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

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

CBSE Class 12 Biology Question Paper (Outside 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.
How many chromosomes do drones of honeybee possess? Name the type of cell division involved in the production of sperms by them.
Answer:
Drones of honey bee possesses 16 chromosomes. Drones produce sperms by mitosis.

Question 2.
What is a cistron?
Answer:
Cistron is that segment of DNA which specifies synthesis of a polypeptide.

Question 3.
Retroviruses have no DNA. However, the DNA of the infected host cell does possess viral DNA. How is it possible?
Answer:
After attacking the host cell, retrovirus enters into macrophages (as in case of HI V) where RNA genome of the virus replicates to form viral DNA with the help of enzyme reverse transcriptase. This viral DNA gets incorporated into the host cell’s DNA and directs the infected cells to produce more viruses. The macrophages continue to produce virus and works as a HIV factory. Hence, the infected host cell possesses viral DNA.

Question 4.
Why do children cured by enzyme-replacement therapy for adenosine deaminase deficiency need periodic treatment?
Answer:
The introduction of genetically engineered lymphocytes into an ADA deficiency patient is not a permanent cure because, the genetically engineered lymphocytes die after somedays. Hence, the patient requires periodic infusion of genetically engineered lymphocytes, so the cure is not permanent.

Question 5.
List two advantages of the use of unleaded petrol in automobiles as fuel.
Answer:
Following are the two advantages of using unleaded petrol as fuel in automobiles:

• The use of unleaded petrol in vehicles fitted with catalytic converters help in reducing emission of poisonous gases.
• As unleaded petrol does not emit harmful compounds, it helps in preventing health diseases like bronchitis, asthma and lung diseases.

Section – B

Question 6.
Why do moss plants produce very large number of male gametes? Provide one reason. What are these gametes called?
Answer:
Mosses are bryophytes and they need water for fertilisation. They lay their flagella male gametes that swim across the water to reach the female gamete. During this process, many of the male gametes are destroyed or lost. Thus, moss plants produce very large number of male gametes so that even if some of the gametes get destroyed, the remaining can fertilise the female gamete. These male gametes are called antherozoids.

Question 7.
(a) Select the homologous structures from the combinations given below:
(i) Forelimbs of whales and bats
(ii) Tuber of potato and sweet potato
(iii) Eyes of octopus and mammals
(iv) Thoms of Bougainvillea and tendrils of Cucurbita
(b) State the kind of evolution they represent.
Answer:
(a) Homologous organs are the organs having similar structure and origin but performing different functions.
From the given options, following are homologous structures:

Forelimbs of whales and bats are similar in structure but perform different functions of swimming and flying, respectively.

Thorns of Bougainvillea and tendrils of Cucurbita are both modifications of a stem arising from axillary bud but perform different functions of protection and climbing, respectively.

Note
Homology indicates common ancestory and is based on divergent evolution.

(b) The evolution represented by homologous organs or structures is divergent evolution as they have common origin but have diverged (became dissimilar) with evolution due to adaptations to different needs.

Question 8.
(a) Why are the plants raised through micropropagation termed as somaclones?
(b) Mention two advantages of this technique.
Answer:
(a) The plants obtained by micropropagation are called somaclones because they are genetically identical
plants developed from any part of a plant by tissue culture.

(b) The advantages of micropropagation are as follows:

1. It helps in the propagation of a large number of plants in a short cpan of time.
2. Healthy plant can be recovered from diseased plant by meristem culture because meristem are free of viruses in diseased plant.

Question 9.
Explain the different steps involved during primary treatment phase of sewage.
Answer:
The steps involved in the primary treatment of sewage involves:

• This involves the physical removal of large and small particles from the sewage through filtration and sedimentation.
• Floating debris are removed by sequential filteration.
• Then the girt (soil and small pebbles) are removed by sedimentation.
• All solids that settle form the primary sludge, and the supenatant forms the effluents.
• The effluent from the primary settling tank is taken for secondary treatment.

Question 10.
What is mutualism? Mention any two examples where the organisms involved are commercially exploited in agriculture.
OR
List any four techniques where the principle of ex-situ conservation of biodiversity has been employed.
Answer:
Mutualism is a type of population interaction between the organisms of two species in which both organisms are benefited from each other.
Examples of the organisms involved that are commercially exploited in agriculture are as follows:

(i) Commercial exploitation of Rhizobium in agriculture: Continuous growth of crops leads to the nutrient deficiency in soil. Farmers, then grow leguminous crops containing Rhizobium in its roots to replenish the lost nutrients (especially nitrogen) in the soil.

(ii) Commercial exploitation of Mycorrhiza in agriculture: Mycorrhiza is an association of the soil fungus with the roots of higher plants. Farmers use mycorrhiza commercially in agriculture as it improves the soil quality and reduces soil erosion by improving plant rooting capacity. The fungal hyphae spread into the root tissues and help the plants to optimally use the soil’s water and minerals.

Thus, to increase the yield of plants and to replenish the soil nutrients. Mycorrhiza is commercially exploited in agriculture.

OR

Four techniques where the principle of ex-situ conservation of biodiversity has been employed are as follows:

• Botanical gardens, zoological parks and wildlife sanctuary are the conventional methods of ex-situ conservation.
• Gametes of threatened species can be preserved in viable and fertile condition for long periods at a low temperature (-196°C) using cryopreservation techniques.
• Eggs can be fertilised Invitro, and plants can be propagated using tissue culture methods (micropropagation).
• Seeds of many different genetic strains of commercially important plants are kept viable for long periods in seed banks.

Note
Ex-situ conservation is a conservation approach in which threatened plants and animals are taken out from their natural habitat and placed in special setting where they can be protected and special care can be given.

Section – C

Question 11.
State what is apomixis. Comment on its significance. How can it be commercially used?
Answer:
Apomixis: It is a form of asexual reproduction that mimics sexual reproduction, and seeds are produced without fertilisation. It is called as apomixis or agamospermy.

Significance: Diploid egg cell is formed without reduction division and develops into embryo without fertilisation, e.g., Asteraceae.

Commercial applications of apomixis :

1. By apomixis, hybrid varieties of seeds can be produced, which will provide higher and better yield.
2. Apomixis prevents the loss of specific characteristics in the hybrid plants.
3. Apomixis is a cost-effective method of producing seeds.

Question 12.
During a monohybrid cross involving at all pea plant with a dwarf pea plant, the offspring populations were tall and dwarf in equal ratio. Work out a cross to show how it is possible.
Answer:
When a cross is made between tall pea plant which is heterozygous and dwarf (small) pea plant which is homozygous. This cross can be represented as follows:

The ratio will be 50% dominant and 50% recessive incase of hybrid or heterozygous individual.

Question 13.
Explain the significance of satellite DNA in DNA fingerprinting technique.
Answer:
Short nucleotide repeats in the DNA are very specific in each individual and vary’ in number from person to person but are inherited. These are called the ‘Variable Number Tandem Repeats’ (VNTRs). These are also called “minisatellites”.

Role of VNTR in DNA fingerprinting: DNA finger-printing technique for identifying individuals generally using repeated sequences in the human genome that produces a pattern of bands that is unique for every individuals. Each individual inherits these repeats from his/her parents which are used as genetic markers in a personal identity test.

For example, a child might inherit a chromosome with six tandem repeats from the mother and the same tandem repeats four times in the homologous chromosome inherited from the father. The half of VNTR alleles of the child resemble that of the mother and half that of the father.

Question 14.
What does the following equation represent? Explain p ² + 2pq + q ² = 1
Answer:
Hardy Weinberg’s principle states that allele frequencies are stable and is constant from generation to generation. The gene pool remains constant called genetic equilibrium. Sum total of all the allele frequencies is one. Suppose there are two alleles M’ and ‘a’ in a population. Their frequencies arep and q, respectively.

The frequency of AA individual in a population is p1. It can be explained that the probability that an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e.,p ² . In the same way, the frequency aa is q ² and for Aa is pq.
p ² + 2pq + q ² = 1
where, p ² represents frequency of homozygous dominant genotype,
2pq represents the frequency of the heterozygous genotype and represents the frequency of homozygous recessive.

Note
VNTRs (Variable Number Tandem Repeats) are repetitive units of 10-60bp and show a higher degree of polymorphism as these base pairs sequences are different in different individuals.

Question 15.
A heavily bleeding and bruised road accident victim was brought to a nursing home. The doctor immediately gave him an injection to protect him against a deadly disease.
(a) Write what did the doctor inject into the patient’s body.
(b) How do you think this injection would protect the patient against the disease?
(c) Name the disease against which this injection was given and the kind of immunity it provides.
Answer:
(a) In the patient’s body, the doctor has injected antiserum containing preformed antibodies against the causative organism or toxin produced by it.
(b) The solution injected by the doctor had antibodies; hence, the injection would protect the patient against the disease and provide him humoral immunity.
(c) The disease against which this injection was given is tetanus caused by Clostridium tetani, which usually exists in environment as spores and may again access to the body through wound.

The kind of immunity that the injection containing antiserum provides is passive immunity as preformed antibodies are used because fast action is required in this emergency case.

Question 16.
Enumerate any six essentials of good, effective Dairy Farm Management Practices.
Answer:
Six important ways of good and effective dairy farm management practices are as follows:
(a) Identification of improved cattle breeds is an important condition of cattle management. Hybrid cattle breeds are essential for the improved productivity. Therefore, it is necessary that hybrid cattle breeds should have a mixture of various desirable genes such as high milk yield and resistance to disease.

(b) Cattle should be fed in scientific manner with healthy and nutritious food consisting of roughage, fibre concentrates and high levels of proteins and other nutrients.

(c) They should be housed-well and kept in ventilated roofs to prevent them from heat, cold and rain.

(d) Animals should be kept in disease-free conditions. Regular bath and brushing should be ensured to control disease. Visit of a veterinary doctor is necessary on regular basis.

(e) The procedure of milking should be hygienic; emphasis should be given to storage and transportation of milk, so that the quality of milk is not affected.

(f) Regular inspection of dairy farms should be done by appointed officials to ensure that all the instructions are being strictly followed.

Question 17.
State the medicinal value and the bioactive molecules produced by Streptococcus, Monascus and Trichoderma.
OR
What are methanogens? How do they help to generate biogas?
Answer:
Streptococcus : The genetically modified Streptococcus produce the enzyme streptokinase which is used as clot- buster for removing clots from blood vessels of patients who have undergone myocardial infraction leading to heart attack.

Monascus purpureus : It produces statins that help in lowering blood cholesterol levels.
Trichoderma polysporum : It produces cyclosporin A that is used as an immunosuppressive agent in organ transplantation.

OR

Methanogens are anaerobic bacteria growing on cellulosic material and produce large amount of methane alongwith CO ₂ and H ₂ gas.

These bacteria are commonly found in the anaerobic sludge during sewage treatment. Examples are: Methano bacterium, Methanococcus.

Methanogens are the bacteria found in cattle dung (gobar) and in anaerobic sludge during sewage treatment. They grow anaerobically on cellulosic material and produce a large amount of methane (main constituent of biogas) alongwith CO ₂ and H ₂ . Thus, methanogens are used in biogas production.

Question 18.
Rearrange the following in the current sequences to accomplish an important biotechnological reaction:
(a) In vitro synthesis of region of DNA of interest
(b) Chemically synthesised oligonucleotides
(c) Enzyme DNA-polymerase
(d) Complementary region of DNA .
(e) Genomic DNA template
(f) Nucleotides provided
(g) Primers
(h) Thermostable DNA-polymerase (from Thermusaquaticus)
(i) Denaturation of ds-DNA
Answer:
The given steps refer to the steps involved in the polymerase chain reaction:
(b) Chemically synthesised oligonucleotides
(f) Nucleotides provided
(h) Thermostable DNA-polymerase (from Thermus aquaticus)
(i) Denaturation of ds DNA
(g) Primers
(e) Genomic DNA template
(c) Enzyme DNA-polymerase
(d) Complementary region of DNA
(a) Invitro synthesis of region of DNA of interest (3 Marks)

Note
Polymerase chain reaction is a molecular biology technique used for the formation of large number of copies of samples produced in small quantities. PCR amplification is commonly used by medical and forensic application.

Question 19.
Describe any three potential applications of genetically modified plants.
Answer:
Three potential applications of genetically modified (GM) plants are as follows:

(i) Pest resistance: Crop losses from insects pests can be incredible, resulting in financial loss for farmers and starvation in developing countries. Growing GM foods such as Bt com, Bt cotton etc. can help eliminate the application of chemical pesticides & reduce the cost of bringing a crop to market.

(ii) Disease resistance: There are many viruses, fungi & bacteria which cause plant diseases. Plant biologists are working to create plants with genetically engineered resistance to these diseases.

(iii) Cold tolerance: Unexpected frost can destroy sensitive seedlings. An antifreeze gene from cold water fish has been introduced into plants such as tobacco and potato.

Note
Generic modification has enhanced nutritional value of food
e.g.. vitamin ‘A’ enriched rice.

Question 20.
How did an American Company, Eli Lilly use the knowledge of r-DNA technology to produce human insulin?
Answer:
Insulin hormone is released as a prohormone, which consists of three peptide chains; A, B and C. This pro-hormone insulin is converted to mature insulin by removal of C peptide. The American company, Eli Lilly, used the knowledge of rDNA technology as follows:

1. DNA sequences corresponding to the two polypeptide;
   A and B of insulin are synthesised Invitro.
2. They are introduced into plasmid DNA of E.coli.
3. This bacterium is cloned under suitable conditions.
4. The transgenes expressed in the form of polypeptides
   A and B, secreted into the tjwdium.
5. They are extracted and combined by creating disulphide bridge to form human insulin.

Question 21.
How do snails, seeds, bears, zooplanktons, fungi and bacteria adapt to conditions favourable for their survival?
Answer:
Snails adapt to unfavourable conditions by producing epiphragm during aestivation that covers the opening of the shell and thus prevents desiccation.

Seeds adapt to unfavourable conditions by getting into the state of dormancy.

Bears adapt to unfavourable conditions by hibernation and reducing their body metabolic activities by 75%.
Zooplanktons adapt to unfavourablc conditions by entering into diapause (stage of suspended development).

Fungi adapt to unfavourabie conditions by reducing their metabolic rate and forming thick-walled spores.
Bacteria adapt to unfavourable conditions by forming endospores.

Note
Adaptation refers to the any attribute of the organism (morphological, physiological and bcha\ioural) that enables the organisms to survive and reproduce in its habitat.

Question 22.
With the help of a flow chart, show the phenomenon of biomagnification of DDT in an aquatic food chain.
Answer:
Biomagnification refers to increase in concentration of the toxicant at successive trophic levels. Biomagnification takes place because a toxic substance is accumulated by an organism cannot be metabolised or exerted and passed on to the next higher trophic level.

This phenomenon is well-known for DDT and mercury the concentration of DDT is increased at successive trophic levels and can ultimately reach 25 ppm in fish-eating birds. It becomes accumulated in birds and cannot be metabolised or excreted out. This high level of DDT disturbs calcium metabolism in birds causing thinning of egg shells and their premature breaking eventually causing decline in bird population.

Section – D

Question 23.
Your school has been selected by the Department of Education to organize and host an interschool seminar on “Reproductive Health-Problems and Practices”. However, many parents are reluctant to permit their wards to attend it. Their argument is that the topic is “too embarrassing.”
Put forth four arguments with appropriate reasons and explanation to justify the topic to be very essential and timely.
Answer:
Reproductive health is the total well-being in all aspects of reproduction. It includes the physical, emotional, behavioural and social well-being of an individual. Therefore, there is an urgent need to educate and discuss topics related to the reproductive health.

Following are the topics about reproductive health that should be discussed with the students:

Sexually transmitted diseases, such as AIDS and Gonorrhoea, are transmitted from one individual to another through sexual contact. Therefore, making the students aware about these diseases will help to prevent their spread. Lack of knowledge about their reproductive status may lead to unwanted pregnancies. Hence, it is necessary to create awareness of reproductve health problems, among people, especially the youth.

Learning about one’s sexuality at a proper age may help the students to know about the different changes happening in their body; thereby, leading to a better mental and physical state of health.

Counselling and creating awareness about reproductive health also helps to solve the problem related to infertility, birth control, mortality, etc.

Note
Infertility refers to the inability or failure of a couples to produce children inspite of unprotected sexual co-habitation.

Section – E

Question 24.
(a) Plan an experiment and prepare a flow chart of the steps that you would follow to ensure that the seeds are formed only from the desired sets of pollen grains. Name the type of experiment that you carried out.
(b) Write the importance of such experiments.
OR
Describe the roles of pituitary and ovarian hormones during the menstrual cycle in a human female.
Answer:
(a) To obtain seeds formed only from the desired sets of pollen grains one can opt for artificial hybridisation. Following are the steps involved:

(b) Artificial hybridisation is important for the following reasons:

• It helps to improve the crop yield.
• It ensures that the crops produced have the desired characteristics.
• It helps to yield commercially superior varieties.

OR

The rhythmic series of changes that occur in the reproductive organs of female primates (monkeys, apes and human beings) is called menstrual cycle.

• It is repeated at an average interval of about 28/29 days.
• The first appearance of menstruation at puberty is called menarche.
• The menstrual cycle has four phases. These are:

(i) Menstrual Phase

• The soft tissue of endometrial lining of the uterus disintegrates causing bleeding.
• The unfertilized egg and soft tissues are discharged.
• It lasts for 3-5days.

(ii) Follicular Phase/Proliferative Phase

• The primary/follicles in the ovary grow and become a fully mature Graafian follicle.
• The endometrium of the uterus is regenerated due to the secretion of LH and FSH from anterior pituitary and ovarian hormone, estrogen.
• It lasts for about 10 to 14 days.

(iii) Ovulatory Phase

• Rapid secretion of LH (LH surge) induced rupture of Graafian follicle, thereby leading to ovulation (released of ovum).
• It lasts for only about 48 hours.

(iv) Luteal Phase/Secretory Phase

• In this phase the ruptured follicle changes into corpus luteum in the ovary and it secrete the hormone progesterone.
• The endometrium thickens further and their glands secrete a fluid into the uterus.
• If ovum is not fertilised, the corpus luteum undergoes degeneration and this causes disintegration of the endometrium leading to menstruation.
• Estrogen and progesterone levels rise during this phase.
• It lasts for only 1 day.

Question 25.
(a) Why are thalassemia and haemophilia categorized as Mendelian disorders? Write the symptoms of these diseases. Explain their pattern of inheritance in humans.
(b) Write the genotypes of the normal parents producing a haemophilic son.
OR
How do m-RNA, t-RNA and ribosomes help in the process of translation?
Answer:
(a) Thalassaemia and haemophilia are categorised as Mendelian disorders because they occur by mutation in a single gene. Their mode of inheritance follows the principles of Mendelian genetics. Mendelian disorders can be autosomal dominant (muscular dystrophy), autosomal recessive (thalassaemia) or sex linked (haemophilia).

Symptoms of Thalassaemia:
Thalassaemia minor results only in mild anaemia, characterised by low haemoglobin level.

Thalassaemia major is also known as Cooley’s anaemia. In this disease, affected infants are normal but as they reach 6 to 9 months of age, they develop severe anaemia, skeletal deformities, jaundice, fatigue, etc.

Symptoms of Hemophilia:
Person suffering from this disease does not develop a proper blood clotting mechanism.

A haemophilic patient suffers from non-stop bleeding even on a simple cut, which may lead to death.

Pattern of Inheritance of Thalassaemia:
Pair of alleles Hb ^A and Hb ^T controls the expression of this disease.

Conditions for thalassemia :
Hb ^A and Hb ^A : Normal
Hb ^A and Hb ^T : Carrier
Hb ^T and Hb ^T : Diseased
Let us assume that both- father and mother are the carriers (Hb ^A Hb ^T ) of beta thalassaemia.

Pattern of Inheritance of Haemophilia:

Haemophilia is an X-linked genetic disorder. Compared to females, males have higher chances of getting affected because females have two XX chromosomes while males have only one X and Y chromosome. Thus, for a female to get affected by one haemophilia, she have the mutant allele on both the X chromosomes while males can b; affected if they carry it on the single X chromosome.

Conditions for haemophilia:
XY; XX: Normal
X ^h Y:
HaemophilicX ^h X:
Carrier
X ^h X ^h :Haemophilic
Let us assume that a carrier female (X ^h X) is married to a normal male.

Note
Hemophilia is caused because of the absence of blood clotting factor VIII (hemophilia-A) and IX (hemophilia-B)

(b) When a normal male marries a carrier female (she is considered normal as she contains the mutant gene on one of her X chromosomes), they can produce a haemophilic son. So, the genotype of the parents would be XY and X ^h X. (2 Marks)

OR

Translation is the process of polymerising amino acid to form a polypeptide chain.
The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain.
The process of translation involves the following three steps:

(i) Initiation
(ii) Elongation
(iii) Termination
During the initiation of the translation, tRNA gets charged when the amino acid binds to it using ATP.
The start (initiation) codon (AUG) present on mRNA is recognised only by the charged tRNA.

The ribosome acts as an actual site for the process of translation and contains two separate sites in a large subunit for the attachment of subsequent aminoacids.

The small subunit of ribosome binds to mRNA at start codon (AUG) followed by the large subunit. Then, it initiates the process of translation.

During the elongation process, the ribosome moves one codon downstream along with mRNA so as to leave the space for binding of another charged tRNA.

The amino acid brought by tRNA gets linked with the previous amino acid through a peptide bond and this process continues result in the formation of a polypeptide chain.

When the ribosome reaches one or more stop codon (UAA, UAG and UGA), the process of translation gets terminated. The polypeptide chain is released and the ribosomes get detached from mRNA. (5 Marks)

Note
Codon AUG codes for amino acid Methionine that serves as an initiation codon in eukaryotes. Whereas GUG codes for aminoacid valine and acts as initiation codon in prokaryotes.

Question 26.
(a) List the different attributes that a population has and not an individual organism.
(b) What is population density? Explain any three different ways the population density can be measured, with the help of an example each.
OR
“It is often said that the pyramid of energy is always upright. On the other hand, the pyramid of biomass can be both upright and inverted.” Explain with the help of examples and sketches.
Answer:
(a) A population certain characteristics that an individual organisms does not have are as follows:

1. Natality refers to the number of birth during a given period in the population that are added to the initial density.
2. Mortality refers to the number of deaths in the population during a given period.
3. Percentage revers to the sex-ratio of male and female.

(b) Population density means number of individuals present per unit area. Population density can be measured by determining the population size. The different methods to study population size are as follows:

1. Quadrat method: It is a method that involves the use of square of particular dimensions to measure the number of organisms.
   Example: The number of Parthenium plants in a given area can be measured using the quadrat method.
2. Direct observation: It involves the counting of organisms in the given area.
   Example: In order to determine the number of bacteria growing in a petridish, their colonies are counted.
3. Indirect method: In this method, there is no need to count the organisms individually.
   Example: The number of fishes caught per trap gives the measure of their total density in a given water body.

OR

Pyramid of energy is always upright because when energy flows from a particular trophic level to the next trophic level, some energy is always lost as heat as well as for performing various activities at each step. Maximum energy is present at the producers level and minimum at level of top carnivores (top consumers). Thus, pyramid of energy is always upright.

(a) The pyramid of biomass in a aquatic ecosystem is inverted. Because, the sum total of the weight of phytoplankton (producer) is far less than a few fishes feeding at higher trophic levels.

(b) Pyramid of biomass in a forest ecosystem is upright because producers are more in biomass than primary consumers. Primary consumers are more than secondary consumers and secondary consumers are more than tertiary consumers (top).