CBSE Sample Papers for Class 12 Biology Set 4 with Solutions

Students can access the CBSE Sample Papers for Class 12 Biology with Solutions and marking scheme Set 4 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Biology Set 4 with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

• All questions are compulsory.
• The question paper has five sections and 33 questions. All questions are compulsory.
• Section – A has 16 questions of 1 mark each; Section – B has 5 questions of 2 marks each; Section – C has 7 questions of 3 marks each; Section – D has 2 case-based questions of 4 marks each; Section – E has 3 questions of 5 marks each.
• There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
• Wherever necessary, neat and properly labeled diagrams should be drawn.

Section – A(16 Marks)

Question 1.
An anther is made up of the following major parts: lobes, theca and sporangia. The theca is cavities in which the sporangia develop.
Which of the following types of anther structure can develop into a tetrasporangiate anther?
(A) Filamentous
(B) Monothecous
(C) Dithecous
(D) Unlobed
Answer:
(C) Dithecous

Explanation: Each anther has two lobes which are attached at the back by a sterile band called connective. When both the anther lobesare present the stamen is called bithecous(or dithecous). A dithecous anther is tetrasporangiate having four microsporangia.

Question 2.
Enclosed within the integument of a typical anatropous ovule is a diploid mass of cellular tissue known as:
(A) Nucellus
(B) Embryo sac
(C) Megaspore mother cell
(D) Synergids
Answer:
(A) Nucellus

Explanation: Nucellus is the main body of the ovule made up of parenchymatous mass. The embryo sac also called female gametophyte is present inside the nucellus.

Question 3.
In the image depicting a level of the packaging of the DNA helix, identify P and Q.

(A) P: Base pairs; Q: DNA
(B) P: Nucleosome; Q: DNA
(C) P: Nucleosome; Q: chromatin
(D) P: Histone; Q: chromosome
Answer:
(B) P: Nucleosome; Q: DNA

Explanation: The label P is nucleosome while label Q is DNA. A nucleosome is a section of DNA that is wrapped around a core of proteins.

Question 4.
The number of gametes that would be produced from a parent with genotype AABBCc is:
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(B) 2

Explanation: The number of types of gametes that would be produced, depends on the number of heterozygous pairs.
No. of types of gametes = 2ⁿ
(Here, n = No. of heterozygous pair)
Therefore, 2¹ = 2
The two types of gametes – ABC and ABc

Question 5.
The process by which organisms with different evolutionary history evolve similar phenotypic adaptation in response to a common environmental challenge, is called:
(A) Natural selection
(B) Convergent evolution
(C) Non-random evolution
(D) Adaptive radiation
Answer:
(B) Convergent evolution

Explanation: Convergent evolution creates analogous structures that have similar forms or function and but were not present in the last common ancestor to those groups.

Question 6.
The presence of NaCl in body fluid shows that life originated in
(A) Salt solution
(B) Primitive oceans
(C) Rivers
(D) All of the above
Answer:
(B) Primitive oceans

Explanation: As marine waters are generally rich in salts like sodium chloride and it is generally accepted that life first evolved in water and the presence of Nad in body fluid proves that life originated in primitive oceans.

Question 7.
A nucleoside differs from a nucleotide. It lacks the:
(A) Base
(B) Sugar
(C) Phosphate group
(D) Hydroxyl group
Answer:
(C) Phosphate group

Explanation: A nitrogenous base is attached to the pentose sugar by an N-glycosidic linkage to form a nucleoside,
i.e., Nucleoside = Nitrogen base + Pentose sugar.
When a phosphate group is attached to the 5′-OH of a nucleoside through phosphodiester linkage, a nucleotide is formed, that is, Nucleotide = Nitrogen base + Pentose sugar + Phosphate (PO₄). So, a nucleoside differs from a nucleotide as it lacks the phosphate group

Question 8.
Consider the pedigree chart given below.

Since only males and off springs from the affected males are getting affected, the gene responsible for the genetic disease is likely to be located on which of the following?
(A) Y chromosome
(B) X chromosome
(C) Autosomal chromosomes
(D) Both sex chromosomes and autosomes
Answer:
(A) Y chromosome

Explanation: The gene responsible for the genetic disease is likely to be located on Y chromosome.

Question 9.
Select the pathogen mismatched with the symptoms of disease caused by it from the list given below:
(A) Entamoeba histolytica: Constipation, abdominal pain.
(B) Epidermophyton: Dry scaly lesions on nail.
(C) Wuchereria bancrofti: Chronic inflammation of lymphatic vessels of lower limb.
(D) Haemophilus influenzae: Blockage of the intestinal passage
Answer:
(D) Haemophilus influenzae: Blockage of the intestinal passage

Explanation: Pneumonia is a bacterial disease caused by Streptococcus pneumonia and Haemophilus influenzae. The disease pneumonia is characterised by the accumulation of fluid in the lungs.

Question 10.
Given below is the restriction site of a restriction endonucleases Pst-I and the cleavage sites on a DNA molecule.

Answer:

Explanation: 5’C-T-G  C-A-G3′
3′-G-A-C-G-7   C-5
This is the resultant fragments by the action of the enzyme Pst-I

Question 11.
Which body of the Government of India regulates GM research and the safety of introducing GM organisms for public services?
(A) Bio-safety committee.
(B) Indian Council of Agricultural Research.
(C) Genetic Engineering Approval Committee.
(D) Research Committee on Genetic Manipulation
Answer:
(C) Genetic Engineering Approval Committee.

Explanation: Indian government has set up an organisation: Genetic Engineering Approval Committee (GEAQ which makes decisions regarding the validity of GM research and its use for public utility.

Question 12.
In the absence of predators, which one of the two curves would appropriately depict the prey population?

(A) Curve a
(B) Curve b
(C) None of the given curve
(D) Date is insufficient to conclude
Answer:
(A) Curve a

Explanation: In the absence of a predator curve (a) would appropriately depict the prey the population because when there is no predator, then the population would exponentially increase.

Question No.13 to 16 consist of two statements- Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

Question 13.
Assertion (A): Geitonogamous flowering plants are cross-pollinated plants.
Reason (R): In Geitonogamous flowering plants, the pollen is transferred to the stigma of another flower of another plant.
Answer:
(C) A is true but R is false.

Explanation: In a geitonogamous flower, the pollen is transferred to the stigma of another flower of the same plant.

Question 14.
Assertion (A): All genetic disorders, Mendelian or chromosomal, are transmitted from one generation to the other.
Reason (R): Genes are located on chromosomes.
Answer:
(D) A is false but R is true.

Explanation: Mendelian disorders are transmitted to offspring on the same lines as in the principles of inheritance. The pattern of inheritance of Mendelian disorders can be traced in a family by the pedigree analysis.

Question 15.
Assertion (A): Ganja is hallucinogen.
Reason (R): It alters perception, causes illusion and damages cardiovascular system.
Answer:
(A) Both A and R are true and R is the correct explanation of A.

Explanation: Ganja is obtained from Cannabis sativa / hemp plant. It is hallucinogen, alters perception, causes illusion and damages cardiovascular system.

Question 16.
Assertion (A): Any fragment of DNA, when linked to the ori region, can be initiated to replicate.
Reason (R): Ori is a genetic sequence that acts as the initiation site for the replication of DNA.
Answer:
(A) Both A and R are true and R is the correct explanation of A.

Explanation: The process of DNA replication begins at the ori sequence. The presence of on in a DNA fragment makes it a self-replicating molecule.

Section – B(10 Marks)

Question 17.
How does zona pellucida of ovum help in preventing polyspermy?
Answer:
Once a sperm comes in contact with the zona pellucida of the ovum, it induces changes in its membrane. These changes prevent the entry of other sperm into the ovum and thus prevent polyspermy.

Question 18.
By using Punnett square depict the genotypes and phenotypes of test crosses (where green pod colour (G) is dominant over yellow pod colour (g) in garden pea with unknown genotype.
Answer:
Case I: When the parent with unknown genotype is homozygous dominant.

This means, if all the F₁ progeny are showing dominant phenotype, the tested parents are homozygous dominant. If 50% F₁ show dominant phenotype and 50% of F₁ show recessive phenotype then tested parent is heterozygous dominant.

Question 19.
In a patient, a mass of cells removed from the liver was found to be producing large amounts of the enzyme pepsin. Iii the same patient, a tumour was found in the stomach.
(a) What property of a tumour can be identified based on the statements above? Give a reason to support your answer.
(b) What are tumours exhibiting the property identified in (a) called?
(c) How will the tumours identified in (b) affect liver cells?
Answer:
(a) Metastasis. Pepsin is produced mainly by the stomach cells, so the tumour may have metastasised from the stomach to the liver.
(b) Malignant tumours
(c) They will disrupt the normal functioning of the liver cells.

Question 20.
Study the figure below and answer the following questions:

(a) Name the process that the figure depicts.
(b) What is the importance of the process depicted by the figure?
(c) Name the structure marked “A” and “B”.
(d) What is “E”? Give an example.
Answer:
(a) ELISA (Enzyme-Linked Immuno Sorbent Assay).
(b) It is a plate-based assay technique designed for detecting and quantifying substance such as peptides, proteins, antibodies and hormones.
(c) “A” is primary antibody and “B” is secondary antibody.
(d) A detection enzyme or other tag that can be linked directly to the primary antibody or introduced through a secondary antibody that recognises the primary antibody. The most commonly used enzyme labels are Horse Radish Peroxidase (HRP) and Alkaline Phosphatase (AP).

Question 21.

Fill up for the tropic levels, labelled 1, 2, 3, 4 in the given figure.
Answer:
1- Producers (First Trophic level- Plants)
2 – Primary Consumers (Second Ttophic level Herbivores)
3 – Secondary Consumers (Third Trophic level Carnivores)
4- Tertiary Consumers (Fourth Hophic level- Top Carnivores).

OR
Humus is formed as a result of the action of decomposers on the organic wastes by the process of humification.
(a) What makes humus a reservoir of nutrients?
(b) Name and write about the process humus undergoes that enriches the soil.
Answer:
(a) Humus is a reservoir of nutrients as it is derived from litter or organic matter scattered over soil surface such as leaves, twigs, dead bodies of organisms and their excretion. Humus is resistant to microbial action and so decomposes very slowly. Being colloidal in nature, it serves as a reservoir of nutrients.
(b) The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is called mineralisation.

Section – C(21 Marks)

Question 22.
Give a graphical representation of hormonal actions during menstrual cycle:

(a) Name the organs related to A, B and C.
(b) What is the name of the hormone D?
(c) What are the types of feedback required as shown in the number1 and 2 of the given figure?
Answer:
(a) A-Hypothalamus
B- Ovary
C- Uterus.
(b) Oestrogen.
(c) 1. Positive 2. Negative

Question 23.
Explain the formation of the placenta after implantation in a human female.
Answer:
After implantation, the cells of inner cell mass of blastocyst differentiate to form the embryo proper. The trophoblast differentiates into two layers, the outer layer secretes enzymes to dissolve the endometrium of uterus. The inner layer grows out as finger like projections called chorionic villi into the uterine stroma. They are surrounded by die uterine tissue and maternal blood vessels. The chorionic villi and; the uterine tissue become interdigitated to form the structural functional unit called placenta. The placenta secretes hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens and progesterone, that are necessary to maintain pregnancy. Placenta also facilitates supply of oxygen and provides nutrients to the embryo through umbilical cord.

Question 24.
Contraception is an artificial method or technique, mainly used to prevent pregnancy as a consequence of sexual intercourse. When a sperm reaches the ova in a woman, she may become pregnant. Contraception is a method that prevents this phenomenon by:
(i) Stopping egg production.
(ii) Keeping the egg distinct from the sperm.
(iii) By stopping the fertilised egg from attaching to the lining of the womb.
(a) Explain the mode of action of Cu⁺⁺ releasing IUDs as a good contraceptive. How is hormone releasing IUD different from it?
(b) Why is ‘Saheli’ a preferred contraceptive by women (any two reasons)?
Answer:
(a) Action of copper releasing IUDs: IUDs increase phagocytosis of sperms. The Cu ions suppress the motility and fertilising capacity of sperms. In contrast to copper releasing IUDs, hormone releasing IUDs (e.g., Progestasert, LNG-2G) make the uterus unsuitable for implantation and the cervix hostile to the sperms.
(b) SaheH is a non-steroidal preparation used as oral contraceptive pills. It is a ‘once a week’ pill with very few side effects and high contraceptive value.

Question 25.
Observe the diagram and answer the questions below:

(a) Identify the types of evolution in the concept diagrams A and B.
(b) Write example pair each for homologous and analogous organs.
Answer:
(a) A- Divergent evolution
B- Convergent evolution
(b) Homologous organs: Cheetah and human bones of forelimbs, thorn of Bougainvillea and tendrils of cucurbits.
Analogous organs: Eye of the octopus and of mammals and flippers of Penguins and Dolphins.

Question 26.
The primary effluent in the treatment of sewage is sent to tanks for secondary treatment in the presence of aerobic bacteria.
(a) How would the BOD of the effluent be affected if anaerobic bacteria are used for secondary treatment?
(b) Name one condition that should be maintained in a sludge digester where biogas is produced.
(c) The slurry formed after biogas production is recommended as manure for plants. Which nutrients will the slurry be rich in and why?
Answer:
(a) Anaerobic bacteria would not use the oxygen in the wastewater; thus, the BOD would reduce.
(b) Anaerobic conditions need to be maintained as biogas-producing bacteria are strict anaerobes.
(c) The slurry will mainly be rich in nitrogen and phosphorus as the carbon in the sludge would be used in the formation of methane and carbon dioxide.

Question 27.
Insulin in the human body is secreted by the pancreas as prohormone/proinsulin. The schematic polypeptide structure of proinsulin is given. This proinsulin needs to undergo processing before it becomes functional in the body. Answer the questions that follow:

(a) State the change the proinsulin undergoes at the time of its processing to become functional.
(b) Name the technique the American company Eli Lilly used for the commercial production of human insulin.
(c) How are the two polypeptides of functional insulin chemically held together?
Answer:
(a) In order to be functional, the C-peptide is removed from the proinsulin.
(b) The American company Eli Lilly used recombinant DNA technology for the commercial production of human insulin.
(c) The two polypeptides of a functional insulin chemically held together by disulphide bonds.

OR

Two children, A and B aged 4 and 5 years respectively visited a hospital with a similar genetic disorder. The Girl A was provided enzyme replacement therapy and was advised to revisit periodically for further treatment. Girl, B was, however, given a therapy that did not require to revisit for further treatment.
(a) Name the ailments the two girls were suffering from?
(b) Why did the treatment provide to girl A required repeated visits?
Answer:
(a) They were suffering from adenosine deaminase (ADA) deficiency.
(b) Girl A was given enzyme replacement therapy, in which lymphocytes isolated from the patient’s blood are cultured in- vitro. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. However, as these cells are not immortal, the patient requires a periodic infusion of such genetically engineered lymphocytes.

Question 28.
Given here is the diagrammatic representation of the global diversity of major taxa of plants. Based on the diagram, answer the following questions:

(a) Which is the most endangered plant among all the categeries?
(b) What is the reason for its endangerment?
(c) How fungi has been able to sustain itself?
Answer:
(a) Mosses are the most endangered plant among all the categories.
(b) Population of mosses and ferns may be less due to environmental pollution, urbanisation, and agriculture expansion or habitat destruction.
(c) Fungi are heterotrophs and lack chlorophyll, they have the ability to reproduce both sexually and a sexually and hence they are able to sustain themselves as a large population.

Section – D(8 Marks)

Question 29.
The clinical gene therapy is given to a four years old patient for an enzyme for the immune system to function. which is crucial for the immune system to function:
Step-1: Lymphocytes of the patient.
Step-2: ⇓
Step-3: Introduction of functional ADA cDNA into lymphocytes.
Step-4: ⇓
Observe the therapeutical flow chart and answer the following question.
(a) Complete the steps at 2 and 4.
(b) Identify the disease to be cured.
(c) Do you think the above method is sufficient to cure the patient? Explain your answer.
OR
(c) At present what method is available to cure this patient permanently?
Answer:
(a) Step-2: Lymphocytes are grown in a culture medium.
Step-4: Infusion of genetically engineered lymphocytes into patients.
(b) Adenosine deaminase deficiency (ADA) disease to be cured by this method.
(c) Genetically engineered lymphocytes are not immortal so the patient requires periodic infusion of the cells.
OR
(c) The gene isolated from the bone marrow cells producing ADA if introduced into the cells at early embryonic stages then it could cure the patient permanently.

Question 30.
The diversity of plants and animals is not uniform throughout the world but shows a rather uneven distribution. For many groups of animals or plants, there are interesting patterns in diversity, the most well-known being the latitudinal gradient in diversity. In general, species diversity decreases as we move away from the equator towards the poles. With very few exceptions, tropics (latitudinal range of 23.5° N to 23.5° S) harbour more species than temperate or polar areas.

Colombia located near the equator has nearly 1,400 species of birds while New York at 41° N has 105 species and Greenland at 71° N only 56 species. India, with much of its land area in the tropical latitudes, has more than 1,200 species of birds. A forest in a tropical region like Ecuador has up to 10 times as many species of vascular plants as a forest of equal area in a temperate region like the Midwest of the USA.
(a) What is called latitudinal gradient of diversity?
(b) Why Colombia has more bird species compared to Greenland?
(c) Why western ghat has more species than eastern ghat in India?
OR
(c) Where could we get the greatest biodiversity on earth?
Answer:
(a) Many groups of animals and plants interestingly diversified along the latitudinal position of their habitat and this creates a gradient of diversity along the latitudes. This is called the latitudinal gradient of diversity.
(b) With increase of latitudes, it has been found the that diversity of flora and fauna become less. As Colombia is located near the equator compared to Greenland so its diversity is more.
(c) Western ghat received more rainfall compared to eastern ghat and due to the presence of tropical rain forest in western ghat its biodiversity is more.
OR
(c) Greatest biodiversity can be observed in Amazon rain forest.

Section – E(15 Marks)

Question 31.
The given structure represents the sectional view of human seminiferous tubule.
(a) Name the cells that undergo spermiogenesis.
(b) Identify the parts labelled as A, B, C and F.
(c) State the ploidy of B and E.
(d) State the function of F cells?

Answer:
(a) Spermatids are the cells that undergo spermiogenesis.
(b) A: Spermatogonium.
B: Primary spermatocyte.
C: Secondary spermatocyte.
D: Sertoli cells.
(c) Primary spermatocytes (B) are diploid while and Spermatids (E) are haploid in nature.
(d) F (Sertoli cells) provides nourishment to germ cells.

OR

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:
Test-tube baby programme Collection of ova and sperm from the donor (Corresponding procedure correctly explained)
Explanation:
(i) IVF: Fertilisation outside the body in almost similar conditions as that in the body.
(ii) ICSI: Sperm is directly injected into the ovum. ET-Embryo is transferred into the reproductive tract/uterus.
(iii) ZIFT: Zygote or early embryos (up to eight blastomeres) transferred into the fallopian tube.
(iv) IUT: Early embryos (with more than eight blastomeres) transferred into the uterus.

Detailed Answer
Test-tube baby should be preferred in the case when both partners are unable to produce functional gametes. In this, the sperms from donor male and ova from donor female are induced to form a zygote in the laboratory (testtube). Then zygote is allowed to divide forming 8 blastomeres. The zygote is transferred into the fallopian tube and then the normal developmental process goes on. It is also called ZIFT.

Question 32.
The chromosome number is fixed for all normal organisms leading to species specification whereas any abnormality in the chromosome number of an organism results in abnormal individuals. For example, in humans 46 is the fixed number of chromosomes both in male and female. In male it is ’44+ XY and in female it is neg 44 + XX’ Thus the human male is heterogametic, in other words produces two different types of gametes one with “22 + X” chromosomes and the other with Y chromosomes respectively. Human female, on the other hand is homogametic i.e., produces only one type of gamete with “22 +X” chromosomes only. Sometimes an error may occur during the meiosis of cell cycle, where the sister chromatids fail to segregate called non disjunction, leading to the production of abnormal gametes with altered chromosome number. On fertilisation such gametes develop into abnormal individuals.
(a) State what is aneuploidy?
(b) If during spermatogenesis, the chromatids of sex chromosomes fail to segregate during meiosis, write only the different types of gametes with altered chromosome number that could possibly be produced.
(c) A normal human sperm (22 + Y) fertilises an ovum with karyotype ’22 + XX’. Name tile disorder the offspring thus produced would suffer from and write any two symptoms of the disorder.
(d) Name a best known and most common autosomal aneuploid abnormality in human and write any two symptoms.
Answer:
(a) Aneuploidyisatypeofchromosomalaberration, where there is one extra chromosome or one missing chromosome.
(b) Due to the non-disjunction of chromatids during spermatogenesis, some sperms will carry both sex chromosomes and some sperms will not carry any sex chromosome.
(c) If a normal human sperm (22 + Y) fertilizes an ovum with karyotype 22 + XX, the zygote will have the karyotype 44 + XXY. This is the trisomy of X-chromosome. It results in disorder called Klinefelter syndrome.
Features of affected individual: Overall masculine development, however, the feminine development is also expressed.
For example:
(i) Development of breast (Gynaecomastia).
(ii) Sterile,
(iii) Mentally retarded.
(d) Down’s syndrome is the best known and most common autosomal aneuploid abnormality.

OR

(a) Explain the process of amino-acylation of tRNA and its role in the process of translation.
(b) How does the initiation of the translation process occur in prokaryotes? Explain.
(c) Where are the untranslated regions located on m-RNA and why?
Answer:
(a) Charging of tRNA (Aminoacylation of tRNA): Here, amino acids are activated (amino acid + ATP) and linked to their cognate tRNA in the presence of aminoacyl tRNA synthetase. This process is commonly known as charging of tRNA or aminoacylation of tRNA. If two such charged tRNAs are brought close enough, the formation of peptide bonds between them would be favoured energetically. This is an essential step as only activated amino acids are carried to the site of protein synthesis by their respective tRNA.

(b) Translation is initiated by formation of an initiation complex consisting of 30S ribosomal subunit, formyl-methionyl (fMet) tRNA, and mRNA. It begins at the 5′-end of mRNA in the presence of an initiation factor. The mRNA binds to the small subunit of ribosome. AUG is recognised by the initiator tRNA. The initiation codon for methionine is AUG. So methionyl tRNA complex would have UAC at the Anticodon site. Now the large subunit (50S) binds to the small subunit to complete the initiation complex. Large subunit (70S) has two binding sites to which tRNA-carrying amino acids can bind. One is called aminoacyl tRNA binding site (A site) and the other is called peptidyl site (P site). There is also a third site called the exit or E site where tRNAs are released.

(c) An mRNA has additional sequences that are not translated (untranslated regions or UTR). UTRs are present at both 5′-end (before start codon) and 3′-end (after stop codon). They are required for an efficient translation process.

Question 33.
(a) Briefly describe the types of cancer associated genes.
(b) Explain Severe combined immunodeficiency (SCID).
Answer:
(a) The cancer associated genes can be divided into three main categories:
(i) Genes that induce cellular proliferation for example, genes encoding growth factors, growth factor receptors, transcription factors, etc.
(ii) Genes that inhibit cellular proliferation like tumour suppressor genes. It has been found that some cancers are not caused by oncogenes but rather by alteration in the genes called anti-oncogenes or tumour suppressing genes. These genes may produce proteins that normally oppose the action of an oncogene.
(iii) Genes that regulate programmed cell death. A mutation in these genes leads to proliferation uncontrollably giving rise to tumour or cancerous growth.

(b) Severe combined immunodeficiency or SCID is a disorder where there is very low number of circulating thrombocytes and both T-cells and B-cells are either absent or negligible at birth. Since, the body lacks the immune system, even a small infection can fatal. The children can be kept alive by keeping them in the isolation suits which resemble the space suits. The air inside the suits is almost sterile, without germs so that the child does not catch any infection, Affected children casually die at an early age.

OR

(a) If a patient is advised anti-retroviral drug, name the possible infection he/ she is likely to be suffering from. Name the causative organism.
(b) How do vaccines prevent subsequent microbial infection by the same pathogen?
(c) How does a cancerous cell differ from a normal cell?
(d) Many microbial pathogens enter the gut of humans along with food. Name the physiological barrier that protects the body from such pathogens.
Answer:
(a) AIDS is caused by the Human Immuno Deficiency Virus.
(b) Vaccines prevent microbial infections by initiating the production of antibodies against these antigens to neutralise the pathogenic agents during later actual infection.
The vaccines also generate memory – B and T-cells that recognise the pathogen quickly on subsequent exposure.
(c) Normal cells show a property called contact inhibition by virtue of which contact with other cells inhibits their uncontrolled growth. Cancer cells appear to have lost this property. These cells grow very rapidly, invading and damaging the surrounding normal tissues. Cells sloughed from such tumours reach distant sites through blood, and wherever they get lodged in the body, they start a new tumours there. This property is called metastasis.
(d) Physiological barriers: Acid in the stomach, saliva in the mouth.