Students must start practicing the questions from RBSE 12th Biology Model Papers Set 6 with Answers in English Medium provided here.
RBSE Class 12 Biology Model Paper Set 6 with Answers in English
Time : 2.45 Hours
Maximum Marks : 56
General Instruction to the Examinee:
- Candidate must write first his/her Roll No. on the question paper compulsorily.
- All the questions are compulsory.
- Write the answer to each question in the given answer book only.
- For questions having more than one part the answers to those parts are to be written together in continuity.
- 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) The ‘eyes’ of the potato tuber are : [1]
(a) root buds
(b) flower buds
(c) axillary buds
(d) shoot buds
Answer:
(c) axillary buds
(ii) When pollination takes place by bats it is called: [1]
(a) omithophily
(b) chiropterophily
(c) entomophily
(d) anemophily
Answer:
(b) chiropterophily
(iii) Haemophilia is a: [1]
(a) X-linked gene
(b) Y-linked gene
(c) XY-linked gene
(d) None of these
Answer:
(a) X-linked gene
(iv) Both deoxyribose and ribose belongs to a class of sugars called: [1]
(a) trioses
(b) tetroses
(c) pentoses
(d) hexoses
Answer:
(c) pentoses
(v) Bacterium, which is concerned with pertussis is : [1]
(a) Bordetella pertussis
(b) Bacillus
(c) Diplococcus
(d) Mycobacterium tuberculum
Answer:
(a) Bordetella pertussis
(vi) Black foot disease is caused due to- [1]
(a) nitrate
(b) fluoride
(c) arsenic
(d) sulphur
Answer:
(c) arsenic
(vii) More than 70 percent of livestock population is in: [1]
(a) Denmark
(b) India
(c) China
(d) India and China
Answer:
(d) India and China
(viii) Palindromic sequence found in E. coli at restriction site is : [1]
(a) GTTAAC
(b) GAATTC
(c) CAATTG
(d) TTAAGC
Answer:
(b) GAATTC
(ix) Transgenic bacteria used to produce insulin: [1]
(a) B. thuringiensis
(b) Brassica napus
(c) Agrobacterium rhizogenes
(d) E. coli
Answer:
(d) E. coli
Question 2.
Fill in the blanks. (1 × 4 = 4)
(i) Anthrax is a ………….. disease of cattle. [1]
(ii) It is estimated that about ………….. plant species are threatened with extinction. [1]
(iii) The central dogma in molecular biology involves ………….. [1]
(iv) In an ecosystem dominated by trees, the pyramid (of numbers) is ………….. type. [1]
Answer:
(i) bacterial
(ii) 25000
(iii) DNA → RNA → Protein
(iv) spindle.
Question 3.
Give the answers of the following questions in one word or one line. (1 × 8 = 8)
(i) What is crossing over? [1]
Answer:
The phenomenon of exchange of chromosomal segments between the homologous chromosomes during meiosis is called crossing over.
(ii) What is meant by linked gene? [1]
Answer:
The genes which tend to transmit together as a unit, as they are located very close on the same chromosome are called linked genes.
(iii) What is disease ? [1]
Answer:
Any condition which interferes with the normal functioning of the body and impairs the health, is called disease.
(iv) Where is the international centre for wheat and maize situated? [1]
Answer:
International centre for wheat and maize is located in maxico.
(v) Who inserted recombinant DNA into host first time? [1]
Answer:
The insertion of recombinant DNA so that the foreign DNA will replicate naturally, as pioneered by Herbert Boyer and Stanley Cohen.
(vi) Define genetically modified organism. [1]
Answer:
Organisms, whose DNA or genes are artificially .designed and altered with desired gene are called Genetically modified organism (GMO).
(vii) Justify the pitcher plant as a producer. [1]
Answer:
Pitcher plant is green and leafy climber and carry out photosynthesis, hence it is a producer.
(viii) Who found that plots with more species showed less year to year variation in total biomass. [1]
Answer:
Tilman.
Section – B
(1.5 Marks)
Question 4.
Write three problems related to reproductive health. [1.5]
Answer:
Problems related to Reproductive Health:
- There is little knowledge of personal hygiene and hygiene of reproductive organs. This causes sexually transmitted diseases.
- Early marriage lead to high maternal and infant mortality rate.
- Due to lack of awareness, there has been a rapid increase in population size.
Question 5.
Write causative organism, symptoms, transmission and treatment of gonorrhoea. [1.5]
Answer:
Gonorrhoea:
Causative Organism : It is caused by diplococcus bacterium Neisseria gonorrhoea.
Symptoms : The bacterium lives in genital tract producing a pus containing discharge, pain over genetalia and burning during urination. It may lead to arthritis, and eye infections.
Transmission : The bacterium transmitted through sexual contact, common toilets and undergarments.
Treatment : For this penicillin and Ampicillin are effective drugs.
Question 6.
(a) Explain what is meant by biofortification, (b) Write down the objectives of plant breeding. [1.5]
Answer:
(a) Biofortification is the process by which the nutritional quality of food crops is improved through agronomic practices, conventional plant breeding or modern biotechnology.
(b) Objectives of plant breeding :
- Higher yield from crops.
- Better quality of plants.
- Disease, insect and pest resistance.
- New varieties adaptive for particular environment.
Question 7.
Which part of the plant is best suited for making virus free plants and why? [1.5]
Answer:
The apical and axillary meristem present in apical and axillary buds respectively are the best plant parts to make-virus free plants. This is because the rate of division of meristematic cells is higher than the rate of multiplication of virus and viruses are unable to invade newly formed meristematic cells.
Question 8.
Differentiate between plasmid DNA and chromosomal DNA. [1.5]
Answer:
Differences between plasmid DNA and chromosomal DNA
Plasmid DNA | Chromosomal DNA | |
1. | Double stranded, circular | Double stranded, circular or linear. |
2. | Not attached with histone. | Attached with histone protein. |
3. | Replicate autonomously. | Replicates under nuclear control. |
Question 9.
Describe the role of DNA ligase in r-DNA technology. [1.5]
Answer:
Role of DNA ligase:
- DNA ligase leads the formation of phosphodiester bond between two DNA fragments.
- Joining of two sticky or blunt ends by the formation of hydrogen bonds.
- Ligation of all purine and pyrimidine bases.
Question 10.
Why is proinsulin so called? How is proinsulin different from functional insulin in humans? [1.5]
Answer:
Pro-insulin is naturally synthesised by human and other mammals contain a c-peptide chain that does not found in mature insulin. Pro-insulin contain three chains A, B and C, when pro-insulin undergoes maturation c-peptide separate out from insulin and fully mature insulin only contain two A and B chains, bonded with di-sulphide bond.
Question 11.
Explain how Eli Lilly, an American company, produced insulin by recombinant DNA technology? [1.5]
Answer:
In 1983, American company Eli Lilly designed two DNA sequences, similar to A and B chain of insulin. E. coli bacteria used to produce A and B chains in culture medium separately. Though disulphide linkage both the chains are exactly like mature insulin. This artificially synthesised insulin is named humulin, which is very safe and cause no allergy.
Question 12.
Explain the pyramid of numbers with example. [1.5]
Answer:
Pyramid of numbers : It gives the graphical representation of the number of individuals found at each trophic level in a food chain of an ecosystem. This pyramid can be inverted or upright depending on the crowd of producers.
Example : In a Grassland ecosystem, this pyramid is upright where in the food chain, the number of producers is followed by the number of herbivores, which in turn is followed by number of secondary and tertiary consumers. Therefore, the number of individuals at the level of producers will be maximum, whereas the number of individuals at the top carnivores will be the least. The pyramid of numbers in a parasitic food chain is inverted, where in the food chain, producers provide food to fruit eating birds which in response support few species of insects.
Question 13.
Differentiate between food chain and food web. [1.5]
Answer:
Difference between food chain and food web
Food chain | Food web | |
1. | Constitutes for a single linear sequence of entities. | Consists of a number of interconnected food chains. |
2. | Members inhabiting higher trophic levels feeds only on one type of entity. | Any given individual has alternate options for food sources. |
Question 14.
What are hot spots? Name one hot spot in India. Name three endangered animal species. [1.5]
Answer:
Hot spots are the richest and most threatened reservoirs of wildlife that need immediate attention for their conservation.
Example : Himalaya.
Name of Endangered Species:
- White lion
- Cheetah
- Rhino.
Question 15.
How co-extinction causes species loss? [1.5]
Answer:
Co-extinction : One species is connected to the other in native habitat in an intricate network. Hence, the extinction of one species causes the extinction of the other wherein they are associated with each other in an obligatory connection. For instance, the extinction of host would cause the extinction of its parasites.
Section – C
(3 Marks)
Question 16.
Define external and internal fertilization. Mention disadvantages of external fertilization. [3]
Or
Explain regeneration process with suitable diagrams. [3]
Answer:
External Fertilization : When the fusion of male and female gametes takes place outside the body of an organism, it is called external fertilization.
Example : Frog
Internal Fertilization : When the fusion of male and female gametes takes place inside the body of female organism is called internal fertilization.
Example : Goat.
Disadvantages of External Fertilization :
- It requires a medium for fusion of gametes.
- It is not ensured and there are chances of wastage and distinction of gametes.
- The young ones are often exposed to predators and no parental care is provided.
Question 17.
What is pedigree analysis? Suggest how such an analysis, can be useful? [3]
Or
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in generation for a dihybrid cross? [3]
Answer:
A pedigree is a diagram of a family tree showing the relationships between individuals together with relevant facts about their medical histories. A pedigree analysis is the interpretation of these data that allows a better understanding of the transmission of genes within the family. In human genetics, pedigree study provides a strong tool, which is utilized to trace the inheritance of a specific trait, abnormality or disease. By carefully observing the position of affected individuals, mutation carriers may be identified. From this data, the risk of carrier status for other family members or the chance that a couple may have an affected child can be estimated.
Question 18.
Differentiate between benign tumors and malignant tumors. [3]
Or
Name the pathogen and their carrier for following diseases :
(i) AIDS, (ii) Leprosy, (iii) Tetanus, (iv) Malaria, (v) Tuberculosis, (vi) Aspergillosis. [3]
Answer:
Differences between benign tumors and malignant tumors.
Benign Tumors | Malignant Tumors | |
1. | Benign tumors are single mass of cells enclosed in capsule. | Malignant tumors are mass of proliferating neoplastic cells without capsule. |
2. | Cells grow slowly. | Cells grow very rapidly. |
3. | Cells are non migratory, hence do not migrate from place of origin. | Cells have pseudopodia like processes, hence are migratory. |
4. | Cells remain combined to the site of their origin. | Cells invade and damage the surrounding normal tissues and lymphatic vessels to be carried to remote areas of body. |
5. | They are non-cancerous. | They are cancerous. |
Section-D
(4 Marks)
Question 19.
Give a diagramatic representation of different stages of development of female gametophyte in angiosperm. [4]
Or
What do you understand by pollination? Write two contrivances for self pollination. [4]
Answer:
Fig. (a) Parts of the ovule showing a large megaspore mother cell, a dyad and a tetrad of megaspores; (b) 2, 4 and 8-nucleate stage of embryo sac; (c) a mature embryo sac;
A diagrammatic representation of the mature embryo sac
Question 20.
How nucleoside and nucleotide are form? Name four nucleotides present in DNA molecules. [4]
Or
A DNA segment has a total of 1000 nucleotides, out of which 240 of them are adenine containing nucleotides. How many pyrimidine bases this DNA segment possesses? [4]
Answer:
Nucleosides : The combination of a molecule of pentose sugar with a nitrogenous base form nucleoside i.e., Pentose sugar + Nitrogenous base = Nucleoside.
Nucleotides : The combination of a nucleoside with a phosphate group form a nucleotide, i.e., Pentose sugar + nitrogenous base + phosphoric acid = Nucleotide.
Nucleotides present in DNA Molecules:
- Deoxyadenylic acid
- Deoxyguanylic acid
- Deoxycytidylic acid
- Deoxythymidylic acid
Leave a Reply