HSC Biology Feb 2026 solve question paper

HSC Biology Feb 2026 solve question paper

BIOLOGY   (56)

Time : 3 Hrs.                                                                                  Max. Marks : 70

General Instructions :

The question paper is divided into four sections.

(1) Section A : Q. 1 contains Ten multiple choice type of questions carrying one mark each. Evaluation will be done for the first attempt only.

Q. No. 2 Contains Eight very short answer type of questions carrying one mark each.

(2) Section B: Q. No. 3 to 14 are short answer type of questions carrying two marks

each. (Attempt any Eight)

(3) Section C: Q. No. 15 to 26 are short answer type of questions carrying three marks

each.   (Attempt any Eight)

(4) Section D : Q. No. 27 to 31 are long answer type of questions carrying four marks

each.   (Attempt any Three)

(5)       Begin the answer of each section on a new page.

 

SECTION – A

Q.1.     Select and write the correct answer for the following multiple-choice type of questions:                                                                                                  [10]

 

(i)

A cell divides mitotically into two. Daughter cells divide and redivide repeatedly. Such growth is called……………….

 

(a)

Arithmetic growth

(b)

Geometric growth

(c)

Growth ratio

(d)

Qualitative growth

Ans

Geometric growth

 

(ii)

The transgenic plant Tomato contains transgene…………… for the production of flavonoids.

 

(a)

Phytoene synthase

(b)

Ferritin

(c)

Chalone isomerase

(d)

Phytase

Ans

Ferritin

 

 

(iii)

The cranial capacity of Neanderthal -man was………………C.C.

 

(a)

650

(b)

900

(c)

1400

(d)

1450

Ans

1400

 

 

(iv)

Identify the organism which meet their energy and nutrient requirements by degrading the detritus:

 

(a)

Omnivores

(b)

Carnivores

(c)

Autotrophs

(d)

Saprotrophs

Ans

Autotrophs

 

 

(v)

Identify the algae among the following:

(a)

Spirulina spp.

(b)

Aspergillus niger

(c)

Candida utilis

(d)

Trichoderma viridi

Ans

Aspergillus niger

 

(vi)

In plants, soluble food is always translocated in the from of …………..

(a)

Glucose

(b)

Starch

(c)

Sucrose

(d)

Fructose

Ans

Glucose


(vii)

Identify invasive plant species from the following :

(a)

Gloriosa

(b)

Michelia

(c)

Sansvieria

(d)

Parthenium

Ans

Parthenium

 

(viii)

The mature plasma cell produces antibodies at the rate of …………….molecules per second.

(a)

2000

(b)

1000

(c)

900

(d)

300

Ans

2000

 

(ix)

Which is the largest WBC among the following?

(a)

Eosinophil

(b)

Monocyte

(c)

Basophil

(d)

Lymphocyte

Ans

Monocyte

 

(x)

The number of deaths when environmental pressures come into play is known as …………..

(a)

Absolute mortality

(b)

Absolute natality

(c)

Realized mortality

(d)

Realized natality

Ans

Absolute mortality

 

Q . 2.Answer the following questions:                                                   [8]

(i)                Give the function of Leydig’s cells.

Ans: Leydig cells, located in the interstitial spaces of the testes, are the primary source of testosterone in males.  Under the regulation of pituitary luteinzing hormone (LH), they produce this androgen to stimulate spermatogenesis (sperm production), drive sexual differentiation in embryos, and maintain male secondary sexual characteristics.

(ii)              Define endangered species.

Ans: An endangered species is a plant or animal population at high risk of extinction due to a sudden, rapid population decrease or loss of critical habit. Listed by organizations like the LUCN red list, these species face a very high risk of vanishing in the near future, largely driven by human activities like habitat destruction, poaching, and pollution.

(iii)           Name the gene which is inherited directly from father to son.

Ans: Y-linked genes  (also known as holandric genes) are inherited directly from father to son. Because fathers have an XY sex chromosome pair, they pass their single Y chromosome exclusively to their male offspring.

(iv)            The producers receive 5000 Joules of light energy, then how much energy will be transferred to the carnivores (secondary consumers)?

Ans: If the producers receive 5000J of energy, then 50J of energy will be transferred to the carnivores (secondary consumers) This calculation is based on Lindeman’s 10% Law, which states that only approximately 10% of the energy from one trophic level is transferred to the next.

(v)              Define the term chlorosis.

Ans: Chlorosis is a condition in plants where leaves turn yellow, white, or pale green due to a lack of chlorophyll, often caused by nutrient deficiencies (iron, magnesium). Poor soil, or disease.

(vi)            Give the name of key factor in DNA profiling.

Ans: The key factor in DNA profiling is variable Number Tandem Repeats (VN TRs) These are specific non-coding regions of DNA where short sequences are repeated, With the number of repeats varying significantly between individuals, allowing for unique identification.

(vii)         Define combined water.

Ans: Combined water is water chemically bound within the crystal structure of minerals, soil, or compounds, rather than existing as free moisture.

(viii)       Give the role of tapetum.

Ans: The tapetum is the innermost nutritive layer of the anther wall, crucial for pollen development. It provides essential nutrients to developing microspores, secretes enzymes (like cells) for microspore separation, produces sporopollenin for the protective exine layer, and contributes pollen kit to assist in insect pollination.

 

SECTION – B

Attempt any EIGHT of the following questions :                                  [16]

Q.3.     Give the features of respiratory surface for gaseous exchange.

Ans: Efficient respiratory surfaces for gaseous exchange, such as alveoli in lungs or gills in fish, must have a large surface area, thin (single-cell thick) membranes to minimize diffusion distance, a moist surface to dissolve gases, and a rich blood supply (high vascularity) to maintain concentration gradients.

Q. 4.   Differentiate between complete sex-linkage and incomplete sex-linkage with reference to location of the gene and inheritance.

Ans: Complete sex linkage involves genes on the non-homologous regions of sex chromosomes inheriting together without recombination. Incomplete sex linkage involves genes located on the homologous (pairing) regions of X and Y chromosomes, allowing for crossing over and recombination during meiosis.

Q. 5.   Explain the properties of nerve fibers with reference to :

(a)       Irritability

(b)       Conductivity

Ans: Nerve fibers are specialized structures possessing key properties, primary irritability (excitability) – the capacity to detect stimuli and generate an action potential – and conductivity – the ability to transmit that electrical impulse along the axon. Irritability, or excitability involves responding to physical, chemical, or electrical stimuli by converting resting potential to an action potential. Conductivity, or action potential propagation, allows the single to travel, often at high speeds and in some cases, bidirectionally.

(a)   Irritability (Excitability)

Definition / Synonyms: Also known as Excitability, this is the ability of a nerve fiber to respond to a stimulus (e.g. electrical, chemical, mechanical) by generating a nerve impulse or action potential.

(b)  Conductivity.

Definition / synonyms: Also known as transmission or propagation, this is the capacity to move a generated nerve impulse or action potential down the axon from the site of stimulation.

Q. 6.   Draw a neat diagram of a nucleosome and label the following:

(i)  histone

(ii) DNA

(iii) Octamer

 

Ans: A nucleosome is the fundamental repeating unit of eukaryotic chromatin, composed of roughly 146-200 base pairs of DNA wrapped around a histone octamer, secured histone.

Here is a diagram resenting the structure :

Text

 

Key components Labeled :

·        [label : DNA] : Represents the double helix DNA wrapping roughly 1.65 times around the core.

·        [label : H1 Histone] : known as the “ linker” or “clamping” histone that secures the DNA in place.

·        Core Octamer : The central protein spool (H2A, H2B, H3, H4X2).


Q. 7.   Give any four advantages of micropropagation.

Ans: Micropropagation allows rapid, year-round production of thousands of uniform, pathogen – free plants from a small amount of parent tissue, regardless of the season.

Four key Advantages of Micropropagation:

  1.  Rapid Multiplication: large numbers of genetically identical plants (clones) are produced within a short period and a limited space.
  2.  Disease-free plants: The tech ique is conducted in sterile conditions, allowing the production and plants, such as virus-free stock.
  3. Year-Round production: production is independent of season, allowing plants to be cultured throughout the year.
  4. Propagation of difficult the year. It allows for the rapid multiplication of rare, endangered, or slow-propagating plant special.

Q. 8.   Name the organism and their role in conversion of rock into soil.

Ans: Organisms contribute to soil formation (pedogenesis) through biological weathering breaking down rock via chemical secretions and physical pressure. Key organisms include lichens (pioneer species releasing acid) mosses (widening cracks), bacteria/ fungi (decomposing material), and earthworms/animals (tunneling and organic mixing), transforming bare rock into fertile soil.

Q. 9.   Explain various harmful effects of noise pollution on human being.

Ans: Harmful Effects of Noise pollution on Humans:

 Hearing impairment: prolonged exposure to sound above 80-90 dB can lead to temporary or permanent hearing loss, along with thinitus ( ringing in ears)

Cardiovascular Diseases : chronic noise triggers stress hormones (like cortisol), leading to high blood pressure (hyper tension) which increases the risk of heart disease and strokes.

Sleep Disturbances and insomnia: Environmental noise from traffic or neighbors disrupts sleep cycles, leading to insomnia, fatigue, and lower work efficiency.

 

Q. 10. Explain skeletal features of Homo erectus.

Ans: Homo erectus skeletons exhibit a maxiture of primitive and primitive and modern traits, characterized by a long, low, and thick-boned skull, a prominent brow ridge, a smaller chinless jaw, and a cranial capacity of 900-1100 cc. They were fully bipedal with shorter arms and longer legs for walking, and possessed robust, thick-walled leg bones. Suited for endurance walking.

Q. 11. Enlist the main objectives for improved animal breeding programs using gene transfer technology.

Ans: The main objectives for improved animal breeding programs using gene transfer technology (genetic engineering) are to enhance production efficiency, quality, and animal heath, key goals include accelerating growth rates, increasing milk, meat, egg, and wool yield, introducing disease resistance, and creating novel, high-value traits (e.g, lower fat, better nutrient profiles), Key objectives of Gene Transfer

Technology in Animal Breeding:

Enhanced production Efficiency: increasing growth rates and improving feed conversion efficiency in livestock.

Q. 12. Mention any four factors affecting water absorption.

Ans: Water absorption, particularly in plants, is crucial for survival and is influenced by several external and internal factors, The key factor include the amount of available soil water, the temperature of the soil, the degree of soil aeration, and the concentration of the soil solution.

Four key factors affecting water absorption

Available soil water: The rate of absorption is directly proportional to the amount of water available in the soil. A higher concentration of available, capillary water in the soil increases the rate of water absorption.

Soil Temperature: Water absorption is most efficient at soil temperatures between 20 C and 30 C. Extremely high or low temperatures decrease the rate of water absorption by affecting metabolism and reducing the permeability of root cell membranes.

Soil  Aeration: Proper soil aeration is necessary. Well-aerated soil contains oxygen, which is necessary for metabolic activity (respiration) of the roots. Poorly aerated soil, which is deficient in oxygen, causes poor absorption rates.

Soil solution Concentration: The concentration of solutes (salts) in the soil affects absorption. High concentrations of salt in the soil increase the osmotic pressure, making it harder for plants to absorb water, thus reducing the rate of absorption.

Q. 13. Write briefly on Alzheimer’s disease.

Ans:   Alzheimer’s disease is a progressive neurodegenerative disorder that acts as the most common cause of dementia, accounting for 60% to 80% of case. It is characterized by the slow destruction of memory, thinking skills, and eventually, the ability to perform simple daily tasks. Alzheimer’s is not a normal part of aging, though the risk increases significantly after age 65.

Key Aspects of Alzheimer’s Disease:

Brain Changes: It is marked by the abnormal buildup of proteins in the brain, forming amyloid plaques between cells and tau tangles within neurons. These structures block communication between nerve cells (neurons) and cause them to die, leading to brain shrinkage (atrophy).

Symptoms: Early signs typically include short-term memory loss (e.g., forgetting recent conversations, repeating questions). As the disease progresses, it causes significant confusion, personality changes, difficulty with language, and inability to recognize family members.

Q. 14. Calculate the amount of ATP required by nitrogen fixer for the formation of 1000 molecules of ammonia Name the specialized cell of cyanobacteria where nitrogen is fixed.

Ans: The formation of 1000 molecules of ammonia ( ) required 8,000 ATP molecules. The specialized cells in cyanobacteria where nitrogen is fixed are called

Key Details:

ATP Calculation : The biological nitrogen fixation reaction (    +  16 ATP ) indicates that 16 ATP are required to produce 2 molecules of ammonia. Therefore, 1000 molecules of ammonia require: 8000 ATP

Specialized Cells: Heterocysts  are thick- walled specialized cells that provide an anaerobic (oxygen-free) environment necessary for the oxygen-sensitive function.

SECTION-C

 

Attempt any EIGHT of the following questions :                                         [24]

Q. 15. Match the respiratory surface to the organism in which it is found :

Organisms

Respiratory surface / organs

a)      Limulus

b)     Protists

c)      Tadpoles of frog

d)     Fishes

e)      Reptiles

f)      Spiders

a)      Lungs

b)     Internal gills

c)      Book lungs

d)     External gills

e)      Plasma membrane

f)      Book gill

g)     Cloaca

 

Ans:

Organisms

Respiratory surface / organs

a)      Limulus

b)     Protists

c)      Tadpoles of frog

d)     Fishes

e)      Reptiles

f)      Spiders

Book gills

Plasma membrane

External gills

Internal gills

Lungs

Book lungs

 

Q. 16. Explain Criss-cross inheritance with reference to bleeder’s disease by suitable charts.

Ans: Criss-cross inheritance (or sex-linked inheritance) is the transmission of a gene from a father to his daughter (F1 carrier and then to her sons (F2 affected). In bleeder’s disease (hemophilia, an X-linked recessive disorder), the gene passes from often skipping a generation in females.

 Key concepts of Criss-cross inheritance in Hemophilia:   

·      Carrier Female:   (Normal but carries one defective gene).

·      Bleeder Male: Y (Affected, as Y carries no matching allele).

·        Pattern: Father to Daughter (Carrier), then Daughter to son (Affected).

2.Example 2: Carrier Female (   ) and Normal Male ( Y)

The carrier female passes the defective gene to 50% of her sons ( Y), who become bleeders, and 50% of her daughters become carriers (

Synonyms and key Features

Synonyms: sex-linked inheritance, X-linked inheritance (specifically related to the X-chromosome).

Bleeder’s Disease: A hereditary disease, officially hemophilia A or B, where blood fails to clot property due to deficient clotting factors (VIII or IX).

Q. 17. (a) Distinguish between hibernation and aestivation.

            (b) Give two behavioral adaptations of desert animals.

Ans: Hibernation is a state of dormancy, or “winter sleep,” where animals conserve energy to survive freezing temperatures and food scarcity. Aestivation is “summer sleep,” a dormant state designed to survive extreme heat and drought by reducing metabolic activity to prevent water loss.

Behavioral Adaptations of Desert Animals

1.      Nocturnal Activity: Many desert animals, such as the kangaroo rat, are activity only at night, hiding in cool burrows during the day to avoid extreme heat and reduce water loss.

2.      Burrowing Behavior: Desert reptiles and small mammals burrow deep into the sand during the afternoon to escape the heat, managing their body temperature by moving between sun and shade.

Q. 18. Define :

(a)       Siphonogamy

(b) Parthenocarpy

(c ) polyembryony 

Ans: (a) siphonogamy: The process of sexual reproduction in seed-bearing plants (spermatophytes) where non-motile male gametes are transported to the female ovule via a hollow pollen tube.

Usage/Synonyms: Often referred to as “pollen tube fertilization”. It ensures that non-motile sperm can reach the egg cell.

       (b) Parthenocarphy: The production of fruit without fertilization of ovules, resulting in seedless (parthenocarpic) fruit.

(c)   Polyembryony: The phenomenon of having more than one embryo in a single seed, which results in the development of multiple seedings from one seed.

 

Q. 19. What is transpiration? Explain stomatal opening and closing mechanism with reference to starch-sugar interconversion theory.

 

Ans: Transpiration is the evaporative loss of water from aerial plant parts, primarily through stomata, facilitating nutrient transport and cooling. Stomatal movement is regulated by turgor changes in guard cells via the starch-sugar interconversion theory: daytime photosynthesis increases sugar levels, driving endosmosis and opening stomata, while night-time starch synthesis increases flaccidity, closing them.   

 

Q. 20. Enlist any six characteristics of genetic code.

Ans: Here are six key characteristics of the genetic code.

1.      Triplet Nature: The code is composed of three consecutive nucleotide bases (codon) that specify a particular amino acid.

2.      Degeneracy (Redundancy): some amino acids are encoded by more than one codon, allowing61 codons to code for only 20 amino acid.

3.      Universality: The genetic code is generally the same across all organisms, from bacteria to humans, with few exceptions.

4.      Non-Overlapping: The code is read sequentially, three bases at a time, and a single base cannot be part of two adjacent codons.

5.      Non-Ambiguous (specific): Each codon codes for one specific amino acid only.

6.      Commas less and polar: There are no commas or spaces between codons in the sequence, and the information is read in a specific 5’  3’ direction.

 

Q. 21. (a) Mention the name of germ layer from which adrenal cortex and pancreas are developed.

(b) Explain the role of alpha ( and delta (  cells of islets of Langerhans.

Ans: The adrenal cortex is derived from the mesoderm, while the pancreas is derive from the endoderm.

Role of lets of Langerhans cells:

Alpha (  cells: Secrete glucagon, which increases blood glucose levels by promoting gluconeogenesis (breakdown of glycogen) and gluconeogenesis in the liver.

Delta (  Cells: Secrete somatostatin, which acts as a regulator by inhibiting the secretion of both insulin (from -cells) and glucagon (from -cells), helping to maintain glucose homeostasis.

 

Q. 22. Define addiction. Give the physiological effects of:

(a) Cocaine

(b) Cannabinoids

Ans: Addiction is defined as a chronic, relapsing brain disease characterized by compulsive drug seeking and use despite harmful, often devastating consequences. It involves functional changes to brain circuits involved in reward, stress, and self-control, leading to an inability to stop using a substance or engaging in a behavior, often accompanied by physical dependence and withdrawal symptoms.

(a)   Cocaine : Cardiovascular system : Acute use increases heart rate and blood pressure, causes constriction of blood vessels, and can lead to arrhythmias, heart attacks, and sudden death.

Behavioral and Mental: It induces euphoria, extreme energy, mental alertness, and with high doses, paranoia, anxiety, and psychosis.

(b)  Cannabinoids : Heart and vascular system: Acute use commonly causes tachycardia (rapid heart rate) and increased blood pressure, which may increase the risk of stroke or heart disease.

Neurological and psychological: Causes relaxation, euphoria (“high”), and altered sensory perception, but can also lead to anxiety, paranoia, and panic attacks.

 

Q. 23. What is infertility? Describe in brief ZIFT and GIFT.

Ans: Infertility is a disease defined as the inability to conceive after one year or more of unprotected intercourse. It affects both men and women due to factors like low sperm count, ovulation issues, or blocked fallopian tubes. ZIFT and GIFT are assisted reproductive technologies (ART) uses to treat infertility, often as alternatives to IVF for patients with at least one functioning fallopian tube, involving the transfer of fertilized eggs (ZIFT) or gametes (GLFT) into the fallopian tube, rather than the uterus.

 

Q. 24. Draw Urey and Miller’s experimental setup and Label it.

Ans: The Urey and Miller experiment (1952) demonstrated that organic molecules, specifically amino acids, could from inorganic constituents under simulated primitive Earth conditions. The apparatus was a closed, sterile glass system combining boiling water, a mixture of methane ( ) ammonia ( ), and hydrogen ( ), and electrical sparks to simulate lightning.

Label Diagram of Miller and Urey Experiment.

 

·        Spark Discharge : Electrodes providing a 60,000 – 100,000 volt electric are to simulate lightning, providing energy for chemical reactions.

·      Gases chamber : Contains the “reducing atmosphere” of methane ( ), ammonia ( ), and hydrogen ( ).

Q. 25. (a) What is grand period of growth?

            (b) Give two examples each of :

            (1) Growth promoters

            (2) Growth inhibitors

Ans: The grand period of growth refers to the specific phase in a plant’s life cycle where rate of growth is at its maximum. It is the period of most rapid growth (exponential / log phase) following the initial slow lag phase and before the growth rate slows down during maturity or senescence.

1)     Growth promoters (stimulate growth cell division, and flowering)

·        Auxins (e.g., IAA – indole – 3- acetic acid) Responsible for cell enlargement and apical dominance.

·        Gibberllins (e.g., GA3 – Gibberellic acid) : Promote stem elongations and seed germination.

·        Other examples: Cytokinins (promote cell division).

2)     Growth inhibitors (inhibit growth, promote dormancy and senescence)

·        Abscisic acid (ABA) : Acts as a stress hormone, inducing stomatal closure and seed dormancy.

·        Ethylene: Promotes senescence and fruit ripening, often acting as a stress- related inhibitor.

·        Other examples : Phenolic inhibitors (e.g., coumarin).

 

Q. 26. (a) What is biofertilizer?

            (b) Draw and Label the T.S. of root nodules.

Ans: Biofertillizers are substance containing living microorganisms-----such as bacteria, fungi, or blue-green algae-that, when applied to seeds, plant surfaces, or soil, promote growth by increasing the supply or availability of essential nutrients enhancers that reduce dependency on chemical fertilizers.

Key Aspects of Biofertillizers:

·      Mechanism: They work by nitrogen fixation (converting atmospheric  to ammonia), solubilizing insoluble phosphorus, or stimulating plant growth through nutrient mobilization.

·        Examples: Rhizobium (symbiotic whit legume roots), Azotobacter (free-living), Azospirillum, and Mycorrhizal fungi (VAM)

( b) T.S. of Root Nodule Labeled Diagram

 

 

SECTION - D

Attempt any THREE of the following questions:                                  [12]

 

Q. 27. With the help of labelled diagram describe the structure of human sperm.

Ans: A human sperm is a microscopic, elongated haploid male gamete. It consists of four main parts:

1.      Head: Contains an elongated haploid nucleus. The anterior portion is covered by a cap-like structure called the acrosome, which contains enzymes (hyaluronidase) that help in fertilizing the ovum.

2.      Neck: A very short region containing two centrioles.

3.      Middle piece: Contains numerous mitochondria coiled around the axial filament. These provide energy (ATP) for sperm motility.

4.      Tail: The longest part, which performs lashing movements to propel the sperm forward.

 

Q. 28. What is cardia cycle? Draw a normal ECG and label it. What do p-wave and QRS complex represent?

 

Ans: Cardiac cycle: It is the sequence of events (electrical and mechanical) that occurs from the beginning of one heartbeat to the beginning of the next.

It typically lasts about 0.8 seconds and includes atrial systole, ventricular systole, and joint diastole.

ECG Representation:

·        P-wave: Represents the electrical excitation or depolarization of the atria, which leads to atrial contraction.

·        QRS complex: Represents the depolarization of the ventricles, which initiates ventricular contraction.

 

Q. 29. Write the name of disorders caused due to hypo and hyper secretion of adrenal corticoids. Mention any 3 symptoms of each of these disorders.

Ans: 1. Hyposecretion: Addison’s Disease symptoms:

·        Extreme fatigue and muscle weakness.

·        Hyperpigmentation (darkening) of the skin.

·        Low blood pressure and salt craving.

 

2.      Hypersecretion: Cushing’s syndrome symptoms:

·        Development of a “moon-shaped” face due fat redistribution.

·        Obesity in the upper body and thin arms/legs.

·        High blood sugar (hyperglycemia) and high blood pressure.

 

Q. 30. Explain nuclear and helobial type of endosperm with suitable examples.

Ans: 1. Nuclear Endosperm: This is the most common type of endosperm development. The primary endosperm nucleus undergoes successive free nuclear divisions without immediate wall formation. This creates a large number of free nuclei distributes in the periphery of the cell around a large central vacuole. Walls are formed much later, starting from the periphery

Example: Coconut water (which is free nuclear endosperm) and Maize.

3.      Helobia Endosperm: This is an intermediate type between nuclear and cellular endosperm. The first division of the primary endosperm nucleus is followed by a wall, dividing the cell into a large micropylar chamber and a small chalazal chamber. In the micropylar chamber, further division are free-nuclear (like nuclear type), while the chalazal chamber may or may not divide further.

Example: Found commonly in the order Helobiales, such as Asphodelus.

 

Q. 31. Describe in brief :

(a) What is germ line gene therapy? Why is it not encouraged?

(b) Explain somatic gene therapy. Mention any two acquired disorders for which it is used.

 

Ans: (a) Definition: This involves inserting functional genes into germ cells (sperm or eggs) or early embryos. These changes are heritable and passed on to future generations.

Why it is not encouraged: It faces major ethical and social concerns because the genetic changes are permanent and affect individuals who cannot give consent (future offspring). There are also risks of unpredictable long-term side effects on the human gene pool.

(b) Definition: This involves the insertion of a therapeutic gene into the somatic (body) cells of a patient to treat a disease. These genetic changes are restricted to the individual patient and are not passed on

·        Acquired disorders it is used for:

1.      Cancer

2.      AIDS (HIV)

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