is a fine powder of recycled modified plastic. The binding property of plastic makes the road last longer besides giving added strength to withstand more loads. This is because: plastic increases the melting point of the bitumen which would prevent it from melting in India’s hot and extremely humid climate, where temperature frequently crosses 50°C. rainwater will not seep through because of the plastic in the blend.

Question 8.
What is the main idea behind ‘Joint Forest Management Concept’ introduced by the Government of India?
[NCERT Exemplar]
Answer:
The main idea behind joint forest management concept introduced by the Government of India was involving the local communities in forest conservation. This concept was adopted considering the extraordinary courage and dedication the local people showed in protecting the wildlife through the movements like Bishnoi’s movement in Jodhpur and Chipko Movement in Garhwal Himalayas.

Long answer type questions

Question 1.
Explain giving reasons the cause of appearance of peaks ‘a’ and ‘b’ in the graph shown below:
img
Answer:
‘a’-High BOD due to sewage discharge.
‘b’-Increase in dissolved oxygen due to sewage decomposition. Micro-organisms involved in biodegradation of organic matter consume a lot of oxygen, therefore, there is a sharp decline in dissolved oxygen. When the sewage is completely degraded, oxygen concentration again increases.

Question 2.
Why is the concentration of toxins found to be more in the organisms occupying the highest trophic level in the food chain in a polluted* water body? Explain with the help of a suitable example.
Answer:
The concentration of toxic materials like heavy metals and pesticides increase at each trophic level of a food chain and is more in organisms of highest trophic level due to their accumulation at each trophic level For example, when DDT was used to control mosquitoes in a lake of USA, 800 times more DDT was found in the phytoplanktons than in the water of the lake. Zooplanktons had about 13 times more DDT than phytoplanktons. It was also observed that the fishes population had 9-40 times more DDT than zooplanktons and fish-eating birds had 25 times more DDT than fish.

Question 3.
Refrigerants arc considered to be a necessity in modem living but are said to be responsible for ozone holes detected in Antarctica. Justify.
Answer:
The widely used refrigerants are CFCs or. chlorofluorocarbons. CFCs discharged in the lower part of atmosphere move upwards to the stratosphere. Here, the UV rays act on them and release chlorine atoms. These free chlorine atoms react with ozone to release molecular oxygen. Chlorine atoms are not consumed in this reaction and hence, these continuously degrade ozone and have resulted in ozone holes.

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.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 14 Respiration in Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 14 Respiration in Plants

Very short answer type questions

Question 1.
Why does anaerobic respiration/fermentation yields less energy than aerobic respiration?
Answer:
It happens due to incomplete oxidation of the substrate.

Question 2.
Write the overall equation of respiration.
Answer:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Energy

Question 3.
Name two openings in plants through which exchange of gases takes place?
Answer:
Stomata and lenticels.

Question 4.
Write the reaction where substrate-level phosphorylation takes place in glycolysis.
Answer:
Substrate level phosphorylation takes place during the following reaction

• When 1, 3-bisphosphoglycerate is converted into 3-phosphoglycerate.
• When phosphoenolpyruvate is converted into pyruvic acid.

Question 5.
List two instances where lactic acid is formed by fermentation.
Answer:
Instances, where lactic acid is formed by fermentation, are given below:

1. During fermentation by lactic acid bacteria.
2. During strenuous exercise, in the striated muscles in humans.

Question 6.
Mention the step of citric acid cycle, which is not mediated by dehydrogenase enzyme.
Answer:
Conversion of oxaloacetic acid to citric acid is not mediated by dehydrogenase enzyme.

Question 7.
At which step of respiration, hydrogen of NADH₂ is used?
Answer:
The hydrogen atoms accepted by NADH₂ during glycolysis are introduced to route I of ETS. In this route 3 ATP molecules are produced.

Question 8.
Mention the number of protons that passes through complex V for the synthesis of 2 molecules of ATP.
Answer:
Two pairs of protons (i.e., 4) passes through complex V for the synthesis of two molecules of ATP.

Question 9.
Name the inhibitor of oxidative phosphorylation.
Answer:
Oligomycin.

Question 10.
Name the unit of oxidative phosphorylation.
Answer:
Oxysomes.

Question 11.
F₀ – F₁ particles participate in the synthesis of ……………………. [NCERT Exemplar]
Answer:
ATP (Adenosine Triphosphate), the energy currency of the cell.

Question 12.
What do you mean by respiratory balance sheet?
Answer:
Respiratory Balance Sheet
The calculations of the net gain of ATP for every glucose molecule oxidized, can be made only on certain assumptions.
But this kind of assumptions are not valid in a living system for the following reasons :

• All pathways work simultaneously and do not take place one after the other.
• Substrates keep entering the pathways and are also withdrawn from the pathways.
• ATP is utilized as and when needed,
• Rates of enzyme actions are controlled by multiple means.

Short answer type questions

Question 1.
Why ATP is called an energy currency?
Answer:
The energy produced during various steps of cellular respiration is stored in the form of ATP. This is later utilized on an SOS basis. So, ATP is also called as energy currency.

Question 2.
There is no special respiratory organ in plants, yet plants efficiently manage exchange of gases. Justify.
Answer:
Every part of plant manages its gas exchange need. There is no exchange of gases between different organs. So unlike animals plants do not need special respiratory organs to facilitate exchange of gases. In leaves the exchange of gases takes place through stomata, while in stems it takes place through lenticels.

Question 3.
How glycolysis takes place in anaerobic environment?
Answer:
Glycolysis is the breakdown of glucose into pyruvic acid and it does not need oxygen. So in all living beings, irrespective of them being either aerobic or anaerobic glycolysis takes place. In fact glycolysis is the first step towards oxidation of glucose and oxidation takes place either during anaerobic respiration or during aerobic respiration.

Question 4.
The maximum concentration of alcohol produced by natural fermentation is 13%. But most of the alcoholic preparations for human consumption contain more than this percentage. How this higher percentage is achieved?
Answer:
The higher percentage of alcohol is achieved through distillation of the liquid, which gives pure alcohol as well. The boiling involved in distillation helps evaporate the liquid part and higher concentration is achieved.

Long answer type questions

Question 1.
Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Answer:

• Oxygen is an essential requirement for aerobic respiration because an element of strong electronegativity to pull the electrons down the chain is needed.
• It ensures that protons are pumped into the outer lumen of the mitochondria, where they can come down their concentration gradient through ATP-synthase making ATP.
• The oxygen picks up electrons and protons, thus forming water. As the electrons in the ETS are used to do work, the electrons lose energy and reach a point at the end of the ETS, where they have to be gotten rid of.
• The scheme the cell uses to do this is to combine the electrons with hydrogen ions and oxygen to produce water.

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 11th Class Biology Important Questions Chapter 15 Plant Growth and Development Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 15 Plant Growth and Development

Very short answer type questions

Question 1.
Identify the actively dividing cells in plants.
Answer:
Meristems are the actively dividing cells present in the plants.

Question 2.
What happens if the meristematic cells ever cease to divide?
Answer:
If meristematic cells cease to divide, the growth of the plant will be hindered and will undergo a period of dormancy depending upon the seasonal changes in the climate. „

Question 3.
Growth is one of the characteristics of all living organisms? Do unicellular organisms also grow? If so, what are the parameters?
Answer:
Yes, unicellular organisms also grow. Their cell size increases up to a certain fixed dimension only.

Question 4.
Mention the name of the internal factors that control development in plants.
Answer:
Internal factors that control development in plants are as follows:

1. Genetic factors (intracellular)
2. Plant growth regulators (intercellular).

Question 5.
Identify the plant hormone-related with intermodal elongation.
Answer:
Gibberellin is the plant hormone-related with internodal elongation.

Question 6.
Mention the name the growth regulator, which was first isolated from endosperm of maize. Give its main biological activity.
Answer:
Zeatin is the growth regulator isolated from endosperm of maize. It controls cell division (cytokinesis) even in non-meristematic tissues.

Question 7.
In most plants, the terminal bud suppresses the development of lateral buds into branches. What is this phenomenon called? Name one phytohormone that can promote this phenomenon.
Answer:
The phenomenon is called apical dominance. Auxin is the phytohormone involved in prompting this phenomenon.

Question 8.
Which air pollutant also acts as a plant hormone?
Answer:
Ethylene.

Question 9.
How do gibberellin help in promoting seed germination?
Answer:
The gibberellin mobilize storage reserves by amylases during germination of seeds.

Question 10.
What are the plant organs responsible for the perception of light variation? What is the pigment responsible for this perception?
Answer:

• Leaves are mainly responsible for perception of light intensity in plants.
• The pigment that performs this perception is called phytochrome.

Question 11.
Name the hormones involved in photoperiodism.
Answer:
Florigen is the hormone involved in photoperiodism.

Question 12.
Beetroot is often known as a long-day plant. Explain why?
Answer:
Beetroot is known as long-day plant because flowering takes place when the plants are exposed to day length longer than a critical period.

Short answer type questions

Question 1.
An owner of an apple orchard wants to get better yield and wants to wait for good market conditions to sell his apples. Which PGR should he use to achieve his goals?
Answer:
He should use Gibberellins. Gibberellins help increase the size of apples. Moreover, they also delay senescence so apples can be left on branches for a longer duration. This will give the orchard owner enough time to wait for good market conditions.

Question 2.
What are plasticity and heterophylly?
Answer:
In some plants, certain structures show different forms, in response to environment or to phases of life. This ability is known as plasticity.

For example, in cotton, coriander, and larkspur, leaves of juvenile plants are different in shape compared to leaves in mature plants. This is called heterophylly. Juvenile In buttercup shape of leaves produced in air is different from that produced in water.

Question 3.
What are the favorable conditions for seed germination?
Answer:
Favorable conditions for seed germination are given below:

• Proper temperature
• Moisture
• Sunlight
• Oxygen.

Question 4.
What are the various man-made meant of overcoming seed dormancy?
Answer:
Man-made means of overcoming seed dormancy are given below :
The seed-coat barrier in some seeds can be broken by mechanical abrasions using knives, sandpaper, etc. or by vigorous shaking.

Effect of inhibitory substances can be removed by subjecting the seeds to chilling conditions or by application of certain chemicals like gibberellic acid and nitrates. Changing environment conditions: like light and temperature.

Question 5.
What do you understand by apical dominance?
Answer:
In most of the higher plants, growing apical bud inhibits the growth of lateral (axillary buds). This phenomenon is called apical dominance. Removal of short tips (decapitation) normally results in the growth of lateral buds. Decapitation is used in tea plantations to get more leaves from a plant.

Long answer type questions

Question 1.
Mention the phenomenon of growth in plants. Explain the phases of growth in detail.
Answer:
Growth is defined as a permanent or irreversible increase in dry weight, mass or volume of cell, organ or organism.
Plant growth takes place in three steps or phases cell, division (meristematic), cell elongation and cell maturation.
(i) Cell Division (Meristematic) Phase:

• It is also called formative phase.
• New cells are produced by mitotic divisions of the pre-existing cells. The meristematic cells have thin cellulose walls with abundant plasmodesma connections, dense protoplasm and conspicuous nuclei.
• In higher plants cell division occurs in meristems or growing points.
• As the formation of new cells requires intense biosynthetic activity, the rate of respiration in the cells of formation phase is very high.

(ii) Cell Elongation Phase:

• It is also called phase of cell enlargement.
• This phase lies just behind the growing points and is mainly responsible for growth of plant parts.
• The newly formed cells, produced informative phase undergo enlargement.
• The cell walls of the enlarging of cell show plastic extension through enzymatic loosening of microfibrils and deposition of new materials.
• The enlarging cell also develops a central vacuole, rate of respiration is high but less than that of the cells in the formative phase.
• Thus, this phase is characterized by cell enlargement, new cell wall deposition and increased vacuolation.

(iii) Cell Maturation Phase

• This phase occurs just behind the phase of elongation.
• The enlarged cells develop into particular type of cells by undergoing structural and physiological differentiation.
• Hence, at this phase, all the diverse tissue types observed in root or stem.