RBSE 12th Biology Model Paper Set 5 with Answers in English

Students must start practicing the questions from RBSE 12th Biology Model Papers Set 5 with Answers in English Medium provided here.

RBSE Class 12 Biology Model Paper Set 5 with Answers in English

Time : 2.45 Hours
Maximum Marks : 56

General Instruction to the Examinee:

1. Candidate must write first his/her Roll No. on the question paper compulsorily.
2. All the questions are compulsory.
3. Write the answer to each question in the given answer book only.
4. For questions having more than one part the answers to those parts are to be written together in continuity.
5. If there is any error/difference/contradiction in Hindi & English version of the question paper, the question of the Hindi version should be treated valid.

Section – A

(1 Mark)

Question 1.
Select the correct option of the following questions and write in notebook. (1 × 9 = 9)
(i) Which of the following structures represent female gametophyte? [1]
(a) Ovule
(b) Embryo
(c) Embryosac
(d) Endosperm
Answer:
(c) Embryosac

(ii) Sometimes pollen grains affect the endosperm in angiosperms. This phenomenon is called: [1]
(a) Xenia
(b) Metaxenia
(c) Protandry
(d) Endrogeny
Answer:
(a) Xenia

(iii) Klinefelter’s syndrome has: [1]
(a) 47 + XXY
(b) 44 + XO
(c) 45 + XY
(d) 66 + XXY
Answer:
(a) 47 + XXY

(iv) In a DNA strand the nucleotides are linked together by: [1]
(a) glycosidic bonds
(b) phosphodiester bonds
(c) peptide bonds
(d) hydrogen bonds
Answer:
(b) phosphodiester bonds

(v) Memory cells are formed from: [1]
(a) monocytes
(b) eosinophils
(c) neutrophils
(d) lymphocytes
Answer:
(d) lymphocytes

(vi) Hepatitis – B is also called- [1]
(a) epidemic jaundice
(b) serum jaundice
(c) catarrhal jaundice
(d) none of these
Answer:
(b) serum jaundice

(vii) Rinderpest disease is the disease of: [1]
(a) honeybee
(b) fishes
(c) poultry
(d) cattle
Answer:
(d) cattle

(viii) In vitro studies are performed in: [1]
(a) nature
(b) human body
(c) lab
(d) plant body
Answer:
(c) lab

(ix) Bt toxin in bacteria do not kill itself because: [1]
(a) toxin enclosed in a poutch
(b) toxin is inactive
(c) bacteria is resistant
(d) toxin separates from body
Answer:
(b) toxin is inactive

Question 2.
Fill in the blanks. (1 × 4 = 4)
(i) Pure line selection is used in ……………… [1]
(ii) IUCN Stands tor ……………… [1]
(iii) ……………… are arranged in linear order on chromosome. [1]
(iv) Plants are called as ……………… because they fix carbon dioxide. [1]
Answer:
(i) self pollination
(ii) International Union for Conservation of Nature and Natural Resources
(iii) Genes
(iv) autotrophs

Question 3.
Give the answers of the following questions in one word or one line. (1 × 8 = 8)
(i) What is a mutagen? Give one example. [1]
Answer:
Substances that cause mutations are called mutagens.
Examples : Mustard gas, UV-radiations.

(ii) A geneticist interested in studying variations and patterns of inheritance in living prefers to choose organisms for experiments with shorter life cycle. Provide a reason. [1]
Answer:
A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms with shorter life cycle, because it enables the geneticist to study many generations of the organisms in a short time period.

(iii) Name two diseases caused by bacteria. [1]
Answer:

• Cholera – Vibrio cholerae
• Pneumonia – Streptococcus pneumoniae.

(iv) Mention the economic value of apis indica. [1]
Answer:
Apis indica is a common source of honey and wax. Honey is used as nutritive, and medicine while wax is used in cosmetics and polish industry.

(v) Give any two uses of genetic engineering. [1]
Answer:

1. Producing a new kind of useful proteins.
2. Produce Genetically Modified Organism (GMO) having advance characters.

(vi) Define biopiracy. [1]
Answer:
Use of biological resources or biochemicals of another country without permission and information. It is a kind of ‘theft’.

(vii) What are detritivores? [1]
Answer:
Organisms that feed on decaying organic matter or dead organisms are called detritivores.

(viii) 22 percent of the total species on earth comprise of. [1]
Answer:
Plants.

Section – B

(1.5 Marks)

Question 4.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement. [1.5]
Answer:
Yes, there has certainly been enormous improvement in reproductive health in the last 50 years. The following examples prove the statement:

• Drastic decreaser in maternal and infant mortality rate due to better postnatal core can be attributed.
• The decrease in the infant mortality rate can be attributed to the massive child immunization programme.
• The awareness created through the family planning programmes has seen people opting for smaller families.

Question 5.
Mention any four characteristics that an ideal contraceptive should have. [1.5]
Answer:

• be easily available.
• be effective and have no side effects.
• not interfere with the sexual derive/desire or the sexual act of the user.
• be user-friendly.

Question 6.
Discuss the role of fishery in enhancement of food production. [1.5]
Answer:
Fishery as an industry that plays a major role in enhancement of food production :

• It has increased the production of fishes, fish products and other aquatic animals, thus meeting the food needs of many.
• It also helps in production of fishes and sea food which are rich in protein and have more nutritional value.

Question 7.
What is plant breeding? Briefly describe various steps involved in plant breeding. [1.5]
Answer:
Plant Breeding: Plant breeding is purposeful manipulation of plant species in order to develop desired plant types that are better suited for cultivation, give high quality and better yields and are disease resistant.
The steps involved in plant breeding are as follows :

• Collection of genetic variation by various means.
• Selection
• Evaluation and release as variety and
• Seed multiplication and distribution among farmers.

Question 8.
Describe selectable marker. [1.5]
Answer:
Antibiotic resistance gene present in vector acts a selectable marker. Selectable marker identifies transformant from non- transformants. The genes encoding resistance to antibiotic such as ampicillin, chloramphenicol, tetracycline are serve as selectable markers.

Addition of selectable marker or antibiotic resistance allows the host organism to survive. Generally, ampicillin, tetracycline etc are the useful selectable marker in E. coli.

Question 9.
Why must a cell be made competent in biotechnology experiments? How does calcium ion help in doing so? [1.5]
Answer:
Competent is the ability to being able to take up the recombinant DNA in it. Recombinant DNA can’t pass the cell membrane directly so first we need to break the cell wall and then cells are treated with cations such as calcium or soaked in CaCl₂, which open the channels on the cell membrane and allow the entry of recombinant DNA into cell.

Question 10.
(a) Mention the cause and the body system affected by ADA deficiency in humans. [1.5]
(b) Why do the toxic insecticidal proteins secreted by bacillus thuringiensis kill the insect and not the bacteria itself?
Answer:
(a) ADA deficiency is caused by the mutation or malfunctioning of ADA gene. Which is responsible for making an enzyme that found in white blood cells (lymphocytes). This enzyme is very crucial for the well functioning of immune system. ADA deficiency affects lymphocytes, lungs, kidney and also affect the skin.

(b) In bacteria, toxic proteins present in inactive form that is protoxic which is not toxic. Once this protein is eaten by insect larvae, toxin get activated in gut of insect and kill the bacteria.

Question 11.
What are the advantages of molecular diagnosis over conventional methods? [1.5]
Answer:
Molecular diagnosis is a great advantage in which you can detect the disease in time before it spreads to the whole body. Initially blood test, urine test, serum test was the conventional methods to detect the infection through virus, bacteria they are very less and small, due to which they are not detected with conventional method.

Molecular diagnosis detect the presence of genetic material or proteins associated with specific health disease, even when they are present in very small number. or non-symptomatic.

Question 12.
Explain phosphorus cycle in brief with flow diagram. [1.5]
Answer:
Phosphorus Cycle Phosphorus is a major constituent of biological membranes, nucleic acids and cellular energy transfer system. Many animals also need phosphorus to make their shells, bones and teeth. The natural reservoir is rock, which contains phosphorus in the form of phosphates. When rocks are weathered, minute amounts of these phosphates dissolve in soil solution and are absorbed by the roots of the plants. Herbivores and other animals obtain this elements from plant source. Organic waste products and the dead organisms are decomposed by the phosphate solubilising bacteria releasing phosphorus.

Fig.: A simplified model of phosphorus cycling in a terrestrial ecosystem

1. Producers — e.g., Grass.
2. Consumers — e.g., Grasshopper.
3. Decomposers — e.g, Bacteria.

Question 13.
Distinguish between litter and detritus. Draw a food chain, starts from detritus. [1.5]
Answer:

Litter	Detritus
Comprises of all kinds of wastes above the ground level.	Comprises of residues of dead animals and plants.
Consists of biodegradable and non-biodegradable substances.	Consists of biodegradable substances only.

Question 14.
How alien species invasions causes of species loss in a geographical area. [1.5]
Answer:
Alien species invasions :
Intentional introduction of non-native species into a particular habitat causes the extinction of indigenous species. Example : Nile perch caused the extinction of more than two hundred species of native fish of Lake Victoria in Kenya when they were introduced in the lake.

Question 15.
Name three methods of in situ conservation of wildlife. [1.5]
Answer:
Biodiversity loss : The loss of different plant and animal species mainly due to four reasons, habitat loss and fragmentation, overexploitation, Alien (exotic) species invasion and coextinction is called biodiversity loss. Methods of insitu conservation of wild life :

1. National parks
2. Sanctuaries
3. Biosphere reserves

Section – C

(3 Marks)

Question 16.
What do you mean by pollination agents? Write the adaptations in flower to anemophily. [3]
Or
What do you understand by budding? Explain budding with example of yeast. [3]
Answer:
Agents of Pollination:
The agents or vectors responsible for cross-pollination in flowering plants have been grouped into two main categories : Abiotic (wind, water) and Biotic (animals such as ants, bees, bat etc.).

1. Wind Pollination or Anemophily : The pollination of flowers by wind is called anemophily. The flowers which are wind pollinated are known as anemophilous. The anemophilous flowers are characterised by the following adaptations :

• Flowers are small, colourless, odourless and nectarless and inconspicious.
• In such flowers, calyx and corolla are either reduced or absent. Anthers are usually versatile and exerted.
• Pollen grains of these flowers are small, light, dry and dusty and sometimes winged, so that they are easily blown by wind over long distance, upto 1300 km.
• Pollen grains are produced in large quantity. For example, a single flower of Cannabis produces 5,00,000 pollens and million by a tassel of Zea mays.
• The stigmas are usually branched, broad, large, well-exposed and hairy to catch pollens from the air.
• When flowers are unisexual, the staminate flowers are much more numerous than the pistillate flowers. In bisexual flowers the stamens are generally numerous.
• In plants like Urtica burst suddenly to throw pollen grains into the air called gunpowder mechanism.
• In certain cases, flowers are produced before the appearance of leaves to increase chances of pollen grains reaching the stigma.

Question 17.
When a cross is made between tall plant with yellow seeds (Tt Yy) and tall plant with green seed (Tt yy), what proportion of phenotype in the offspring could be expected to be :
(a) tall and green, (b) dwarf and green. [3]
Or
Write a brief note on Klinefelter’s syndrome. [3]
Answer:
When a cross is made between tall plant with yellow seeds (Tt Yy) and tall plant with green seeds (Tt Yy), the phenotypic proportion in the offspring could be expected are : three tall and green, one dwarf and green.

Phenotypic ratio = 3 : 3 : 1 : 1
Tall yellow = 3, tall green = 3, dwarf yellow = 1, dwarf green = 1

Question 18.
What are communicable and non-communicable diseases? Write any two ways for direct transmission of communicable diseases. [3]
Or
The diesease which are caused by pathogen, and are transmitted from infected person to a healthy person are called communicable diseases. [3]
Answer:
Communicable Diseases : The diseases which are caused by pathogen and are transmitted from infected person to a healthy person are called communicable diseases. Non

Communicable Diseases : The diseases which are not caused by pathogen and do not communicate from infected person to healthy person.

Ways for direct transmission:

• Droplet Infection: Diseases like diphtheria, influenza, common cold etc. are spread by droplet infection (germs in tiny droplets of mucus) from sneezing, coughing, spiting or even taking of infected reason.
• Direct Contact : Some diseases like chicken pox, leprosy, ring worm etc. are spread by physical contact between a infected person and a healthy person.

Section – D

(4 Marks)

Question 19.
With a neat, labelled diagram, describe the part of a typical angiosperm ovule. [4]
Or
Explain about any four types of ovule. [4]
Answer:
An ovule is a female megasporangium where the formation of megaspores takes place.

Various parts of ovule are as follows :

• Funicle : It is a stalk-like structure which represent the point of attachment, of the ovule to the placenta of the ovary.
• Hilum : It is the point where the body of the ovule is attached to the funiculus.
• Integuments : They are the outer layers surrounding the ovule that provide protection to the developing embryo.
• Micropyle : It is a narrow pore formed by the projection of integuments. It marks the point where the pollen tube enters the ovule at the time of fertilization.
• Nueellus : It is a mass of the parenchymatous tissue surrounded by the integuments from the outside. The nueellus provides nutrition to the developing embryo. The embryo sac is located inside the nueellus.
• Chalaza : It is the basal swollen part of the nueellus from where the integuments originate.

Question 20.
(a) Write four goals of human genome project:
(b) Briefly describe the following:
(i) Transcription
(ii) Polymorphism
(iii) Translation
(iv) Bioinformatics. [4]
Or
‘DNA replicates semiconservatively’. Prove it with the help of Mathew Meselson and Franklin Stahl experiment. Draw the linear diagram of the experiment. [4]
Answer:
(a) Goals of human genome project:

• Optimization of the data analysis.
• Sequencing the entire genome.
• Identification of the complete human genome.
• Creating genome sequence databases to store the data.

(b) (i) Transcription : It is a DNA directed synthesis of RNA in which the RNA transcribed on 3′ → 5′ template strand of DNA in 5′ → 3′ direction.
(ii) Polymorphism : Variation of genetic level occurred due to mutation, is called polymorphism. Such variations are unique at particular site of DNA, forming satellite DNA. The polymorphism in DNA sequences is the basis of genetic mapping and DNA finger printing.
(iii) Translation : The process of protein synthesis from mRNA, tRNA, rRNA is called translation. This process occurs on ribosomes with the help of aminoacids.
(iv) Bioinformatics: Computational method of handling and analysing biological databases.