Please refer to Class 12 Biology Sample Paper With Solutions Set A provided below. The Sample Papers for Class 12 Biology have been prepared based on the latest pattern issued by CBSE. Students should practice these guess papers for class 12 Biology to gain more practice and get better marks in examinations. The Sample Papers for Biology Standard 12 will help you to understand the type of questions which can be asked in upcoming examinations.
Section-A
Q1. Why does endosperm development precede embryo development?
Answer : The cells of endosperm are filled with reserve food materials and are used for the nutrition of the developing embryo.
Q2. How many meiotic divisions are required to produce 76 seeds in a Guava fruit?
Answer : 95
Q3. How does pollination take place in water hyacinth and water lily?
Answer : In water hyacinth and water lily, the flowers emerge above the level of water and are pollinated by insects or wind as in most of the land plants.
Q4. Name the glands that contribute to human seminal plasma.
Answer : Prostate, Seminal vesicle and Bulbourethral gland. (any two)
Q5. A snapdragon plant with violet flowers was crossed with another such plant with white flowers.
The F1 progeny obtained had pink flowers. Explain, in brief, the inheritance pattern seen in offsprings of F1 generation?
Answer : The inheritance is incomplete dominance. In this, a new intermediate phenotype between the two original phenotypes is obtained. One allele for a specific trait is not completely expressed over the other allele for the same trait.
Q6. Differentiate between aneuploidy and polyploidy.
Answer : Aneuploidy is chromosomal abnormality in which one or more chromosomes are gained or lost. Polyploidy is when an entire extra set of chromosomes is added. (It may be triploid orte traploid.)
Q7. Predict the effect if, the codon UAU coding for an amino acid at the 25th position of a polypeptide of 50 amino acids, is mutated to UAA.
Answer : A polypeptide of 24 amino acids will be formed as UAA is a stop codon which will prevent further translation.
Q8. Differentiate between pro-insulin and mature insulin.
Answer : Pro- insulin contains an extra stretch called the C peptide which is not present in the mature insulin.
Q9. Name the commonly used vector for cloning genes into higher organisms.
Answer : Retrovirus/ Adenoviruses/Papilloma virus/Cauliflower mosaic virus/Tobacco mosaic virus.
Q10. Which of the three forests- Temperate, Mangroves and Tropical Evergreen is more vulnerable to invasion by outside animals and plants?
Answer : Tropical Evergreen Forests
Q11. Assertion (A): Primary transcripts in eukaryotes are nonfunctional.
Reason (R): In a nonuniform electric field, a dipole experiences a force as well as torque.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : B
OR
Assertion (A): An organism with lethal mutation may not even develop beyond the zygote stage.
Reason (R): All types of gene mutations are lethal.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : B
Q12. Assertion (A): E. coli having pBR322 with DNA insert at BamHI site cannot grow in medium containing tetracycline. 1
Reason (R): Recognition site for Bam HI is present in tetR region of pBR322.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : A
Q13. Assertion (A): A community with more species is more stable than that with less species. 1
Reason (R): More the number of species, lesser the variation in the total biomass production year after year.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : A
Q14. Assertion (A): In Ophrys one petal of the flower bears an uncanny resemblance to the female bee.
Reason (R): Two closely related species competing for the same resource can coexist simultaneously. 1
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : C
Read the following and answer any four questions from 15(i) to 15(v) given below:
Q15. Ecological Indicators: The presence of dragonflies can reveal changes in the water ecosystems more quickly than studying other animals or plants. In fact, from the nymph to the adult stage, the dragonfly has a significant, positive ecological impact. Dragonfly eggs are laid and hatched in or near water, so their lives impact both water and land ecosystems. Once hatched, dragonfly nymphs can breathe underwater which enables them to eat mosquito larvae, other aquatic insects and worms, and even small aquatic vertebrates like tadpoles and small fish and in the air. Adult dragonflies capture and eat adult mosquitoes. Community wide mosquito control programs that spray insecticides to kill adult mosquitoes also kill dragonflies. 4
(i) The approach to biological control includes:
(a) Import and release of an insect pest to a new area to provide hosts for natural enemies
(b) Import and release of natural enemies from the native home of an alien insect pest that has invaded a new area
(c) Preservation of natural enemies (predators and parasitoids) that are already established in an area
(d) Use of insecticides to reduce alien insect pests to establish new equilibrium position.
Answer : A
(ii) Two diseases less likely to occur in a region with plenty of dragonflies are_____
(a) Yellow fever and amoebic dysentery
(b) Malaria and Yellow fever
(c) Anthrax and typhoid
(d) Cholera and typhoid
Answer : B
(iii) Dragonflies indicate positive ecological impact as
(a) The presence of dragonflies indicates polluted water.
(b) Dragonfly nymphs selectively eat mosquito larvae.
(c) They help to decrease the probability of diseases spread by vectors.
(d) Dragonfly do not cause any harm to beneficial species.
Answer : C
(iv) The most effective stages in the life cycle of dragonfly that eradicate mosquitoes area.
(a) Larvae and Adult
(b) Caterpillar and Adult
(c) Nymph and Adult
(d) Pupa and Adult
Answer : C
(v) Assertion (A): Releasing dragonflies in areas where there is an outbreak of malarial diseases can be an environment friendly method of control.
Reason (R): Dragon flies are dominant species and will not allow mosquitoes to reproduce (a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : C
Read the following and answer any four questions from 16(i) to 16(v) given below:
Q16. Sickle cell anemia is a genetic disorder where the body produces an abnormal hemoglobin called hemoglobin S. Red blood cells are normally flexible and round, but when the hemoglobin is defective, blood cells take on a “sickle” or crescent shape. Sickle cell anemia is caused by mutations in a gene called HBB. It is an inherited blood disorder that occurs if both the maternal and paternal copies of the HBB gene are defective. In other words, if an individual receives just one copy of the defective HBB gene, either from mother or father, then the individual has no sickle cell anemia but has what is called “sickle cell trait”. People with sickle cell trait usually do not have any symptoms or problems but they can pass the mutated gene onto their children. There are three inheritance scenarios that can lead to a child having sickle cell anemia:
– Both parents have sickle cell trait
– One parent has sickle cell anemia and the other has sickle cell trait
– Both parents have sickle cell anemia 4
(i) Sickle cell anemia is a/ an ______________________ disease.
(a) X linked
(b) autosomal dominant
(c) autosomal recessive
(d) Y linked
Answer : C
(ii) I f both parents have sickle cell trait, then there is _______________of the child having sickle cell anemia.
(a) 25% risk
(b) 50% risk
(c) 75% risk
(d) No risk
Answer : A
(iii) If both parents have sickle cell trait, then there is _______________of the child having sickle cell trait.
(a) 25% risk
(b) 50% risk
(c) 75% risk
(d) No risk
Answer : B
(iv) If one parent has sickle cell anemia and the other has sickle cell trait, there is __________ that their children will have sickle cell anemia and ___________will have sickle cell trait.
(a) 25% risk, 75% risk
(b) 50% risk, 50% risk
(c) 75% risk, 25% risk
(d) No risk
Answer : B
(v) The following statements are drawn as conclusions from the above data (Kenya). (Image 447)
I. Patients with SCD (Sickle Cell Disease) are less likely to be infected with malaria.
II. Patients with SCD (Sickle Cell Disease) are more likely to be infected with malaria.
III. Over the years the percentage of people infected with malaria has been decreasing.
IV. Year 2000 saw the largest percentage difference between malaria patients with and without SCD.
Choose from below the correct alternative.
(a) only I is true
(b) I and IV are true
(c) III and II are true
(d) I and III are true
Answer : D
Section-B
Q17. State the composition and principle of oral pills as a contraceptive measure taking the example of Saheli.
Answer : The composition of oral pills comprises: Either progestogens alone or progestogen—estrogen combination Saheli is a Non-steroidal preparation. It inhibits ovulation and implantation. It also alters the quality of cervical mucus to prevent/ retard the entry of sperms.
Q18. Karyotype of a child shows trisomy of chromosome number 21. Identify the disorder and state the symptoms which are likely to be exhibited in this case.
Answer : Disorder- Down’s Syndrome
Symptoms: The affected individual is short statured with small round head; has furrowed tongue; partially open mouth; Palm is broad with characteristic palm crease; Physical,
psychomotor and mental development is retarded (any three symptoms)
Q19. Explain four advantages of mycorrhizal association to plants.
Answer : The fungal symbiontin in mycorrhizal associations with plants:
(i) absorbs phosphorus from soil and passes it to the plant.
(ii) provides resistance to root-borne pathogens,
(iii) enhances tolerance to salinity and drought,
(iv) induces an overall increase in plant growth and development.
Q20. Explain the method to increase the competency of the bacterial cell membrane to take up recombinant DNA?
Answer : The Recombinant DNA can be forced into the bacterial cell treated with divalent cations and incubating it with recombinant DNA on ice. This is to be followed by placing it briefly at 420C (heat shock), and then putting it back on ice. This process would enable the bacteria to take up the recombinant DNA.
OR
What are bioreactors? How are large volumes of cultures maintained and processed in them?
Answer : Bioreactors are vessels in which raw materials are biologically converted into specific products such as enzymes using microbial, plant, animal or human cells. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions like temperature, pH, substrate, salts, vitamins and oxygen.
Q21. Explain the role of enzymes in the extraction of DNA from Rhizopus in its purest form.
Answer : The extraction of DNA from Rhizopus in its purest form can be done by treating the fungal cells with enzymes such as Chitinase which will dissolve the cell wall. The RNA can be removed by treatment with ribonuclease whereas proteins can be removed by treatment with protease.
Other molecules can be removed by appropriate treatments thereby purifying DNA.
Q22. What are sticky ends? State their significance in recombination DNA technology.
Answer : ● Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands. This leaves single stranded portions at the ends. These overhanging stretches on each strand are called sticky ends .
● They form hydrogen bonds with their complimentary counterparts and facilitate the action of DNA ligase enzyme.
OR
Explain the procedure by which PCR aids in early detection of cancer.
Answer : ● A single stranded DNA or RNA is tagged with a radioactive molecule(probe)
● It is allowed to hybridize to its complementary DNA in a clone of cells followed by detection using autoradiography.
● The clone having the mutated gene will hence not appear on the photographic film,
● because the probe will not have complementarity with the mutated gene. Hence, cancer induced mutation can be detected.
Q23. Explain how advanced ex-situ conservation techniques assist in preserving threatened species of plants and animals.
Answer : ● Advanced techniques are being used now for ex situ conservation. Gametes of threatened species can be preserved in viable and fertile condition for long periods using
cryopreservation techniques. Eggs can, thus, be fertilized invitro.
● In plants, the explants can be propagated using tissue culture methods and can be kept for long periods in seed banks.
Q24. Define interference competition. Give one example that supports competitive exclusion occurring in nature.
Answer : Interference competition is the feeding efficiency of one species which might be reduced due to the interfering and inhibitory presence of the other species, even if resources (food and space) are abundant. Examples that support competitive exclusion occurring in nature are:
● The Abingdon tortoise became extinct within a decade after goats were introduced on the island, apparently due to the greater browsing efficiency of the goats.
● The larger and competitively superior barnacle Balanus dominates the intertidal area and excludes the smaller barnacle Chathamalus from that zone.
Q25. The Tropical regions are likely to have more biological diversity than the Temperate ones. Give two reasons to justify the statement.
Answer : Some possible reasons are:
● Speciation is generally a function of time, unlike temperate regions subjected to frequent glaciations in the past, tropical latitudes have remained relatively undisturbed for millions
of years and thus, had a long evolutionary time for species diversification.
● Tropical environments, unlike temperate ones, are less seasonal, relatively more constant and predictable. Such constant environments promote niche specialisation and lead to a
greater species diversity.
● There is more solar energy available in the tropics, which contributes to higher productivity; this in turn might contribute indirectly to greater diversity.
(Any two reasons)
Section-C
Q26. A fully developed foetus initiates its delivery from the mother’s womb. Justify the statement.
Answer : ● The signals for parturition originate from the fully developed foetus and the placenta which induce mild uterine contractions called foetal ejection reflex.
● This triggers the release of oxytocin from the maternal pituitary.
● Oxytocin acts on the uterine muscle and causes stronger uterine contractions, which in turn stimulates further secretion of oxytocin.
● The stimulatory reflex between the uterine contraction and oxytocin secretion continues resulting in stronger and stronger contractions.
● Parturition is induced by a complex neuroendocrine mechanism involving cortisol, oestrogens and oxytocin.
● This leads to expulsion of the baby out of the uterus through the birth canal–parturition.
Q27. How would you find out the genotype of a pea plant with violet flowers? Explain with the help of Punnets’ square showing crosses.
Answer : (Image 452)
Q28. Define flocs and state their importance in biological treatment of waste water.
Answer : ● Flocs are masses of semi – decayed organic matter along with decomposer microbes which are surrounded by slime. They separate the organic matter from waste water.
● Flocs settle down in secondary tanks and take part in the formation of sludge.
● They can be used as inoculum in biological treatment of waste water as well as source of biogas and manure.
Q29. A farmer noticed that nematode infection in tobacco plants has resulted in the reduction in the yield. Suggest a strategy which provides cellular defense for providing resistance to this pest. Explain the technique.
Answer : ● Strategy based on the process of RNA interference (RNAi) – as a method of cellular defence can be used.
● This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRNA (silencing).
● The source of this complementary RNA can be from an infection by viruses having RNA genomes or mobile genetic elements (transposons) that replicate via an RNA intermediate.
● Using Agrobacterium vectors, nematode-specific genes are introduced into the host plant.
The introduction of DNA produces both sense and anti-sense RNA in the host cells.
● Two RNA’s being complementary to each other form a double stranded (dsRNA) that initiateRNAi and thus, silence the specific mRNA of the nematode.
● As a consequence, the parasite cannot survive in a transgenic host expressing specific interfering RNA. The transgenic plant therefore gets protected from the parasite.
Q30. The graph given below represents three categories of organismic responses – L, M and N to cope with stressful conditions. Identify the categories L and M. (Image 448)
Given below are examples of some of the activities performed by animals. Categorise these activities into the appropriate kind of the organismic response (L, M or N) as shown in the
graph with reasons.
(i) In summers we sweat profusely.
(ii) Sometimes desert lizards bask in the sun and sometimes they move into shade.
Answer : L: Conformers, M: Regulators
(i) To regulate the body temperature – M/Regulators
(ii) To keep their body temperature constant by behavioural response for coping with variations in environment– L/Conformers
OR
Give reasons for the following:
(a) Very small animals are rarely found in polar regions.
(b) Mammals from colder climate generally have shorter ear and limbs.
(c) Initially we feel nausea and fatigue when we reach a high altitude such as Rohtang Pass and then, gradually, we feel normal.
Answer : (a) Since small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside; then they have to expend much energy to generate body heat through metabolism. This is the main reason why very small animals are rarely found in polar regions.
(b) Mammals from colder climates generally have shorter ears and limbs to minimise heat loss. (This is called the Allen’s Rule.)
(c) This is because in the low atmospheric pressure of high altitudes, the body does not get enough oxygen. But gradually we get acclimatised and stop experiencing altitude sickness.
Section-D
Q31. Study the graph given below related with menstrual cycle in females: 5
(a) Identify ovarian hormones X and Y mentioned in the graph and specify their source. (Image 449)
(b) Corelate and describe the uterine events that take place according to the ovarian hormone levels X and Y mentioned in the graph on –
(i) 6 – 15 days
(ii) 16 – 25 days
(iii) 26 – 28 days (when ovum is not fertilized)
Answer : (a) X- Estrogen secreted by growing follicles;
Y – Progesterone secreted by corpus luteum
(b) Uterine events that take place according to the ovarian hormone levels X and Y on –
(i) 6 – 15 days: Endometrium of the uterus regenerates by proliferation under the influence of estrogen.
(ii) 16 – 25 days: Under the influence of Progesterone the endometrium of the uterus is maintained for implantation of fertilised ovum and other events of pregnancy.
(iii) 26 – 28 days (when ovum is not fertilized): in the absence of fertilisation, corpus luteumdegenerates which causes disintegration of endometrium leading to
menstruation, marking a new cycle.
OR
The following figure shows a foetus within the uterus. On the basis of the given figure, answer the questions that follow: (Image 449)
(a) In the above figure, choose and name the correct part (A, B, C or D) that act as a temporary endocrine gland and substantiate your answer. Why is it also called the functional junction?
(b) Mention the role of B in the development of the embryo.
(c) Name the fluid surrounding the developing embryo. How is it misused forsexdetermination?
Answer : (a) Part labeled A -Placenta. It acts as an endocrine tissue as it produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens, progestogens, etc. It is also called the functional junction because it facilitates the supply of oxygen and nutrients to the embryo and removes carbon dioxide and excretory/waste materials produced by the embryo.
(b) The placenta is connected to the embryo through an umbilical cord which helps in the transport of substances to and from the embryo.
(c) Amniotic fluid; a foetal sex determination test is based on the chromosomal pattern of the cells in the amniotic fluid surrounding the developing embryo.
Q32. Evaluate the suitability of DNA and RNA as genetic material and justify the suitability of the one that is preferred as an ideal genetic material.
Answer : Evaluation of DNA and RNA on the basis of the properties of the genetic material:
1. It should be able to generate its replica (Replication): As per the rule of base pairing and complementarity, both the nucleic acids (DNA and RNA) have the ability to direct their duplications.
2. The genetic material should be chemically and structurally stable enough not to change with different stages of life cycle, age or with change in physiology of the organism.
Presence of 2’-OH group and uracil make RNA more reactive and structurally less stable than DNA. Therefore, DNA is a better genetic material than RNA.
3. It should provide the scope for slow changes (mutation) that are required for evolution: Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutates at a faster
rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.
4. It should be able to express itself in the form of ‘Mendelian Characters’: RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA, however,
is dependent on RNA for synthesis of proteins. The protein synthesising machinery has evolved around RNA.
5. The above discussion indicates that both RNA and DNA can function as genetic material, but DNA being more stable is preferred for storage of genetic information.
OR
Explain the mechanism of DNA replication as suggested by Watson and Crick.
Answer : Mechanism of Replication of DNA suggested by Watson and Crick
● The two strands of DNA would separate and act as a template for the synthesis of new complementary strands. After the completion of replication, each DNA molecule would have one parental and one newly synthesised strand. This scheme was termed as semiconservative replication of DNA.
● In living cells, such as E. coli, the process of replication requires a set of catalysts (enzymes). The main enzyme is referred to as DNA-dependent DNA polymerase, since it uses a DNA template to catalyse the polymerisation of deoxynucleotides.
● Furthermore, energetically replication is a very expensive process. Deoxyribonucleoside triphosphates serve dual purposes. In addition to acting as substrates, they provide energy
for polymerisation reaction.
● For long DNA molecules, since the two strands of DNA cannot be separated in its entire length (due to very high energyrequirement), the replication occurs within a small opening
of the DNA helix, referred to as replication fork.
● The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5′ → 3′.
● Consequently, on one strand (the template with polarity 5′ → 3′), the replication isconti nuous, while on the other (the template with polarity 5′ → 3′), it is discontinuous.
The discontinuously synthesised fragments are later joined by the enzyme DNA ligase. (Image 454)
● The DNA polymerases on their own cannot initiate the process of replication.
● There is a definite region in E. coli DNA where the replicationoriginates, such regions are termed as origin of replication.
● 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. A failure in cell division after DNA replication results into polyploidy.
Q33. Identify and name the disease in which the patient’s cells lose the property of contact inhibition.
State its possible causes and explain any three methods to accurately detect the pathological and physiological changes that take place due to the disease in living tissues.
Answer : Disease: Cancer
Probable Causes:
Physical/ Environmental- Exposure to X – rays/ gamma rays/ UV rays;
Chemicals/Nicotine in tobacco/ other carcinogens
Biological- Viral oncogenes/ Mutations
Detection and diagnosis:
1. Cancer detection is based on biopsy and histopathological studies of the tissue; blood and bone marrow tests for increased cell counts in the case of leukemias. In biopsy, a piece
of the suspected tissue cut into thin sections is stained and examined under microscope (histopathological studies) by a pathologist.
2. Techniques like radiography (use of X-rays), CT (computed tomography) and MRI (magnetic resonance imaging) are very useful to detect cancers of the internal organs.
Computed tomography uses X-rays to generate a threedimensional image of the internals of an object. MRI uses strong magnetic fields and non-ionising radiations to accurately
detect pathological and physiological changes in the living tissue.
3. Antibodies against cancer-specific antigens are also used for detection of certain cancers.
4. Techniques of molecular biology can be applied to detect genes in individuals with inherited susceptibility to certain cancers. (any three methods)
OR
A patient had tested positive to ELISA Test. Identify the disease and the pathogen responsible, give reasons for the reduced/ weak immunity of the patient and trace the path, spread and effects of this pathogen in the human body.
Answer : Disease: AIDS (Acquired Immuno Defeciency Syndrome)
Pathogen: Human Immuno deficiency virus (HIV).
Reason:
Due to decrease in the number of helper T lymphocytes, the person starts suffering from infections that could have been otherwise overcome such as those due to bacteria especially
Mycobacterium, viruses, fungi and even parasites like Toxoplasma.
The path of this pathogen and its spread andeffect on the human body:
● After getting into the body of the person, the virus enters into macrophages where RNA genome of the virus replicates to form viral DNA with the help of the enzyme reverse transcriptase.
● This viral DNA gets incorporated into host cell’s DNA and directs the infectedcells to produce virus particles.
● The macrophages continue to produce virus and in this way acts like a HIV factory.
● Simultaneously, HIV enters into helper T-lymphocytes (TH), replicates and produce progeny viruses.
● The progeny viruses released in the blood attack other helper T-lymphocytes. This is repeated leading to a progressive decrease in the number of helper T-lymphocytes in the body of the infected person.
● During this period, the person suffers from bouts of fever, diarrhoea and weight loss.
