CBSE Class 12 Biology 2023 : 50 Most Important (2 Marks) Question with Answers
BRIGHT CBSE MASTER
In this article, we have compiled and presented the 50 most Important Questions (2 Marks) and answers for Class 12 Biology board examinations of CBSE in the 2023 session. These important questions have been designedby subject experts focusing on the latest changes in the syllabus, sample papers and previous year questions.
Biology
Most Important (2 Marks) Questions,
Q1. When are the non-flowering plants said to be homothallic and monoecious and heterothallic and dioecious? Give an example of each.
Ans. Non-flowering plants are said to be homothallic and monoecious when they bear both male and female reproductive parts on the same plant. They are said to be heterothallic and dioecious when they have male and female reproductive parts on separate plants, i.e. male and female plants.
Q2. Cucurbits and papaya plants bear staminate and pistillate flowers. Mention the categories they are put under separately on the basis of the type of flowers they bear.
Ans. Cucurbits are monoecious plants i.e., they bear both male and female flower on the same plant. On the other hand, papaya plants are dioecious, i.e., both male and female flowers are present on separate plant.
Q3. Mention the unique flowering phenomenon exhibited by Strobilanthes kunthiana (neelakurinji).
Ans.Strobilanthes kunthiana produce flowers once in twelve years.
Q4. Name the mode of reproduction that ensures the creation of new variants.
Ans. Sexual reproduction ensures creation of new variants. It involves, fusion of gametes from different parents which results in genetic recombination that cause variations.
Q5. Name the phase all organisms have to pass through before they can reproduce sexually.
Ans. All organisms have to pass through the period of growth known as juvenile or vegetative phase before they start reproducing sexually. The period of growth is between the birth of an individual upto reproductive maturity.
Q6. Name the phenomenon and the cell responsible for the development of a new individual without fertilisation as seen in honey bees.
Ans. Female gamete undergoes development to form new individual without fertilisation. This phenomenon is known as parthenogenesis.
Q7. How are Cucurbita plants different from papaya plants with reference to the flowers they bear?
Ans. Cucurbits are monoecious plants i.e., they bear both male and female flower on the same plant. On the other hand, papaya plants are dioecious, i.e., both male and female flowers are present on separate plant.
Q8. Mention the advantages of emasculation and bagging in artificial hybridisation in plants bearing unisexual and bisexual flowers.
Ans.Emasculation is removal of stamens from floral buds of female parents. A breeder needs to emasculate a flower to eliminate the chances of self pollination.
Q9. Express the process of pollination in Vallisneria.
Ans. Pollination in Vallisneria is accomplished through water. The female flower reach the surface of water by the long stalk and the male flowers or pollen grains are released on to the surface of water. They are carried passively by water currents and some of them eventually reach the female flowers and the stigma. Pollen grains are covered by mucilage which helps them in sticking to stigma as well as protects them from wetting by water. After pollination, the female flower is pulled inside water by the coiling of its stalk.
Q10. Explain the functions of myometrium and endometrium in human females.
Ans. Myometrium is middle thick layer of smooth muscles fibres that brings about contraction of the uterus during the delivery of the baby. The endometrium is the inner glandular layer that undergoes cyclical changes during the menstrual cycle.
Q11. Mention the names and the characteristics of different uterine wall layers in humans. Which one of them undergoes cyclic changes during menstrual cycles?
Ans. The wall of the uterus is composed of three layers of tissues. The perimetrium is an outer thin covering of peritoneum. The myometrium is a middle thick layer of smooth muscle fibres which shows strong contraction during delivery of the baby. The endometrium is inner glandular layer that lines the uterine cavity. The endometrium undergoes cyclical changes during menstrual cycle.
Q12. Why are copper containing intrauterine devices considered an ideal contraceptive for human females?
Ans. Copper containing intrauterine devices (CuT, Cu7, etc) are considered an effective contraceptives for human females as the Cu ions released by them suppress sperm motility and fertilising capacity of the sperms. Hence, they act as effective birth control method.
Q13. What do oral pills contain and how do they act as effective contraceptives?
Ans. Oral pills contain either progestin (progestogen) alone or a combination of progestogen and estrogen both. Oral pills inhibit ovulation motility and secretory activity of oviducts and changes the cervical mucus that impairs transport of sperms and also alter the uterine endometrium and makes it unsuitable for implantation. Hence, they act as effective contraceptives for human females.
Q14. When does a geneticist need to carry a test cross? How is it carried?
Ans. To determine the genotype of a plant i.e., whether the individual is exhibiting dominant character is homozygous or heterozygous, a test cross is carried out by a geneticist. The individual having dominant phenotype is crossed with its homozygous recessive parent. If heterozygous tall is crossed with homozygous recessive parent, tall and dwarf will be produced, in equal proportion while if homozygous tall is crossed with homozygous recessive, the upcoming progenies will contain all tall plant.
Q15. State and explain the law of segregation as proposed by Mendel in a monohybrid cross?
Ans. Principle of segregation states that, “when a pair of contrasting factor or gene are brought together in a hybrid; these factors do not blend or mix up but simply associate themselves and remain together and separate at the time of gamete formation”, i.e, allele pairs segregate during gamete formation and the paired condition is restored by random fusion of gametes during fertilisation. The above law is also known as “law of purity of gametes” because each gamete is pure in itself.
Q16. Give an example of a gene responsible for multiple phenotypic expressions. What are such genes called ? State the cause that is responsible for such an effect.
Ans. Multiple alleles is the presence of more than two alleles of a gene. They are produced due to repeated mutation of the same gene but in different directions and show meristic type of germinal variations, e.g., eye colour in Drosophila. Multiple alleles occur on the same gene locus of the same chromosome or its homologue and are responsible for multiple phenotypic expression. For example, the wild type of allele for red eye colour (w+ or W) in Drosophila melanogaster mutated to form allele for white eye (w). Further, mutations in both have produced as much as 15 alleles which are recessive to wild type and dominant over white eye (w) but have incomplete intermediate dominance over one another.
Q17. Although a prokaryotic cell has no defined nucleus, yet DNA is not scattered throughout the cell. Explain.
Ans. In prokaryotes, DNA lies in the cytoplasm which is supercoiled (coiled and recoiled) with the help of RNAs and non-histone basic proteins like polyamines. DNA being negatively charged is held in place with the help of these proteins that have positive charges in a region termed as nucleoid. The DNA in nucleoid is organised in large loops held by proteins.
Q18. Describe the structure of a nucleosome.
Ans. DNA packing in eukaryotes is carried out with help of lysine and arginine rich basic proteins called histones. The unit of compaction is called nucleosome. There are five types of histone proteins - H1, H2A, H2B, H3 and H4. Four of them (H2A, H2B, H3 and H4) occur in pairs to produce histone octamer, called nu body or core of nucleosome. Their positively charged ends are towards the outside. They attract negatively charged strands of DNA. DNA over nu body forms 1¾ turns to form nucleosome core. A typical nucleosome contains 200 bp of DNA helix. DNA connecting two adjacent nucleosomes is called interbead or linker DNA. It bears H1 histone protein.Nucleosome chain gives a beadson-string appearance.
Q19. What is central dogma ? Who proposed it?
Ans. Concept of central dogma was proposed by Crick in 1958. It refers to the flow of information from DNA to mRNA (transcription) and then decoding the information present in mRNA in the formation of polypeptide chain or protein (translation).
Q20. What was proposed by Oparin and Haldane on origin of life? How did S.L. Miller’s experiment support their proposal?
Ans. Oparin of Russia and Haldane of England proposed that the first form of life could have originated from pre-existing non-living organic molecules (e.g., RNA, protein, etc.) and that formation of life was preceded by chemical evolution, i.e., formation of diverse organic molecules from inorganic constituents. The Oparin-Haldane theory (also called chemical theory or naturalistic theory) was experimentally.
Q21. Mention the contribution of S.L. Miller’s experiment on origin of life.
Ans. Oparin of Russia and Haldane of England proposed that the first form of life could have originated from pre-existing non-living organic molecules (e.g., RNA, protein, etc.) and that formation of life was preceded by chemical evolution, i.e., formation of diverse organic molecules from inorganic constituents. The Oparin-Haldane theory (also called chemical theory or naturalistic theory) was experimentally.supported by Stanley Miller. Miller's experiment supported chemical evolution of life. He created laboratory conditions similar to primitive earth. He used mixture of methane, ammonia, hydrogen and water in an air tight apparatus and passed electrical discharge from electrodes at 800 °C. He passed the mixture through a condenser. He circulated the gases in same way for about a week and then analysed the contents. Formation of simple organic compounds supported chemical evolution of life.
Q22. Write the Oparin and Haldane’s hypothesis about the origin of life on Earth. How does meteorite analysis favour this hypothesis?
Ans. Oparin and Haldane proposed that life originates from pre-existing, non-living organic molecules, such as RNA, proteins, etc., and formation of life was preceded by chemical evolution. Meteroite analysis confirmed presence of similar compounds elsewhere in space, maintaining that, life had reached earth in the form of spores from other heavenly bodies.
Q23. Name the scientist who has used the set-up shown. Write the purpose of ‘a’ in the set-up and the conclusion the scientist arrived at.
Ans. Stanley Miller had used the set up shown. In the given figure, ‘a’ represents electrodes, used for simulation of lightning. From the experiment, Miller concluded that complex organic molecules were formed from simple inorganic molecules.
Q24. Name one air borne and a water borne disease in humans. List one specific symptom of each one of them.
Ans. Influenza (flu) caused by Orthomyxovirus is an air borne disease. It is caused by droplet infection. Person suffering from influenza experiences headache and fever. Typhoid caused by bacterium Salmonella typhi is
a water borne disease. It is transmitted through faecal oral route. Patient suffering from typhoid has high fever and feels abdominal pain.
Q25. List the symptoms of ascariasis. How does a healthy person acquire this infection?
Ans. Ascariasis is caused by the common round worm Ascaris lumbricoides, a giant intestinal worm. Symptoms of this disease include internal bleeding, muscular pain, fever, anemia and blockage of the intestinal assage. A healthy person acquires infection through contaminated water, vegetables, fruits, etc.
Q26. A patient showed symptoms of sustained high fever, stomach pain and constipation, but no blood clot in stools. Name the disease and its pathogen. Write the diagnostic test for the disease. How does the disease get
transmitted?
Ans. Bacterial disease typhoid, caused by Salmonella typhi is characterised by sustained high fever (39° - 40°C), stomach pain and constipation. Typhoid fever is diagnosed by Widal test and is transmitted through contaminated food and water.
Q27. Name the hormone with which a cow is administered using MOET technology. State the function of this hormone.
Ans. In MOET, genetic mother is the biological mother of the offspring. Hormones with FSH like activity are given to genetic mothers for inducing follicular maturation and superovulation. As a result, 6-8 eggs are produced instead of 1 egg per cycle. She is then mated with a superior bull or artificially inseminated. Embryos at 8-32 cell stage are recovered and transferred to surrogate mother and the genetic mother is again available for super ovulation.
Q28. Why is cross-breeding in animals practiced? How is a breed Hisardale developed?
Ans. Cross-breeding allows the desirable qualities of two different breeds to be combined. In this method, superior males of one breed are mated with superior females of other breed. Hisardale is a new breed of sheep developed in Punjab by cross-breeding the Bikaneri ewes and Merino rams.
Q29. “Artificial insemination helps overcome several problems of normal mating in cattle”. Do you agree? Support your answer with any three reasons.
Ans. Yes, artificial insemination helps us overcome several problems of normal mating in cattle. It involves the insemination of the semen of superior bulls of exotic or indigenous breeds into the reproductive tract of the selective female.
The three reasons are as follows–
(i) It increases the rate of conception and considerably fewer sperms are required to achieve conception.
(ii) It controls the spread of certain diseases and also ensures good quality of progeny.
(iii) It is very economical as it excludes the need for importing and maintaining the bulls.
Q30. Differentiate between outbreeding and outcrossing.
Ans. Outbreeding is breeding of unrelated animals which may be between individuals of same breed or between different breeds or different species. Outcrossing is the practice of mating of animals within the same breed but having no common ancestors. This is the best breeding method for animals that are below average in milk production, growth rate in beef cattle, etc. A single outcross often helps to overcome inbreeding depression.
Q31. Explain giving reasons, the need to keep the bee-hives in the fields during flowering season.
Ans. Bees are the pollinators of many of our crop species such as sunflower, Brassica, apple and pear. Keeping beehives in crop fields during flowering period increases pollination efficiency and improves the crop yield. Also bees collect nectar from flowers of these crop plants to make honey. Hence, honey yield also increases.
Q32. How is a pureline in an animal raised? Explain.
Ans. A pureline in an animal is raised through inbreeding. In this method, superior males and females of the same breed are selected and mated. The progeny obtained from such matings are evaluated and superior males and females among them are identified for further mating. Inbreeding increases homozygosity. It exposes harmful recessive genes that are eliminated by selection.
Q33. How does inbreeding depression set in? Mention the procedure you would suggest to reverse this.
Ans. Continued inbreeding (mating between animals of same breed for 4-6 generations) reduces fertility and productivity of animals, also known as inbreeding depression. This could be the reason of reduced fertility and productivity in the herd of cattle. In this condition, the selected animal should be mated with superior animals of the same breed but unrelated to the breeding population, e.g., outcrossing. This helps in restoring fertility and yield.
34. Write the basis of naming the restriction endonuclease EcoRI.
Ans. Type II restriction enzymes are named for the bacterium from which they have been isolated. The first letter used for the enzyme is the first letter of the bacterium’s genus (in italics). Then comes the first two letters of the species (in italics). EcoRI is obtained from bacterium Escherichia coli RY13. The capital letter E comes from genus Escherichia. The letter Co are from species Coli. The letter R is from RY13 (strain). The roman number I indicates that it was the first enzyme isolated from bacterium E.coli RY13.
35. All cloning vectors do have a ‘selectable marker’. Describe its role in recombinant DNA technology.
Ans. Selectable markers are the gene sequence present in cloning vectors that help in selecting those host cells which contain vectors (transformants) and eliminating the non-transformants. Generally, the genes encoding resistance to antibioticssuch as tetracycline, ampicillin, kanamycin or chloramphenicol are useful selectable markers for E.coli. Plasmid pBR322 has two resistance genes -ampicillin resistance (ampR) and tetracycline resistance (tetR) which are considered useful for selectable markers.
36. b-galactosidase enzyme is considered a better selectable marker. Justify the statement.
Ans. The gene for the enzyme b-galactosidase is an alternative selectable marker. Alternative selectable marker is developed to differentiate recombinants and non-recombinants on the basis of their ability to produce colour in the presence of a chromogenic substance. Now a recombinant DNA is inserted in the coding sequence of an enzyme b-galactosidase. Presence of insert results into insertional inactivation of the b-galactosidase and therefore the colonies do not produce any colour and are marked as recombinant colonies. Hence, b-galactosidase enzyme is considered a better selectable marker.
37. Why does Bt toxin not kill the bacterium that produces it, but kill the insect that ingests it?
Ans. Soil bacterium Bacillus thuringiensis produces proteins that kill certain insects like lepidopterans (tobacco budworm, armyworm), coleopterans (beetles) and dipterans (flies, mosquitoes), etc. B. thuringiensis forms some protein crystals. These crystals contain a toxic insecticidal protein. This toxin does not kill the Bacillus (bacterium) because it exists as inactive protoxins in them. But, once an insect ingests the crystals, it is converted into an active form of toxin due to the alkaline pH of the alimentary canal that solubilises the crystals. The activated toxin binds to the surface of midgut epithelial cells and creates pores which cause cell swelling and lysis and finally cause death of the insect.
38. What does ‘cry’ genes in Bacillus thuringiensis code for? State its importance in cotton crop.
Ans. cry genes code for certain crystal (cry) proteins that are toxic to insect larvae. The genes cryIAc and cryIIAb control cotton bollworm. When these genes are introduced into cotton plants through genetic engineering, these plants become resistant to the attack of cotton bollworm.
39. How does ‘RNA interference’ take place in eukaryotes? Mention its importance.
Ans. RNA interference (RNAi) is the phenomenon of inhibiting activity of a gene through production of sense and antisense RNA. RNAi takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRNA (silencing). The source of this complementary RNA could be from an infection by viruses having RNA genomes or mobile genetic elements (transposons) that replicate via an RNA intermediate.
40. How is normal human body temperature of 37°C maintained during (i) summer and (ii) winter? Explain.
Ans. Human beings are able to maintain a constant body temperature at about ~37°C.
(i) During summers the external temperature may rise upto 45°C. Humans begin to sweat profusely when external temperature rises above 37°C. Cooling of the body occurs as sweat evaporates.
(ii) During winter, when external temperature is low, our body inadvertently starts shivering. It is an exercise that raises body temperature.
41. Why the plants that inhabit a desert are not found in a mangrove? Give reasons.
Ans. Plants inhabiting desert (xerophytes) are not found in mangroves, because xerophytic plants are adapted to dry and hot environment. They possess various physical modifications to tolerate extreme water scarcity and heat, like extensive root system, succulent organs, leaf reduced to spines, etc. Mangrove swamp is a region of vegetation where soil is highly saline and water logged. Only halophytes can survive in such regions as they possess aerial roots called pneumatophores through which gaseous exchange occurs. Roots of xerophytes are positively geotropic and will suffocate and die in such badly aerated soil ultimately leading the whole plant to death.
42. Plants that inhabit a rainforest are not found in a wetland. Explain.
Ans. Plants that inhabit rainforest are not able to germinate in wetland due to presence of excess water and anaerobic conditions (due to water logging). Wetlands are marshy areas and plants growing there have negatively geotropic roots, called pneumatophores which help in gaseous exchange. Pneumatophores are not present in plants inhabiting rainforests.
43. Many freshwater animals cannot survive in marine environment. Explain
Ans. If a freshwater animal is placed in marine environment, then it will not be able to survive because of osmoregulation problem. The freshwater animal is adapted to live in fresh environment, so, if it is kept in saline water, it will not be able to cope with outside hypertonic environment and it would face death.
44. How is humus formed? Mention any three characteristics of humus.
Ans. Humus is dark coloured, amorphous organic matter formed by decomposition of detritus. Humus is rich in lignin and cellulose. It is quite resistant to microbial action. It is a reservoir of nutrients and is helpful in maintenance of soil moisture as well as aeration.
45. How does the dead organic matter get decomposed in nature? Explain.
Ans. In nature, dead organic matter gets decomposed by decomposer microorganisms like bacteria, fungi through the process of decomposition. Decomposition is the process of breaking down of complex organic matter into inorganic substances like carbon dioxide, water and nutrients.
46. Describe how do oxygen and chemical composition of detritus control decomposition.
Ans. Decomposition of detritus is an oxygen requiring process, i.e., aerobic conditions are essential for the activity of decomposer organisms. Chemical composition of detritus effects its decomposition. Chitin and lignin decompose at a very slow rate. Cellulose decomposition also takes time whereas, detritus which is rich in nitrogen and water soluble substances (like sugars) decomposes rapidly.
47. Substantiate with the help of one example that in an ecosystem mutualists (i) tend to co-evolve and (ii) are also one of the major causes of biodiversity loss.
Ans. (i) An ecosystem mutualists often involve co-evolution, i.e., the evolution of the flower and its pollinator species are tightly linked with one another. For example, fig species can be pollinated only by its partner wasp species and no other species.
(ii) Elimination of one invariably leads to the extinction of the other.
48. Mention the kind of biodiversity more than a thousand varieties of mangoes in India represent. How is it possible?
Ans. More than a thousand varieties of mango in India exhibit genetic diversity. Genetic diversity represents the diversity in number and types of genes as well as chromosomes and variations in the genes and their alleles in the same species. The reason for this enetic diversity is the occurrence of variations in environmental parameters and use of horticulture techniques like grafting, breeding, etc.
49. How did a citizen group called Friends of Arcata Marsh, Arcata, California, USA, help to improve water quality of the marshland using Integrated Waste Water Treatment? Explain in four steps.
Ans. In the town of Arcata situated along the Northern Coast of California, an integrated waste water treatment process was developed with the help of biologist from Humboldt State University. Waste water is treated in the following manner :
(i) Waste water is filtered to remove floating and large suspended solids.
(ii) Precipitated solids (sediments) are removed in the next step.
(iii) The clear water is now chlorinated with chlorine or perchlorate salt. This treated water contains lots of heavy metals and other dangerous pollutants.
(iv) In this step, a series of six connected marshes in 60 hectares of marshland seeded with bacteria, algae, fungi and plants, which absorb, assimilate and neutralise the pollutants. The naturally purified water is then allowed to flow out.
50. Plenty of algal bloom is observed in a pond in your locality.
Ans. (a) Algal bloom is the excess growth of planktonic algae that causes characteristic colouration of water. It may have caused due to passage of sewage and run off from the fields into the ponds. Nutrients present in sewage and run off cause nutrient enrichment or eutrophication particularly with nitrogen and phosphorus. Algal blooms cut off light for submerged plants. Water quality is greatly affected as there is decreased oxygen replenishment inside water. Decreased oxygen level is quiet harmful for aquatic life of the pond.
(b) Organic waste and other types of waste materials should not be dumped into the pond. Domestic wastes with organic nutrients must be treated before passing into it.
(c) Write what has caused this bloom and how does it affect the quality of water.
(d) Suggest a preventive measure.