RBSE 12th Biology Model Paper Set 9 with Answers in English

Students must start practicing the questions from RBSE 12th Biology Model Papers Set 9 with Answers in English Medium provided here.

RBSE Class 12 Biology Model Paper Set 9 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 note book. (1 × 9 = 9)
(i) Gametes are always : [1]
(a) haploid
(b) diploid
(c) triploid
(d) haploid or diploid
Answer:
(a) haploid

(ii) Through which cell of the embryo sac does the pollen tube enter the embryo sac? [1]
(a) Egg cell
(b) Central cell
(c) Persistent Synergid
(d) Degenerated Synergid
Answer:
(d) Degenerated Synergid

(iii) How many pairs of characters are taken by Mendel in his experiments? [1]
(a) 2 pairs
(b) 4 pairs
(c) 5 pairs
(d) 7 pairs
Answer:
(d) 7 pairs

(iv) A complex of ribosomes attached to a single strand of RNA is known as : [1]
(a) polypeptide
(b) Okazaki fragment
(c) polysome
(d) polymer
Answer:
(c) polysome

(v) Cancer cells are- [1]
(a) HeLa cells
(b) CD₄ cells
(c) plasma cells
(d) Memory cells
Answer:
(a) HeLa cells

(vi) In AIDS, which of these cells are most affected? [1]
(a) B-cells
(b) T-cells
(c) Monocytes
(d) Neutrophils
Answer:
(b) T-cells

(vii) Which one of the following is a viral disease of chickens? [1]
(a) Pullorum disease
(b) Ranikhet
(c) Mycotic disease
(d) Fowl spirochaetosis
Answer:
(b) Ranikhet

(viii) Covid-19 vaccine is a type of: [1]
(a) t-RNA
(b) m-RNA
(c) s-RNA
(d) r-RNA
Answer:
(b) m-RNA

(ix) Nematode specific genes are introduced into the tobacco host plant using a vector [1]
(a) pBR322
(b) Plasmid
(c) Bacteriophage
(d) Agrobacterium
Answer:
(d) Agrobacterium

Question 2.
Fill in the blanks. (1 × 4 = 4)
(i) The plant part that used for tissue culture is called …………….. [1]
(ii) Biodiversity Act of India was passed by the parliament in the year …………….. [1]
(iii) Extranuclear DNA found in bacterial cells called …………….. [1]
(iv) The major reservoir of carbon dioxide on earth is …………….. [1]
Answer:
(i) callus
(ii) 2002
(iii) Plasmid
(iv) oceans.

Question 3.
Give the answers of the following questions in one word or one line. (1 × 8 = 8)
(i) Define the term ‘homozygous’ and ‘heterozygous’. [1]
Answer:
When similar pair of alleles are present for a character, it is called homogyzous. e.g., TT. When dissimilar or different pairs of alleles are present for a character, it is called heterozygous condition, e.g., Tt.

(ii) What are alleles? [1]
Answer:
The alternative forms of a gene are called alleles.

(iii) Name two fungal diseases. [1]
Answer:

1. Ringworm – Trichophyton
2. Athlete’s foot – Tinea rubrum.

(iv) Give the nutritive importance of honey. [1]
Answer:
Honey contains sugars like levulose, dextrose and maltose, enzymes, pigments, minerals, vitamins and water.

(v) Which is the largest vaccine maker institute in the world? [1]
Answer:
The largest vaccine maker institute in the world, by volume is Serum Institute of India in Pune.

(vi) Which American company designed insulin? [1]
Answer:
In 1983, American companies, Eli-Lilly, designed two DNA sequences, similar to A and B chain of insulin.

(vii) How does energy flow in an ecosystem? [1]
Answer:
It is unidirectional. It flows through producers to herbivores to carnivores.

(viii) Name two causes of biodiversity losses. [1]
Answer:

1. Habitat loss
2. Natural calamities.

Section – B

(1.5 Mark)

Question 4.
Explan the Zygote Intra Fallopian Transfer Technique (ZIFT). How is intra Uterine Transfer (IUT) Technique different from it. [1.5]
Answer:
Zygote Intra Fallopian Transfer (ZIFT) is the technique in which zygote or early embryo (produced by in vitro fertilization) with up to 8 blastomere is transferred into the fallopian tube of female.

On the other hand in Intra Uterine Transfer (IUT), embryo with more than 8 blastomeres is transferred into the uterus.
These are two principal procedures adopted for test tube bahy programme.

Question 5.
What is tubectomy? Why is tubectomy considered as a contraceptive method? [1.5]
Answer:
Tubectomy : Tubectomy is the surgical removal of a small portion of fallopian tubes in a female. So that fusion of ovum with sperm is chocked. In tubectomy, a small part of fallopian tube is cut and tied up to block the passage of ova from ovary to the site of fertilization in fallopian tube. It prevents fertilization so, it is considered as a contraceptive method.

Question 6.
Explain in brief the role of animal husbandary in human welfare. [1.5]
Answer:
Rearing and management of domesticated animals is called animal husbandry. Animal husbandry plays a major role in the human welfare in the following ways :

• It increases food production and caters to the increasing food demand.
• It provides various animal products like milk, eggs, meat, wool, silk, leather, honey etc., used by the humans.
• On the commercial aspects, it also fetches a lot of money from export of the animal products.

Question 7.
If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production. [1.5]
Answer:
Following measures should be undertaken for improving the quality and quantity of milk production.

• Selection of good breed of cattle.
• Proper nutrition of animals.
• Living place of animals should be clean, hygienic and free from dirt and pollution.

Question 8.
Explain the significance of palendrome nucleotide sequenses in r-DNA technology. [1.5]
Answer:
Palindromic nucleotide sequences on DNA read same either we read from 5′ to 3′ or 3′ to 5′ direction. This means restriction enzymes read them from both sides which increases the chance of action of restriction enzyme to cut DNA strand.

The recognition site of restriction endonuclease is a far away from the center of palindromic sites, but they cut b/w the same base/pairs of DNA strand and leaves sticky overhang end for the action of DNA ligase so presence of palindromic site promote the action of restriction endonuclease for the formation of r-DNA.

Question 9.
Name any two natural cloning vector. Give reasons that make them act as cloning vector. Give reasons that make them act as cloning vector. Write two charecteristics the engineered vectors are made to possess. [1.5]
Answer:
Plasmids derived from the environment directly are called natural plasmids. Two natural cloning vectors :

1. Plasmid (e.g. Ti Plasmid)
2. Bacteriophage

They act as a cloning vector because they have ability to replicate it self into the host cell very fast and make their protein independently in host cell.

Question 10.
Write three differences between RNA and DNA. [1.5]
Answer:

	RNA	DNA
1.	Contains ribose pentose sugar.	Contains deoxyribose pentose sugar.
2.	Single stranded.	Double stranded.
3.	Four bases : thymine, adenine, cytosine and guanine.	thymine replaced with uracil.

Question 11.
What is transgenic bacteria? Illustrate using one example. [1.5]
Answer:
Transgenic bacteria are those in which external genes are inserted into their DNA. These foreign gene contain desired characters which use for human welfare, like production of food, medicines, biochemical etc.

The most commonly used transgenic bacteria is E.coli which is highly efficient to produce desired protein e.g., insulin.

E.coli in insulin production :
E.coli bacteria through rDNA technology used to grow in culture medium with amino acid that make up insulin. Both the chain A and B are synthesised separately and later bonded together to form insulin.

Question 12.
Construct a pyramid of biomass starting with phytoplanktons. Label 3 trophic levels. Is the pyramid upright or inverted? Why? [1.5]
Answer:

The pyramid is inverted because the biomass of fishes is much more than that of the phytaplanktons.

Question 13.
What is primary productivity? Why does it vary in different types of ecosystems? [1.5]
Answer:
Primary productivity is defined as the amount of biomass (energy produced per unit area in a given time) by plants during photosynthesis.

It depends upon plant species inhabiting in a particular area, environmental factors, availability of nutrients and photosynthetic capacity of plants which vary in different types of ecosystems.

Question 14.
What is species diversity? [1.5]
Answer:
Species Diversity:
It is the diversity at the species level. For example, the Western ghats have a greater amphibian species diversity than the eastern ghats. Number of species present in a unit area at given time is known as species richness of the area.

Question 15.
How habitat loss and fragmentation is a cause of loss of biodiversity? [1.5]
Answer:
Habitat loss and Fragmentation : This is the most important cause of plants and animals extinction. Tropical rainforests once covering more than 14% of the earth’s land surface, now cover only 6%, they are being destroyed fast. The amazon rain forest (called lungs of the planet) is being cut and cleared for cultivating soya beans or for conversion to grasslands for raising beef cattle.

In case when large habitats are broken up into small fragments due to various human induced activities, mammals and birds requiring large territories and certain animals with migratory habits are badly affected leading to their population decline.

Section – C

(3 Marks)

Question 16.
Why are offsprings of oviparous animals at a greater risk as compared to offsprings of viviparous animals? Give one example of each oviparous and viviparous animals. [3]
Or
What do you understand by multiple fission? Explain.
Answer:
Oviparous animals lay their eggs outside the body. As a result, the eggs of these animals are under continuous threat from various environmental conditions. On the other hand, in viviparous animals, the development of the eggs takes place inside the body of female. Hence, the offspring of an egg laying or oviparous animal is at greater risk as compared to the offspring of a viviparous animal. Which gives birth to its young ones.
Example : Viviparous animal : Cow Oviparous animal: Pigeon.

Question 17.
Explain XX-XO type of sex determination.
Or
What is crossing over. Write significance of crossing over.
Answer:
XX-XO type of Sex-determination : It is found mainly in insects. In them female is homogametic having AA + XX chromosomes. All eggs produced are of one type with A + X chromosomes. The male is heterogametic having AA + XO chromosomes, i.e., it contains only one sex-chromosome (X). It produces sperms of two types : 50% having A + X chromosomes and other 50% having only autosomes (A) and no sex-chromosome.

Question 18.
Write the prevention and control measures of alcohol abuse. [3]
Or
(i) Differentiate between benign and malignant tumors.
(ii) Why is colostrum a boon to the newborn baby? [3]
Answer:
As adolescence is the most sensitive and vulnerable period of an individual life as it is accompanied with several biological and behavioural changes, so adolescents are most likely to be attracted towards drugs or alcohol abuse and addiction. A control of addiction is a time-taker and associated with certain serious withdrawl symptoms, so management of addiction lays more stress on prevention. Main preventive measures are :

• Alcohol also has a direct depressant effect so the cases of suicides are more in alcoholics than in rest society.
• An alcoholic spends huge money on costly alcohol which causes deficiency of basic needs and malnutrition of his family members. Which leads to deficiency of vitamins, especially thiamine (B₁).
• The children of an alcoholic suffer from anxiety, frustation and insecurity.
• Alcoholism also causes certain other social problems like absenteeism from work, unemployment, child abuse, violence and traffic offences.
• Reckless behaviour, vandalism and violence.
• Common warning signs among youth are drug in academic performance, unexplained absenteeism, lack of interest in personal hygiene, aggressive behaviour, deteriorating relationship, change in sleeping and eating habits, stealing of money etc.
• Alcohol causes visual problems like blurred vision, double vision and improper judgement of distance so increase the chances of road and industrial accidents.
• Alcoholism breeds social crimes like corruption, illegal distillation and sale of illicit liquor, etc.

Section – D

(4 Marks)

Question 19.
Write four differences between male gametophyte and female gametophyte. [4]
Or
Write short notes on the following: [4]
(i) Autogamy
(i) Geitonogamy
(iii) Homogamy and
(iv) Herkogamy.
Answer:

	Male Gametophyte	Female Gametophyte
1.	In angiosperms, the male gametophyte is derived from the microspore which is produced inside the pollen sac of anther.	In angiosperms, the female gametophyte is derived from the megaspore which is formed inside the nucellus of ovule.
2.	Usually the microspores are shed of maturity and transferred from anther to the stigma by the process of pollination.	The megaspores are not shed from the ovule.
3.	The mature male gametophyte is 3-celled (one vegetative and two male gametes).	The mature female gametophyte is 7-celled and 8-nucleated (an egg, 2-synergids, 3-antipodals, 2-polar nuclei).
4.	Pollen tube is formed in male gametophyte.	Pollen tube is not formed.

Question 20.
Explain Meselson and Stahl’s experiment to demonstrate semiconservative replication of DNA. [4]
Or
Differentiate between leading strand and lagging stand of DNA. Draw a diagrammatic representation of Semidiscontinuous replication of DNA synthesis. [4]
Answer:
Meselson and Stahl’s Experiment:
Methew Mesolson and Franklin Stahl (1958) carried out experiment on Escherichia coli and demonstrated that DNA replication is semiconservative.
(i) E. coli bacteria were grown in a culture medium containing a heavy isotope of nitrogen ¹⁵N using ¹⁵NH₄CL Many generation of this bacteria were cultured in the same medium. This produced a population of E. coli with both the strands of their DNA containing ¹⁵N. They represented the parental generation and their DNA with ¹⁵N in both the strand was the heaviest.

(ii) These bacteria with ¹⁵N were transferred in culture medium containing ¹⁴N isotope (lighter isotope of N) and allowed to multiply only for one generation. The DNA from these first generation of bacteria which was isolated after about 20 minutes was found to be hybrid having one strand with ¹⁵N (heavier) and the other with ¹⁴N (lighter). The heavier strand represents the parental strand and lighter one is the new one indicating semi-conservative method of DNA replication.

(iii) DNA extracted from the culture after another generation (that is after 40 minutes, second generation) was composed of equal amounts of this hybrid DNA (¹⁵N-¹⁴N) and normal light DNA (¹⁵N¹⁴N).

(iv) When DNA was extracted from E. coli cells after 60 minutes, it belonged to the third generation and only 25% DNA molecules was hybrid, the remaining 75% was normal DNA molecule with ¹⁴N¹⁴N.