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Page 1
1
MARKING SCHEME
SAMPLE QUESTION PAPER
2019-20
CLASS XII (BIOLOGY)
TIME 3 HOURS MM 70
Section – A
1. b) Leydig cells
OR
b)Amniocentesis
1
2. d) Cell-mediated immune response
OR
d) ii and iv
1
3. d) P enzyme is Restriction endonuclease and Q enzyme is ligase 1
4. a) Sal I 1
5. b) Habitat loss and fragmentation 1
Section B
6. Encysted Amoeba divides by multiple fission / produces amoeba or pseudopodiospores /cyst wall
bursts out/spores are liberated to grow as amoebae(sporulation)
(½X4=2 Marks)
OR
Gemmule-asexual reproductive structure in sponges ( ½+½=1Mark)
Conidia-asexual reproductive structure in Penicillium.(or any other correct example)
(½+½ = 1Mark)
2
7. CuT,Cu7,Multiload 375 (Any two) (½ and ½ =1Mark )
Cu ions released suppresses sperm motility and the fertilizing capacity of sperms.
(½ +½ =1 Mark )
2
8. Control crosses cannot be performed in human beings, Alternate method-Pedigree analysis (study
of the traits in several generations of a family). (1+1=2 Marks)
2
9. A is more reactive ½ Mark
2'-OH group present in the pentose sugar ½ Mark
Makes it more labile/ catalytic and easily degradable. ½+½ =1 Mark
2
10. • Tissue culture ½ Mark
• Meristem apical or axillary is excised. ½ Mark
• Explant grown in a test tube under sterile condition/special nutrient medium
½+½ = 1 Mark
2
Page 2
1
MARKING SCHEME
SAMPLE QUESTION PAPER
2019-20
CLASS XII (BIOLOGY)
TIME 3 HOURS MM 70
Section – A
1. b) Leydig cells
OR
b)Amniocentesis
1
2. d) Cell-mediated immune response
OR
d) ii and iv
1
3. d) P enzyme is Restriction endonuclease and Q enzyme is ligase 1
4. a) Sal I 1
5. b) Habitat loss and fragmentation 1
Section B
6. Encysted Amoeba divides by multiple fission / produces amoeba or pseudopodiospores /cyst wall
bursts out/spores are liberated to grow as amoebae(sporulation)
(½X4=2 Marks)
OR
Gemmule-asexual reproductive structure in sponges ( ½+½=1Mark)
Conidia-asexual reproductive structure in Penicillium.(or any other correct example)
(½+½ = 1Mark)
2
7. CuT,Cu7,Multiload 375 (Any two) (½ and ½ =1Mark )
Cu ions released suppresses sperm motility and the fertilizing capacity of sperms.
(½ +½ =1 Mark )
2
8. Control crosses cannot be performed in human beings, Alternate method-Pedigree analysis (study
of the traits in several generations of a family). (1+1=2 Marks)
2
9. A is more reactive ½ Mark
2'-OH group present in the pentose sugar ½ Mark
Makes it more labile/ catalytic and easily degradable. ½+½ =1 Mark
2
10. • Tissue culture ½ Mark
• Meristem apical or axillary is excised. ½ Mark
• Explant grown in a test tube under sterile condition/special nutrient medium
½+½ = 1 Mark
2
2
11. • RNA interference ½ Mark
• silencing of a specific mRNA due to a complementary RNA ½ Mark
• dsRNA/Introduction of DNA was such that it produced both sense/ and anti-sense RNA in
the host cells/these two RNAs formed dsRNA that initiated RNAi 1 Mark
2
12.
The pyramid is inverted because the biomass of fishes is much more than that of the zooplankton
and phytoplankton. 1+1= 2 Marks
2
Section C
13.
(Diagram =1 Mark )
(Any four labellings ½ x 4=2)
3
14. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a
chromosome(s) ( aneuploidy) (1 Mark)
Autosomes:-
Down’s Syndrome: The cause is the presence of an additional copy of the chromosome number
21 (trisomy of 21). (½ Mark)
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.
(Any one symptom ½ Mark)
Sex chromosomes:-
Klinefelter’s Syndrome : This is caused due to the presence of an additional copy of X-
chromosome resulting into a karyotype of 47, XXY. ½ Mark
3
Page 3
1
MARKING SCHEME
SAMPLE QUESTION PAPER
2019-20
CLASS XII (BIOLOGY)
TIME 3 HOURS MM 70
Section – A
1. b) Leydig cells
OR
b)Amniocentesis
1
2. d) Cell-mediated immune response
OR
d) ii and iv
1
3. d) P enzyme is Restriction endonuclease and Q enzyme is ligase 1
4. a) Sal I 1
5. b) Habitat loss and fragmentation 1
Section B
6. Encysted Amoeba divides by multiple fission / produces amoeba or pseudopodiospores /cyst wall
bursts out/spores are liberated to grow as amoebae(sporulation)
(½X4=2 Marks)
OR
Gemmule-asexual reproductive structure in sponges ( ½+½=1Mark)
Conidia-asexual reproductive structure in Penicillium.(or any other correct example)
(½+½ = 1Mark)
2
7. CuT,Cu7,Multiload 375 (Any two) (½ and ½ =1Mark )
Cu ions released suppresses sperm motility and the fertilizing capacity of sperms.
(½ +½ =1 Mark )
2
8. Control crosses cannot be performed in human beings, Alternate method-Pedigree analysis (study
of the traits in several generations of a family). (1+1=2 Marks)
2
9. A is more reactive ½ Mark
2'-OH group present in the pentose sugar ½ Mark
Makes it more labile/ catalytic and easily degradable. ½+½ =1 Mark
2
10. • Tissue culture ½ Mark
• Meristem apical or axillary is excised. ½ Mark
• Explant grown in a test tube under sterile condition/special nutrient medium
½+½ = 1 Mark
2
2
11. • RNA interference ½ Mark
• silencing of a specific mRNA due to a complementary RNA ½ Mark
• dsRNA/Introduction of DNA was such that it produced both sense/ and anti-sense RNA in
the host cells/these two RNAs formed dsRNA that initiated RNAi 1 Mark
2
12.
The pyramid is inverted because the biomass of fishes is much more than that of the zooplankton
and phytoplankton. 1+1= 2 Marks
2
Section C
13.
(Diagram =1 Mark )
(Any four labellings ½ x 4=2)
3
14. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a
chromosome(s) ( aneuploidy) (1 Mark)
Autosomes:-
Down’s Syndrome: The cause is the presence of an additional copy of the chromosome number
21 (trisomy of 21). (½ Mark)
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.
(Any one symptom ½ Mark)
Sex chromosomes:-
Klinefelter’s Syndrome : This is caused due to the presence of an additional copy of X-
chromosome resulting into a karyotype of 47, XXY. ½ Mark
3
3
Such an individual has overall masculine development
• has overall masculine development
• feminine development is also expressed by the development of breast/ Gynaecomastia).
Such individuals are sterile.
(Any one symptom ½ Mark)
If students give the example of Turner’s Syndrome, it should be considered and marks given.
OR
a) i. point mutation/ single base substitution ½ Mark
ii. point mutation/ single base deletion ½ Mark
b) i 4 aminoacids 1 Mark
ii 4 aminoacids 1 Mark
15. In some species, the diploid egg cell is formed without reduction division and develops into the
embryo without fertilization. 1 Mark
In many Citrus and Mango varieties some of the nucellar cells surrounding the embryo sac start
dividing, protrudes into the embryo sac and develops into the embryos. In such species each ovule
contains many embryos. 2 Mark
3
16. a.) Chemical evolution – First form of life originated from pre-existing non-living organic
molecules.
b.) Amino acids
c.) H
2
1x3 =3 Mark
3
17. a.)
Amino acid Phe Val
DNA Code in Gene AAA CAC
Codon in mRNA i)UUU ii)GUG
Anticodon in tRNA iii)AAA iv)CAC
1Mark
b.)
i) A polypeptide containing 14 different amino acid = 14x3=42 base pairs. 1Mark
ii) 14 different types of RNA are needed for the synthesis of polypeptide. 1Mark
3
18. Advantages:-Inbreeding is necessary if we want to evolve a pure line in any animal.
• It helps in accumulation of superior genes and elimination of less desirable genes
• Inbreeding exposes harmful recessive genes that are to be eliminated by selection.
• Where there is selection at each step, it increases the productivity of inbred population.
(Any two 1 Mark each)
Disadvantages:-
• reduces fertility
• decreases productivity.
(Any two ½ x2=1 Mark)
3
19. Specific Bt toxin genes isolated from Bacillus thuringiensis is incorporated into cotton is coded
by the genes cryIAc and cryIIAb that control the cotton bollworms (½ + ½ = 1 Mark)
• Bacillus forms protein crystals that contain a toxic insecticidal protein.
• once an insect ingest the inactive toxin, it is converted into an active form
• The toxin in the form of crystals gets solubilised due to alkaline pH in the gut
• The activated toxin binds to the surface of gut epithelial cells and perforate the walls
causing the death of insect larva (½ x2=2 Marks)
3
Page 4
1
MARKING SCHEME
SAMPLE QUESTION PAPER
2019-20
CLASS XII (BIOLOGY)
TIME 3 HOURS MM 70
Section – A
1. b) Leydig cells
OR
b)Amniocentesis
1
2. d) Cell-mediated immune response
OR
d) ii and iv
1
3. d) P enzyme is Restriction endonuclease and Q enzyme is ligase 1
4. a) Sal I 1
5. b) Habitat loss and fragmentation 1
Section B
6. Encysted Amoeba divides by multiple fission / produces amoeba or pseudopodiospores /cyst wall
bursts out/spores are liberated to grow as amoebae(sporulation)
(½X4=2 Marks)
OR
Gemmule-asexual reproductive structure in sponges ( ½+½=1Mark)
Conidia-asexual reproductive structure in Penicillium.(or any other correct example)
(½+½ = 1Mark)
2
7. CuT,Cu7,Multiload 375 (Any two) (½ and ½ =1Mark )
Cu ions released suppresses sperm motility and the fertilizing capacity of sperms.
(½ +½ =1 Mark )
2
8. Control crosses cannot be performed in human beings, Alternate method-Pedigree analysis (study
of the traits in several generations of a family). (1+1=2 Marks)
2
9. A is more reactive ½ Mark
2'-OH group present in the pentose sugar ½ Mark
Makes it more labile/ catalytic and easily degradable. ½+½ =1 Mark
2
10. • Tissue culture ½ Mark
• Meristem apical or axillary is excised. ½ Mark
• Explant grown in a test tube under sterile condition/special nutrient medium
½+½ = 1 Mark
2
2
11. • RNA interference ½ Mark
• silencing of a specific mRNA due to a complementary RNA ½ Mark
• dsRNA/Introduction of DNA was such that it produced both sense/ and anti-sense RNA in
the host cells/these two RNAs formed dsRNA that initiated RNAi 1 Mark
2
12.
The pyramid is inverted because the biomass of fishes is much more than that of the zooplankton
and phytoplankton. 1+1= 2 Marks
2
Section C
13.
(Diagram =1 Mark )
(Any four labellings ½ x 4=2)
3
14. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a
chromosome(s) ( aneuploidy) (1 Mark)
Autosomes:-
Down’s Syndrome: The cause is the presence of an additional copy of the chromosome number
21 (trisomy of 21). (½ Mark)
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.
(Any one symptom ½ Mark)
Sex chromosomes:-
Klinefelter’s Syndrome : This is caused due to the presence of an additional copy of X-
chromosome resulting into a karyotype of 47, XXY. ½ Mark
3
3
Such an individual has overall masculine development
• has overall masculine development
• feminine development is also expressed by the development of breast/ Gynaecomastia).
Such individuals are sterile.
(Any one symptom ½ Mark)
If students give the example of Turner’s Syndrome, it should be considered and marks given.
OR
a) i. point mutation/ single base substitution ½ Mark
ii. point mutation/ single base deletion ½ Mark
b) i 4 aminoacids 1 Mark
ii 4 aminoacids 1 Mark
15. In some species, the diploid egg cell is formed without reduction division and develops into the
embryo without fertilization. 1 Mark
In many Citrus and Mango varieties some of the nucellar cells surrounding the embryo sac start
dividing, protrudes into the embryo sac and develops into the embryos. In such species each ovule
contains many embryos. 2 Mark
3
16. a.) Chemical evolution – First form of life originated from pre-existing non-living organic
molecules.
b.) Amino acids
c.) H
2
1x3 =3 Mark
3
17. a.)
Amino acid Phe Val
DNA Code in Gene AAA CAC
Codon in mRNA i)UUU ii)GUG
Anticodon in tRNA iii)AAA iv)CAC
1Mark
b.)
i) A polypeptide containing 14 different amino acid = 14x3=42 base pairs. 1Mark
ii) 14 different types of RNA are needed for the synthesis of polypeptide. 1Mark
3
18. Advantages:-Inbreeding is necessary if we want to evolve a pure line in any animal.
• It helps in accumulation of superior genes and elimination of less desirable genes
• Inbreeding exposes harmful recessive genes that are to be eliminated by selection.
• Where there is selection at each step, it increases the productivity of inbred population.
(Any two 1 Mark each)
Disadvantages:-
• reduces fertility
• decreases productivity.
(Any two ½ x2=1 Mark)
3
19. Specific Bt toxin genes isolated from Bacillus thuringiensis is incorporated into cotton is coded
by the genes cryIAc and cryIIAb that control the cotton bollworms (½ + ½ = 1 Mark)
• Bacillus forms protein crystals that contain a toxic insecticidal protein.
• once an insect ingest the inactive toxin, it is converted into an active form
• The toxin in the form of crystals gets solubilised due to alkaline pH in the gut
• The activated toxin binds to the surface of gut epithelial cells and perforate the walls
causing the death of insect larva (½ x2=2 Marks)
3
4
20. criteria for determining biodiversity hot spots are: –
• high levels of species richness (1 Mark)
• High degree of endemism. (1 Mark)
hotspots In India - Western Ghats, Himalaya (Indo-Burma/Sunderland to be accepted)
(Any 2) (½+½ = 1Mark)
OR
In-situ Conservation– Threatened /endangered plants and animals are provided with urgent
measures to save from extinction within their natural habitat and they are protected and
allowed to grow naturally.
Example- wildlife sanctuaries/ national parks /biosphere reserves/ sacred groves
(Any one example) (½ Mark, 1 Mark for difference)
Ex-situ Conservation –Threatened animals and plants are taken out from their natural
habitat and placed in a setting where they can be protected and given care
Example- in botanical gardens/ zoological gardens/ seed/pollen/gene banks
(Any one example) (½ Mark, 1 Mark for difference )
3
21. (a) To maintain the cells in their physiologically most active log/exponential phase. 1 Mark
(b) Temperature, pH, substrate, salts, vitamins, oxygen (Any 4) (½ x4 =2 Mark)
3
Section D
22. a.) Each primary spermatocyte will undergo meiosis-I and meiosis-2 which will result in 4
spermatozoa
300 million/4=75 million 1 Mark
b) Since replication has occurred by this stage
46x2 = 92 chromatids 1 Mark
Meiosis –I is completed by this time 92/2 =46 chromatids - 1 Mark
3
23. a) Vigorous growth of useful aerobic microbes into flocs. 1 Mark
b) Activated sludge – some of it is pumped back into the aeration tank to serve as the inoculum
½ + ½ Mark
c) During this digestion, a mixture of gases such as methane, hyrogensulphide is made and carbon
dioxide. These gases form biogas. 1 Mark
3
24. Platinum-pallidium Rhodium (Any two ½ +½ = 1Mark)
CO
2
,H
2
0 and CO [any 2] ½ + ½=1 Mark
Nitric oxide 1 Mark
3
Section E
25. Polygenic inheritance 1 Mark
• If we assume skin colour is controlled by three genes A, B, C
• Dominant forms (A,B,C) are responsible for dark skin colour and recessive form (a, b, c) for
light skin colour 1 Mark
• The genotype with all dominant alleles (AABBCC) will be darkest skin colour and with
recessive alleles will be light test skin colour (aabbcc) (1+1=2 Marks)
5
Page 5
1
MARKING SCHEME
SAMPLE QUESTION PAPER
2019-20
CLASS XII (BIOLOGY)
TIME 3 HOURS MM 70
Section – A
1. b) Leydig cells
OR
b)Amniocentesis
1
2. d) Cell-mediated immune response
OR
d) ii and iv
1
3. d) P enzyme is Restriction endonuclease and Q enzyme is ligase 1
4. a) Sal I 1
5. b) Habitat loss and fragmentation 1
Section B
6. Encysted Amoeba divides by multiple fission / produces amoeba or pseudopodiospores /cyst wall
bursts out/spores are liberated to grow as amoebae(sporulation)
(½X4=2 Marks)
OR
Gemmule-asexual reproductive structure in sponges ( ½+½=1Mark)
Conidia-asexual reproductive structure in Penicillium.(or any other correct example)
(½+½ = 1Mark)
2
7. CuT,Cu7,Multiload 375 (Any two) (½ and ½ =1Mark )
Cu ions released suppresses sperm motility and the fertilizing capacity of sperms.
(½ +½ =1 Mark )
2
8. Control crosses cannot be performed in human beings, Alternate method-Pedigree analysis (study
of the traits in several generations of a family). (1+1=2 Marks)
2
9. A is more reactive ½ Mark
2'-OH group present in the pentose sugar ½ Mark
Makes it more labile/ catalytic and easily degradable. ½+½ =1 Mark
2
10. • Tissue culture ½ Mark
• Meristem apical or axillary is excised. ½ Mark
• Explant grown in a test tube under sterile condition/special nutrient medium
½+½ = 1 Mark
2
2
11. • RNA interference ½ Mark
• silencing of a specific mRNA due to a complementary RNA ½ Mark
• dsRNA/Introduction of DNA was such that it produced both sense/ and anti-sense RNA in
the host cells/these two RNAs formed dsRNA that initiated RNAi 1 Mark
2
12.
The pyramid is inverted because the biomass of fishes is much more than that of the zooplankton
and phytoplankton. 1+1= 2 Marks
2
Section C
13.
(Diagram =1 Mark )
(Any four labellings ½ x 4=2)
3
14. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a
chromosome(s) ( aneuploidy) (1 Mark)
Autosomes:-
Down’s Syndrome: The cause is the presence of an additional copy of the chromosome number
21 (trisomy of 21). (½ Mark)
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.
(Any one symptom ½ Mark)
Sex chromosomes:-
Klinefelter’s Syndrome : This is caused due to the presence of an additional copy of X-
chromosome resulting into a karyotype of 47, XXY. ½ Mark
3
3
Such an individual has overall masculine development
• has overall masculine development
• feminine development is also expressed by the development of breast/ Gynaecomastia).
Such individuals are sterile.
(Any one symptom ½ Mark)
If students give the example of Turner’s Syndrome, it should be considered and marks given.
OR
a) i. point mutation/ single base substitution ½ Mark
ii. point mutation/ single base deletion ½ Mark
b) i 4 aminoacids 1 Mark
ii 4 aminoacids 1 Mark
15. In some species, the diploid egg cell is formed without reduction division and develops into the
embryo without fertilization. 1 Mark
In many Citrus and Mango varieties some of the nucellar cells surrounding the embryo sac start
dividing, protrudes into the embryo sac and develops into the embryos. In such species each ovule
contains many embryos. 2 Mark
3
16. a.) Chemical evolution – First form of life originated from pre-existing non-living organic
molecules.
b.) Amino acids
c.) H
2
1x3 =3 Mark
3
17. a.)
Amino acid Phe Val
DNA Code in Gene AAA CAC
Codon in mRNA i)UUU ii)GUG
Anticodon in tRNA iii)AAA iv)CAC
1Mark
b.)
i) A polypeptide containing 14 different amino acid = 14x3=42 base pairs. 1Mark
ii) 14 different types of RNA are needed for the synthesis of polypeptide. 1Mark
3
18. Advantages:-Inbreeding is necessary if we want to evolve a pure line in any animal.
• It helps in accumulation of superior genes and elimination of less desirable genes
• Inbreeding exposes harmful recessive genes that are to be eliminated by selection.
• Where there is selection at each step, it increases the productivity of inbred population.
(Any two 1 Mark each)
Disadvantages:-
• reduces fertility
• decreases productivity.
(Any two ½ x2=1 Mark)
3
19. Specific Bt toxin genes isolated from Bacillus thuringiensis is incorporated into cotton is coded
by the genes cryIAc and cryIIAb that control the cotton bollworms (½ + ½ = 1 Mark)
• Bacillus forms protein crystals that contain a toxic insecticidal protein.
• once an insect ingest the inactive toxin, it is converted into an active form
• The toxin in the form of crystals gets solubilised due to alkaline pH in the gut
• The activated toxin binds to the surface of gut epithelial cells and perforate the walls
causing the death of insect larva (½ x2=2 Marks)
3
4
20. criteria for determining biodiversity hot spots are: –
• high levels of species richness (1 Mark)
• High degree of endemism. (1 Mark)
hotspots In India - Western Ghats, Himalaya (Indo-Burma/Sunderland to be accepted)
(Any 2) (½+½ = 1Mark)
OR
In-situ Conservation– Threatened /endangered plants and animals are provided with urgent
measures to save from extinction within their natural habitat and they are protected and
allowed to grow naturally.
Example- wildlife sanctuaries/ national parks /biosphere reserves/ sacred groves
(Any one example) (½ Mark, 1 Mark for difference)
Ex-situ Conservation –Threatened animals and plants are taken out from their natural
habitat and placed in a setting where they can be protected and given care
Example- in botanical gardens/ zoological gardens/ seed/pollen/gene banks
(Any one example) (½ Mark, 1 Mark for difference )
3
21. (a) To maintain the cells in their physiologically most active log/exponential phase. 1 Mark
(b) Temperature, pH, substrate, salts, vitamins, oxygen (Any 4) (½ x4 =2 Mark)
3
Section D
22. a.) Each primary spermatocyte will undergo meiosis-I and meiosis-2 which will result in 4
spermatozoa
300 million/4=75 million 1 Mark
b) Since replication has occurred by this stage
46x2 = 92 chromatids 1 Mark
Meiosis –I is completed by this time 92/2 =46 chromatids - 1 Mark
3
23. a) Vigorous growth of useful aerobic microbes into flocs. 1 Mark
b) Activated sludge – some of it is pumped back into the aeration tank to serve as the inoculum
½ + ½ Mark
c) During this digestion, a mixture of gases such as methane, hyrogensulphide is made and carbon
dioxide. These gases form biogas. 1 Mark
3
24. Platinum-pallidium Rhodium (Any two ½ +½ = 1Mark)
CO
2
,H
2
0 and CO [any 2] ½ + ½=1 Mark
Nitric oxide 1 Mark
3
Section E
25. Polygenic inheritance 1 Mark
• If we assume skin colour is controlled by three genes A, B, C
• Dominant forms (A,B,C) are responsible for dark skin colour and recessive form (a, b, c) for
light skin colour 1 Mark
• The genotype with all dominant alleles (AABBCC) will be darkest skin colour and with
recessive alleles will be light test skin colour (aabbcc) (1+1=2 Marks)
5
5
• The genotypes (AaBbCc) will be of intermediate skin colour i.e. with three dominant alleles
and three recessive alleles 1 Mark
OR
• The sequences were arranged based on some overlapping regions present in them (Alignment
of these sequences was not humanly possible) 1Mark
• Therefore, specialized computer based programme was developed. 1Mark
• These sequences were subsequently annotated and were assigned to each chromosome-1Mark
• Chromosome 1 1Mark
• Caenorhabditis elegans 1Mark
26. a) Inducing mutation artificially using chemicals /radiations /and selecting plants with
desirable characters ½ x 2 = 1Mark
Mung Bean 1Mark
Yellow mosaic virus 1Mark
b) AQUACULTURE PISCICULTURE
1. It involves production and culturing of all
types of aquatic organisms in water bodies.
Production and culturing of fishes is called
pisciculture. 1x2= 2 Mark
OR
a) AIDS caused by the Human Immuno deficiency Virus (½+½= 1 Mark)
b) Vaccines prevent microbial infections by initiating production of antibodies against these
antigens to neutralise the pathogenic agents during later actual infection. (1/2)
The vaccines also generate memory – B and T-cells that recognize the pathogen quickly on
subsequent exposure. (1/2) 1 Mark
c) Normal cells show a property called contact inhibition by virtue of which contact with other
cells inhibits their uncontrolled growth. Cancer cells appear to have lost this property.(1)
These cells grow very rapidly, invading and damaging the surrounding normal tissues.
Cells sloughed from such tumors reach distant sites through blood, and wherever they get
lodged in the body, they start a new tumor there. This property called metastasis. (1)
2 Marks
d) Physiological barriers : Acid in the stomach and saliva in the mouth. ½ Mark
5
27.
(Marks to be given only if relative contribution is correct) (½ x 4 = 2 Marks )
5
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FAQs on Biology: CBSE Marking Scheme with Solution (2019-20) - Sample Papers for Class 12 Medical and Non-Medical
| 1. What is the CBSE marking scheme for Biology for the academic year 2019-20? |  |
Ans. The CBSE marking scheme for Biology for the academic year 2019-20 is a guideline provided by the Central Board of Secondary Education (CBSE) that outlines the distribution of marks for different sections and types of questions in the Biology exam. It helps students understand the weightage of each topic and the criteria for awarding marks.
| 2. What is the importance of the CBSE marking scheme in Biology exams? | |
Ans. The CBSE marking scheme in Biology exams is important as it provides clarity to students regarding the marking criteria. It helps them understand the expectations of examiners and enables them to prepare accordingly. The marking scheme also ensures fairness and consistency in evaluating students' performance.
| 3. How can students use the CBSE marking scheme to prepare for Biology exams? | |
Ans. Students can use the CBSE marking scheme to prepare for Biology exams by analyzing the weightage given to different topics and types of questions. They can focus more on topics with higher marks weightage and practice answering different types of questions as per the marking scheme. This will help them allocate their study time effectively and improve their chances of scoring well in the exam.
| 4. Can the CBSE marking scheme help in identifying common mistakes made by students in Biology exams? | |
Ans. Yes, the CBSE marking scheme can help in identifying common mistakes made by students in Biology exams. By referring to the marking scheme, students can understand the specific requirements for each question and the expected answer format. This can help them identify any misconceptions or errors they might have in their understanding of certain concepts. By practicing with the marking scheme, students can rectify their mistakes and improve their performance.
| 5. Is the CBSE marking scheme for Biology exams subject to change every year? | |
Ans. Yes, the CBSE marking scheme for Biology exams may be subject to change every year. The marking scheme is designed by the board based on the curriculum and guidelines for that particular academic year. It may be revised to align with any changes in the syllabus or evaluation pattern. Therefore, it is important for students to refer to the marking scheme specific to their academic year while preparing for the Biology exam.
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Document Description: Biology: CBSE Marking Scheme with Solution (2019-20) for Class 12 2024 is part of Sample Papers for Class 12 Medical and Non-Medical preparation.
The notes and questions for Biology: CBSE Marking Scheme with Solution (2019-20) have been prepared according to the Class 12 exam syllabus. Information about Biology: CBSE Marking Scheme with Solution (2019-20) covers topics
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