CBSE Class 12 Biology Question Paper (Delhi 2014) with Solutions

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CBSE Class 12 Biology Question Paper (Delhi 2014) with Solutions

Allowed: 3 Hours
Maximum Marks : 70

General Instructions:

1. All questions are compulsory.
2. This question paper consists of four Sections A, B, C and D. Section A contains 8 questions of one mark each, Section B is of 10 questions of two marks each, Section C is of 9 questions of three marks each and Section D is of 3 questions of five marks each.
3. There is no overall choice. However, an internal choice has been provided in one question of 2 marks, one question of 3 marks and all the questions of 5 Marks weightage. A student has to attempt only one of the alternatives in such questions.
4. Wherever necessary, the diagrams drawn should be neat and properly labelled.

Section – A

Question 1.
Write the name of the organism that is referred to as the ‘Terror of Bengal’.
Answer:
An organism that is referred to as ‘Terror of Bengal’ is called water hyacinth or Eicchornia crassipes.

Note
Water hyacinth is called ‘terror of Bengal’because it is an alien species which was introduced in India for it’s lovely flowers and it causes havoc by it’s excessive growth. It grew abundantly in eutrophic water bodies and leads to an imbalance in die ecosystem dynamics of the water body.

Question 2.
What are ‘true breeding lines’ that are used to study inheritance pattern of traits in plants ?
Answer:
True breeding lines are those plants that have been generated through repeated self-pollination and have become homozygous for particular trait. This trait is then passed onto the future generations if bred with another true breeding plant.

Question 3.
Name any two types of cells which act as ‘Cellular barriers’ to provide Innate Immunity in humans.
Answer:
Certain types of leukocytes (WBCs) such as Polymorpho-Nuclear Leukocytes (PMNL-neutrophils), monocytes and natural killer cells in the blood as well as macrophages in the tissues that can phagocytise and destroy microbes. These cells act as cellular barriers and provide innate immunity in humans.

Note
Innate immunity is non-specific type of defence which is present at the time of birth.

Question 4.
Mention the type of host cells suitable for the gene guns to introduce an alien DNA.
Answer:
An undifferentiated plant cells are the most suitable host cells for the gene guns or biolistics gun. Plant cells are used because they have rigid cell wall that can be broken easily by bombarding them with high velocity micro-particles of gold or tungsten coated with DNA in a gene gun.

Question 5.
How is ‘stratification’ represented in a forest ecosystem ?
Answer:
Stratification refers to the vertical distribution of different species occupying different levels. It involves trees occupy top vertical strata or layer of a forest, shrubs t(ie second and herbs and grasses occupy the bottom layers.

Question 6.
Give an example of an organism that enters ‘diapause’ and why.
Answer:
Under unfavourable-conditions several zooplankton species in lakes and ponds are enter into a diapause, a stage of suspended development.

Note
Diapause takes place in adverse conditions such as drought, extreme temperature, and reduced food availability. The physiological and metabolic activities are inhibited during this phase.

Question 7.
Identify ‘a and ‘b’ in the figure given below representing proportionate number of major vertebrate taxa.

Answer:
In the given figure, a represents mammals and b represents amphibians.

Question 8.
State the cause of Accelerated Eutrophication.
Answer:
Accelerated eutrophication is the ageing of a water body because of nutrient enrichment of water. This phenomenon can either be natural or artificial.

Causes of accelerated eutrophication:

• Fertilisers runoff from the field to the water body results in algal bloom that leads to cause eutrophication.
• Industrial and sewage effluent also lead to increase in the temperature and biochemical oxygen demand of the water body. It results in increased biological activity leads to cause algal blooms.

Section – B

Question 9.
Why do algae and fungi shift to sexual mode of reproduction just before the onset of adverse conditions ?
Answer:
Organisms such as fungi and algae that switch to sexual mode of reproduction during adverse conditions because sexual reproduction brings variation into the individuals. It helps individuals to adapt the changing environmental conditions and promotes survival in unfavourable conditions. This process helps in the continuity of the species.

Note
When two parents (male and female) participate in the reproductive process and the fusion of male and female gametes is called sexual reproduction.

Question 10.
A cross was carried out between two pea plants showing the contrasting traits of height of the plant. The result of the cross showed 50% of parental characters.
(i) Work out the cross with the help of a punnett square.
(ii) Name the type of the cross carried out.
Answer:
(i) Diagrammatic Representation of Cross:

In the above cross, 50% progeny is tall (dominant) and 50% progeny is dwarf (recessive).

(ii) The type of cross carried out is called test cross. In test cross, an unknown dominant phenotype is crossed with an individual homozygous recessive for that specific trait.

Question 11.
How does the gene ‘I’ control ABO blood groups in humans? Write the effect the gene has on the structure of red blood cells.
OR
Write the types of sex-determination mechanisms the following crosses show. Give an example of each type.
(i) Female XX with Male XO
(ii) Female ZW with Male ZZ
Answer:
The inheritance of human blood group is an example of codominance and multiple alleles. ABO blood grouping in human beings are controlled by I gene. The plasma membrane of the red blood cells has sugar polymers that are found on the surface of RBCs and is controlled by this gene.

The I gene has three alleles I ^A , I ^B and i. The gene I ^A and I ^B are dominant over i and both I ^A and I ^B express their own types of sugars. This phenomenon is called co-dominance. Hence, red blood cells have both A and B types of sugars. While i allele do not produce any sugar. There are three different alleles and there are six different genotypes of the human ABO blood types.

Tabular representation of genetic basis of Blood Groups in Human population:

OR

(i) The type of sex determination mechanism shown in female XX with male XO is male heterogamety such as in humans.
(ii) The type of sex determination mechanism shown in female ZW with male ZZ is female heterogamety such as in birds.

Question 12.
(i) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion.
Answer:
(i) George Gamow was scientist who suggested that the genetic code should be made of a combination of three nucleotides.

(ii) According to George Gamow, there are only 4 nitrogenous bases and if they have to code for 20 amino acids, the code should constitute a combination of bases. He also suggested that in order to code for all the 20 amino acids, the code should be made up of three nucleotides. This was a very bold proposition, because a permutation combination of 4 ³ (4 × 4 × 4) that would generate 64 codons. He provide an evidence that the codon was triplet in nature.

Note
A codon is a sequence of three DNA or RNA nucleotides. There are 64 codon out which 61 codes for 2C aminoacides and 3 codons are stop codon that terminates the process of translation.

Question 13.
State the disadvantage of inbreeding among cattle. How it can be overcome ?
Answer:
Disadvantage of inbreeding is inbreeding depression. Inbreeding reduces fertility and productivity of an animal. It can be overcome by using different techniques such as:
Outbreeding in which breeding of the unrelated animals that may be between individuals of the same breed but no common ancestors.

Outcrossing is a technique in which mating of animals within the same but having no common ancestors on either side of their pedigree upto 4-6 generations. Cross-breeding is a technique in which superior males of one breed are mated with superior females of another breed. Interspecific hybridisation involves the mating of male and female animals of two different species.

Note
Inbreeding refers to the mating of more closely related individuals within the same breed for 4-6 generations.

Question 14.
Explain with the help of a suitable example the naming of a restriction endonuclease.
Answer:
The convention for naming enzyme restriction endonucleases such as EcoRI contains the first letter of the name that comes from the genes and the second two letters comes from the species of the prokaryotic cell from which they were isolated.

For example: In case of EcoRI comes from Escherichia coli RY 13. In this, the letter ‘R’ is derived from the name of strain. Roman numbers following the names indicate the order in which the enzymes were isolated from that strain of bacteria.

Question 15.
State how has Agrobacterium tumifaciens been made a useful cloning vector to transfer DNA to plant cells.
Answer:
Agrobacterium tumifaciens, is a pathogen of several dicot plants is able to deliver a piece of DNA known as ‘T-DNA’ to transform normal plant cells into tumor.

• To direct these tumor cells to produce the chemical required by the pathogen.
• Retroviruses in animals have ability to transform normal cells into cancerous cells.
• The tumor inducing (Ti) plasmid of Agrobacterium tumifaciens has modified into a cloning vector and is no more pathogenic to the plants and is used to deliver genes of interest into a variety of plants.
• The Ti plasmid contains genes that codes for the synthesis of auxin and cytokinin hormone and its introduction in a plant helps to produce its own nutrient machinery.

Question 16.
Construct an age pyramid which reflects a stable growth status of human population.
Answer:
Diagrammatic Representation of Stable Age pyramid:

Stable age pyramid indicates equal proportion of population in each age group. (2 Marks)

Note
Age pyramids represent age distribution of males and females in a combined diagram. As the shape of the pyramid reflects the growth status to the population.

Question 17.
Apart from being part of the food chain, predators play other important roles. Mention any two such roles supported by examples.
Answer:
Important roles played by predators apart from being a part of the food chain:

• Predators check the prey-population.
• They prevent the over-population of prey and
• Predators also help in maintaining the biodiversity in an ecosystem.

Question 18.
How are ‘sticky ends’ formed on a DNA strand ? Why are they so called ?
Answer:
The sticky ends are produced by the restriction enzymes. The restriction enzymes cut the strand of the DNA a little away from the centre of the palindromic sites, but between the same two bases on the opposite strands. This leaves single stranded portions at the ends. There are overhanging stretches called sticky ends on each strand.

They are called sticky ends because they form hydrogen bonds with their complementary cuts. This stickiness of the ends facilitates the action of the enzyme DNA ligase.

Diagrammatic Representation of action of Restriction Enzymes:

Section – C

Question 19.
Explain any three advantages the seeds offer to angiosperms.
Answer:
The three advantages that seeds offer to angiosperms are as follows:

• The seeds of angiosperms provide protection to the embryo from harsh environmental conditions.
• It provides nourishment and parental care to the developing embryo.
• The dispersal of the seeds to far-off places prevents competition among the members of the same species and prevents their extinction.

Question 20.
Name and explain the role of inner and middle walls of the human uterus.
Answer:
The middle wall of the uterus is called myometrium. It consists of smooth muscles that bring about contraction of uterine muscles during delivery of the baby.

The inner wall of the uterus is the inner glandular layer called endometrium and it plays an essential role during menstrual cycle. As it undergoes cyclic changes during menstrual cycle. Endometrium is necessary for implantation of the fertilised ovum and other events of pregnancy.

Question 21.
A colourblind child is bom to a normal couple. Work out a cross to show how it is possible. Mention the sex of this child.
OR
Mendel published his work on inheritance of characters in 1865, but it remained unrecognized till 1900. Give three reasons for the delay in accepting his work.
Answer:
Colour blindness is a sex-linked disease. The gene for this disorder is present on X chromosome. As males have only one X chromosome while females have two X chromosomes. If a colour blind child is bom from a normal couple then in this case, the mother will be carrier for colour blindness and colour blind child will be male.
Diagrammatic Representation Cross of Colourblindness:

OR

Three reasons that are responsible for the delay in accepting Mendel’s work:

• Lack of communication and publicity.
• His concept of factors or gene as discrete units that did not blend with each other was not accepted in the terms of variations that occur naturally in nature.
• Mendel’s approach to explain biological phenomenon with the help of mathematics was also not accepted.

Question 22.
Women are often blamed for producing female children. Consequently, they are ill-treated and ostracized. How will ytra address this issue scientifically if you were to conduct an awareness programme to highlight the values involved?
Answer:
The sex of the child is determined by father as the type gametes child receives from the father. As, the father contains two types of chromosomes such as XY whereas the mothers contains only XX chromosomes. When X chromosome from father fuses with the X chromosome from mother then the child will be girl while if the Y chromosome from father fuses with X chromosome of mother then the child will be boy.

So, the female or mother is not responsible for the sex determination of child. Women should not be ill-treated for giving birth to a girl child, as both males and females are equally important for the balance of nature and continuity of species.

Question 23.
(a) Name the tropical sugar cane variety grown in South India. How has it helped in improving the sugar cane quality grown in North India ?
(b) Identify ‘a’, ‘b’ and ‘c’ in the following table :

No.	Crop	Variety	Insect Pests
1.	Brassica	Pusa Gaurav	(a)
2.	Flat bean	Pusa Sem 2	(b)
3.	(c)	Pusa Sawani Pusa A-4	Shoot and fruit borer

Answer:
(a) Saccharum officinarum is the tropical variety of sugar cane grown in South India that had higher stems and higher sugar contents. Saccharum barberi was grown in North India and had poor sugar content and yield. So these two species were successfully crossed to get sugar varieties that combining the desirable qualities of high yield, thick stems, high sugar content and ability to grow in the sugar cane areas of North India.

Question 24.
Why are beehives kept in crop field during flowering period? Name any two crop fields where this is practiced.
Answer:
Beehives are kept in the crop fields during the flowering period to increase efficiency of pollination and also improves yield. Bees collect huge amount of nectar and produces more honey and in return they pollinate the flowers from which they collect nectar. This technique is practiced in the crop fields such as sunflower and apple.

Note
Bee-keeping or agiculture is the maintenance of hives of honeybees for the production of honey.

Question 25.
How did the process of RNA interference help to control the nematode from infecting roots of tobacco plants ? Explain.
Answer:
Meloidogyne incognitia is a nematode that causes infections in the root of tobacco plants. This reduces the yield of tobacco plants. In order to protect the tobacco plants from infection, a process called RNA interference occurs in all eukaryotic organisms as a method of cellular defense. The process of RNA interference involves mRNA silencing because of complementary dsRNA molecule that get binds to and prevents the translation of mRNA.

The complementary RNA is obtained due to the infection by viruses that contain an RNA genome or mobile genetic elements (transposons). It replicate via an RNA intermediate.

By using Agrobacterium vectors, nematode-specific gene were used to introduced into the host plant. After the introduction of DNA into the host, it produces both sense and anti-sense RNA in the host cells.

The two RNA’s are complementary to each other formed a double strand (dsRNA) that initiates the process of RNAi and silenced the specific mRNA of the nematode. After that, the nematode is not able to survive in a transgenic host expressing specific interfering RNA.

In this way, the transgenic plant remain protected from the parasite.

Question 26.
Study the graph given below and answer the questions that follow :

(i) Write the status of food and space in the curves (a) and (b).
(ii) In the absence of predators, which one of the two curves would appropriately depict the prey population?
(iii) Time has been shown on X-axis and there is a parallel dotted line above it. Give the significance of this dotted line.
Answer:
(i) The curve (a) represents exponential growth of the population when the responses are not limiting the growth. Whereas curve (b) represents logistic growth of the population when responses are limiting the growth.

(ii) In the absence of predator, the curve (a) is increasing as the responses are not limiting the growth. So in the absence of predator, prey population continuously increases.

(iii) The parallel dotted line represents K (carrying capacity). Carrying capacity is defined as in nature; a given habitat has enough resources to support a maximum possible number beyond which no further growth is possible.

Question 27.
(i) What is primary productivity? Why does it vary in different types of eco-systems?
(ii) State the relation between gross and net primary productivity.
Answer:
(i) Primary productivity is defined as the amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis. The primary productivity depends on different plant species present in a given ecosystem and each of their photosynthetic efficiency. The environmental factors, availability of various nutrients vary in different ecosystems results in variations in primary productivity.

(ii) Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of Gross primary productivity is utilized by the plants in respiration. Net primary productivity or NPP is the Gross primary productivity minus the respiratory losses (R).
NPP = GPP – R

Note
The rate of biomass production is called Productivity and it is expressed in terms of gm ^-2 yr ^-1 or (kcal m ^-2 ) yr ^-1 .

Section – D

Question 28.
(a) Coconut palm is monoecious, while date palm is dioecious. Why are they so called?
(b) Draw a labelled diagram of sectional view of a mature embryo sac of an angiosperm.
OR
(a) How ‘oogenesis’ markedly different from ‘spermatogenesis’ with respect to the growth till puberty in the humans?
(b) Draw a sectional view of human ovary and label the different follicular, stages, ovum and Corpus luteum.
Answer:
(a) Coconut palm is monoecious because both male and female gametes are borne on the same plant. Whereas in case of date palm, it is dioecious because both male and female flowers are borne on different plants. A single plant has either male flower or female flower. (2 Marks)

(b) Diagrammatic Representation of Mature Embryo Sac of an Angiosperm:

OR

(a) The process of formation of a mature female gamete is called oogenesis and is different from spermatogenesis. Oogenesis is initiated during the embryonic development stage when a couple of millions of oogonia or gamete mother cells are formed within each fetal ovary. These cells start division and enter into prophase-I of the mciotic division and get temporarily arrested at that stage called primary oocytes.

Each primary oocyte then gets surrounded by a layer of granulosa cells and then called the primary follicle. A large number of these follicles degenerate during the phase, from birth to puberty. So, at puberty only 60,000-80,000 primary follicles are left in each ovary.

The primary follicles get surrounded by more layers of granulosa cells and a new theca and this is called secondary follicles. Then secondary follicles transforms into a tertiary follicle that is characterised by a fluid filled cavity called antrum. The theca layer is organised into an inner theca interna and an outer theca externa.

The primary oocyte within the tertiary follicle grows in size and completes its first meiotic division. It is an unequal division that results in the formation of a large haploid secondary oocyte and a tiny first polar body. The secondary oocyte retains bulk of the nutrient rich cytoplasm of the primary oocyte.

The tertiary follicle further changes into the mature follicle or Graafian follicle. The secondary oocyte forms a new membrane called zona pellucida surrounding it. Then the Graafian follicle ruptures to release the secondary oocyte from the ovary by the process called ovulation.

Whereas in the process of spermatogenesis, the immature male germ cells or spermatogonia produce sperms in the testis that begins at the puberty.

Note
Spermatogenesis starts at the age of puberty due to significant increase in the secretion of gonadotropin releasing hormone (GnRH).

(b) Diagrammatic Representation of sectional view of ovary and follicular stages of ovum as well as corpus luteum:

Question 29.
(a) Explain the process of DNA replication with the help of a schematic diagram.
(b) In which phase of the cell cycle does replication occur in Eukaryotes ? What would happen if cell-division is not followed after DNA replication ?
OR
(a) Explain Darwinian theory of evolution with the help of one suitable example. State the two key concepts of the theory.
(b) Mention any three characteristics of Neanderthal man that lived in near east and central Asia.
Answer:
(a) DNA replication is the biological phenomenon in which a duplicate copy of DNA is synthesised. It involves following steps:

• The process of DNA replication takes place in the S-phase of the cell cycle.
• DNA dependent DNA polymerase enzyme is required for the process of replication.
• Deoxyribonucleoside triphosphate (DNTPs) plays dual role such as it acts as substrate as well as provides energy for polymerisation reaction.
• It originates at specific regions in DNA called the origin of replication because of the requirement of the origin of replication that a piece of DNA if needed to be propagated during recombinant DNA that requires a vector.
• DNA polymerase enzyme polymerises a large number of nucleotides in a very short time.
• For long DNA molecules, since the two strands of DNA cannot be separated in its entire length because of very high energy requirement.
• So the replication takes place within a small opening of the DNA helix called replication fork. The DNA-dependent DNA polymerase catalyse the polymerisation only in one direction such as 5’→3′
• One strand called as template strand having polarity 3’→5’, so the replication is continuous while the other strand having polarity 5’→3′ so the replication is discontinuous.
• The discontinuously synthesised fragments are called okazaki fragment are later joined by the DNA ligase enzyme.

Diagrammatic Representation of replication fork during replication:

(b) In eukaryotes, the replication of DNA takes place at S-phase of the cell-cycle. The replication of DNA and cell division cycle should be highly coordinated. If cell division is not followed after DNA replication then the replicated chromosome would not be distributed to daughter nuclei. A repeated replication of DNA without any cell division results in the accumulation of DNA inside the cell. This would increase the volume of the cell nucleus causing cell expansion.

OR

(a) Branching descent and natural selection are the two key concepts of Darwinian theory of Evolution.

According to the concept of Branching descent, various species have come into existence from a common ancestor.

In Natural selection, the nature selects the individuals that are most fit to adapt to their environment. It occurs because of availability of limited natural resources, variations in the characters of the members of population and inheritance of variations to next generation.

According to the Darwin theory of evolution:

• New forms keep on gradually evolving with time in the history.
• Those organisms that adapt themselves better to the surrounding survive and reproduce whereas other die and promote the survival of the fittest one.
• There is always variation in characteristics of populations that help them to adapt better to the surroundings.

Example of natural selection is industrial melanism in which before industrialization, it was observed that there were more white-winged moths or melanised moths were found on the trees than dark-winged moths. But after industrialization in 1920s, it was observed that there were more dark-winged moths were found in the same area. It occurs because during post-industrialisation period, the tree trunks became dark due to industrial smoke and soots. In this condition, the white-winged moth did not survive because of predators. Hence, dark-winged or melanised moth survived.

Note
Evolution is defined as the change in the characteristics of a species over several generations and is responsible for natural selection.

(b) The three characteristics of Neanderthal man are as follows:

• They have brain capacity 1400cc and lived near east and central Asia between 1,00,000-40,000 years back.
• They used hides to protect their body.
• They buried their dead.

Question 30.
(a) Name the technology that has helped the scientists to propagate on large scale the desired crops in short duration. List the steps carried out to propagate the crops by the said technique.
(b) How are somatic hybrids obtained ?
OR
(a) Cancer is one of the most dreaded diseases of humans. Explain ‘Contact inhibition’ and ‘Metastasis’ with respect to the disease.
(b) Name the group of genes which have been identified in normal cells that could lead to cancer and how they do so ?
(c) Name any two techniques which are useful to detect cancers of internal organs.
(d) Why are cancer patients often given a-interferon as part of the treatment ?
Answer:
(a) Tissue culture refers to the technique that helps scientists to propagate the desired crops on a large scale in a short duration. Hence, a large number of plants are propagated through this culture and this process is also referred to as micropropagation.

The following steps are carried out to propagate crops by tissue culture:

• Explants are derived from any part of the plant to be propagated.
• The explants are grown in sterile conditions in special nutrient media to regenerate complete plants.
• The nutrient medium must contain a carbon source such as sucrose, organic salts, vitamins, amino acids and phytohormones such as auxin and cytokinin.
• All the plants obtained by tissue culture technique are called somaclones as they are genetically identical to each other as well as to the parent plant. Many important food plants such as tomato, banana, apple and so on have been produced on commercial scale by using plant tissue culture technique.

(b) Somatic hybrids are hybrids produced by somatic hybridisation technique. In this technique, isolated single cells from plants are digested by scientists. Further they have digesting the cell walls to isolate naked protoplasts. Isolated protoplasts from two different varieties of plants having a desirable character that can be fused to get hybrid protoplasts that can be further grown to form a new plant.

OR

(a) Normal cells show a property called contact inhibition by virtue of which contact with other cells that inhibits their uncontrolled growth. Cancer cells lack the property of contact inhibition because of this the cancerous cells divides continuously and give rise to masses of cells called tumors.

There are two types of tumors such as benign and malignant. Benign tumors normally remain at their original location and do not spread to the other parts of the body results in little damage. Whereas the malignant tumors are a mass of proliferating cells called neoplastic or tumor cells. Such cells grow rapidly, invading and also damaging the surrounding normal tissue. All the tumor cells divide actively and grow.

These cells also starve the normal cells by competing for vital nutrients. The cells sloughed from such tumors reach distant sites through blood and they get lodged in the body. Then they start a new tumor at new location and this property is called metastasis which is the most feared property of malignant tumors.

(b) Certain genes called cellular oncogene or proto oncogene that have been identified in normal cells that are activated under several conditions that leads to cause oncogenic transformation of the cells.

(c) Techniques such as techniques as radiotherapy, CT (Computer Tomography) and MRI (Magnetic Resonance Imaging) are useful for the detection of cancers in the internal organ.

(d) Cancer patients are treated with substances called biological response modifiers called alpha- interferon. It activates their immune system and help in destroying the tumor.