Bihar Board 12th Biology Important Questions Long Answer Type Part 1

Bihar Board 12th Biology Important Questions Long Answer Type Part 1 are the best resource for students which helps in revision.

Bihar Board 12th Biology Important Questions Long Answer Type Part 1

Question 1.
Write short notes on the following:
(a) Nuclear type endosperm
(b) Dicot non-endospermic seed
(c) Seed dormancy
(d) Geitonogamy.
Answer:
(a) Nuclear type endosperm: During the development of nuclear type endosperm, the primary endosperm nucleus divides mitotically into free nuclei divisions without wall formation. It results into a large number of free nuclei in the central cell of embryo sac. A big central vacuole is developed into embryo sac pushing all the nuclei to the periphery. In the second phase the cleavage of multinucleate peripheral cytoplasm results in centripetal cell wall formation. Finally, all the endosperm is converted into a cellular tissue; e.g. Coconut.

Fig.: A-C. Stages in the development of nuclear endosperm in Acalypha indica

(b) Dicot non-endospermic seed: The seeds which store their food in cotyledons are called on endospermic seed or exalbuminous seeds, e.g. bean, gram etc. The seeds of bean are formed within pod, which is a ripened ovary, The seed is attached by funiculus or seed stalk. When seeds shed the funiculus breaks off leaving a prominent stalk, hilum. Below the hilum lies micropyle. The seed coats have characteristic colours.

In these seeds there is no endosperm, its already consumed by developing embryo. Most of the food is stored in cotyledons which never function as true leaves.

Fig.: Structure of Bean seed (non-endospermic) A. Entire seed in side view B. Entire seed in micropylar view

(c) Seed dormancy: The dormancy of seed is the condition of seed when it fails to germinate even though the environmental conditions which are considered favourable for active growth. Seed dormancy may be due to seed coats, condition of embryo, due to specific light requirement, due to germination inhibitors.

Seed dormancy is beneficial in various ways as it is a adaptation to ensure germination only under favourable conditions. It enables the seeds to be disseminated.in time and space. Seed dormancy can be broken by softening or breaking of seed coats (scarification), by shaking (impaction), altering the temperature, treatment of seeds with growth promoters.

(d) Geitonogamy: Geitonogamy is the kind of self-pollination. It takes place when pollen grains are transferred from anther of a flower to the stigma of the same flower. The flowers are essentially bisexual, e.g. Pea, Wheat, rice etc.

Question 2.
What is menstrual cycle? Describe its various phases with hormonal regulation?
Answer:
The gamete formation in females in a cyclic activity that takes about 28 days and involes in the structure and function the entire reproductive system. It is called menstrual cycle.

Menstural cycle is divided into four phases:
(i) Prolife erative phases:
It follows the menstrual phases an last four about 10 – 20 days (from 5th – 14 day of menstural cycle). It involves the following changes

Under the stimulation of FSH-RF, there in increased secretion of FSH.
FSH stimulates the change of primary follicle of the ovary into a Graffian follides
Graffian follicle secretes estrogen
Estrogen stimualte the grwoth, maintenance and normal functioning of secondary sex organs.
Uterine endomentrium become thick, more glandular & mone vasuclar.
Estrogen inhibit the secretion of FSH and stimulate the secretion of LH.

(ii) Ovulatory phase: It involves the ovulation from the Graffian follicle. It occurs between two menstrual cycle on 14th day of the onset of the menstrual cycle.

Ovulation is controlled by the increased level of LH in blood. LH also starts the change of empty Graffian-follicle into Corpus-Luteum and secretion of progesterone from corpus.

(iii) Secretary phase: It lasts for about 12 – 14 days and extends from 16th – 28th day of menstrual cycle. Carpus Luteum formed from Graffian follice, increase in size, so it is called Luteal phase.

Carpus Luteum begins to secrete progesterone hormone. Progesterone decreases the secretion of FSH and LH, so inhibits the maturation of follicle and ovulation. Progesterone reduces uterine movement.

In the obsence of fertilization, carpus Luterum degeneration LH level falls, Progesterone level is reduced. Reduced level of both progesterone anu estrogen causes mensus.

(iv) Menstrual phase: It last for about 3 days, and extends from 1st – 4th days of the Menstrual cycle. The level of progesterone in the blood is declines. The uterine tissue fails to be maintained. Then the unfertilized ovum along with ruptured uterine epithelium, asoul 50 – 100 ml of blood and some neucus is discharged out through vaginal orifice and is called menstrual flow or menstrualtion. Decrease the level of progesterone and estrogen in the blood stimulates the hypothalamus and anterior pituitary to release FSH-RH and FSH respectively. FSH starts the follicular phase of next menstrual cycle.

Question 3.
(a) Describe with the help of a sketch the structure of a maize grain.
(b) How does it differ from been seed?
Answer:
(a) Structure of maize grain: Maize grain is endospermic grain, it is a one seeded fruit called caryopsis. The pericarp is fused with testa. It consists of following parts:
(i) Seed coat: The outer brownish layer. In this, seed and fruit are fused together.
(ii) Endosperm: Its the major part of grain and full with reserve food. Its made up of two regions. (1) Aleurone layer: outer single layer made up of aleurOne proteins, (2) Inner starcy endosperm. Endosperm is separated from embryo by epithelium.
(iii) Embryo: It is a single cotyledonous embryo called scutellum. An embryo axis with plumule and radical are present at its two ends. At the tip of radicle is present root cap, radical is surrounded by a protective layer known as coleorhiza. Plumule is protected by covered sheath called coleoptile.

(b) Difference between maize grain and bean seed:

Question 4.
Explain female mammary glands with the help of diagram.
Answer:
The mammary glands in female animals are paired structures that contain glandular tissue and variable amount of fat. The glandular tissue of each breast (mammary gland) is divided into 15 – 20 mammary lobes containing clusters of cells called alveoli (Fig.).

The cells of alveoli secrete milk, which is stored in the cavities (lumens) of alveoli. The alveoli open into mammary tubules. The tubules of each to be join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to lactiferous duct through which milk is sucked out.

Question 5.
Describe briefly:
(i) Which modes of contraception are 100% reliable?
(ii) Where does the test tube baby develop?
(iii) Is there any non-invasive method of determining the condition of the foetus before birth?
Answer:
(i) Sterilization is at present the most effective means of birth control. It involves vasectomy and tubectomy. These are permanent methods of birth control. Vasectomy, is done in males and tubectomy is done in females. It involves the removal of a short segment of each vas defense or oviduct and tying up of the remaining ends tightly with surgical thread. These operations are minor and do not affect the normal healthy life.

(ii) Test tube babies develop by invitro fertilization and embryo transfer. Ovum from the wife or donor female and sperms from the husband or donor male are collected and are induced to form zygote under simulated conditions in the laboratory. The zygote or embryo is then transfered into fallopian tube (ZIFT) or uterus (IUT). Thus test tube babies develop as normal embryo, inside the mother’s womb.

(iii) Ultrasound technique or sonography is a non-invasive method of determining the condition of the foetus. In it high frequency sound waves are passed through the body. These waves have the property to pass unimpeded through the homogenous tissue till they strike another tissue or organ. Another related technique is called Doppler Ultrasound Scanning. It is based on the fact that a few foetal blood cells leak across the placent into the mother’s blood stream. A blood sample from the mother provides enough foetal cells that can be tested for genetic disorders.

Question 6.
Write the cause and the symptoms of the following:
(a) Down’s Syndrome
(b) Klinefelter’s Syndrome
(c) Turner’s Syndrome.
Answer:
(a) Down’s Syndrome: The cause of this genetic disorder the presence of an additional copy of the chromosome number 21. The affected individual is short statured with shall round head, furrowed tongue and partially open mouth. Physical, psychomotor and mental development is retarded.

(b) Klinefelter’s Syndrome: This genetic disorder is caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY. An individual suffering from such syndrome has masculine development, however, the feminine development (development of breast) is also expressed. Such individuals are sterile,

(c) Turner’s Syndrome: Such disorder is caused due to the absence of one of the X-chromosomes. Such females are sterile. The ovaries are rudimentary besides other features including lack of other secondary sexual characters.

Question 7.
Discuss the mechanism of protein synthesis.
Answer:
Translation (protein synthesis) takes place in several steps:
(i) Activation of amino acids: It is carried out by aminoacyl tRNA synthetases. In the presence of ATP and mg2+, an amino acid combines with specific aminoacyl t-RNA synthetase. It produces aminoacyl-adenylate-enzyme complex.

This complex, reacts with tRNA specific for the amino acid.
AA ~ AMP-E + tRNA → AA – tRNA + AMP + E

(ii) Initiation: It requires initiation factors. Procaryotes have three initiation factors IF3, IF2, and IFI. Eukaryotes have nine – eIF2, eIF3,eIFl, eIF4, eIF4B, eIF4C, eIF4D, eIF5, eIF6.

For initiation the ribosome binds to the mRNA at start codon AUG. which is recognised by initiator tRNA for methionine. It takes place in following steps:

(iii) Elongation: During this phase, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along with the mRNA. Amino acids are added one by one, translated into polypeptide seequences dictated by DNA and represented by mRNA. A peptide bond (-CO – NH-) is established between the carboxyl group (-COOH) of amino acid attached to tRNA at P-site. Peptidyl transferase enzyme catalyses the formation of peptide bond.

For every single amino acid incorporated in peptide chain 1 ATP and 2 GTP molecules are used.

(iv) Termination: Polypeptide synthesis terminates when a non-sense codon of mRNA reaches A-site. there are three non-sense codons -UAA, UAG and UGA. These are not recognised by any tRNA. The polypeptide is released in the presence of release factor on encounter of any non-sense codons.

Question 8.
Write some comparison between chromosomes and genes.
Answer:
Chromosomes:

There occur in pairs
At the time of gamete formation they segregate.
One of each pair is transmitted to a gamete.
Independent pairs segregate independently of each other.

Genes:

There also occur in pairs.
At the time of gamete also segregate.
Only one of each pair is transmitted to a gamete.
One pair segregates independently of another pair.

Question 9.
(i) Briefly describe the process of transcription.
(ii) Distinguish amongst structural gene, regulator gene and operator gene.
Answer:
(i) Transcription is the formation of RNA over DNA template. It requires enzyme RNA polymerase and takes place in several steps:
(a) Activation of ribonucleotides: All the four types of nucleotides are activated through phosphorylation. Enzyme phosphorylase and energy is required for the process.

(b) DNA template: Each DNA segment has a promoter region and a terminator region. A promoter region has RNA polymerase recognition site. RNA polymerase binds at promoter site, unwind the DNA helix with the help of helicases and gyrases. One strand act as template strand. Transcription occurs at 5′ → 3′ direction.

(c) Base pairing: Ribonucleoside triphosphates form complementary pairs, with the nitrogen bases of DNA template using pyrophosphatase enzyme and releasing energy. U pairs with A, A with T, C with G and G with C.

(d) Chain formation: With the help of RNA polymerase the adjacent ribonucleotides join to form RNA chain. The rate of chain formation is 30 nucleotides per second.
RNA synthesis stops with polymerase reaches terminator region.

(e) Separation of RNA: Terminator factor of Rho(P) factor has ATP – as activity, it helps in the release of completed RNA chain.

(ii) Differences amongst regulator gene, operator gene, promoter gene and structural gene:

Question 10.
What are the evidences for evolution?
Answer:
Following are the evidences for evolution:
Some fossils appear similar to modern organisms. They represent extinct organisms. Study of fossils in different sedimentary layers indicates the geological period in which they existed.

The studies showed that the life forms varied our time and certain life forms are restricted to certain geological time-spans. Then, new forms have arisen at different times in the history of earth.

Question 11.
Explain with an example evolution by natural selection.
Answer:
In 1850s in England, in a collection of moths, before industrialisation was sent in, it was observed that there were more white-winged moths on trees than dark winged or melanised moths. However, after industrialisation, there were more dark-winged moths in the same area i.e., the proportion was reversed.

The reason for this was, during post-industrialisation, period the tree trunks became dark due to industrial smoke and soots. Under this condition the white winged moth did not survive due to predators, dark-winged or melanised moth survived.

Before industrialisation, thick growth of white-coloured lichen covered the trees-in that background the white-winged moth survived but the dark-colored moth were picked out by predators. Hence, moths that were able to camouflage themselves i.e., hide in the background, survived.

Question 12.
Describe the causes, symptoms and prevention of AIDS.
Answer:
Causes of AIDS: AIDS is caused by Human Immuno Deficiency Virus (HIV), a member of group of retroviruses.

Symptoms: Symptoms of AIDS are headache, dizziness, unexplained weight loss, purple patches on the skin and mucous membrane inside the mouth, unexplained bleeding and prolonged diarroea A person may suffer from any are all of these symptoms.

Causative agents in transmitted to the healthy person by the following ways:

Sexual contacts with infected person.
Transfusion of contaminated blood and blood products.
Sharing of injected needles as in the case of I.V. drugs abusers.
Infected mother to her foetus through placenta.

Preventions:
Preventive measures can be

Clean or safe sex habits using condoms regularly
Do not shave items that could become contained by blood like razor, tooth brush, or any skin, piercing instruments.
Checking of blood before transfusion.
Infected mother should not bear a child.

Question 13.
Write short notes on following:
(i) Cloning vectors
(ii) Production of monoclonal antibodies
(iii) Applications of biotechnology
Answer:
(i) The cloning of a foreign fragment of DNA, in bacteria is made possible with the help of cloning vectors or carriers, which are additional sequences of DNA and do not affect the normal functioning of the cell. These are various kinds:

(a) Plasmids: Plasmids are most commonly used vectors for genetic engineering. They are circular, double stranded extra chromosomal DNA with self replicating system. A cut is made in plasmid DNA by restriction enzyme, the desired segment of DNA is introduced and joined by DNA ligases. This reforms a circular hybrid or chimeric plasmid. The chimeric plasmid is transferred to a bacteria like E. Coli where it replicates and expresses itself.

(b) Phages are viruses which infact bacteria. For example X (Lamda) phage. They have sites for restriction enzymes and can easily enter inside bacteria and insert their genome.

(c) Costnid and phasmid vectors: Cosmids are plasmid vectors containing a fragment of X phage DNA. They provide an efficient means of cloning large fragments of foreign DN A. They are used in constructing cDNA libraries of eukaryotes.

Phasmids are also plasmid vectors containing a fragment of phage DNA. They have the advantages of both plasmids and phages. Plasmid may insert into phage DNA in the same may by which phage DNA enters in bacterial chromosome during lysogenic cycle.

(ii) Production of monoclonal antibodies:
It involves following steps

First a mouse or other animal is injected with specific antigen.
The animal will develop an immune response and produces antibodies against the antigen in B-lymphocyte cells in spleen.
The spleen of the animal is taken out and its B-lymphocyte cells are isolated.
In the same way, the cells producing bone marrow cancer (myeloma cells) are isolated. These cells should not be able to synthesise their own nutrients.
Both the myeloma cells and antibody-producing cells are made to fuse together in cultures. The fused cells are named hybridomas.
The unfused myeloma cells are separated from fused myeloma cells by growing them on nutrient deficient medium (needed by myeloma cells)
The surviving fused myeloma (hybridoma) cells are grown separately and each clone is tested for production of antibody.
The clones which show positive results are isolated and cultured for large scale production of antibody.

(iii) Applications of biotechnology:

Food: The important foods employing micro-organisms during their preparation are Bread, Idli, Dosa, Acid Porridge, Shoyu, Soy Sauce, Miso. Koji Tempeh, Ontjom, etc.
Dairy Products: Yogurt (=yoghurt), Curd, Kefir (Russian), Cheese and butter require specific strains of micro-organisms during their preparation or maturation.
Alcoholic Beverages: Beers, Wines, Whisky, etc. are produced through fermentation of different foods by suitable micro-organisms.
Nonalcoholic Beverages: Curing of coffee beans and Tea leaves is a microbial process.
Sewage Treatment: Different types of bacteria, fungi and algae take part in removal of organic matter in the sewage.
Biogas (Gobar gas): Cowdung, farm refuse, garbage, etc. are placed in biogas plants where anaerobic conditions allow methane bacteria to’produce methane and other fuel gases. The organic remains of the biogas plants are used as manure.
Biofertilizers: They are mostly nitrogen fixing microorganisms which may live free in the soil or form associations with plants. Special strains of these organisms are now inoculated to soil or seeds.
Organic Acid: A number of organic acids (acetic acid, lactic acid, citric acid, gluconic acid) and amino acids are obtained through biotechnology.
Enzymes: Enzymes required for industrial and medicinal use are available from microbial processes.
Vitamins: Some vitamins are still manufactured with the help of microorganisms. Others are synthesised chemically. Food yiest is rich in both proteins and vitamins.
Antibiotics: Only a few are synthetic. All others are products of micro-organisms;
Vaccines: They are employed for providing immunity against important diseases. Vaccines contain attenuated or killed pathogens or their antigens.
Monoclonal Antibodies: Antibodies against pathogens can now be obtained in form clonal cultures.
Hormones: Insulin, growth hormone and other hormones are presently synthesised through the use of microbes and genetic engineering.
Tissue culture: This is an important tool for improvement of agriculture, forestry introduction of mutations and resistance, synthesis of specific biochemicals, etc.
Genetic engineering: Recombinant DNA technology is applied to several biotechnological processes in obtaining particular biochemicals, improvement of genetic make-up of organism and fighting genetic defects.
Transplants: Test tube babies and embryo transplants are now being routinely undertaken for production of offspring of desired parents.
Steroids: Micro-organisms are employed for transformation of one type of steroids with other types. Steroids and some other products are regularly employed in antifertility formations.

Question 14.
What is basis of classifying cancer? Name and explain the different categories of cancer. Mention any two approaches for cancer treatment.
Answer:
Cancers are classified according to organs and tissues.
The different categories of cancers are
(a) Sarcomas: It affects the mesodermal tissue like connective lymphoid tissue etc. They are rare in humans. It is of three types-

Lymphomes – Cancer of lymphatic tissue. They appear in the lymph nodes. Spleen and reticuloendothelial tissues.
Lipomas – Tumours of adipose tissue
Osteoma – Cancers of bones.

(b) Carcinomas: It affects the epithelial and glandular tissue they include breast cancer, lung cancer, stomach and pancreatic cancer.

(c) Leukaemia: Cancer of bone marrow results form excessive formation of WBCs and Lymphatic nodes thereby increasing their number in the blood.

Cancer Treatment.

Chemotherapy – It is administration of certain anticancer drugs.
Immunotherapy – In this administration of certain anticancer drugs.

Question 15.
Write notes of vaccination and immunisation?
Answer:
Vaccination is the process of development of immunisation against a particular disease by inoculation of harmless antigenic materials like attenuated pathogen into a healthy person. Immunisation develops due to formation of memory cells by the immune system. When a vaccinated person receives an injected dose of the pathogen, the existing memory Tor B cells recoginise the antigen and induce a massive formation of antibodies for elimination of invaders. Edward Jenner cesned the term vaccine. Vaccine is a expression of killed or attenuated pathogenic microorganisms. In some case a single dose in enough for life time whereas other’s 2 – 3 boosters doses are required to develope life time immunity.

Question 16.
(i) Give a schematic representation of construction of cDNA library, (ii) Role of biotechnology in diagnosis of diseases.
Answer:
(i) Construction of cDNA library:

(ii) Now-a-days biotechnology is contributing a lot to medicines, both in the diagnosis of diseases and in the development of pharmaceutical products. Its main purpose is to identify the genes whose mutation causes genetic diseases. All diseases involve changes in gene expression within the affected cells, and within the patient’s immune system. By comparing gene expression in healthy and diseased cells, biotechnologists may identify genes which are turned on or off in particular diseases.

(a) Polymerase chain reaction (PCR) s a technique by which any piece of DNA can be copied many times in short period. If the source of DNA is impure it can be amplified to using PCR technique. Amplification makes DNA identification much easier. For example the sequence of HIV DNA is known, its amplification by PCR can help to detect HIV DNA in blood or tissue samples.

DNA from single embryonic Cells is amplified by PCR for rapid parental (before birth) diagnosis of genetic disorders. DNA technology can help to identify individuals with genetic disorders before the appearance of symptoms or carriers of potentially harmful recessive alleles.

(b) Pharmaceutical products: With the help of genetic engineering the gene for a desired protein is transferred into bacteria or yeast and a large amount of protein can be produced in short time or the gene is directly inserted inside the host cells to avoid purifications. Some pharmaceutical products like:

Hormones: Mammalian hormones are produced by DNA technology. e.g. Human growth hormone (HGH), insulin etc.
Immunomodulation: These fight with certain diseases where tradition drug therapy does not work.
Antisense nucleic acids: These are single stranded DNA or RNA with pairs with mRNA and block translation and inhibit the propagation of diseases like cancer.
Genetically engineered proteins: They mimic or block the surface receptors of cell membranes. For example, one such protein mimics the HIV receptor protein as a result the HIV virus binds to this protein and fails to enter the white blood cell. Recombinant DNA technology is also helpful in the production of vaccines.
Interferons: These are proteins which limit the cell-to-cell spread of viruses inside the host. Interferons can be produced in large amounts by recombinant DNA technology.

(c) Gene therapy: It may help scientists to replace the defective genes with normal genes. This new system of medicine is called gene therapy.

Question 17.
Write briefly about recombinant DNA technology?
Answer:
Recombinant DNA Technology is engaged in production of newer and safer therapeutic drugs, called recombinant drugs. There drugs do not induce immunological reactions as some other drugs obtained from non human sources, i.e. insulin form animals. About 30 recombinant drugs have been approved for hurrhjn use. In India 12 of them are marketed. Recombinant DNA can then be forced into such cells by incubatting the cells with recombinant DNA on ice followed by placing then briefly at 42° C (heat shock) an then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

The joining of DNA involves several processes. After having cut the source DNA as well as the vector DNA with a specific restriction enzyme, the cut out ‘gene of interest’ from the source DNA and the ligase in added.

Question 18.
What are genetically modified foods? What are the advantages of GM foods? What are their disadvantages?
Answer:
Genetically modified (GM foods) crop plants are created for human and animal consumption using the latest molecular biology techniques.

The plants are created with the exact desired trait very rapidly and with great accuracy. For example, Bt cotton is a genetically modified cotton variety.

Advantages of GM foods:

Pest resistance: Chemical pesticides and fertilizers can poison the soil, ground water and cause harm to the environment. Growing genetically modified foods can help to eliminate the application of chemical pesticides and reduce the cost of production.
Herbicide tolerance: Farmers destroy weeds by chemical herbicides which is time consuming and expensive. Crop plants are genetically engineered to resist the herbicide to retain the crop yield.
Disease resistance: Many viruses, bacteria and fungi cause various plant diseases. Genetic engineers are working to create plants with resistance to these diseases.
Cold tolerance: Unexpected frost can destroy sensitive seedling. An antifreeze gene from cold water fish has been introduced into tobacco and potato plants. With this modification the plants can tolerate cold temperature.
Drought tolerance and salinity tolerance: Genetically modified plants which can tolerate high salt content and drought will help people to grow crops in formerly inhospitable places.
Nutrition: Plants can be genetically modified to contain extra vitamins, for e.g., golden rice contains high content of beta-carotene (vitamin-A).
Pharmaceuticals: Researches are working to develop edible vaccines in tomato and potatoes. They will not need any special conditions for storage and transportation.
Phytoremediation: Plants have been genetically engineered to clean up heavy metal pollution from contaminated soil.

Disadvantages: There are some major concerns about genetically modified foods, These are as follows:
(i) Environmental hazards: According to study published in nature the pollen from Bt cotton caused high mortality rates in monarch butterfly caterpillars.

Many people are concerned that over a period of time, insects will become resistant to B.t. or other crops which are genetically modified to produce their own pesticides.

Another area of concern is that crops are modified for herbicide tolerance, if these plants will crossbreed, with weeds they will produce superweeds which will also become herbicide tolerant.

(ii) Human health risks: Many children in US and Europe have developed allergies to some genetically modified food products. People are concerned that introducing foreign genes into food plants may have unexpected and negative impact on human health, whereas scientists believe that these foods do not present a risk to human risk.

(iii) Economic concerns: Producing genetically modified crops is a lengthy and costly process. Many of these GM foods are patented that will raise the cost of the seed so that small farmers and third world countries will not be able to afford these seeds.