Biology: CBSE Marking Scheme with Solution (2019-20) Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

Class 12: Biology: CBSE Marking Scheme with Solution (2019-20) Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

The document Biology: CBSE Marking Scheme with Solution (2019-20) Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12 is a part of the Class 12 Course Sample Papers for Class 12 Medical and Non-Medical.
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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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