CBSE Class 12 Biology Question Paper (Delhi 2015) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Delhi 2015) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Delhi 2015) with Solutions

Time Allowed: 3 Hours
Maximum Marks : 70

General Instructions:

1. There are a total of 26 questions and five sections in the question paper. All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3. Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5. Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examinee is to attempt any one of the questions out of the two given in the question paper with the same question number.

Section – A

Question 1.
A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms for experiments with shorter life cycle. Provide a reason.
Answer:
A geneticist prefers to choose organisms for experiments with shorter life cycle because it helps the geneticist to study many generations in the shorter life spans of those organisms.

Question 2.
Name the transcriptionally active region of chromatin in a nucleus.
Answer:
Euchromatin is the transcriptionally active region of chromatin in a nucleus.

Note.
Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin that appears as a thread-like structure. Some region of chromatin are loosely packed (and stains light) and are called euchromatin and the chromatin that is more densely packed and stains dark are called as Heterochromatin.

Question 3.
State a reason for the increased population of dark coloured moths coinciding with the loss of lichens (on tree barks) during industrialization period in England.
Answer:
Before industrialisation, 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
Before indus trialisation, the free trunks are covered by white-coloured lichen and in the background the white winged moth survived.

Question 4.
Indiscrimate diagnostic practices using X-rays etc.. should be avoided, Give one reason.
Answer:
Indiscrimate diagnostic practices using X-rays should be avoided because excessive use of X-rays leads to cause mutation and cancer.

Note
X-rays are ionizíng radiation that can affect the atoms in living things. Cells on exposure to ionizing radiation leads to cause tissue and DNA damage.

Question 5.
What is Biopiracy?
Answer:
The term biopiracy is defined as the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries as well as people concerned without compensatory payment.

Section – B

Question 6.
After a brief medical examination a healthy couple came to know that both of them are unable to produce functional gametes and should look for an ‘ART’ (Assisted Reproductive Technique). Name the ‘ART’ and the procedure involved that you can suggest to them to help them bear a child.
Answer:
The doctor suggests ZIFT (Zygote intra fallopian transfer) to those couples who are not able to bear a child. In this procedure, the sperm is collected either from the husband or donor and an ovum is collected from wife or donor. Then sperm and ova are induced to form zygote under controlled conditions in the laboratory. Then the zygote or early embryos upto 8 blastomeres stages are then transferred into the fallopian tube of the female for further development.

Note
ART (Assisted reproductive technology) involves the treatments and procedures for those couples that are nor able to bear child.

Question 7.
Differentiate between male and female heterogamety.
Answer:
The difference between male and female heterogamety are as follows:

Question 8.
How has mutation breeding helped in improving the production of mung bean crop?
Answer:
Mutation breeding is a process in which mutation is induced artificially by the use of chemicals or radiations such as gamma radiations. Then selection and use of such plants that contains desirable character as a source in breeding. In case of mung beans, mutation breeding is used to induce resistance against yellow mosaic virus and powdery mildew.

Question 9.
Mention a product of human welfare obtained with the help of each one of the following microbes:
(a) LAB
(b) Succharomyces cerevisiae
(c) Propionibacteriuni sharmanii
(d) Aspergillus niger
Answer:
(a) LAB (Lactic acid bacteria): LAB are commonly grow in milk and helps in the conversion of milk into curd. It produces acid that helps in coagulation and partial digestion of milk proteins. LAB also plays essential role in checking disease causing microbes.
(b) Saccharomyces cerevisiae: It is also called baker’s yeast. It is used for making bread and also used in beverage industry.
(c) Propionibacterium sharmanii: It is used for the production of large holes in ‘Swiss cheese’. Large holes are produced in the cheese because of the production of large amount of carbon dioxide by bacteria.
(d) Aspergillus niger: It is a fungus used for the production of citric acid.

Question 10.
Many fresh water animals can not survive in marine environment. Explain.
OR
How are productivity gross productivity, net primary productivity and secondary productivity interrelated’
Answer:
Many fresh water fishes cannot survive in the marine environment because their bodies are not adapted to the marine environment. As freshwater fishes lose body water because the surrounding water has higher salt concentration (hypertonic solution) and this makes the survival of fresh water fishes difficult in marine water.

OR

Productivity: Productivity is defined as the rate of biomass production.
Gross primary productivity: Gross primary productivity is defined as the rate of organic matter production during photosynthesis.
Net primary productivity: Net primary productivity involves the gross-productivity minus respiratory losses (R).
So, NPP = GPP-R
Secondary productivity: Secondary productivity is defined as the rate of formation of new organic matter by consumers.
As per their definitions, all the terms are interrelated to each other.

Section – C

Question 11.
Double fertilisation is reported in plants of both. castor and groundnut. However, the mature seeds of groundnut are non-albuminous and castor are albuminous. Explain the post fertilization events that are responsible for it. (3 Marks)
Answer:
The post fertilisation events that are responsible for the formation of non-albuminous mature seeds of groundnut and albuminous seeds of castor are as follows:
The primary endosperm nucleus divides repeatedly to give rise to free nuclei and this stage is called free nuclear endosperm. The cell wall is formed after the formation of cellular endosperm.

Hence, if the endosperm is consumed fully by the developing embryo before seed maturation results in the formation of non-albuminous seeds such as in groundnut. Whereas if the endosperm persists in the mature seed and can be used up during seed germination then is called as albuminous such as in castor.

Note
The process of double fertilisation involves the fusion of one gamete with nucleus of egg cell results in the formation of zygote whereas other male gamete move towards the two polar nuclei that is located in the central cell and fuses with polar nuclei to give rise to primary endosperm nucleus (PEN). It is also called triple fusion as it involves the fusion of three haploid nuclei.

Question 12.
Describe the process of Parturition in humans.
Answer:
The process of parturition involves following events:

• The average duration of human pregnancy is about 9 months and this period is called gestation period.
• The process of parturition involves vigorous contraction of the uterus at the end of pregnancy results in the expulsion or delivery of the foetus. So, parturition is defined as the process of delivery of the foetus and is also called childbirth.
• It is induced by a complex neuroendocrine mechanism. As the signals for parturition are originated from the fully developed foetus and the placenta results in induction of mild uterine contractions and this process is called foetal ejection reflex.
• Foetal ejection reflex triggers the release of oxytocin from the maternal pituitary.
• Oxytocin hormone acts on the uterine muscle and causes stronger uterine contraction that stimulates the secretion of oxytocin.
• The stimulatory reflex between the uterine contraction and oxytocin secretion that leads to cause stronger and stronger contractions.
• It leads to cause expulsion of the baby out of the uterus through the birth canal.

Question 13.
A teacher wants his/her students to find the genotype of pea plants bearing purple coloured flowers in their school garden. Name and explain the cross that will make it possible.
Answer:
Purple colour is a dominant phenotype in pea plant and the genotype of the pea plant with purple flowers can be determined by test cross.

In test cross, the pea plant whose genotype is to be determined is crossed with a homozygous recessive parent (ww) having white flowers. It helps the students to determine whether the genotype of the plant is heterozygous dominant (Ww) or homozygous dominant (WW).

Diagrammatic representation of a test cross is as follows:

Interpretation: Unknown flower is homozygous dominant

Question 14.
(a) 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?
(b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer:
(a) According to chargaff’s rule, the ratio of purine (adenine and guanine) and pyrimidine is equal.
As, the ratio of adenine + thymine and guanine + cytosine = 1
If a DNA contains 1000 nucleotides and out of this, 240 are adenine.
Then, the ratio of adenine and thymine are same = 240
A + T = 240 + 240 = 480
So, number of pyrimidine bases = 1000 – 480
= 520
Ratio of cytosine = \(\frac{520}{2}\) = 260
Ratio of guanine = \(\frac{520}{2}\) = 260
Hence, the number of pyrimidine nitrogenous bases = Cytosine + Thymine
= 260 + 240 = 500 (2 Marks)

(b) Diagrammatic representation of structure of DNA:

Question 15.
Explain adaptive radiation with the help of a suitable example.
Answer:
Adaptive radiation is defined as the process of evolution of different species in a specific geographical area starting from a point and literally radiating to another areas of geography or habitat. Darwin finches and Australian marsupials are examples of adaptive radiation.

Question 16.
A team of students are preparing to participate in the interschool sports meet. During a practice session you find some vials with labels of certain cannabionoids.
(a) Will you report to the authorities ? Why?
(b) Name a plant from which such chemicals are obtained.
(c) Write the effect of these chemicals on human body.
Answer:
(a) Yes, I would like to report the matter to the higher sports authorities because cannabinoids are classified under drugs and its abuse is considered as an illegal practice.
(b) Natural cannabinoids are obtained from the inflorescences of the plant Cannabis sativa.
(c) Cannabinoids are the group of chemicals that interact with the cannabinoids receptors present in the brain. It affects the cardiovascular system of the body.

Question 17.
Enlist the steps involved in inbreeding of cattle. Suggest two disadvantages of this practice.
Answer:
Inbreeding refers to the mating of more closely related individuals within the same breed for 4-6 generations. It involves the mating of superior males and superior females of the same breed that are identified in pairs. The progeny obtained after mating are evaluated and superior males and females are identified for further mating. The animals having high yielding capacity of the same breed are mated.

Disadvantages of inbreeding:

• Inbreeding increases homozygosity.
• It also increases inbreeding depression as continued inbreeding reduces fertility and productivity.

Question 18.
Choose any three microbes, from the following which are suited for organic farming which is in great demand these days for various reasons. Mention one application of each one chosen. Mycorrhiza; Monascus; Anabaena; Rhizobium; Methanobacterium; Trichoderina.
Answer:
(a) Mycorrhiza: It is an fungi that plays an essential role in the symbiotic association with plants and involves in nitrogen fixation in plants. It is also involved in the absorption of phosphorus from soil and passes it to the plants.
(b) Monascus purpureus: It is yeast that is used for the commercial production statins used as a blood- cholesterol lowering agents.
(c) Anabaena: It is a cyanobacteria that is involved in nitrogen fixation. They are autotrophic and free-living. They are also used as a good source of biofertiliser
(d) Rhizobium: It forms symbiotic association with the root nodules of the leguminous plant. They fix atmospheric nitrogen into organic and absorbable forms for plants that can be used as a source of nutrient.
(e) Methanobacterium: It is an bacteria used for the biological generation of methane by anaerobic processes.
(f) Trichoderma polysporum: It is a fungus used for the production of bioactive molecule called cyclosporine A that is used as an immunosuppressive agent in organ-transplant patients.

Question 19.
Recombinant DNA-technology is of great importance in the field of medicine. With the help of a flow chart, show how this technology has been used in preparing genetically engineered human insulin.
Answer:
Representation of flow chart for the production of recombinant insulin:

Diagrammatic representation of recombinant insulin:

Question 20.
Draw a labelled sketch of sparged-stirred-tank bioreactor. Write its application.
Answer:
Diagrammatic representation of sparged-stirred-tank bioreactors:

Bioreactors refers to an apparatus in which a biological reaction is carried out.

Application of sparged-stirred-tank bioreactor:

• Sparged-stirred tank bioreactors are used for the production of large amount of proteins.
• It is also used for the large-scale production of alcohol.

Question 21.
Following the collision of two trains a large number of passengers are killed. A majority of them are beyond recognition. Authorities want to hand over the dead to their relatives. Name a modem scientific method and write the procedure that would help in the identification of kinship.
Answer:
DNA fingerprinting is the molecular biology technique and a modern scientific method which is used for the kinship analysis.

Procedure used in DNA fingerprinting:

• Variable Number Tandem Repeats (VNTRs) are satellite DNAs that show higher degree of polymorphism.
• In DNA fingerprinting, VNTRs probes are used.
• DNA from an individual is collected from every tissue such as blood, hair-follicle, skin, bone and so on.
• Isolated DNA sample from an individual is then cut with the help of restriction endonucleases.
• Then, the fragments are separated through gel electrophoresis on the basis of their size.
• Then, the separated DNA fragments are immobilised on a synthetic nylon or nitrocellulose membrane.
• Immobilised DNA fragments are hybridised with the help of VNTRs probe.
• The hybridised DNA fragments can be detected by autoradiography.
• VNTRs probes vary in sizes from 0.1-20kb.
• So, in the autoradiogram, a band of different sizes will be obtained.
• Such bands serve as the characteristics of an individual.
• VNTRs are different in every individual.

Note
DNA fingerprinting technique was developed by Alec Jeffreys and he used satellite DNA as probe that represents very higher degree of polymorphism called VNTRS (Variable Number of Tandem Repeats). VNTRs are repetitive units of 10-60pb and show a higher degree of polymorphism as these base pair sequences are different in different individuals.

Question 22.
Many plant and animal species are on the verge of their extinction because of loss of forest land by indiscriminate use by the humans. As a biology student what method would you suggest along with its advantages that can protect such threatened species from getting extinct?
OR
‘Determination of Biological Oxygen Demand (BOD) can help in suggesting the quality of a water body.’ Explain.
Answer:
As a biology student, following method would be suggested in order to protect the threatened species from getting extinct are:

• Ex-situ conservation: In this, the threatened species of plants and animals are taken out of their habitats. Such species of both plants and animals are kept in special habitat such as zoological parks, botanical gardens and wildlife parks to provide natural environment for their survival.
• Gametes of several endangered plants and animal species can be preserved by methods involves cryopreservation. It can be fertilised In vitro followed by propagation through tissue culture methods.
• Seeds are preserved in seed banks and this method is called off-site conservation method.

OR

BOD (Biological oxygen demand) refers to the amount of oxygen consumed if all the organic matter in one litre of water were oxidised by bacteria.

BOD of the waterbody helps in the determination of quality of water body. Presence of more organic waste increases the BOD that consumes a large amount of oxygen from water. Greater the BOD of waste water indicates that water is more polluting by microorganisms.

Note
BOD test helps in the measurement of oxygen uptake by micro-organisms in a sample of water.

Section – D

Question 23.
Since October 02, 2014 ‘Swachh Bharat Abhiyan” has been launched in our country.
(a) Write your views on this initiative giving justification.
(b) As a biologist name TWO problems that you may face while implementing the programme in your locality.
Answer:
“Swachh Bharat Abhiyan” was started by India’s current Prime Minister Narendra Modi on October 02, 2014.
(a) I am completely in support for the movement of “Swachh Bharat Abhiyan”. It is our primary duty to clean our nation because waste materials cause biggest disaster as it make difficulty in the progress and development of a country. The waste material results in unclean surroundings that leads to cause health hazards. Wastes support the growth of mosquitoes and flies that are responsible for spread of various infectious diseases. Waste material also pollute air and water results in pollution. It effects the development and economic growth of the country.

(b) Two problems arises while implementing the programme in my locality such as:

• Improper sanitation management.
• Failure in the separation of biodegradable and non- biodegradable wastes.

(c) The methods used to overcome these problems are as follows:

• Maintain proper awareness regarding sanitation among people and also encourage them to maintain proper sanitation in their locality in order to overcome sanitation problem.
• Separate dustbins are placed everywhere for degradable and biodegradable wastes so that those wastes can be used and recycled accordingly.

Note
Placenta acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG) human placental lactogen (hPL), estrogens, progestogens during pregnancy, a hormone called relaxin is also secreted by the ovary hCG, hPL and relaxin are produced in women only during pregnancy.

Section E

Question 24.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds.
Answer the following questions giving reasons:
(a) What is the minimum number of pollen grains that must have been involved in the pollination of its pistil?
(b) What would have been the minimum number of ovules present in the ovary?
(c) How many megaspore mother cells were involved?
(d) What is the minimum number of microspore mother cells involved in the above case?
(e) How many male gametes were involved in this case?
OR
During the reproductive cycle of a human female, when, where and how does a placenta develop ? What is the function of placenta during pregnancy and embryo development?
Answer:
The number of viable seeds that are produced by the tomato plant by sexual reproduction is 240.
(a) The minimum number of pollen grains that is involved in the pollination of its pistil is 240 because each pollen grain contains two male gametes. One male gamete fuses with polar nuclei and forms endosperm whereas the other male gamete fuses with the egg cell to form zygote and it is further developed into seeds. So, 240 pollen grains are required to obtain 240 seeds.
(b) 240 ovules present in the ovary and are involved in the process of fertilisation as the 240 seeds are viable. After fertilisation, the ovary develops into fruit and ovules are developed into seeds. So, the numbers of ovules are further developed into seeds.
(c) 240 megaspore mother cells are involved in the process of gametogenesis. As, in this process only one megaspore of the tetrad becomes functional and further developed. Whereas the other three megaspores are degenerated. (1 Mark)
(d) 60 microspore mother cells are involved in the process of gametogenesis. They have undergone meiotic division prior to dehiscence of anther. As each microspore mother cell gives rise to 4 microspores and ! microspores mother cell would produce 4 microspores. So, 60 microspore mother cells are required to obtain 240 microspores. (1 Mark)
(e) 240 male gametes are required for the process of seed formation as each male gamete will fuse with one egg to form zygote that will be further developed into the seed.

OR

The placenta develops after the implantation of zygote in the uterus during the reproductive phase of human female.

After implantation, numerous finger-like projections called chorionic villi are formed on the trophoblast. The chorionic villi are surrounded by uterine tissues and maternal blood. Placenta is surrounded by uterine tissues.

• Placenta plays an essential role during embryonic development as it facilitates the supply of oxygen and nutrients to me embryo.
• It is also involved in the removal of carbon dioxide and other excretory waste materials by the embryo.
• The placenta is connected to the embryo through an umbilical cord that helps in the transportation of
  substances in and out from the embryo.

Question 25.
Explain the genetic basis of blood grouping in human population.
OR
How did Hershey and Chase established that DNA is transferred from virus to bacteria?
Answer:
The inheritance of human blood group is an example of codominance and multiple alleles. ABO blood grouping in human beings are controlled by l 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 l 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. 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

To proof that DNA is the genetic material an experiment was performed by Alfred Hershey and Martha Chase in 1952. They worked with viruses that infect bacteria called bacteriophage.

Following steps are involved in Hershey and Chase experiment:

• They grow viruses on a different medium containing radioactive phosphorus and radioactive sulphur.
• Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not protein because DNA contains phosphorus but is absent in protein.
• Whereas virus grown in a medium containing radioactive sulphur contained radioactive protein but not radioactive DNA because sulphur is absent is DNA.
• Radioactive phages were-allowed to attach to E.coli bacteria.
• As the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender.
• Then the virus particles were separated from the bacteria by spinning them in a centrifuge.
• So the bacteria that infected with viruses that contain radioactive DNA were radioactive.
• This indicates that DNA was the material that can be passed from the virus to bacteria.
• Bacteria that were infected with viruses that contain radioactive proteins were not radioactive.
• This indicates that proteins did not enter the bacteria from the viruses.
• Hence, it is proved that DNA is the genetic material that is passed from virus to bacteria.

Diagrammatic representation of Hershey and Chase experiment:

Question 26.
“Analysis of age-pyramids for human population can provide important inputs for long term planning strategies.” Explain.
OR
Describe the advantages for keeping the ecosystems healthy.
Answer:
Age pyramid is defined as a way for representing the age- sex structure of a population. There are three types of age distribution pyramids such as expanding, stable and declining. A population is composed of individuals of different age groups.

The analysis of age pyramid helps in the determination of growth status of the population. It involves three types of structure that represents that whether the population is expanding, stable or declining. The structure of age pyramids for human population emphasis on providing food to population, development of proper health care facilities and so on.

Diagrammatic representation of age pyramids for human population:

OR

The advantages of maintaining an ecosystem healthy are as follows:

• Healthy forest ecosystem helps in purification of air and water.
• It is important to maintain the biodiversity for the maintenance of a healthy ecosystem.
• Healthy ecosystem controls drought and flood conditions and also maintains nutrient cycle.
• Healthy ecosystem provides aesthetic, spiritual and cultural values.
• It is also responsible for generation of fertile soil and also provide safe habitat for wildlife.