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CBSE Sample Papers for Class 12 Physical Education Paper 1 are part of CBSE Sample Papers for Class 12 Physical Education. Here we have given CBSE Sample Papers for Class 12 Physical Education Paper 1.

CBSE Sample Papers for Class 12 Physical Education Paper 1

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 1 of Solved CBSE Sample Paper for Class 12 Physical Education is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 70

General Instructions:

• All questions are compulsory.
• Answers to questions carrying 1 mark should approximately 10-20 words.
• Answers to questions carrying 3 marks should approximately 30-50 words.
• Answers to questions carrying 5 marks should approximately 75-100 words.

Questions.

Question 1.
Define Planning? 1

Question 2.
List down the components of Barrow General Motor Ability Test? 1

Question 3.
What do you understand by Balance Diet? 1

Question 4.
Define disability and disorder among children? 1

Question 5.
Define Posture? 1

Question 6.
What do you mean by Motor Development? 1

Question 7.
List down the physiological factors determining Physical Fitness? 1

Question 8.
Define Vital Capacity? 1

Question 9.
What do you understand by Sports Medicine? 1

Question 10.
Define Soft Tissue Injury? 1

Question 11.
List Newton’s Law of motion? 1

Question 12.
Describe the Axis and Plane used in movements? 3

Question 13.
Describe the importance of intramural in physical education? 3

Question 14.
Describe various types of biomechanical movements? 3

Question 15.
Differentiate between Micro and Macro Nutrient? 3

Question 16.
Describe the benefits of yoga for healthy life? 1

Question 17.
Explain the factors effecting motor development? 3

Question 18.
Children of a particular residential school were very indisciplined, fighting, and creating problems around, resulting in poor academic achievement. The new school principal asked the teachers to start inter-section sports program for children along with extra PE classes at weekends. After the year end, students attended for assessment test related to psychomotor skills. The analysis of report highlighted increase in academics performance and various other 1 parameters.
(i) What would those various parameters which developed among the participating students?
(ii) Which type of sports competition was organized by the school?
(iii) What psychomotor test could be planned for testing fitness of school children? 3

Question 19.
List down the test components of Rikli and Jones for elderly people? 3

Question 20.
Appraise coping strategies for stress management among athletes? 5

Question 21.
Analyze the muscles involved in Throwing a cricket ball? 5

Question 22.
Differentiate Physical and Physiological differences between male and female athletes? 5

Question 23.
Categorize various types of sports injuries and explain the preventive measures? 5

Question 24.
Explain corrective measures through physical activity and exercise for improving posture? 5

Question 25.
Draw a fixture of 13 Football teams participating in a Tournament on the basis of knock out? 5

Question 26.
Describe advantages of physical activities for children with special needs? 5

Answers.

Answer 1.
Planning is basic management function involving formulation of detailed plans to achieve optimum balance of need or demands with available resources.

Answer 2.
The Barrow Motor Ability Test consisting of three items designed to test the motor ability of high school boys.
Test 1 – Standing Jump Test – It is used as a measure of explosive leg power.
Test 2 – Zigzag Run Test – Its objective is to measure agility.
Test 3 – Medicine Ball Put Test – Its objective is to measure arm and shoulder girdle strength.

Answer 3.
A diet which consists of all the essential food constituents which are necessary for growth and maintenance of the body. Protein, carbohydrates, fat, minerals, vitamins are essential micro and macro nutrients which form part of balance diet.

Answer 4.
Disability: Any degree of physical disability, malformation or disfigurement that is caused by bodily injury, birth defect or illness. Disorder to is the disturbance which affects normal functions performed by an individual.

Answer 5.
Posture is the alignment of the body part for producing various task. Posture involves of muscles, ligaments and tendons along with joints to work together for performing any action, it also effects the functioning of the organic systems.

Answer 6.
Motor development refers to the development of a child’s bones, muscles and ability to move around and manipulate his or her environment. Motor development can be divided into two sections: gross motor development and fine motor development.

• Gross motor development involves the development of the large muscles in the child’s body. These muscles allow us to sit, stand, walk and run, among other activities.
• Fine motor development involves the small muscles of the body, especially in the hand.

Answer 7.
Physiological factors determining physical fitness are:

• Cardiovascular Endurance
• Lung Capacity
• Muscle Composition
• Muscle Fibre size
• Somatotype
• Muscular Strength
• Muscular Speed

Answer 8.
Vital Capacity (VC) is the maximum amount of air a person can expel from lungs after a maximum inhalation.

Answer 9.
A field of medicine concerned with the functioning of the human body during physical activity and with the prevention and treatment of athletic injuries.

Answer 10.
Soft tissue injuries are the most common injury to tissues that connect, support, or surround other structures and organs of the body. Soft tissue includes muscles, tendons, ligaments, fascia, nerves, fibrous tissues, fat, blood vessels, and synovial membranes.

Answer 11.
There are three Newton’s Law of motion.

1. The Law of Inertia: A body at rest tends to remain at rest. A body in motion tends to continue in motion with consistent speed and in the same direction unless acted upon by an outside force.
2. The Law of Acceleration: The velocity of a body is changed only when acted upon by an additional force.
3. The Law of Counterforce: The production of any force will create another force opposite and equal to the first force.

Answer 12.
Human movements are described in three dimensions based on a series of planes and axis,

(i) A plane is the surface on which movement take place. There are three planes of motion that pass through the human body.

• The sagital plane lies vertically and divides the body into right and left parts.
• The frontal plane also lies vertically and divides the body into anterior and posterior parts.
• The transverse plane lies horizontally and divides the body into superior and inferior parts.

(ii) An axis is a straight line around which an object rotates. Movement at a joint takes place in a plane about an axis. There are three axis of rotation.

• The sagital axis passes horizontally from posterior to anterior and is formed by the intersection of the sagital and transverse planes.
• The frontal axis passes horizontally from left to right and is formed by the intersection of the frontal and transverse planes.
• The vertical axis passes vertically from inferior to superior and is formed by the intersection of the sagital and frontal planes.

Answer 13.
Intramural activities are performed within the institution, where outside students can’t participate and the main focus is on maximal participation. Following are the importance of intramurals:

• Opportunities for physical, mental, emotional and social development of students
• Inculcation of moral and ethical values through sports
• Awareness of health and wellness among children
• Source of enjoyment, fun and recreation among children.
• Opportunities for maximization of sports participation among students. Develop Leadership, group cohesion among students.

Answer 14.
There are various types of biomechanical movements:

• Flexion
• Adducation
• Extension
• Ulnar deviation
• Abduction

Extension – It is a straightening movement that increases the angle between body parts. When a joint can move forward and backward, such as the neck and trunk, extension refers to movement in the posterior direction. For example, when standing up, the knees are extended. Extension of the hip or shoulder moves the arm or leg backward. When the chin is against the chest, the head is flexed, and the trunk is flexed when a person leans forward.

Abduction – It refers to a motion that pulls a structure or part away from the midline of the body. In the case of fingers and toes, it refers to spreading the digits apart, away from the centerline of the hand or foot. Abduction of the wrist is also called radial deviation For example, raising the arms up, such as when tightrope-walking, is an example of abduction at the shoulder. When the legs are splayed at the hip, such as when doing a star jump or doing a split, the legs are abducted at the hip.

Adduction – It refers to a motion that pulls a structure or part toward the midline of the body, or towards the midline of a limb. In the case of fingers and toes, it refers to bringing the digits together, towards the centerline of the hand or foot. Adduction of the wrist is also called ulnar deviation. Dropping the arms to the sides, and bringing the knees together, are examples of adduction.

Answer 15.

Answer 16.
More importantly, yoga is extremely effective in:

1. Increasing Flexibility – Yoga has positions that act upon the various joints of the body including those joints that are never really on the ‘radar screen’ let alone exercised.

2. Increasing lubrication of the joints, ligaments and tendons – Yoga positions exercise the different tendons and ligaments of the body. Surprisingly it has been found that the body which may have been quite rigid starts experiencing a remarkable flexibility in even those > parts which have not been consciously work upon. Why? It is here that the remarkable research behind yoga positions proves its mettle. Seemingly unrelated “non strenuous” yoga positions act upon certain parts of the body in an interrelated manner. When done together, they work in harmony to create a situation where flexibility is attained relatively easily.

3. Massaging of ALL Organs of the Body – Yoga is perhaps the only form of activity which massages all the internal glands and organs of the body in a thorough manner, including those – such as the prostate – that hardly get externally stimulated during our entire lifetime. Yoga acts in a wholesome manner on the various body parts. This stimulation and massage of the organs in turn benefits us by keeping away disease and providing a forewarning the first possible instance of a likely onset of disease or disorder.

4. Complete Detoxification – By gently stretching muscles and joints as well as massaging the various organs, yoga ensures the optimum blood supply to various parts of the body.This helps in the flushing out of toxins from every nook and cranny as well as providing nourishment up to the last point. This leads to benefits such as delayed ageing, energy and a remarkable zest for life.

5. Excellent toning of the muscles – Muscles that have become flaccid, weak or slothy are
stimulated repeatedly to shed excess flab and flaccidity.

Answer 17.
There are many factors that effect motor development:

1. Nutrition: Nutritious food promotes good motor development. Sensory motor development is dependent upon nutrition that the child gets to a great extent. Children get stronger and development is good if they get nutritious food.
2. Immunisation: If mother and child both are immunized at a proper time it leads to good sensory motor development.
3. Environment: Encouragement, love and security help the child to take risk to explore fearlessly and to know more about environment which leads to a better sensory development.

Answer 18.
(i) Parameters developed are:

• Leadership
• Motivation
• Personality
• Posture
• Fitness

(ii) Intramural competitions
(iii) Test for School Children:

• AAPHER physical fitness test

Answer 19.
Rikli and Jones is the senior citizen test. It was developed by Rikli and Jones in 2001. This test is beneficial for various senior citizens. It helps the early identification of at-risk participants. It is significant to plan safe and effective physical exercise programmes for senior citizens because individual’s health and fitness level can be known better with the help of this test.

1. Chair Stand Test – This test measures the lower body strength, particularly legs.
2. Arm Curl Test – The arm curl test is a test for upper body strength and endurance which is required for performing household activities.
3. Chair sit and reach test: It assess the lower body flexibility, which is important for good posture.

Answer 20.
Coping refers to the thoughts and actions which we usually use to deal with a threatening situation. Lazarus and Folkman said “Coping is a process of constantly changing cognitive and behavioral efforts to manage specific external and/or internal demands or conflicts appraised as taxing or exceeding one’s resources”. These are the following types of strategies:
(i) Problem focused coping strategies – They aim at changing or eliminating the authentic source of stress by:

• analyze the stressful situation
• Seek professional help
• reframing
• slow down pace
• stay focused
•  plan properly

(ii) Emotion focused coping strategies – tried to reduce the negative emotional responses linked with stress by:

• denial of reality
• blaming
• mentally disengaging from stressful situation or people
• humorous attitude
• seek support
• be positive

Answer 21.
Throwing comprises of two phases, the preparatory phase and the throwing phase.
Most actions are rotational in the transverse plane and longitudinal axis and the two joints primarily involved are the elbow and shoulder.
The elbow is a hinge joint formed by the humerus and ulna.
The shoulder is a ball and socket joint formed between the humerus and the scapula.

Answer 22.
The male and female athletes have various physiological differences. These differences generally result in difference in performance ouputs interms of strength, speed, endurance.

1. Cardiovascular Fitness: Athletes’ cardiovascular fitness is measured by their maximum oxygen consumption, also known as VO2 max, which measures their capacity to transport and use oxygen during exercise. This is measured by calculating the point at which an athlete’s oxygen consumption remains steady despite an increase in an exercise intensity. Elite male athletes have a higher oxygen carrying capacity than women, which allows them to reach their maximum training peak earlier. According ACSM’s Primary Care Sports Medicine reference book, this is probably due to women’s lower hemoglobin levels and men’s larger body size. Maximum oxygen consumption is directly related to body size,

2. Bones and Ligaments: Male athletes have longer and larger bones, which provide a clear mechanical advantage over female athletes. The increased articular surface and larger structure of male bones provide them with a greater leverage and a wider frame on which to support muscle. Similarly, the ligaments of female athletes are generally more lax and fragile than those of their male counterparts. This gives male athletes an advantage in sports that involve throwing, kicking and hitting, and explains the higher incidence of musculoskeletal injuries among female athletes. On the other hand, female athletes have a wider pelvis and a lower center of gravity, which provides excellent balance.

3. Strength: Male athletes have a higher ratio of muscle mass to body weight, which allows for greater speed and acceleration. This explains why female speed records in running and swimming are consistently 10 percent slower than men’s, and why, on average, they have two thirds of the strength of men. However, when you factor out the larger muscle mass in men and compare muscular strength relative to cross-section area of muscle, the strength of male and female athletes is nearly equal.

4. Endurance: Endurance is largely determined by a body’s efficiency when converting calories into energy. Female athletes are more efficient than male athletes at converting glycogen to energy. Glycogen is a secondary source of fuel you use when glucose levels drop. This is why female athletes excel in ultra-long-distance sports and rarely hit the wall during long races. It also explains why ultra-running, which includes races longer than a marathon, is one of the few sports where elite female and male athletes regularly compete together, and in which female athletes sometimes win.

Answer 23.
Classification of injury:
(i) Soft Tissue
(ii) Bone Injury
(iii) Joint Injury

(i) Soft Tissue Injury:

• Contusion: Direct impact with blunt object which causes bleeding deep with muscles due to damage in capillaries.
• Strain: Injury to musculo-tendon injury
• Sprain: Ligament injury.
• Abrasion: Loss of epidermis (Outer layer of skin) Superficial injury with loss of skin
• Incision: Cut on arteries, tendon, veins, nerves due to sharp objects
• Laceration: Irregular tear in skin, cut in epidermis and dermis with blunt edge objects

(ii) Bone Injury-Fracture:
A.Close Fracture

• Transverse Fracture
• Oblique Fracture
• Spiral Fracture
• Comminute Fracture
• Impact Fracture

B.Open Fracture

• Compound Fracture

(iii) Joint Injury: Joint injuries often occur as a result of bicycle wrecks, falling is contact sports, and car accidents. They can range from sprains to fractures and dislocations. Common joints injuries include:

• Runners knee
• Plica syndrome of the knee
• Rotator cut injury (shoulder)
• Sprained Ankle

Preventive Measures:

• Wear protective gear such as helmets, protective pads, etc.
• Warm up and cool down
• Keep in mind the rules of the game.

Answer 24.
Exercise for Posture Correction:

1. Back exercises – to put shoulder at original place, some workouts of back strengthening are required.
2. Trunk exercises – twisting of trunk often strengthens the back and abdomen.
3. Leg press, hamstring, curls and leg extensions.
4. Aerobics
5. Walking, running, jumping, etc.

Answer 25.
Total number of teams =13
Upper half = n + 1/2 = 13 + 1/2 = 7
Lower half = n – 1/2 = 13 – 1/2 = 6
Power of two’s = 24 n. of teams
= 16 -13 = 3 No. of Byes = 3
No. of Byes in Lower Half =nb + 1/2 = 3 + 1/2 = 2
No. of Byes in Upper Half = nb – 1/2 = 3 – 1/2 = 1

Answer 26.
Scientific research has demonstrated repeatedly that physical education can enhance academic performance and cognitive function. However, for children with special needs, it’s valuable for so many reasons, from providing an opportunity to build collaborative and social skills, to teaching individuals how to focus on specific goals and overcome obstacles.

When students with special needs participate in physical activity and sports, they see improvements in everything from their hand-eye coordination and flexibility, to their muscle strength, endurance, and even cardiovascular efficiency. These are all simply the natural benefits of exercise — a development of better motor skills and enhanced physical health that helps individuals to fight back against problems such as obesity, and the health complications that follow.

1. Mental Improvements in Confidence and Well-Being: Regular exposure to sports through physical education classes isn’t only good for a child’s body — it’s beneficial to their mind, too. Physical activity improves general mood and wellness in psychiatric patients suffering from anxiety and depressive disorders. What’s more, regular fitness links to improvements in self-esteem, social awareness, and self-confidence — all essential for empowering the lives of young people with special needs.

2. Reduces stress and anxiety: Providing a physical outlet may help students reduce or cope with anxiety, stress and depression — while interaction and involvement with other students will help to give children a sense of accomplishment and confidence. Their physical education teachers to involve them in environments where they can feel as though they’re successfully contributing to a group and their abilities in other areas will improve according to their positive self-image and confidence.

3. Behavioral Improvements in Attention, Relationships, and Academics: Physical education is about a lot more than simply learning how to engage in a particular sport — it teaches children a range of skills, from how to work as a team, to how to solve problems, increase attention span, and focus on task-based behavior. Eventually, those skills can transfer into other classroom settings too, so that students with special needs have a greater ability to leam and engage with their peers outside of physical education.

4. Self-Esteem: Developing a sense of self-esteem and confidence is an extremely important part of special education. These children need to be involved in environments where they feel that they are contributing successfully to a group. Their abilities in all other skill areas will improve as a result of a positive self-image and confidence.

5. Cognitive Benefits: The hands-on nature of physical education leads to cognitive improvements in children with special needs, allowing them to access skills that they couldn’t challenge within a traditional classroom setting. The structure of sport – which comes with a set of rules and organization, can be a learning tool that helps children to practice self-regulation and enhance their decision making skills. On top of that, children with special needs can leam to focus on specific goals, and work on their verbal communication by interacting with peers through sport.

6. Improves appetite

7. Improves quality of sleep
As these children often feel isolated, they

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These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 6.

CBSE Sample Papers for Class 12 Biology Paper 6

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 6 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
What is coleorhiza?

Question 2.
What is the advantage of use of biotechnology in molecular biology over traditional pathological tests?

Question 3.
What are cleistogamous flowers?

Question 4.
How is Agrobacterium tumefaciens considered useful?

Question 5.
Which attribute of human population do the following figures represent?

SECTION-B

Question 6.
What are the basic steps involved in genetically modifying an organism?

Question 7.
Explain the four types of barriers of Innate immunity.

Question 8.
What do you mean by inbreeding depression? How this problem should be solved during animal breeding?

OR

Write with examples, how use of microbes helps us to make different types of cheese with specific texture & flavors?

Question 9.
S strain → Inject into mice→ ……
….. → Inject into mice → Mice live
Complete the diagram above. What was this experiment about and who performed it?

Question 10.
What may be the reasons for low productivity of ocean?

SECTION-C

Question 11.
Discuss the barrier methods for contraception.

Question 12.
An individual has genotype with an extra chromosome 21.
(a) What is this disorder called?
(b) What will be the physical appearance?

Question 13.
Discuss the role of microbes in sewage treatment

Question 14.
What is DNA finger printing? On what principle does it work? Mention its two applications.

OR

Explain Miller’s experiment to prove the ‘theory of chemical origin of life’ as proposed by Oparin and Haldane.

Question 15.
Differentiate between spermatogenesis and oogenesis with a diagram.

Question 16.
What are the advantages of GM plants?

Question 17.
What do you understand by the term bio-pesticide? Name and explain the mode of action of a popular bio-pesticide

Question 18.
Represent schematically the life cycle of malarial parasite.

Question 19.
Compare and contrast: isogamy and anisogamy. With examples.

Question 20.
Answer the following:  
(a) Expand IUT.
(b) In which part of the female reproductive system the 8-celled embryo will be transferred during test tube baby programme.

Question 21.
Halpoid content of human DNA is 3.3 x 10⁹ bp and the distance between 2 consecutive bp is 0.34x 10^-9. What is the length of the DNA molecule?

Question 22.
Explain convergent evolution with examples.

SECTION-D

Question 23.
Rakhi and her parents were watching a TV serial in the evening. During a commercial break, an advertisement flashed on the screen which was promoting use of sanitary napkins. Rakhi was still watching the TV. The parents got embarrassed and changed the channel. Rakhi objected to her parent’s behaviour and explained the need for these advertisements.
(a) What values did the parents show?
(b) Briefly describe the phases of a menstrual cycle.

SECTION-E

Question 24.
Plant breeding programmes are carried out in a systematic way worldwide. Explain the five main steps in breeding a new genetic variety.

OR

Why are CO₂, CH₄, N₂Oetc known as greenhouse gases?
Why is CNG better than Diesel?

Question 25.
What are chromosomal disorders?

OR

(a) Explain primary productivity and the factors that influence it. 5
(b) Describe how oxygen and chemical composition of detritus control decomposition.

Question 26.
Draw schematically a single polynucleotide strand (with at least three nucleotides). Provide labels and directions.

OR

What are the post pollination events? Explain it.

Answers

SECTION-A

Answer 1.
In embryos of monocots the root cap and radicle are enclosed in an undifferentiated sheath called coleorhiza.

Answer 2.
Biotechnological methods make early diagnosis possible to detect a disease and start treatment at an early stage rather conventional methods of diagnosis where it is not possible.

Answer 3.
Self pollinating flowers in which stamens and pistil are in close proximity.

Answer 4.
Agrobacterium tumefaciens help a tumor causing gene in the bacteria to be substituted with a gene of interest and its introduction into plants.

Answer 5.
The figures represent the age pyramids of human population as Expanding Stable and Declining

SECTION-B

Answer 6.
Three basic steps involved in genetically modifying an organism are:

1. Identification of DNA with desirable genes.
2. Introduction of the identified DNA into the host.
3. Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

Answer 7.
Innate immunity consists of four types of barriers. These are:

1. Physical barriers: Skin on our body is the main barrier which prevents entry of the micro­organisms.
2. Physiological barriers: Acid in the stomach, saliva in the mouth, tears from eyes-all prevent microbial growth.
3. Cellular barriers: Certain types of leukocytes (WBC) of our body like polymorpho-nuclear leukocytes (PMNL-neutrophils) and monocyte can phagocytose and destroy microbes.
4. Cytokine barriers: Virus-infected cells secrete proteins called interferons which protect non-infected cells from further viral infection.

Answer 8.
Continued inbreeding, especially close inbreeding, usually reduces fertility and even productivity. This is called inbreeding depression. This problem can be solved by selecting animals of the breeding population to be mated with unrelated superior animals of the same breed which usually helps restore fertility and yield.

OR

The specificity of texture, flavour and taste in different variety of cheese comes from microbes used. For example,

1. The large holes in ‘Swiss Cheese’ are due to production of a large amount of CO, by Propionibacterium sharmanii.
2. The ‘Roquefort cheese’ are ripended by growing a specific fungi on them, which gives them a particular flavor

Answer 9.
S strain → Inject into mice → Mice die
R strain → Inject into mice →Mice live

Answer 10.
Low productivity of ocean is due to:

1. Lack of light.
2. High salinity.
3. High pressure and
4. Waves and tides

Answer 11.
Various barrier methods available for both males and females are:

(1) Condoms: Barriers made of thin rubber/latex sheath that are used to cover the penis in the male or vagina and cervix in the female, just before coitus so that the ejaculated semen would not enter into the female reproductive tract.

(2) Diaphragms, Cervical Caps and Vaults: Barriers made of rubber that are inserted into the female reproductive tract to cover the cervix during coitus. They prevent conception by blocking the entry of sperms through the cervix. They are reusable. Spermicidal creams, jellies and foams are usually used along with these barriers to increase their contraceptive efficiency.

(3) Intra Uterine Devices (IUDs): These devices are inserted by doctor or expert nurses in the uterus through vagina. These are presently available as the non-medicated IUDs (g., Lippes loop), copper releasing IUDs (CuT, Cu7, Multiload 375) and the hormone releasing IUDs (Progestasert, LNG-20). IUDs increase phagocytosis of sperms within the uterus and the Cu ions released suppress sperm motility and the fertilising capacity of sperms. The hormone releasing IUDs, in addition, make the uterus unsuitable for implantation and the cervix hostile to the sperms.

Answer 12.
An individual has genotype with an extra chromosome 21.
(a) The disorder is called Down’s syndrome or trisomy of 21.
(b) The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease. Physical, psychomotor and mental development is retarded.

Answer 13.
Heterotrophic microbes naturally present in the sewage plays a major role in the treatment of sewage which is carried out in two stages:

Primary Treatment: This initially involves removal of floating debris by sequential filtration followed by grit (soil and small pebbles) removal by sedimentation. All solids that settle down form the primary sludge, and the supernatant forms the effluent. The effluent from the primary settling tank is taken for secondary treatment.

Secondary Treatment or Biological Treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful aerobic microbes into floes. The significantly reduced BOD (biochemical oxygen demand) of the effluent is then passed into a setting tank where the bacterial ‘floes’ are allowed to form activiated sludge. A small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum. The remaining major part of the sludge is pumped into anaerobic sludge digesters, where, other kinds of bacteria digest the bacteria and the fungi in the sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. These gases form biogas and can be used as source of energy as it is inflammable.

Answer 14.
DNA finger printing is a genetic molecular method that identifies and evaluates an individual from another individual on the basis of unique patterns (polymorphisms) in their DNA. DNA finger printing works on a principle of repetitive DNA sequences that gives a unique identity to an individual. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation and categories, into micro-satellites, mini-satellites. These sequence normally do not code for any proteins, but they form a large portion of human genome. These sequence show high degree of polymorphism and form the basis of DNA finger printing. Its two applications are:

1. As an identification tool in forensics.
2. Paternity testing, in case of disputes.

OR

Oparin and Haldane proposed that the first form of life could have come from pre-existing non-living organic molecules (e.g. RNA, protein, etc.) and that formation of life was preceded by chemical evolution, i.e. formation of diverse organic molecules from inorganic constitutents. The conditions on earth were—high temperature, volcanic storms, reducing atmosphere containing CH₄, NH_r Miller created similar conditions in a laboratory scale. He created electric discharge in a closed flask containing CH₄, H₂, NH₃ and water vapour at 8000° C. He observed formation of amino acids. In similar experiments others observed, formation of sugars, nitrogen bases, pigment and fats. Analysis of meteorite content also revealed similar compounds indicating that similar processes are occurring else where in space.

Answer 15.

Answer 16.
Genetically Modified crops/plants have the following advantages:

1. Crops are more tolerant to abiotic stresses (cold, drought, salt, heat).
2. Have reduced reliance on chemical pesticides (pest-resistant crops).
3. Helpful in reduction of post harvest losses.
4. Plants have increased efficiency of mineral usage to prevent early exhaustion of fertility of soil.
5. Crops yield enhanced nutritional value of food, e.g., Vitamin ‘A’ enriched rice.
6. Tailor-made plants have created alternative resources for industries, in the form of starches, fuels and pharmaceuticals.

Answer 17.
Bio-pesticides are derived from natural materials such as animals, plants, bacteria and certain minerals. This microbial bio-control agents that is used to control butterfly caterpillars is the bacteria Bacillus thuringiensis (often written as Bt). These are mixed with water and sprayed onto vulnerable plants such as brasssicas and fruit trees, where these are eaten by the insect larvae. In the gut of the larvae, the toxin is released and the larvae get killed. The bacterial disease will kill the caterpillars, but leave other insects unharmed.method of controlling pests relies on natural predation rather than introduced chemicals. An example of B. thuringiensis toxin genes has also been introduced with the help of genetic engineering into plants which offers resistance to insect pests. Bt-cotton is one such example

Answer 18.

Answer 19.

Answer 20.
(a) IUT stands for Intra Uterine transfer. The 8-celled embryo developed by In-vitro fertilization or ICSI is transferred in this technique.
(b) The 8-celled embryo is transferred to the uterus.

Answer 21.
Haploid content is 3.3 x 10⁹
Therefore, diploid content is 6.6 x 10⁹ Distance between bp is 0.34 x 10^-9
Therefore length is diploid content x distance between bp = 6.6 x 10⁹ x 0.34 x 10 ⁹ = 2.24 m

Answer 22.
Convergent evolution occurs when more than one adaptive radiation appear to have occurred in an isolated geographical area of different habitats. The analogous organs have almost similar appearance and perform the same function but develop in different groups and are totally different in their basic structure and development origin. For example, Darwins’ Finches and Australian Marsupials where each marsupial differ from each other.

SECTION-D

Answer 23.
(a) The parents were traditional but understood the need for such advertisements. They showed maturity and openness later.
(b)

1. Menstrual phase
2. Proliferative phase
3. Ovulatory phase
4. Secretory phase.

Answer 24.
The main steps in breeding a new genetic variety of a crop are as follows:

(1) Collection of Variability: The entire collection of plants/seeds having all the diverse alleles for all genes including all the different wild varieties, species and relative of the cultivated species in a given crop called as germplasm collection, is collected.

(2) Evaluation and Selection of Parents: The germplasm is evaluated so as to identity plants with desirable combination of characters. The selected plants are multiplied and used in the process of hybridization.

(3) Cross Hybridization among the Selected Parents: The desired characters are combined from two different plants (parents), to produce hybrids that genetically combine the desired characters in one plant. This is a very time-consuming and tedious process since the pollen grains from the desirable plant chosen as male parent have to be collected and placed on the stigma of the flowers selected as female parent. Usually only one in few hundred to a thousand crosses shows the desirable combination, for example high protein quality of one parent may need to be combined with disease resistance from another parent.

(4) Selection and Testing of Superior Recombinants: Progency plants that are superior to both of the parents are selected. The selection process is crucial to the success of breeding objective and requires careful scientific evaluation of the progency.

(5) Testing, Release and Commercialization of new Cultivars: The newly selected lines are evaluated for their yield and other agronomic traits of quality, disease resistance etc. The material is then evaluated in comparison to the best available local crop cultivar usually by a check or reference cultivar.

Answer 25.

OR 

(a) Primary productivity is defined as the amount of biomass or organic matter produced per unit area
over a time period by plants during photosynthesis. It is expressed in terms of weight (g ²) or energy
(kcal nr²). It is also expressed in terms of g ² yr¹ or (kcal nr²) yr¹ to compare the productivity of different ecosystems.

It is divided into gross primary productivity (GPP), net primary productivity (NPP) and secondary productivity.

Gross primary productivity of an ecosystem is the rate of production of organic matter ditfing photosynthesis.

Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores and decomposers), i.e., NPP = GPP – R.

Secondary productivity is defined as the rate of-formation of new organic matter by consumers.

The factors on which primary productivity depends are: (z) Plant species inhabiting a particular area, (ii) On a variety of environmental factors such as availability of nutrients. (iii) Photosynthetic capacity of plants.

(b) Decomposition is an oxygen-requiring process. The rate of decomposition is controlled by both chemical composition of detritus and climatic factors. In a particular climatic condition, decomposition rate is slower if detritus is rich in lignin and chitin, and quicker, if detritus is rich in nitrogen and water-soluble substances like sugars. Temperature and soil moisture are the most important climatic factors that regulate decomposition through their effects on the activities of soil microbes. Warm and moist environment favour decomposition whereas low temperature and anaerobiosis inhibit decomposition resulting in build up of organic matter.

Answer 26.

The post pollination events are:
1. The pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores.
2. The contents of the pollen grain move into the pollen tube. Pollen tube grows through the tissues of the stigma and style and reaches the ovary.
3. The generative cell divides and forms the two male gametes during the growth of pollen tube in the stigma.
4. Pollen tube, after reaching the ovary, enters the ovule through the micropyle and then enters one of the synergids through the filiform apparatus.
All these events are together referred to as pollen-pistil interaction also.

We hope the CBSE Sample Papers for Class 12 Biology Paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 12 Biology Paper 6, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 5.

CBSE Sample Papers for Class 12 Biology Paper 5

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 5 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
What is the number of chromosome in human zygote?

Question 2.
What is totipotency?

Question 3.
What are palindromic sequences?

Question 4.
What is Allen’s rule?

Question 5.
How do you define NPP?

SECTION-B

Question 6.
Write the transcription product sequence for
(a) 5 ‘ATGCACTGATCCAA 3’
(b) 3 ‘ GTACGTACGTAC 5’

Question 7.
Complete the table:

 Question 8.
What are the types of acquired immunity?

Question 9.
Which microbe converts milk to curd?

Question 10.
Give some examples of diseases and their insect vector.

OR

What are the different methods of breeding?

SECTION-C

Question 11.
List the salient features of DNA double helix model.

Question 12.
What is the fate of the product of fertilization in humans?

Question 13.
How was the genetic code elucidated?

Question 14.
p² + 2pq + q² = 1.Explain this equation.

Question 15.
What are the different levels at which gene regulation can be achieved?

Question 16.
What are the primary lymphoid organs?

Question 17.
Explain gene therapy with an example?

Question 18.
Diagrammatically represent the replication of retrovirus.

Question 19.
How have cry proteins been utilized?

OR

Explain carbon cycle with diagram.

Question 20.
Explain two reasons for loss of biodiversity.

Question 21.
Give some adaptations of desert plants to survive the heat.

Question 22.
What does the picture represent?

SECTION-D

Question 23.
In art class the teacher asked Sunita to mix green and red paint and report on the combined colour formed. Sunita could not find red colour in her box and was scolded by the teacher who found it lying right in front. Suddenly Vijay realized that Sunita was not able to identify red colour and reported the matter to the teacher who was of the opinion that she lacked colour concept. After school was over, Vijay reported this matter to Sunita’s parents.                                                                                                (a) What values did Vijay possess?
(b) Did Sunita lack knowledge of colours? If not, give the biological reason for the same.
(c) Give the technical term for this type of inheritance. Explain with a typical example

SECTION-E

Question 24.
Explain with diagram the experiment that proved that DNA is a genetic material.

OR

Explain pollination by wind and water.

Question 25.
Give the journey of sperm formation with diagram. What are the hormones involved?

OR

Explain the technique of fingerprinting with diagram.

Question 26.
What is parasitism? What are the types?

OR

What are ecosystem services?

Answers

SECTION-A

Answer 1.
46

Answer 2.
Capacity of generating a whole plant from cell/explant is called totipotency.

Answer 3.
The palindrome in DNA is a sequence of base pairs on the two strands that reads the same when orientation of reading is kept same

Answer 4.
Mammals from colder climates generally have shorter ears and limbs to minimise heat loss. This is known as Allen’s rule.

Answer 5.
NPP stands for Net Primary Productivity. It is the available biomass for the consumption by heterotrophs (herbivores and decomposers) NPP = GPP – R. Where ‘GPP’ is Gross Primary Productivity, ‘R’ is Respiration losses.

SECTION-B

Answer 6.
(a) 3 ‘TACGTGACTAGGTT 5’                                      ‘
(b) 5 ‘AUGCACUGAUCCAA 3 ’

Answer 7.

Answer 8.
1. Antibody mediated immunity or humoral immune response by the production of antibodies against antigens.
2. Cell mediated immunity initiated by T lymphocytes.

Answer 9.
Micro-organisms such as Lactobacillus and others commonly called lactic acid bacteria (LAB) grow in milk and convert it to curd. During growth, the LAB produce acids that coagulate and partially digest the milk proteins. A small amount of curd added to the fresh milk as inoculum or starter contain millions of LAB, which at suitable temperatures multiply, thus converting milk to curd.

Answer 10.
(1) Malaria – female Anopheles mosquito
(2) Dengue or Chikungunya – Aedes mosquito.

OR

Different methods of breeding are:
(a) Inbreeding;
(b) Out breeding;
(c) Out crossing;
(d) Cross breeding; and
(e) Interspecific hybridisation

SECTION-C

Answer 11.
The salient features of the Double-helix structure of DNA are as follows:

1. It is made of two polynucleotide chains, where the backbone is constituted by sugar phosphate, and the bases project inside.
2. The two chains have anti-parallel polarity. It means, if one chain has the polarity 5’→3′, the other has
   3′ → 5′.
3. The base in two strands are paired through hydrogen bond (H-bonds) forming base pairs (bp). Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa. Similarly, Guanine is bonded with Cytosine with three H-bonds. As a result, always a purine comes opposite to a pyrimidine. This generates approximately uniform between the two strands of the helix.
4. The plane of one base pair stacks over the other in double helix. This, in addition to H-bonds, confersstability of the helical structure.

Answer 12.
The product of fertilization is the zygote. The mitotic division starts as the zygote moves through the isthmus of the oviduct called cleavage towards the uterus and forms 2,4,8,16 daughter cells called blastomeres. The embryo with 8 to 16 blastomeres is called a morula. The morula continues to divide and transforms into blastocyst as it moves further into the uterus. The blastomeres in the blastocyst are arranged into an outer layer called trophoblast and an inner group of cells attached to trophoblast called the inner cell mass. The trophoblast layer then gets attached to the endometrium and the inner cell mass gets differentiated as the embryo. After attachment the uterine cells divide rapidly and cover the blastocyst. As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is called implantation and it leads to pregnancy.

Answer 13.
Har Govind Khorana, Marshall Nirenberg and Severo Ochoa elucidated the genetic code. Genetic Code directs the sequence of amino acids during protein synthesis of proteins. The chemical method was developed by Har Gobind Khorana to synthesise RNA molecules with defined combinations of bases (homopolymers and copolymers). Marshall Nirenberg’s cell-free system for protein synthesis finally helped the code to be deciphered. Further, Severo Ochoa enzyme (polynucleotide phosphoryalse) was also helpful in polymerizing RNA with defined sequences in a template independent manner (enzymatic synthesis of RNA). This finally gave rise to the checker-board, for genetic code. The salient features of genetic code are as follows:
(a) The codon is triplet 4³ = 64. (61 codons code for amino and 3 codons do not code for any amino acids, hence they function as stop codons.)
(b) One codon codes for only one amino acid, hence, it is unambiguous and specific.
(c) Some amino acids are coded by more than one codon, hence the code is degenerate.
(d) The codon is read in mRNA in a contiguous fashion. There are no punctuations.
(e) The code is nearly universal: for example, from bacteria to human UUU would code for Phenylalanine (phe).
(f) AUG has dual functions. It codes for Methionine (met), and it also as initiator codon

Answer 14.
p² + 2pq + q² = 1, equation represents the gene frequency of a given population as stated by Hardy-Weinberg law. It states that the frequencies of allele in a given population are stable and constant.
Where p² = frequency of homozygous dominant alleles (AA).
q² = frequency of homozygous recessive alleles (aa).
2pq = frequency of heterozygous alleles (Aa).
This is also referred to as genetic equilibrium and is a binomial expansion of (p + q)².

Answer 15.
Gene regulation could be exerted at

1. Transcriptional level (formation of primary transcript).
2. Processing level (regulation of splicing).
3. Transport of mRNA from nucleus to the cytoplasm.
4. Translational Level.

Answer 16.
The primary lymphoid organs are bone marrow and thymus where immature lymphocytes differentiate into antigen-sensitive lymphocytes. Both provide micro-environments for the development and maturation of T-lymphocytes.The bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.

The thymus is a lobed organ located near the heart and beneath the breastbone. The thymus is quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained, it reduces to a very small size.

Answer 17.
Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in child/embryo. Here, genes are inserted into a person’s cells and tissues to treat a disease. Correction of a genetic defect involves delivery of a normal gene into the individual or embryo to take over the function of and compensate for the non-functional gene.

The first clinical gene therapy was given to a 4-year-old girl with adenosine deaminase (ADA) deficiency. This disorder is caused due to the deletion of the gene for adenosine deaminase. In some children, ADA deficiency can be cured by bone marrow transplantation; in others it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. Both of these approaches are not completely curative.

Answer 18.

Answer 19.
The Bt toxin is coded by a gene named cry protein. The proteins encoded by the genes crylAc and crylAb control the cotton bollworms that of crylAb control com borer. B. thuringiensis  forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. The Bt toxin protein exist as inactive protoxin but once an insect ingest 1 the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut which solubilise the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually cause death of the insect.

OR

Carbon cycling occurs through atmosphere, ocean and through living and dead organisms. A considerable amount of carbon returns to the atmosphere as CO₂ through respiratory activities of the producers and consumers. Decomposers also contribute substantially to CO₂ pool by their processing of waste materials and dead organic matter of land oceans. Some amount of the fixed carbon is lost to sediments and removed from circulation. Burning of wood, forest fire and combustion of organic matter, fossil fuel, volcanic activity are additional sources for releasing CO₂ in the atmosphere.

Answer 20.
Two reasons for the loss of biodiversity are:
1. Habital Loss and Fragmentation: The degradation of many habitats by pollution also threatens the survival of many species. When large habitats are broken up into small fragments due to various human activities, mammals and birds requiring large territories and certain animals with migratory habits are badly affected, leading to population declines.

2. Co-extinctions: When a species becomes extinct, the plant and animal species associated with it in an obligatory way also become extinct. When a host fish species becomes extinct, its unique assemblage of parasites also meets the same fate. Another example is the case of a coevolved plant-pollinator mutualism where Extinction of one invariably leads to the extinction of the other.

Answer 21.
Some adaptive features that the desert plants have are:

1. Thick cuticle on their leaf surfaces.
2. Their stomata arranged in deep pits to minimize water loss through transpiration.
3. Special photosynthetic pathway (CAM) enables their stomata to remain closed during day time.
4. Some plants like opuntia, have no leaves-they are reduced to spines-and the photosynthetic function is taken over by the flattened stems

Answer 22.
The following picture represents the global biodiversity

Fishes have the maximum global biodiversity in the phylum vertebrates followed by birds, reptiles, amphibians and mammals.

SECTION-D

Answer 23.
(a) Vijay was alert, curious, clever and a responsible child.
(b) According to the teacher, Sunita lacks the concept of colours. When she could not identify red colour, it proved to be a case of colour blindness. It is a sex linked inherited disorder.
(c) This is a human disease which causes the loss of ability to differentiate between red and green colour. The gene for this red-green colour blindness is present on X chromosome. Colour blindness is recessive to normal vision. If a colour blind man (XcY) marries a girl with normal vision (XX), the daughters would have normal vision but would be a carrier, while sons would also be normal. If the carrier girl (heterozygous for colour blindness, XcX) now marries a colour blind XcY the off spring would show 50% females and 50% males. Of the females, 50% would be the carrier for colour blindness and the rest 50% would be colour blind. Of the males, 50% would have normal vision and the 50% would be colour blind.

SECTION-E

Answer 24.
Hershey and Chase grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but not protein. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur. Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge. Bacteria which were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.

Pollination By Wind:
(a) Wind pollination requires that the pollen grains are light and non-sticky so that they can be transported in wind currents.
(b) Possess well-exposed stamens so that the pollens are easily dispersed into wind currents.
(c) Ltirgc and feathery stigma to easily trap air-bome pollen grains.
(d) Have a single ovule in each ovary and numerous flowers packed into an inflorescence.
(e) Quite common in grasses.

Pollination By Water: 
(a) Water pollination is quite rare in flowering plants and is limited to mostly monocotyledons.
(b) Water is a regular mode of transport for the male gametes among the lower plant groups such as algae, bryophytes and pteridophytes.
(c) Some examples of water pollinated plants are Vallisneria and Hydrilla which grow in fresh water and several marine sea-grasses such as Zostera.
(d) In Vallisneria, the female flower reach the surface of water by the long stalk and the male flowers or pollen grains are released on the surface of water. They are carried passively by water currents some of them eventually reach the female flowers and the stigma.
(e) In sea grasses, female flowers remain submerged in water and the pollen grains are released inside the water.
(f) Pollen grains in many such species are long, ribbon like and they are carried passively inside the water; some of them reach the stigma and achieve pollination. In most of the water- pollinated species, pollen grains are protected from wetting by a mucilaginous covering.

Answer 25.
In testis, the immature male germ cells (spermatogonia) produce sperms by spermatogenesis that begins at puberty. The spermatogonia (sing, spermatogonium) present on the inside wall of seminiferous tubules multiply by mitotic division and increase in numbers. Each spermatogonium is diploid and contains 46 chromosomes. Some of the spermatogonia called primary spermatocytes periodically undergo meiosis. A primary spermatocyte completes the first meiotic division (reduction division) leading to formation of two equal, haploid cells called secondary spermatocytes, which have only 23 chromosomes each. The secondary spermatocytes undergo the second meiotic division to produce four equal, haploid spermatids. The spermatids are transformed into spermatozoa (sperms) by the process called spermiogenesis. After spermiogenesis, sperm heads become embedded in Sertoli cells, and are finally released from the seminiferous tubules by the process called sperimation.

The hormones involved are:

1. GnRH – Gonadotropin Releasing Hormone
2. LH – Luteinising Hormone
3. FSH – Follicle Stimulating Hormone.

OR

DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times.

• These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.

• The bulk DNA forms a major peak and the other small peaks are reffered to as satellite DNA.
• Depending on base composition (A : T rich or G:C rich), length of segment, and number of repetitive units, the satellite DNA is classified into many categories, such as micro- satellities, mini-satellites etc.
• These sequences normally do not code for any proteins, but they form a large portion of human genome.
• These sequence show high degree of polymorphism and form the basis of DNA fingerprinting.

The technique of DNA fingerprinting involves the following steps: 

1. Isolation of DNA, i.e., extraction from the nuclei of the different possible cells.
2. The DNA molecules are digested with the help of enzyme restriction endonuclease (called chemical knife) that cuts them into fragments. The fragments of DNA also contains the VNTRs.
3. The fragments are separated according to the size by the gel electrophoresis.
4. Multiplication of fragments of a particular size having VNTRs through PCR technique.
   Here they are treated with alkaline chemicals to split them into single stranded DNAs.
5. Blotting/transferring of separated fragments of a single stranded DNA to a nylon membrane.
   
6. Radioactive DNA probes having repeated base sequences complementary to possible VNTRs are poured over the nylon membrane. Some of them will bind to the single stranded VNTRs. Hybridisation of DNA with probes is called Southern Blotting. The nylon membrane is washed to remove extra probes.
7. Autoradiography: An X-ray film is exposed to the nylon membrane to mark the place where the radioactive DNA probes have bound to the DNA fragments. These places are marked as dark bands when X-ray film is developed.
8. The dark bands so formed on X-ray film represent the DNA fingerprints (= DNA profiles).

Answer 26.
Parasitism is an interspecies relationship in which one organism gets benefitted and the other is harmed. Parasites evolve special adaptations such as the loss of unnecessary sense organs, presence of adhesive organs or suckers to cling on to the host, loss of digestive system and high reproductive capacity. The life cycles of parasistes are often complex, involving one or two intermediate hosts or vectors to facilitate parasitisation of its primary host. The human liver fluke (a trematode parasite) depends on two intermediate hosts (a snail and a fish) to complete its life cycle. The malarial parasite needs a vector (mosquito) to spread to other hosts.

There are three types of parasitism:

(1) Ecoparasitsm: Parasites feeding on the external surface of the host organism. For example the lice on humans and ticks on dogs, marine fish are infested with ectoparasitic copepods and, cuscuta found growing on hedge plants, to derive its nutrition from the host plant which it parasitizes.

(2) Endoparasitism: Parasites living inside the host body at different sites, /.<?., liver, kidney, lungs, red blood cells etc. The life cycles of endoparasites are more complex because of their extreme specialisation. Their morphological and anatomical features are greatly simplified while emphasising their reproductive potential.

(3) Brood parasitism: Parasitisism in which birds lays its eggs in the nest of its host and lets the host incubate them. The eggs of the parasitic bird have evolved to resemble the host’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.

OR

The benefits that people obtain from ecosystems are termed as ecosystem services. The four categories of ecosystem services are supporting, provisioning, regulating and cultural. These services forms a base for a wide range of economic, environmental and aesthetic goods and services. Out of the total cost of various ecosystem services, the soil formation accounts for about 50 percent, and contributions of other services like recreation and nutrient cycling, are less than 10 per cent each. The cost of climate regulation and habitat for wildlife are about 6 per cent each. Examples of such services are as follows:
(a) Healthy Forest ecosystems.
(b) Purify air and water.
(c) Mitigate droughts and floods.
(d) Cycle nutrients.
(e) Generate fertile soils.
(f) Provide wildlife habitat.
(g) Maintain biodiversity.
(h) Pollinate crops.
(i) Provide storage site for carbon.

We hope the CBSE Sample Papers for Class 12 Biology Paper 5 help you. If you have any query regarding CBSE Sample Papers for Class 12 Biology Paper 5, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 12 Physics Paper 5 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 5.

CBSE Sample Papers for Class 12 Physics Paper 5

Time Allowed : 3 Hours

Max. Marks : 70

General Instructions 

• All questions are compulsory. There are 26 questions in all.
• This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
• Section A contains five questions of 1 mark each. Section B contains five questions of 2 marks each. Section C contains twelve questions of 3 marks each. Section D contains one value based question of 4 marks and Section E contains three questions of 5 marks each.
• There is no overall choice. However, an internal choice has been provided in 1 question of 2 marks, 1 question of 3 marks and all the 3 questions of 5 marks weightage. You have to attempt only 1 of the choices in such questions.
• You may use the following values of physical constants wherever necessary :

Questions
SECTION : A

Question 1.
The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator?

Question 2.
A heating element is marked 210 V, 630 W. What is the value of current drawn by the element when connected to a 210 V dc source?

Question 3.
Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.

Question 4.
Show on a graph, the variation of resistivity with temperature for a typical semiconductor.

Question 5.
Why should electrostatic field be zero inside a conductor?

SECTION : B

Question 6.
Draw a plot showing the variation of
(i) electric field (E) and
(ii) distance r due to a point charge Q.

Question 7.
Distinguish between ‘Analog and Digital signals’

OR

Mention the functions of any two of the following used in communication system :
(a) Transducer
(b) Repeater
(c) Transmitter
(d) Bandpass Filter

Question 8.
A ray of light incident on an equilateral prism (μ_g = \(\sqrt { 3 } \)) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.

Question 9.
The susceptibility of a magnetic material is -2.6 x 10^-5. Identify the type of magnetic material and state its two properties.

Question 10.
Two identical circular wires P and Q each of radius R and carrying current ‘I’  are kept in perpendicular planes such that they have a common centre. Find the magnitude and direction of the net magnetic field at the common centre of the two coils.

SECTiON : C

Question 11.
A cell of emf E and internal resistance r is connected to two external resistance R₁ and R₂ and a perfect ammeter. The current in the circuit is measured in four different situations :
(i) without any external resistance in the circuit
(ii) with resistance R₁only
(iii) with R₁ and R₂ in series combination
(iv) with R₁ and R₂ in parallel combination.
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order.Identify the currents corresponding to the four cases mentioned above.

Question 12.
Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self inductance L is given by 1/2 LI²

Question 13.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
(b) In Young’s double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3.

Question 14.
When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used the current flows continuously. How does one explain this based on the concept of displacement current?

Question 15.

A rectangular loop of wire of size 4 cm x 10 cm carries a steady current of 2 A. A straight long wire carrying 5 A current is kept near the loop as shown. If the loop and the wire are coplanar,
find
(i) the torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current carrying wire.

Question 16.
In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs ε₁ and ε₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A.
Find
(i) ε₁/ε₂ and
(ii) position of null point for the cell ε₁

How is the sensitivity of a potentiometer increased?

OR

Using Kirchhoffs rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Q resistance. Also find the potential between A and D.

Question 17.
Write any two factors which justify the need for modulating a signal. Draw a diagram showing an amplitude modulated wave by superimposing a modulating signal over a sinusoidal carrier wave.

Question 18.
The current in the forward bias is known to be more (~ mA) than the current in the reverse bias (~ μA). What is the reason, then, to operate the photodiode in reverse bias?

Question 19.
A metallic rod of length ‘L’ is rotated with angular frequency of ‘ω’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the induced em/between the centre and the metallic ring.

Question 20.
Write Einstein’s photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation. Write the three salient features observed in photoelectric effect which can be explained using this equation.

Question 21.

The figure shows a series LCR circuit with L = 5 H, C = 80 μF, R = 40 Ω connected to a variable frequency 240 V source, calculate
(i) the angular frequency of the source which drives the circuit at resonance,
(ii) the current at the resonating frequency,
(iii) the rmspotential drop across the inductor at resonance.

Question 22.
(i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
(ii) Show that the density of nucleus over a wide range of nuclei is constant independent of mass number A.

SECTION : D

Questoin 23.
Renu and her friend went to see an exhibition. There, security guard standing on the entrance gate, asked them to come through a metal detector gate. Her friend was scared of it. But Renu convinced her and explained the purpose and working of a metal detector.
(i) What values Renu possess?
(ii) What is a metal detector and how it works?

SECTION : E

Question 24.
A parallel plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change;
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor

OR

(a) Define electric flux. Write its S.I. units.
(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(c) How is the field directed if
(i) the sheet is positively charged,
(ii) negatively charged?

Question 25.
Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 150 cm and an eye-piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye-piece.

OR

How is the working of a telescope different from that of a microscope? The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.

Question 26.
Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain A_v of the amplifier is given by

is the current again ; R_L is the load resistance and r. is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?

OR

(a) Draw the circuit diagram of a full wave rectifier using p-n junction diode. Explain its working and show the output, input waveforms.
(b) Show the output waveforms (Y) for the following inputs A and B of
(i) OR gate
(ii) NAND gate.

 Answers :
SECTION : A

Answer 1.
On the equator, the value of both angle of dip (5) and vertical component of earth’s magnetic field is zero. So, in this case,
B_v = 0.

Answer 2.
In dc source, P = VI
Given that P = 630 W and V = 210 V
So, I = P/V = 630/210 = 3 A.

Answer 3.
Using Lenz’s law we can predict the direction of induced current in both the rings. Induced current oppose the cause of increase of magnetic flux. So, it will be clockwise in ring 1 and anticlockwise in ring 2.

Answer 4.
The following curve shows the variation of resistivity with temperature for a typical semiconductor.

This is because for a typical semiconductor resistivity decreases rapidly with increasing temperature.

Answer 5.
Charge on conductor resides on its surface. So if we consider a Gaussian surface inside the conductor to find the electrostatic field,

Where, q = charge enclosed in Gaussian surface. q = 0, inside the conductor, hence the electrostatic field inside the conductor is zero.

SECTION : B

These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 4.

CBSE Sample Papers for Class 12 Biology Paper 4

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
What are cleistogamous flowers?

Question 2.
What is coleorrhiza?

Question 3.
What are alleles?

Question 4.
What are introns?

Question 5.
What is apiculture?

SECTION-B

Question 6.
Explain MOET.

Question 7.
Write a note on discovery of penicillin.

Question 8.
What does the diagram below signify?

Question 9.
What are the basic steps involved in genetically modifying an organism?

OR

What is humus?

Question 10.
Represent the age pyramids for human population.

SECTION-C

Question 11.
Explain the parts of an ovule with a diagram

Question 12.
How is RNA synthesized in bacteria? Illustrate

Question 13.
Discuss the barrier methods for contraception

Question 14.
Explain convergent evolution with examples.

Question 15.
What are the major causes of cancer?

Question 16.
Is it possible to obtain large quantities of DNA from a single cell?

Question 17.
What are the advantages of GM plants?

Question 18.
Explain T_i plasmid of Agrobacterium tumefaciens.

Question 19.
Give the equations of both exponential and logistic growth curves. Also represent them graphically.

OR

Present a case study for remedy of plastic waste

Question 20.
Complete the diagram below:

Question 21.
Haploid content of human DNA is 3.3 x 10⁹ bp and the distance between two consecutive
bp is 0.34 x 10 ⁹. What is the length of the DNA molecule?

Question 22.
Label the diagram given below:

SECTION-D

Question 23.
Rita and her parents were watching a TV serial in the evening. During a commercial break
an advertisement flashed on the screen which was promoting use of sanitary napkins. Rita
was still watching the TV when the parents got embarrassed and changed the channel. Rita
objected to her parent’s behaviour and explained the need for these advertisements.
(a) What values did the parents show?
(b) Briefly describe the phases of a menstrual cycle.

SECTION-E

Question 24.
What are the post pollination events?

OR

Explain endosperm development.

Question 25.
What are chromosomal disorders?

OR

List the observations of Human Genome Project.

Question 26.
Explain some interspecific relationship where no species is harmed.

OR

Explain ecological succession.

Answers

SECTION-A

Answer 1.
Self pollinating flowers in which stamens and pistil are in close proximity.

Answer 2.
In embryos of monocots the root cap and radicle are enclosed in an undifferentiated sheath j called coleorhiza.

Answer 3.
Genes which code for a pair of contrasting traits.

Answer 4.
Intervening sequences in DNA which are not expressed in mature or processed RNA.

Answer 5.
Apiculture is the maintenance of hives of honeybees for the production of honey.

SECTION-B

Answer 6.
Multiple Ovulation Embryo Transfer Technology (MOET) is a programme for herd improvement. In this method, a cow is administered hormones, with FSH-like activity, to induce follicular maturation and super ovulation – instead of one egg, which they normally yield per cycle, they produce 6-8 eggs. The animal is either mated with an elite bull or artificially inseminated. The fertilised eggs at 8-32 cells stages, are recovered non-surgically and transferred to surrogate mothers. The genetic mother is available for another round of super ovulation.

Answer 7.
Alexander Fleming while working on Staphylococci bacteria, once observed a mould growing in one of his unwashed culture plates around which Staphylococci could not grow. He found out that it was due to a chemical produced by the mould and he named it Penicillin after the mould Penicillium notatum.

Answer 8.
The first diagram is that of the normal cells which show controlled growth due to property of contact inhibition.
The second diagram shows loss of contact inhibition by cancer cells. This uncontrolled cell growth leads to tumor

Answer 9.
Three basic steps in genetically modifying an organism are:

• Identification of DNA with desirable genes;
• Introduction of the identified DNA into the host;
• Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

OR

Humus is a dark coloured amorphous substance which is highly resistant to microbial action and undergoes decomposition at an extremely slow rate. Being colloidal in nature it serves as a reservoir of nutrients. The process of humus formation is called humification.

Answer 10.

SECTION-C

Answer 11.

The ovule is a small structure attached to the placenta by means of a stalk called funicle.The body of the ovule fuses with funicle in the region called hilum. Each ovule has one or two protective envelopes called integuments. Integuments encircle the ovule except at the tip where a small opening called the micropyle is organised. Opposite the micropylar end, is the chalaza, representing the basal part of the ovule. Enclosed within the integuments is a mass of cells called the nucellus. Cells of the nucellus have abundant reserve food materials. Located in the embryo sac or female gametophyte.

Answer 12.

In a bacteria, there are three major types of RNAs: mRNA (messenger RNA), tRNA (transfer RNA), and rRNA (ribosomal RNA). The mRNA provides the template, tRNA brings aminoacids and reads the genetic code, and rRNAs play structural and catalytic role during translation. There is single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria. RNA polymerase binds to promoter and initiates transcription (Initiation). It uses nucleoside triphosphates as substrate and polymerises in a template depended fashion following the rule of complementarity. It also facilitates opening of the helix and continues elongation. Only a short stretch of RNA remains bound to the enzyme. Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription.

Answer 13.
In barrier methods, ovum and sperms are prevented from physically meeting. Such methods are available for both males and females.
(1) Condoms are made of thin rubber/latex sheath that are used to cover the penis in the male or vagina and cervix in the female, just before coitus so that the ejaculated semen would not enter into the female reproductive tract. This can prevent conception.

(2) Diaphragms, cervical caps and vaults are also barriers made of rubber that are inserted into the female reproductive tract to cover the cervix during coitus. They prevent conception by blocking the entry of sperms through the cervix. They are reusable.

(3) Spermicidal creams, jellies and foams are usually used along with these barriers to increase their contraceptive efficiency.

(4) Intra Uterine Devices (IUDs): These devices are inserted by doctors or expert nurses in the uterus through vagina. These Intra Uterine Devices are presently available as the non- medicated IUDs (e.g., Lippes loop), copper releasing IUDs (CuT, Cu7, Multiload 375) and the hormone releasing IUDs (Progestasert, LNG-20). IUDs increase phagocytosis of sperms within the uterus and the Cu ions released suppress sperm motility and the fertilising capacity of sperms. The hormone releasing IUDs, in addition, make the uterus unsuitable for implantation and the cervix hostile to the sperms.

Answer 14.
Convergent evolution is a evolution of different structures for the same function and hence having similarity. It is the similar habitat that has resulted in selection of similar adaptive features in different groups of organisms but toward the same function.
For example the eye of the octopus and of mammals, the flippers of Penguins and Dolphins, sweet potato (root modification) and potato (stem modification) similarities in proteins and genes performing a given function among diverse organisms give clues to common ancestory.

Answer 15.
The major causes of cancer are:

1. Transformation of normal cells into cancerous neoplastic cells which may be induced by physical, chemical or biological agents called carcinogens.
2. Ionising radiations like X-rays and gamma rays and non-ionizing radiations like UV cause DNA damage leading to neoplastic transformation.
3. The chemical carcinogens present in tobacco smoke have been identified as a major cause of lung cancer.
4. Cancer causing viruses called oncogenic viruses have genes called viral oncogenes. Cellular oncogenes (cone) or proto oncogenes have been identified in normal cells which,when activated under certain conditions, could lead to oncogenic transformation of the cells.
   

Answer 16.
Yes, it is possible to make large quantities of DNA from a single cell by

PCR stands for Polymerase Chain Reaction. In this reaction, multiple copies of the gene (or DNA) of interest is synthesised in vitro using two sets of primers (small chemically synthesised oligonucleotides that are complementary to the regions of DNA) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as template. If the process of replication of DNA is repeated many times, the segment of DNA can be amplified to approximately billion times, i.e., 1 billion copies are made. Such repeated amplification is achieved by the use of thermostable DNA polymerase (isolated form a bacterium, Thermus aquaticus), which remain active during the high temperature induced denaturation of double stranded DNA. The amplified fragment if desired can now be used to ligate with a vector for further cloning.

Answer 17.
Advantages of GM plants are:

• GM crops are more tolerant to abiotic stresses (cold, drought, salt, heat).
• These have reduced reliance on chemical pesticides (pest-resistant crops).
• These crops helped to reduce post harvest losses.
• GM crops have increased efficiency of mineral usage which prevents early exhaustion of fertility of soil.
• These crops have enhanced nutritional value of food, e.g., Vitamin ‘A’ enriched rice.
• These crops have also helped in creation of tailor-made plants to supply alternative resources to industries, in the form of starches, fuels and pharmaceuticals.

Answer 18.
Agrobacterium tumifaciens is a pathogen of several dicot plants which has an ability to deliver ‘T-DNA’ that further transforms normal plant cells into a tumor and direct these tumor cells to produce the chemicals required by the pathogen. The tumor inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified into a cloning vector which is no more pathogenic to the plants but is still able to use the mechanisms to deliver genes of our interest into a variety of plants.

Answer 19.

Answer 20.

Answer 21.
Haploid content = 3.3 x 10⁹.
Therefore, diploid content = 6.6 x 10⁹
Distance between bp = 0.34 x 10 ⁹
Therefore length = diploid content x distance between bp
⇒ 6.6 x 10⁹ x 0.34 x 10~⁹ = 2.24 m

Answer 22.

SECTION-D

Answer 23.
(a) The parents were traditional but understood the need for such advertisements.
(b)

SECTION-E

Answer 24.

Following compatible pollination, the pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores. The contents of the pollen grain move into the pollen tube. Pollen tube grows through the tissues of the stigma and style and reaches the ovary. In plants, where the pollen grains are shed at the two celled stage, the generative cell divides and forms the two male gametes during the growth of pollen tube in the stigma. In plants which shed pollen in the three-celled condition, pollen tubes carry the two male gametes from the beginning. Pollen tube after reaching the ovary, enters the ovule through the micropyle and then enters one of the synergids through the filiform apparatus. Filiform apparatus present at the micropylar part of the synergids guides the entry of pollen tube. All these events are together referred to as pollen-pistil interaction.

OR

Endosperm development precedes embryo development. The primary endosperm cell divides repeatedly and forms a triploid endosperm tissue. The cells of this tissue are filled with reverse food materials and are used for the nutrition of the developing embryo. In the most common type of endosperm development, the PEN undergoes successive nuclear divisions to give rise to free nuclei. This stage of endosperm development is called free-nuclear endosperm. Subsequently cell wall formation occurs and the endosperm becomes cellular. The number of free nuclei formed before cellularisation varies greatly

Answer 25.
The chromosomal disorders are caused due to absence or excess or abnormal arrangement of one or more chromosomes. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome), called aneuploidy.

Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism and is called polyploidy.

Sometimes, either an additional copy of a chromosome may be included in an individual or an individual may lack one of any one pair of chromosomes: These situations are known as trisomy or monosomy of a chromosome, respectively.

Common examples of chromosomal disorders are Down’s syndrome, Turner’s syndrome, Klinefelter’s syndrome.

Down’s Syndrome: Caused by trisomy of 21 chromosome number.

Klinefelter’s Syndrome: Caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 447, XXY.

Turner’s Syndrome: Caused due to the absence of one of the X chromosome, i.e. 45 with XO.

OR

The salient observations drawn from human genome project are:

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.
3. The total number of genes is estimated at 30,000-much lower than previous estimates of 80,000 to 1,40,000 genes. Almost all (99.9 per cent) nucleotide bases are exactly the same in all people.
4. The functions are unknown for over 50 per cent of the discovered genes.
5. Less than 2 per cent of the genome codes for proteins.
6. Repeated sequences make up very large portion of the human genome.
7. Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.
8. Chromosome 1 has most genes (2968), and the Y has the fewest (231).
9. Scientists have identified about 1.4 million location where single base DNA differences (SNPs- single nucleotide polymorphism, pronounced as ‘snips’) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequences and tracing human history.

Answer 26.
Some interspecific relationship where no species is harmed are:
(1) Commonalism: This is the interaction in which one species benefits and the other is neither harmed nor benefited. For example: an orchid growing as an epiphyte on a mango branch, barnacles growing on the back of a whale, the cattle egret and grazing cattle, and Sea anemones stinging tentacles protect the clown fish from predators that lives among them.

(2) Mutualism: This interaction confers benefits on both interacting species. For example, Lichens represent an intimate mutualjstic relationship between a fungus and photosynthesising algae or cyanobacteria.

(3) Plant-Animal Relationships: Plants need the help of animals for pollinating their flowers and dispersing their seeds; in return plants offer rewards or fees in the form of pollen and nectar for pollinators and juicy and nutritious fruits for seed dispersers.

The gradual and fairly predictable change in the species composition of a given area is called Ecological succession. During succession some species colonise an area and their populations become more numerous, whereas populations of other species decline and even disappear. The entire sequence of communities that successively change in a given area are called sere(s). The individual transitional communities are termed serai stages or serai communities. In the successive serai stages there is a change in the diversity of species of organism, increase in the number of species and organisms as well as an increase in the total biomass. Succession is a process that starts where no living organisms are there-these could be areas where no living organisms ever existed, i.e., bare rock; or in areas that somehow, lost all the living organisms that existed there. The former is called primary succession, while the latter is termed secondary succession.

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