CBSE Class 12 Biology Question Paper (Outside Delhi 2018) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Outside Delhi 2018) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Outside Delhi 2018) with Solutions

Time Allowed: 3 Hours
Maximum Marks : 70

General Instructions:

1. There are a total of 26 questions and five sections in the question paper. All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3. Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5. Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examinee is to attempt any one of the questions out of the two given in the question paper with the same question number.

Section – A

Question 1.
How do cytokine barriers provide innate immunity in humans?
Answer:
Cytokines play a vital role in the innate immune response by means of direct mechanisms which inhibit viral replication by secreting interferon. Interferons are the proteins secreted by viral-infected cells that provide protection to the non-infected cells from further viral infections.

Question 2.
Write the dual purpose served by Deoxyribonucleoside triphosphates in polymerisation.
Answer:
Deoxyribonucleoside triphosphates (DTPs) serves as substrates i.e. nucleotides during replication and also supply energy for polymerisation reaction by breaking of high energy terminal phosphates bond. In PCR, deoxynucleotide triphosphates (dNTPs) serve as building blocks for new DNA strands.

Note
DNA replication is the process by which a double stranded DNA molecule is copied to produce two identical DNA molecules. DNA replication takes place in the cytoplasm of prokaryotes and in the nucleus of eukaryotes.

Question 3.
Write the names of the following :
(a) A 15 mya primate that was ape-like
(b) A 2 mya primate that lived in East African grasslands
Answer:
(a) The names of a 15 mya primate that was ape-like is Dryopithecus.
(b) The name of a 2 mya primate that lived in East African grasslands is Australopithecus.

Question 4.
Mention the chemical change that proinsulin undergoes, to be able to act as mature insulin.
Answer:
Mature functional protein is produced by the processing or pro-hormone insulin which contains an extra peptide called C-peptide or connecting peptide. This C-peptide is removed during the maturation of proinsulin, and A & B chains are linked by disulphide linkage.

Note
An American company Eli Lilly in 1983 prepared two DNA sequences corresponds to A and B, chains of human insulin and introduced them in plasmids of E.coli to produce insulin chains.

Question 5.
Name two diseases whose spread can be controlled by the eradication of Aedes mosquitoes.
Answer:
The two diseases whose spread can be controlled by the eradication of Aedes mosquitoes are chikungunya and dengue.

Section – B

Question 6.
How did a citizen group called Friends of Areata Marsh, Areata, California, USA, help to improve water quality of the marshland using Integrated Waste Water Treatment ? Explain in four steps.
Answer:
A citizen group called Friends of the Areata Marsh (FOAM), Areata, California, USA, helps to improve water quality of the marshland using Integrated Waste Water Treatment by following ways:

(a) The conventional sedimentation, filtering and chlorine treatments are given. After this stage, lots of dangerous pollutants like dissolved heavy metals still remain.
(b) To combat this, an innovative approach was taken and the biologists developed a series of six connected marshes over 60 hectares of marshland.
(c) Appropriate plants, algae, fungi and bacteria were seeded into this area, which neutralise, absorb and assimilate the pollutants. Hence, as the water flows through the marshes, it gets purified naturally.
(d) The marshes also constitute a sanctuary, with a high level of biodiversity in the form of fishes, animals and birds.

Question 7.
Your advice is sought to improve the nitrogen content of the soil to be used for cultivation of a non-leguminous terrestrial crop.
(a) Recommend two microbes that can enrich the soil with nitrogen.
(b) Why do leguminous crops not require such enrichment of the soil?
Answer:
(a) Azospirillum, Azotobacter, Anabaena, Oscillatoria, etc.
(b) Leguminous plants have symbiotic association with Rhizobium bacteria which traps N ₂ which is present in atmosphere and provides it to the plant, and in turn gets food and shelter.

Question 8.
You have obtained a high yielding variety of tomato. Name and explain the procedure that ensures retention of the desired characteristics repeatedly in large populations of future generations of the tomato crop.
Answer:
Micropropagation (type of vegetative propagation) ensures retention of the desired characteristics repeatedly in large populations of future generations of the tomato crop. A small part of plant called explant, is excised and grown under sterile condition in special nutrient medium to obtain such plants that would be genetically identical to the original plants.

Note
The method of producing thousands of plants through tissue culture from a single explants is called micropropagation.

Question 9.
(a) Name the source plant of heroin drug. How is it obtained from plant?
(b) Write the effects of heroin on the human body?
Answer:
(a) Papaver somniferum is the source plant for heroin drug. This is obtained by the acetylation of morphine, which is extracted from the latex of poppy plant.
(b) Heroin is a depressant and used to slow down body functions.

Note
Heroin is a opioid. Opioids are the drugs, which bind to specific opioid receptors that are present in central nervous system and gastrointestinal tract.

Question 10.
With the help of an algebraic equation, how did Hardy- Weinberg explain that in a given population the frequency of occurrence of alleles of a gene is supposed to remain the same through generations?
OR
Although a prokaryotic cell has no defined nucleus, yet DNA is not scattered throughout the cell. Explain.
Answer:
In a given population, one can find out the frequency of occurrence of alleles of a gene or a locus. This frequency is supposed to remain fixed and even remain the same through generations. Hardy-Weinberg principle stated it using algebraic equations.

This principle states that the allele frequencies in a population are stable and is constant from generation to generation. The gene pool (total genes and their alleles in a population) remains constant. This is called genetic equilibrium. Individual frequencies, for example, can 12. be named p, q, etc. In a diploid, p and q represents the frequency of allele A and allele a. The frequency of AA individuals in a population is simply p ² .

This is simply stated in another ways, i.e., the probability that an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e., p ² . Similarly for aa is q ² , Aa is 2pq.

Hence, p ² + 2pq + q ² = 1. This is a binomial expansion of (p + q) ² . When frequency measured, differs from expected values, the difference (direction) indicates the extent of evolutionary change. Disturbance in genetic equilibrium, or Hardy-Weinberg Equilibrium, i.e., change of frequency of alleles in a population would then be interpreted as resulting in evolution.

OR

In prokaryotes, the DNA (negatively charged) is scattered in the cytoplasm means that it is naked and is not covered by any membrane. The prokaryotes use an arrangement that helps to pack genetic material tightly into a specific region, positively charged protein hold it in large loops known as nucleoid because prokaryote does not have a well defined nucleus. So, the DNA is not scattered but present in the form of membrane less structure called nucleoid. This nucleoid floats in the cytoplasm and can be found anywhere in the cytoplasm. Also the DNA in form of single chromosomes is attached to mesosome at a point.

Section – C

Question 11.
(a) Differentiate between analogous and homologous structures.
(b) Select and write analogous structures from the list given below:
(i) Wings of butterfly and birds
(ii) Vertebrate hearts
(iii) Tendrils of Bougainvillea and Cucurbita
(iv) Tubers of sweet potato and potato
Answer:
(a) Difference between homologous and analogous organs:

Homologous organ	Analogous organ
(i) Homology is based on divergent evolution.	(i) Analogy is based on convergent evolution.
(ii) In homology, the structures are evolved from the same origin and have common ancestors but they have different functions.	(ii) In analogy, the structures are evolved from different origin and have different ancestors but have similar functions.
Eg. Wings of birds and forelimbs of human.	Eg. Wing’s of birds and wings of insects.

(b) These are the analogous structure from the given list:
(i) Wings of butterfly and birds.
(ii) Tubers of sweet potato and potato.

Question 12.
How has the use of Agrobacterium as vectors helped in controlling Meloidegyne incognita infestation in tobacco plants ? Explain in correct sequence.
Answer:
A nematode Meloidegyne incognita infects the roots of tobacco plants and causes a great reduction in yield. A novel strategy was adopted to prevent this infestation which was based on the process of RNA inteference (RNAi). Using Agrobacterium vectors, nematode-specific genes were introduced into the host plants. The introduction of DNA was such that it produced both sense and anti-sense RNA in the host cells.

These two RNA’s being complementary . to each other formed a double stranded (dsRNA) that initiated RNAi and thus, silenced specific mRNA of the nematode. The consequence was that the parasite could not survive in a transgenic host expressing specific interfering RNA. The transgenic plant therefore got itself protected from the parasite.

Question 13.
(a) “India has greater ecosystem diversity than Norway.” Do you agree with the statement ? Give reasons in support of your answer.
(b) Write the difference between genetic biodiversity and species biodiversity that exists at all the levels of biological organisation.
OR
Explain the effect on the characteristics of a river when urban sewage is discharged into it.
Answer:
(a) Yes, India has greater ecosystem diversity than Norway as India has deserts, rain forests, mangroves, coral reefs, wetlands, estuaries, and alpine meadows.

Note
Ecosystem diversity refers to the variations in ecosystems within a geographical location and its overall impact on existance of human population in an environment.

(b) Genetic diversity:

1. It is the total number of genetic characteristics in the genetic makeup of a species.
2. A single species might show high diversity at the genetic level (E.g. Man : Chinese, Indian, American, African etc.). India has more than 50,000 genetically different strains of rice, and 1,000 varieties of mango.
3. It allows species to adapt to changing environments. This diversity aims to ensure that some species survive drastic changes and thus carry on desirable genes.

Species diversity:

1. It is the ratio of one species population over total number of organisms across all species in the given biome. ‘Zero’ would be infinite diversity, and ‘one’ represents only one species present.
2. It is a measure of the diversity within an ecological community that incorporates both species richness (the number of species in a community) and the evenness of species.

Question 14.
Explain the mechanism of ‘sex determination’ in birds. How does it differ from that of human beings?
Answer:
In birds, sex determination is of ZW – ZZ type.
In this type the males are homogametic and have ZZ sex chromosomes, and females are heterogametic with ZW pair of sex chromosomes.

whereas, in human beings, the chromosomal mechanism of sex determination is of XX – XY type. The human male is heterogametic and have XY sex chromosomes and human female is homogametic with XX sex chromosomes.

Question 15.
Explain out-breeding, out-crossing and cross-breeding husbandry.
Answer:
Out-breeding : Out-breeding is the breeding of the unrelated animals, which may be between individuals of the same breed but having no common ancestors for 4-6 generations (out-breeding) or between different breeds (cross-breeding) or different species (inter-specific hybridization).

Out-crossing: This is the practice of mating of animals within the same breed, but having no common ancestors on either side of their pedigree up to 4-6 generations.

The offspring of such a mating is known as an outcross. It is the best breeding method for animals that are below average in productivity in milk production, growth rate in beef cattle, etc. A single outcross often helps to overcome inbreeding depression.

Cross-breeding: In this method, superior males of one breed are mated with superior females of another breed. Cross – breeding allows the desirable qualities of two different breeds to be combined. The progeny hybrid animals may themselves be used for commercial production. Alternatively, they may be subjected to some form of inbreeding and selection to develop new stable breeds that may be superior to the existing breeds. Many new animal breeds have been developed by this approach. Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri ewes and Marino rams.

Question 16.
(a) Organic farmers prefer biological control of diseases and pests to the use of chemicals for the same purpose. Justify.
(b) Give an example of a bacterium, a fungus and an insect that are used as biocontrol agents.
Answer:
(a) Biological pest control has important advantages compared to chemical pest control, such as being safer for humans and the- environment. Chemical methods often kills both useful and harmful life forms indiscriminately. Eradication of the creatures that are often described as pests in not only possible, but also undesirable, for without them the beneficial predatory and parasitic insects which depend upon them as food or hosts would not be able to survive.

(b) Bacterium, a fungus and an insect that are used as biocontrol agents are:
Insects = Ladybird and Dragonflies.
Bacteria = Bacillus thuringiensis.
Fungus = Trichoderma

Question 17.
(a) How has the development of bioreactor in biotechnology?
(b) Name the most commonly used bioreactor and describe its working.
Answer:
(a) Small volume cultures cannot yield appreciable quantities of products. In order to produce in large quantities, bioreactor are developed in which, large volumes (100 – 1000 litres) of culture can be processed. Thus, bioreactors can be thought of as vessels in which raw materials are biologically converted into specific products, individual enzymes, etc., using microbial plant, animal or human cells.

Note
Bioreactor is defined as a vessel that carries out a biological reaction and is used to culture aerobic cells for conducting cellular or enzymatic immobilization.

(b) The most commonly used bioreactors are of stirring type. A stirred – tank reactors is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents. The stirrer facilitates even mixing and oxygen availability throughout the bioreactor. The bioreactor has an agitator system, an oxygen delivery system and a foam control system, a temperature control system. pH control system and sampling ports so that small volumes of the culture can be withdrawn periodically.

Question 18.
Explain the roles of the following with the help of an example each in recombinant DNA technology.
(a) Restriction Enzymes
(b) Plasmids
Answer:
(a) Restriction enzymes :

1. Restriction enzymes belongs to class of enzymes nucleases which breaks nucleic acids by cleaving their phosphodiester bonds.
2. Since restriction endonucleases cuts DNA at specific recognition site, they are used to cut the donor DNA to isolate the desired gene.
3. The desired gene has sticky ends which can be easily ligated to cloning vector which was cut by same restriction enzymes having complementary sticky ends to form recombinant DNA.
4. An example is EcoRI which is obtained from E.coli bacteria “R” strain which cuts DNA at specific palindromic recognition site.

Note
Pallindromes are the group of letters that form the same words when read both forward and backward for e.g. MALAYALAM.

(b) Plasmids :

1. Plasmids are autonomous, extra chromosomal circular double stranded DNA of bacteria.
2. They are small and self-replicating, and they are used as cloning vectors in genetic engineering.
3. Some plasmids have antibiotic resistance genes which can be used as marker genes to identify recombinant plasmids from non-recombinant ones.
4. The plasmids are cut and ligated with desired genes and transformed into host cell for amplification to obtain the desired products.
5. An example of artificial modified plasmids is pBR322 or pUC.

Question 19.
Differentiate between Parthenocarpy and Parthenogenesis. Give one example of each.
Answer:
Parthenogenesis and parthenocarpy are two such processes that results in fruits and individuals from unfertilized ovules or eggs prior to fertilisation.

In most plants, flowers need to be pollinated and fertilized to produce fruits. However, some plants can produce fruits before fertilisation or without fertilisation.

Parthenocarpy is the process which produces fruits from unfertilised ovules in plants. Unfertilised ovules develop into fruits prior to fertilisation. These fruits do not contain seeds. E.g. banana and grapes.

Parthenogenesis is a type of reproduction commonly shown in organisms mainly by some invertebrates and lower plants. It can be described as a process in which unfertilised ovum develops into an individual (virgin birth) without fertilisation. Therefore, it can be considered as a method of asexual reproduction.

The key difference between parthenogenesis and parthenocarpy is, parthenogenesis is shown by animals and plants while parthenocarpy is shown only by plants. Parthenogenesis is seen in organism like rotifers, honeybees and even some lizards and birds (turkey).

Question 20.
Medically it is advised to all young mothers that breastfeeding is the best for their newborn babies. Do you agree? Give reasons in support of your answer.
Answer:
Yes, I do agree with the fact that breastfeeding is the best for newborn babies. Mammary glands of the female undergo differentiation during preganancy and starts producing milk. The milk produced during the initial few days and lactation is called colostrum which contains IgAantibodies. It helps in developing resistance for newborn baby by providing innate immunity to the developing infant. It helps the baby fight to against viruses and bacteria. Thus breast milk is packed with disease fighting substances that protect the baby from illness. It naturally contains many of the vitamins and minerals that are required by newborn.

Question 21.
Draw a diagram of a mature human sperm. Label any three parts and write their functions.
Answer:
Acrosome : It is a cap like structure, filled with hydrolytic enzymes that help in the fertilisation of the ovum.
Middle piece : Possesses numerous mitochondria, which produces energy for the movement of tail and facilitate the movement of tail that facilitates sperm motality essential for fertilisation.

Tail : Facilities sperm motility essential for fertilisation

Note
The seminal plasma is rich in fructose, calcium and certain enzymes and seminal plasma alongwith sperms constitute the semen. The secretions of bulbourethral glands also helps in the lubrication of the penis.

Question 22.
(a) Expand VNTR and describe its role in DNA fingerprinting.
(b) List any two applications of DNA fingerprinting technique.
Answer:
(a) VNTR stands for “Variable Number of Tandem Repeats”. The VNTR belongs to a class of satellite DNA referred to as mini-satellite. A small DNA sequence is arranged tandemly in many copy numbers. The copy number varies from chromosome to chromosome in an individual. The numbers of repeat show very high degree of polymorphism. As a result the size of VNTR varies in size from 0.1 to 20 kb. Consequently, after hybridization with VNTR probe, the autoradiogram gives many bands of differing sizes. These bands give characteristic pattern for an individual DNA which is used to identify individuals.

(b) Since DNA from every tissue (such as blood, hair – follicle, skin, bone, saliva, sperm etc.), from an individual show the same degree of polymorphism, they become very useful identification tool in forensic applications to identify criminals. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basic of paternity testing, in case of disputes.

Section – D

Question 23.
Looking at the deteriorating air quality because of air pollution in many cities of the country, the citizens are very much worried and concerned about their health. The doctors have declared health emergency in the cities where the air quality is very severely poor.
(a) Mention any two major causes of air pollution.
(b) Write any two harmful effects of air pollution to plants and humans.
(c) As a captain of your school Eco-club, suggest any two programmes you would plan to organize in the school so as to bring awareness among the students on how to check air pollution in and around the school.
Answer:
(a) Causes of air pollution are:

1. Industrial effluents
2. Smoke released from vehicles.
3. Burning of fossil fuels
4. Smoke stacks of thermal power plants.

(b) Harmful effects of air pollution.

1. It affects respiratory system of animals as well as human beings.
2. It also reduces growth and yield of crops & cause premature death of plants.

(c)

1. To boost public transport i.e. buses should use CNG instead of diesel fuel.
2. Planting more trees to reduce pollution.

Note
Motor vehicles are equipped with catalytic converter should use unleaded petrol because lead in the petrol inactivates the catalyst

Section – E

Question 24.
(a) Write the scientific name of the organism Thomas Hunt Morgan and his colleagues worked with for their experiments. Explain the correlation between linkage and recombination with respect to genes as studied by them.
(b) How did Sturtevant explain gene mapping while working with Morgan ?
OR
(a) State the ‘Central dogma’ as proposed by Francis Crick. Are there any exceptions to it ? Support your answer with a reason and an example.
(b) Explain how the biochemical characterization (nature) of “Transforming Principle” was determined, which was not defined from Griffith’s experiments.
Answer:
(a) Drosophila melanogaster : Morgan carried out several dihybrid crosses in Drosophila to study gens that were sex-lined. Morgan and his group knew that the genes were located on the X chromosome and saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type.

Morgan attributed this due to the physical association or linkage of the two genes and coined the term linkage to describe this physical association of genes on a chromosome and the term recombination to describe the generation of non- parental gene combination. Morgan and his group also found that even when genes were grouped on the same chromosome, some genes were very tightly linked (showed very low recombination) while others were loosely linked.

Morgans student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome.

Genetic maps are extensively used as a starting point in the sequencing of whole genomes.

(b) Alfred Sturtevant expressed the frequency of recombination between gene pairs present on the same chromosome as the distance between those genes. He then mapped the positions of the genes on the chromosome. Today, gene maps are used as a starting point in the genome sequencing.

OR

(a) Francis Crick proposed the central dogma, which states that the genetic information flows from DNA to RNA and then to protein, replication

There are few exceptions to this as observed in the case of some viruses (retrovirus). In retrovirus, DNA is synthesized from RNA with the help of reverse transcriptase which is known as reverse transcription.

(b) Transforming Principle : In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumonia), witnessed a transformation in the bacteria. During the cource of his experiment, a living organism (bacteria) had changed in physical form.

He concluded that the R strain bacteria had somehow been tranformed by the heat – killed S strain bacteria. Some ‘transforming principle’, transferred from the heat – killed S strain, had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material.

However, the biochemical nature of genetic material was not defined from his experiments. Oswald Avery, Colin MacLeod and Maclyn McCarty worked to determine the biochemical nature of ‘tranforming principle’ in Griffith’s experiment. They purified biochemicals (proteins , DNA, RNA, etc.) from the heat – killed S cells to see which ones could transform live R cells into S cells. They discoveredf that DNA alone from S bacteria caused R bacteria to become transformed.

They also discovered that protein – digesting enzymes (proteases) and RNA – digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material, but not all biologist were convinced.

Question 25.
(a) Following are the responses of different animals to various abiotic factors. Describe each one with the help of an example.
(i) Regulate
(ii) Conform
(iii) Migrate
(iv) Suspend
(b) If 8 individuals in a population of 80 butterflies die in a week, calculate the death rate of population of butterflies during that period.
OR
(a) What is a trophic level in an ecosystem? What is ‘standing crop’ with reference to it?
(b) Explain the role of the ‘first trophic level’ in an ecosystem.
(c) How is the detritus food chain connected with the grazing food chain in a natural ecosystem?
Answer:
(a) (i) Regulate: Some organisms are able to maintain nomeostatis by regulating their body temperatures. The mechanisms used by most mammals to regulate their body temperature are similar to what we humans use. For example: Body temperature remains constant at 37°C. In summer, when outside temperature is more than our body temperature, we sweat profusely and when the temperature is much lower than 37°C, we shiver thus body temperature remains constant. E.g. Birds and mammals.

(ii) Conform: Many animals, cannot maintain a constant internal environment. Their body temperature changes with the ambient temperature.

These are conformers. Heat loss or heat gain is a function of surface area. Since small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside. Eg. Shrews and humming birds.

(iii) Migrate: The organism can move away temporarly from the stressful habitat to a more hospitable area & return when stressful period is over. Eg: Every winter, the famous Keolado National Park in Bharatpur host thousands of migratory birds coming from Siberia & other northern regions.

(iv) Suspend: In animals, if migration is not possible, they might avoid the stress by escaping in time. Eg:

1. Bears go into hibernation during winter.
2. Fishes go into aestivation to avoid summer related problems heat & dessication.
3. Under unfavourable conditions many zooplankton species in lakes and ponds are known to enter diapause, a stage of suspended development.

(b) Death Rate : Number of deaths per 1000 individuals of a population.
Death Rate = 8/80 = 0.1

OR

(a) Tropic level: Organisms occupy a place in the natural surroundings or in a community according to their feeding relationship with other organisms. Based on the source of their nutrition or food, organisms occupy a specific place in the food chain and this is known as trophic level.
Standing crop: It is measured as the mass of living organisms (biomass) or the number in a unit area.

(b) First trophic level is formed by producers. This is the basic unit. These organisms can live without feeding on any another level. The only thing that these organisms need to survive is sunlight and water which they can turn into energy themselves. All other trophic levels depend on this level for energy.

(c) GFC is Grazing Food Chain: It is depicted as below : Producers, Primary consumers, Secondary consumers DFC is Detritus Food Chain : It begins with dead organic matter. It is made up of decomposers which are heterotrophic organisms like fungi, bacteria etc. GFC is the major conduct for energy flow. DFC may be connected with GFC at some levels : Some of the organisms of DFC are prey to the GFC animals.
These natural inter connection of food chains forms food web.

Note
Food chain refers to the a linear sequence of transfer of food and energy from which represents the flow of energy from one trophic level to another.

Question 26.
(a) Describe any two devices in a flowering plant which prevent both autogamy and geitonogamy.
(b) Explain the events upto double fertilisation after the pollen tube enters one of the synergids in an ovule of an angiosperm.
OR
(a) Explain menstrual cycle in human females.
(b) How can the scientific understanding of the menstrual cycle of human females help as a contraceptive measure?
Answer:
(a) Autogamy : Transfer of pollen grains from anther to the stigma of same flower. It is a type of self-pollination.
Geitonogamy : Transfer of pollen grains from anther to the stigma of another flower of same plant. Two devices that prevent both autogamy and geitonogamy are :

1. Self- incompatibility: This is a genetic mechanism & prevents self-pollen from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.
2. Dioecious plants : Male and female flowers are present on different plants, that is each plant is either male orfemale.

Note
Geitonogamy is functionally cross-pollination that involves a pollinating agent, genetically it is similar to autogamy since the pollen grains come from the same plant.

(b) The events seen after the pollen tube enters one of the synergids in an ovule are as follows:

1. Pollen tube, after reaching the ovary, enters the ovule through the micropyle and thus enters one of the synergids through filiform apparatus.
2. After entering one of synergids, the pollen tube releases the two male gametes into the cytoplasm of the synergid.
3. One of the male gametes move towards the egg cell and fuses with its nucleus thus results in formation of zygote (diploid cell). This is called syngainy.
4. The other male gamete move towards the two polar nuclei located in the central cell and fuses to form triploid primary endosperm nucleus (PEN). This involves fusion of three haploid nuclei & hence termed as triple fusion.
5. Two types of fusions, syngamy & triple fusion takes place in an embryosac and hence the phenomenon is termed as double fertilisation.
6. After fertilisation, PEN becomes the primary endosperm cell (PEC) & develops into endosperm while zygote develops into an embyo.

OR

(a) Menstrual Cycle:

(i) The reproductive cycle in the female primates (e.g. monkeys, apes and humans) is called menstrual cycle.

(ii) The first menstruation begins at puberty and is called menarche.

(iii) In human females, menstruation is repeated at an average interval of about 28/29 days and the cycle of events starting from one menstruation till the next one is menstrual cycle.

(iv) The phases of menstrual cycle are as follows: Menstrual phase : It lasts for 3-5 days. The menstrual flow results due to breakdown of endometrial lining of the uterus and its blood vessels which forms liquid that comes out through vagina. Menstruation only occurs if the released ovum is not fertilised.

Follicular phase : It lasts for 8-10 days. During this phase, the primary follicles in the ovary grow to become a fully mature Grafian follicle and simultaneously the endometrium of uterus regerates through proliferation. The secretion of LH and FSH increases gradually.

Ovulatory phase : It lasts for 1 day. There is release of ovum.

Luteal phase : It lasts for 13 days. There is LH surge. These induces the remaining parts of Grafian follicle to tranform as corpus luteum and its secretes progesterone for maintenance of the endometriun.

(v) If fertilisation occurs, endomentrium starts preparing for implantation. In the absence of fertilisation, corpus luteum degenerate.

Note
In human beings, menstrual cycles ceases around 50 years of age and is termed as menopause. Cyclic menstruation is an indicator of normal reproductive phase and extends between menarche and menopause.

(b) Scientific understanding of menstrual cycle of human females are very important as a contraceptive measures. It helps in following ways:

(i) Safe period (Rhythm method)
A week before and a week after menstrual bleeding is considered as safe period for sexual intercourse. The idea is based on following facts:

(ii) Ovulation occurs on 14^ day of cycle and ovum survives for about 2 days.

(iii) Pills used by females are also dependent on menstrual cycle. The pills have to be taken daily for a period of 21 days starting preferably within first five days of menstrual cycle. It is repeated again after period of 7 days. These inhibit ovulation and implantation as well as alter the quality of cervical mucus to prevent/retard entry of sperms.