RBSE 12th Biology Model Paper Set 3 with Answers in English

Students must start practicing the questions from RBSE 12th Biology Model Papers Set 3 with Answers in English provided here.

RBSE Class 12 Biology Model Paper Set 3 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) Vegetative propagation in Bryophyllum mainly occurs by: [1]
(a) bulbils
(b) rhizome
(c) leaf bud
(d) runner
Answer:
(c) leaf bud

(ii) Embryo sac is found in:
(a) endosperm
(b) embryo
(c) ovule
(d) seed
Answer:
(c) ovule

(iii) Mendel’s law of independent assortment is applicable for:
(a) all non-linked genes only
(b) all genes in all organisms
(c) all genes of pea plant only
(d) all linked genes only
Answer:
(a) all non-linked genes only

(iv) DNA multiplication is called:
(a) translation
(b) replication
(c) transduction
(d) transcription
Answer:
(a) translation

(v) Antigens are present
(a) inside nucleus
(b) on the cell surface
(c) inside cytoplasm
(d) on nuclear membrane
Answer:
(b) on the cell surface

(vi) Which of the following is not a hereditary disease?
(a) Cretinism
(b) Cystic fibrosis
(c) Thalassemia
(d) Haemophilia
Answer:
(a) Cretinism

(vii) Man-made crop Triticale is a hybrid between:
(a) wheat and rye
(b) rice and barley
(c) maize and barley
(d) rice and maize
Answer:
(a) wheat and rye

(viii) Which purine is found in RNA?
(a) Cytosine
(b) Thymine
(c) Guanine
(d) Uracil
Answer:
(d) Uracil

(ix) Midgut epithelium breakdown can be done by:
(a) protoxic
(b) activated protein crystals
(c) protoplast fusion
(d) dehydration
Answer:
(b) activated protein crystals

Question 2.
Fill in the blanks. (1 × 4 = 4)
(i) ……………. is always the main objective of plant breeding. [1]
(ii) Desert National park (Jaisalmer) is famous for ……………. [1]
(iii) New strands of DNA are formed only in the ……………. direction. [1]
(iv) A saprophytic Food chain proceeds from ……………. animals to microorganisms. [1]
Answer:
(i) Higher yield
(ii) Great Indian Bustard
(iii) 5′ → 3′
(iv) dead

Question 3.
Give the answers of the following questions in one word or one line. (1 × 8 = 8)
(i) Give an example of an organism that exhibits haplo-diploidy sex-determination system. [1]
Answer:
Haplo-diploidy sex-determination system is seen in honeybees.

(ii) Give an example of polygenic trait in humans. [1]
Answer:
Human skin colour.

(iii) Name the vector ofkalaazar and cholera. [1]
Answer:
Kala azar — sandfly (Phlebotomus)
Cholera — Housefly (Musca domestica).

(iv) Define animal husbandry. [1]
Answer:
The science of rearing, feeding, caring, breeding and management of domesticated animals is called animal husbandry.

(v) What is sonification? How can you break DNA into smaller fragments using other than restriction enzymes? [1]
Answer:
Bonification is another method used to generate small DNA fragments by using mechanical shearing. It is done with the help of ultrasound blenders.

(vi) Name the source organism of the gene Cry IAc and its target pest. [1]
Answer:
Cry IAc gene controls the cotton bollworm and is isolated from bacteria Bacillus thuringiensis.

(vii) Define ecological succession. [1]
Answer:
An orderly and progressive replacement of one community by another till the development of a stable or climax community in a particular area is termed as ecological succession.

(viii) More than 25 percent of the drugs currently sold in the market worldwide are derived from. [1]
Answer:
Plants.

Section – B

(1.5 Marks)

Question 4.
Differentiate between in-vitro and in-vivo fertilization. [1.5]
Answer:
The fertilization of ovum outside the body of female almost in the similar conditions as those in the body is known as in-vitro fertilization. in-vivo fertilization is the method of sexual reproduction where the fusion of sperm and egg takes place inside the body of the female.

Question 5.
Describe the technique by which chromosomal disorder in a developing foetus can be detected. [1.5]
Answer:
Amniocentesis is a technique by which chromosomal or genetic disorder in a developing foetus can be detected. This is based on the chromosomal/karyotype pattern in the cells found the amniotic fluid surrounding the developing foetus. Amniotic fluid contains cells and molecules shed by the foetus. The chromosomes of foetal cells can also be used to find out the sex of the foetus and abnormalities if any. So, if an abnormality is found, the mother can start the treatment or get the foetus aborted.

Question 6.
Suggest four important Steps to produce a disease resistant plant through conventional plant breeding technology. [1.5]
Answer:
The four important steps to produce a disease resistant plant through conventional plant breeding technology are :

1. Screening of germplasm from resistant sources.
2. Hybridization of selected parents.
3. Selection and evaluation of the hybrids.
4. Testing of new variety.

Question 7.
“Large scale cultivation of Spirullina is highly advantageous for human population”. Explain giving two reasons. [1.5]
Answer:

• Spirullina is a very fast, easily growing alga that can produce huge quantity of protein even in a small area of land.
• These microbes grow on material like wastewater from potato processing plants, straw, molasses, animal mannure and even sewage.

Utilisation of these may reduce environmental pollution.

Question 8.
Given below is the diagram representing a typical gel electrophoresis technique, Observe the illustration and answer the following questions. [1.5]

(a) What will be the direction of movement of DNA fragments and why?
(b) What will be the sites of largest and smallest molecules of DNA and why?
(c) Which medium is used to separate DNA fragments?
Answer:
(a) The direction of movement of DNA fragments will be Anode (A) to cathode (B).
Because DNA have overall negative charge and so they move at B site from A site:
(b) Largest molecules can be seen toward A site or positively charged Anode site and smallest size of molecules of DNA found at the negatively charged B site because they move faster than large molecules.
(c) Agarose gel or polyacrylamide (see weed) medium is used to separate DNA fragments.

Question 9.
Give the reason, if Hind II enzyme was first isolate than Hind I, then why it named so? [1.5]
Answer:
There are some specific reasons :

1. Well Hind I was firstly discover enzyme but then, it was not isolated, Hind II was the first isolated enzyme.
2. Hind I was exonuclease, so the first discovered endonuclease enzyme was Hind II.
3. There are many types of restriction enzyme Hind II was the type 2 enzyme, that why it named so.

Question 10.
Explain the role of biotechnology in agriculture by giving an example. [1.5]
Answer:
Agriculture biotechnology has been producing countless new products, that have change our live for the better role of biotechnology which are followed to increase the food production.

• Agriculture based on Agrochemical.
• Organic agriculture.
• Agriculture based on genetically engineered crop.

Question 11.
What are GMOs? List any four ways by which GMO’s have been useful for enhanced crop output. [1.5]
Answer:
GMOs : Any cell of plants, bacteria, algae fungi and animals, whose DNA or genes are artificially designed and altered with desired gene are called Genetically modified organisms (GMOs).

• GMO’s are resistant to pest so reduce the dependency on agrochemicals.
• GMO’s have high yield so due to resistancy post harvest losses are greatly reduced.
• GM crops are able to use maximum minerals of soil.

Question 12.
Write important features of a sedimentary cycle in an ecosystem. [1.5]
Answer:
Sedimentary cycles have reservoirs in the crust of earth or rocks Nutrient elements such as phosphorus, sulphur, potassium and calcium are present in the sediments of the earth. These sedimentary cycles are unhurried and slow. It takes a long time to complete their circulation and are deemed to be less perfect cycles. It is because when recycling, the nutrient elements may get sealed in the reservoir pool taking a long time to come out and continue circulation. Hence, they typically go out of circulation for a long period of time.

Question 13.
Write the difference between upright pyramid and inverted pyramid. [1.5]
Answer:
Differences Between Upright and inverted pyramid

Upright pyramid	Inverted pyramid
Pyramid of energy is always upright.	Pyramid of numbers and biomass can be inverted.
At the producer level of an ecosystem, this pyramid has the highest number and biomass of organisms which declines at each trophic level in a food chain.	At the producer level of an ecosystem, this pyramid has the lowest number and biomass of organisms which rises at each increasing tropic level in a food chain.

Question 14.
How habitat loss and fragmentation causes of species losses in a geographical area. [1.5]
Answer:
Habitat loss and fragmentation:
Uncontrolled and unsustainable human activities such as slash, deforestation, mining, burn agriculture and urbanization causes habitats of different entities to be destructed or altered. This leads to the breakup of habitat into smaller pieces affecting the migration of animals and also a decline in the genetic exchange between populations resulting in a decline in the species.

Question 15.
(a) What are sacred groves? What is their role in conservation?
(b) Why should biodiversity be conserved, list any three ethical arguments in its support? [1.5]
Answer:
(a) Sacred Groves : They are referred to the tracts of forests that are restored in the places of worship. These are found to be in Western Ghats of Karnataka, Rajasthan, Maharashtra, Madhya Pradesh and Meghalaya. These groves help in safeguarding threatened, rare and endemic species of animals and plants present in a region. The act of deforestation is sternly banned in tribal regions. Thus sacred groves are rich regions.

(b) (i) Narrwly utilitarion arguments.
(ii) Broadly utilitarian arguments
(iii) Ethical reasons.

Section – C

(3 Marks)

Question 16.
Explain:
(a) Juvenile phase.
(b) Reproductive phase.
(c) Senescent phase. [3]
Or
Name the vegetative reproductive structure of the following :
(i) Ginger (ii) Potato (iii) Onion (iv) Sweet potato (v) Colocasia (vi) Eichornia.
Answer:
(1) Juvenile or Vegetative Phase : All the organisms grow to a certain maturity in their life before they start reproducing sexually. The period of growth between their birth upto their reproductive maturity is called the juvenile phase. In plants, the period of growth between seed germination up to initiation of flowering is called vegetative phase. This phase varies from organisms to organisms.

(2) Reproductive Phase : The later part, when the organisms start reproducing sexually is called reproductive phase. In higher plants (seed plants), the flowering phase is called as their reproductive phase. In animals the end of juvenile phase is indicated by appearance of certain morphological and physiological changes prior to reproductive phase. This phase also shows a great variation among different organisms.

(3) Senescence or Old age : The third or last phase of life cycle is known as ‘old age’ or ‘senescence’. During this phase, the end of reproductive phase takes place and ultimately the organism leads towards death. In annuals, biennial plants, vegetative, reproductive and senescent phases can be clearly seen. Whereas in the case of trees which are perennials it is very difficult to clearly define phases. Senescence carries finally to death.

Question 17.
Explain the low of dominance using a monohybrid cross. [3]
Or
What is pedigree analysis? Suggest how such an analysis, can be useful? [3]
Answer:
When two different factors (genes) or a pair of contrasting forms of a character are present in an organism, only one expresses itself in the F₁ generation and is termed as dominant while the other remains unexpressed and called recessive factor (gene).

Example : when a monohybrid cross between tall (TT) and dwarf (tt) pea plants was carried out, all the plants in F₁ generation were observed to be tall (Tt). Here tallness is dominant over dwarfness which expressed itself and the dwarfness got suppressed.

Question 18.
(a) It is generally observed that the children who had suffered from chickenpox in their childhood may not contract the same disease in their adulthood. Explain giving reasons the basis of such an immunity in a individual. Name this kind of immunity.
(b) What are interferons ? Mention their role. [3]
Or
What measure would you take to prevent waterborne disease? 3
Answer:
(a) A child who had suffered from chickenpox in childhood may not contract disease in his/her adulthood. It is because during the first encounter with pathogen (chickenpox virus) specific antibodies (By humoral immune response) are produced to counter the attack.

During this attack, memory cells are also produced. Due to this, on subsequent exposure to the same pathogen, the immune response is more rapid and intense. That is why, second exposure to the chickenpox does not cause disease. It is known as acquired immunity of the body.

(b) Interferons are special type of proteins secreted by virus infected cells. They protect the healthy cells from the virus attacks.

Section – D

(4 Marks)

Question 19.
Draw a well labelled diagram of T.S. of mature anther. Write the functions of the following:
(i) Tapetum (ii) Sporogenous tissue (iii) Endothecium (iv) Stomium. [4]
Or
Explain microsporogenesis. [4]
Answer:

Functions :
(i) Tapetum : It nourishes the developing microspores.
(ii) Sporogenous tissue : It gives rise microspores.
(iii) Endothecium : The endothecium cells are hygroscopic in nature, hence play an important role in dehiscence of anthers.
(iv) Stomium : It helps in dispersal of pollengrain or dehiscence of pollen sacs.

Question 20.
Difference between mRNA and tRNA. Write short notes on the following :
(i) Small nuclear RNA
(ii) Genetic RNA
(iii) Enzymatic RNA [4]
Or
Answer the following questions based on Meselson and Stahl’s experiment on E.coli.
(i) Write the name of the chemical substance used as the only source of nitrogen in the experiment.
(ii) Why did they allow the synthesis of the light and the heavy DNA molecules in the organism?
(iii) How did they distinguish the heavy DNA molecules from the light DNA molecules?
(iv) Write the conclusion the scientists arrived at the end of the experiment. [4]
Answer:

	mRNA	tRNA
1.	It is linear.	It is clover-leaf shaped.
2.	It carries coded, informations.	It carries information for association with an amino acid and an anticode for its incorporation in a polypeptide.
3.	mRNA undergoes additional processing, i.e., capping tailing and splicing.	It does not require any processing.
4.	Nitrogen bases are unmodified.	Nitrogen bases may be modified.

1. Small Nuclear RNAs (snRNAs) :
These are found in the nucleus of eukaryotic cells. snRNA has 100 to 215 nucleotides. These are important for DNA replication and processing of long, newly transcribed hnRNA molecules.

2. Genetic RNA (gRNA) : In some viruses, in which, RNA is the hereditary material, gRNA is found. It may be single stranded or double stranded.

3. Enzymatic RNAs : Certain RNAs act as biological catalysts. These are called ribozymes. These are associated with the maturation of hnRNAs. 23S and 28S ribosomal RNAs of ribosomes catalyse formation of peptide bonds during synthesis of polypeptide chains.