Class 12 Biology: CBSE Marking Scheme (2023-24) | Sample Papers for Class 12 Medical and Non-Medical PDF Download

 Page 1


1 
 
MARKING SCHEME (2023-24 
Class XII  
Biology (Subject Code-044) 
Q. 
No. 
Answer 
Marks 
Section - A 
1 d) black pepper 1 
2 d) tapetum 1 
3 d)  7 1 
4 a) males and females, respectively 1 
5 a) 0.32 1 
6 a) random and directionless 1 
7  d) Nucleotide 1 
8 d) aa 1 
9 c) Cyclosporin A produced from Trichoderma polysporum 1 
10 d) 1 
11 b) Reduce pesticide accumulation in food chain 1 
12 d) Soil Sample C 1 
13 d) A is false but R is true 1 
14 c) A is true but R is false  1 
15 a) Both A and R are true and R is the correct explanation of A. 1 
16 a) Both A and R are true and R is the correct explanation of A. 1 
Section – B 
17 Spermatogenesis starts at the age of puberty due to significant increase in the secretion of 
gonadotropin releasing hormone (GnRH). This is a hypothalamic hormone.                           [0.5]                                                                                                                                 
The increased levels of GnRH then act at the anterior pituitary gland and stimulate secretion of 
two gonadotropins – luteinising hormone (LH) and follicle stimulating hormone (FSH).          [0.5]                                                                                                                                 
LH acts at the Leydig cells and stimulates synthesis and secretion of androgens. Androgens, in 
turn, stimulate the process of spermatogenesis.                                                                   [0.5]                                                                                                                                 
FSH acts on the Sertoli cells and stimulates secretion of some factors which help in the process of 
spermiogenesis.                                                                                                                 [0.5]                                                                                                                                 
2 
18 a) CTT would become CAT which codes for valine. Thus, valine would replace glutamic acid at 
that point.                                                                                                                     [0.5]                                                                                                                                 
b) Sickle cell anaemia [0.5], the mutant haemoglobin molecule undergoes polymerization [0.5] 
leading to the change in the shape of the RBC from biconcave disc to elongated sickle like 
structure.                                                                                                                      [0.5] 
2 
Page 2


1 
 
MARKING SCHEME (2023-24 
Class XII  
Biology (Subject Code-044) 
Q. 
No. 
Answer 
Marks 
Section - A 
1 d) black pepper 1 
2 d) tapetum 1 
3 d)  7 1 
4 a) males and females, respectively 1 
5 a) 0.32 1 
6 a) random and directionless 1 
7  d) Nucleotide 1 
8 d) aa 1 
9 c) Cyclosporin A produced from Trichoderma polysporum 1 
10 d) 1 
11 b) Reduce pesticide accumulation in food chain 1 
12 d) Soil Sample C 1 
13 d) A is false but R is true 1 
14 c) A is true but R is false  1 
15 a) Both A and R are true and R is the correct explanation of A. 1 
16 a) Both A and R are true and R is the correct explanation of A. 1 
Section – B 
17 Spermatogenesis starts at the age of puberty due to significant increase in the secretion of 
gonadotropin releasing hormone (GnRH). This is a hypothalamic hormone.                           [0.5]                                                                                                                                 
The increased levels of GnRH then act at the anterior pituitary gland and stimulate secretion of 
two gonadotropins – luteinising hormone (LH) and follicle stimulating hormone (FSH).          [0.5]                                                                                                                                 
LH acts at the Leydig cells and stimulates synthesis and secretion of androgens. Androgens, in 
turn, stimulate the process of spermatogenesis.                                                                   [0.5]                                                                                                                                 
FSH acts on the Sertoli cells and stimulates secretion of some factors which help in the process of 
spermiogenesis.                                                                                                                 [0.5]                                                                                                                                 
2 
18 a) CTT would become CAT which codes for valine. Thus, valine would replace glutamic acid at 
that point.                                                                                                                     [0.5]                                                                                                                                 
b) Sickle cell anaemia [0.5], the mutant haemoglobin molecule undergoes polymerization [0.5] 
leading to the change in the shape of the RBC from biconcave disc to elongated sickle like 
structure.                                                                                                                      [0.5] 
2 
2 
 
19 On administration of the first dose of  the vaccine (L), the body shows a response of low 
intensity (X) as the immune system comes in contact with the  antigenic protein of the 
weakened/inactivated pathogen for the first time. This is called primary immune response. [1] 
On subsequent encounter with the same antigenic protein in the second dose (M), the body 
elicits a highly intensified secondary response (Y). Because of the memory of the first contact 
with the antigen, the secondary immune response is faster and stronger, leading to more 
effective pathogen elimination in comparison to the primary immune response.                        [1] 
2 
20 a) Plate 1, b-galactosidase enzyme is responsible for blue colour. Gene is inserted in the 
b-galactosidase site of the plasmid thereby causing insertional inactivation of the enzyme, 
so no blue colour is made.                                                                                               [1] 
b) Plate II - Gene of interest not inserted in the plasmid                                                      [0.5] 
     Plate III - No plasmid                                                                                                     [0.5] 
2 
21 
 
OR 
a) Gross Primary Productivity is  45000 + 40367 = 85367 KJm
-2
y
-1                                                              
[1] 
b) Net production is gradually reducing as we move from producers to consumers due to heat 
loss/respiration /10% law.                                                                                               [1] 
2 
Section – C 
22 a) Sperm A                                                                                                                [0.5] 
b) In the figure given, Sperm ‘A’has come in contact with the zona pellucida layer (P) of the 
ovum (Q), it will induce changes in the membrane that will block the entry of additional 
sperms (B and C). Thus, it ensures that only one sperm can fertilise the ovum.          [0.5] 
? The secretions of the acrosome of sperm A will help it to enter into the cytoplasm of the 
ovum (Q) through the zona pellucid (P) and the plasma membrane, this will induce the 
completion of the meiotic division of the secondary oocyte (Q).                                  [1] 
? The second meiotic division in Q being unequal will result in the formation of a second 
polar body and a haploid ovum. Then, the haploid nucleus of the sperm ‘A’ and that of the 
ovum (Q) will fuse together to form a diploid zygote.                                                  [1] 
3 
23 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 covers the blastocyst. 
? As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is 
called implantation and it leads to pregnancy.                                                         [0.5X6=3] 
3 
Page 3


1 
 
MARKING SCHEME (2023-24 
Class XII  
Biology (Subject Code-044) 
Q. 
No. 
Answer 
Marks 
Section - A 
1 d) black pepper 1 
2 d) tapetum 1 
3 d)  7 1 
4 a) males and females, respectively 1 
5 a) 0.32 1 
6 a) random and directionless 1 
7  d) Nucleotide 1 
8 d) aa 1 
9 c) Cyclosporin A produced from Trichoderma polysporum 1 
10 d) 1 
11 b) Reduce pesticide accumulation in food chain 1 
12 d) Soil Sample C 1 
13 d) A is false but R is true 1 
14 c) A is true but R is false  1 
15 a) Both A and R are true and R is the correct explanation of A. 1 
16 a) Both A and R are true and R is the correct explanation of A. 1 
Section – B 
17 Spermatogenesis starts at the age of puberty due to significant increase in the secretion of 
gonadotropin releasing hormone (GnRH). This is a hypothalamic hormone.                           [0.5]                                                                                                                                 
The increased levels of GnRH then act at the anterior pituitary gland and stimulate secretion of 
two gonadotropins – luteinising hormone (LH) and follicle stimulating hormone (FSH).          [0.5]                                                                                                                                 
LH acts at the Leydig cells and stimulates synthesis and secretion of androgens. Androgens, in 
turn, stimulate the process of spermatogenesis.                                                                   [0.5]                                                                                                                                 
FSH acts on the Sertoli cells and stimulates secretion of some factors which help in the process of 
spermiogenesis.                                                                                                                 [0.5]                                                                                                                                 
2 
18 a) CTT would become CAT which codes for valine. Thus, valine would replace glutamic acid at 
that point.                                                                                                                     [0.5]                                                                                                                                 
b) Sickle cell anaemia [0.5], the mutant haemoglobin molecule undergoes polymerization [0.5] 
leading to the change in the shape of the RBC from biconcave disc to elongated sickle like 
structure.                                                                                                                      [0.5] 
2 
2 
 
19 On administration of the first dose of  the vaccine (L), the body shows a response of low 
intensity (X) as the immune system comes in contact with the  antigenic protein of the 
weakened/inactivated pathogen for the first time. This is called primary immune response. [1] 
On subsequent encounter with the same antigenic protein in the second dose (M), the body 
elicits a highly intensified secondary response (Y). Because of the memory of the first contact 
with the antigen, the secondary immune response is faster and stronger, leading to more 
effective pathogen elimination in comparison to the primary immune response.                        [1] 
2 
20 a) Plate 1, b-galactosidase enzyme is responsible for blue colour. Gene is inserted in the 
b-galactosidase site of the plasmid thereby causing insertional inactivation of the enzyme, 
so no blue colour is made.                                                                                               [1] 
b) Plate II - Gene of interest not inserted in the plasmid                                                      [0.5] 
     Plate III - No plasmid                                                                                                     [0.5] 
2 
21 
 
OR 
a) Gross Primary Productivity is  45000 + 40367 = 85367 KJm
-2
y
-1                                                              
[1] 
b) Net production is gradually reducing as we move from producers to consumers due to heat 
loss/respiration /10% law.                                                                                               [1] 
2 
Section – C 
22 a) Sperm A                                                                                                                [0.5] 
b) In the figure given, Sperm ‘A’has come in contact with the zona pellucida layer (P) of the 
ovum (Q), it will induce changes in the membrane that will block the entry of additional 
sperms (B and C). Thus, it ensures that only one sperm can fertilise the ovum.          [0.5] 
? The secretions of the acrosome of sperm A will help it to enter into the cytoplasm of the 
ovum (Q) through the zona pellucid (P) and the plasma membrane, this will induce the 
completion of the meiotic division of the secondary oocyte (Q).                                  [1] 
? The second meiotic division in Q being unequal will result in the formation of a second 
polar body and a haploid ovum. Then, the haploid nucleus of the sperm ‘A’ and that of the 
ovum (Q) will fuse together to form a diploid zygote.                                                  [1] 
3 
23 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 covers the blastocyst. 
? As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is 
called implantation and it leads to pregnancy.                                                         [0.5X6=3] 
3 
3 
 
 
Fig : Fertilisation and passage of growing embryo in fallopian tube 
24 a) The embryo has Turner’s Syndrome [0.5] due to aneuploidy of the sex chromosome. Such a 
disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with XO.      [0.5] 
 
b)   She was advised MTP as the child will have the following problems: 
· rudimentary ovaries 
· poorly developed breasts 
· lack of other secondary sexual characters 
· delayed or no onset of the menstrual cycle and infertile.                           [Any 2; 2 marks]   
3 
25 a) A -stabilising; B - directional; C - disruptive;                                                         [1.5] 
b) Graph A – Stabilising 
 Graph B – Directional  
 Graph C – Disruptive  
                            [1.5] 
3 
26 
? It will adversely affect the secondary treatment or biological treatment of sewage. 
? When the aeration tank is not functional, the air will not be pumped into it. 
? This will not allow the vigorous growth of useful aerobic microbes into flocs (masses 
of bacteria associated with fungal filaments to form mesh like structures). 
? Thus, the major part of the organic matter in the effluent will not be consumed by 
these bacteria. 
? The BOD (biochemical oxygen demand) of the effluent will not be reduced. BOD 
refers to the amount of the oxygen that would be consumed if all the organic matter 
in one liter of water were oxidised by bacteria. 
3 
Page 4


1 
 
MARKING SCHEME (2023-24 
Class XII  
Biology (Subject Code-044) 
Q. 
No. 
Answer 
Marks 
Section - A 
1 d) black pepper 1 
2 d) tapetum 1 
3 d)  7 1 
4 a) males and females, respectively 1 
5 a) 0.32 1 
6 a) random and directionless 1 
7  d) Nucleotide 1 
8 d) aa 1 
9 c) Cyclosporin A produced from Trichoderma polysporum 1 
10 d) 1 
11 b) Reduce pesticide accumulation in food chain 1 
12 d) Soil Sample C 1 
13 d) A is false but R is true 1 
14 c) A is true but R is false  1 
15 a) Both A and R are true and R is the correct explanation of A. 1 
16 a) Both A and R are true and R is the correct explanation of A. 1 
Section – B 
17 Spermatogenesis starts at the age of puberty due to significant increase in the secretion of 
gonadotropin releasing hormone (GnRH). This is a hypothalamic hormone.                           [0.5]                                                                                                                                 
The increased levels of GnRH then act at the anterior pituitary gland and stimulate secretion of 
two gonadotropins – luteinising hormone (LH) and follicle stimulating hormone (FSH).          [0.5]                                                                                                                                 
LH acts at the Leydig cells and stimulates synthesis and secretion of androgens. Androgens, in 
turn, stimulate the process of spermatogenesis.                                                                   [0.5]                                                                                                                                 
FSH acts on the Sertoli cells and stimulates secretion of some factors which help in the process of 
spermiogenesis.                                                                                                                 [0.5]                                                                                                                                 
2 
18 a) CTT would become CAT which codes for valine. Thus, valine would replace glutamic acid at 
that point.                                                                                                                     [0.5]                                                                                                                                 
b) Sickle cell anaemia [0.5], the mutant haemoglobin molecule undergoes polymerization [0.5] 
leading to the change in the shape of the RBC from biconcave disc to elongated sickle like 
structure.                                                                                                                      [0.5] 
2 
2 
 
19 On administration of the first dose of  the vaccine (L), the body shows a response of low 
intensity (X) as the immune system comes in contact with the  antigenic protein of the 
weakened/inactivated pathogen for the first time. This is called primary immune response. [1] 
On subsequent encounter with the same antigenic protein in the second dose (M), the body 
elicits a highly intensified secondary response (Y). Because of the memory of the first contact 
with the antigen, the secondary immune response is faster and stronger, leading to more 
effective pathogen elimination in comparison to the primary immune response.                        [1] 
2 
20 a) Plate 1, b-galactosidase enzyme is responsible for blue colour. Gene is inserted in the 
b-galactosidase site of the plasmid thereby causing insertional inactivation of the enzyme, 
so no blue colour is made.                                                                                               [1] 
b) Plate II - Gene of interest not inserted in the plasmid                                                      [0.5] 
     Plate III - No plasmid                                                                                                     [0.5] 
2 
21 
 
OR 
a) Gross Primary Productivity is  45000 + 40367 = 85367 KJm
-2
y
-1                                                              
[1] 
b) Net production is gradually reducing as we move from producers to consumers due to heat 
loss/respiration /10% law.                                                                                               [1] 
2 
Section – C 
22 a) Sperm A                                                                                                                [0.5] 
b) In the figure given, Sperm ‘A’has come in contact with the zona pellucida layer (P) of the 
ovum (Q), it will induce changes in the membrane that will block the entry of additional 
sperms (B and C). Thus, it ensures that only one sperm can fertilise the ovum.          [0.5] 
? The secretions of the acrosome of sperm A will help it to enter into the cytoplasm of the 
ovum (Q) through the zona pellucid (P) and the plasma membrane, this will induce the 
completion of the meiotic division of the secondary oocyte (Q).                                  [1] 
? The second meiotic division in Q being unequal will result in the formation of a second 
polar body and a haploid ovum. Then, the haploid nucleus of the sperm ‘A’ and that of the 
ovum (Q) will fuse together to form a diploid zygote.                                                  [1] 
3 
23 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 covers the blastocyst. 
? As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is 
called implantation and it leads to pregnancy.                                                         [0.5X6=3] 
3 
3 
 
 
Fig : Fertilisation and passage of growing embryo in fallopian tube 
24 a) The embryo has Turner’s Syndrome [0.5] due to aneuploidy of the sex chromosome. Such a 
disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with XO.      [0.5] 
 
b)   She was advised MTP as the child will have the following problems: 
· rudimentary ovaries 
· poorly developed breasts 
· lack of other secondary sexual characters 
· delayed or no onset of the menstrual cycle and infertile.                           [Any 2; 2 marks]   
3 
25 a) A -stabilising; B - directional; C - disruptive;                                                         [1.5] 
b) Graph A – Stabilising 
 Graph B – Directional  
 Graph C – Disruptive  
                            [1.5] 
3 
26 
? It will adversely affect the secondary treatment or biological treatment of sewage. 
? When the aeration tank is not functional, the air will not be pumped into it. 
? This will not allow the vigorous growth of useful aerobic microbes into flocs (masses 
of bacteria associated with fungal filaments to form mesh like structures). 
? Thus, the major part of the organic matter in the effluent will not be consumed by 
these bacteria. 
? The BOD (biochemical oxygen demand) of the effluent will not be reduced. BOD 
refers to the amount of the oxygen that would be consumed if all the organic matter 
in one liter of water were oxidised by bacteria. 
3 
4 
 
? The greater the BOD of waste water, more is its polluting potential. Thus, the effluent 
will remain polluted with high amount of organic matter and high BOD.        [0.5X6=3] 
27 a) Cry I Ab [0.5] 
b)  The spores of Bt contain crystalline toxin which is inactive [0.5]; for this crystalline toxin 
protein to become active it needs alkaline pH, which is present in insect  gut [0.5] The gut 
lining is broken down/mid gut epithelial cells become porous/swollen/cell lysis.                 [0.5] 
c) The Bt-toxin gene is cloned and inserted into the plant genome by recombinant 
DNA technology. These genetically modified (GM) plants express the Bt-toxin genes and 
become pest-resistant.                                                                                                    [1] 
OR 
a)    (i) Functional enzyme lipase is given to the patient by injection.                                       [0.5] 
(ii) This procedure is not completely curative.                                                                [0.5] 
b)  
? The disease can be treated by using Gene therapy.                                                   [0.5] 
? Gene therapy is a collection of methods that allows correction of a gene defect that has 
been diagnosed in a child/embryo.                                                                           [0.5] 
? 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.     [1] 
3 
28 Prokaryotic organisms’ diversity is not given any figures by ecologist because of following reasons. 
? Classification and identification of vast diversity of microbes is very difficult and cannot be 
efficiently done with use of currently available methods. 
? For many microorganisms, it is difficult to culture them under laboratory condition. 
? According to current biochemical and molecular techniques, it is estimated that microbes 
diversity can range in billions with microbes inhabiting diverse habitat on earth, with 
enormous diversity present in air, water and soil. Hence, more advanced molecular and 
biochemical techniques are needed to classify and identify this enormous diversity of 
microbes. 
3 
Section – D 
29 a) Plasmids which can be used to insert the geneof interest from a desired organism into a 
host/ they act as vectors to transfer gene of interest into the host.                             [1] 
OR  
Ori- Origin of replication (ori) - No replication will take place resulting in no copies of 
linked DNA.  
b)   i)   5’... ATC GTA/AAG CTT /CAT…3’ 
           3’... TAG CAT/TTC GAA /GTA…5’                                          [1 mark for both strand ] 
OR 
       5’... AAG CTT  …3’ 
           3’... TTC GAA …5’ ’                                                                 [1 mark for both strand ] 
 
ii)   No, as the restriction enzymes need to be the same which cut the     
           DNA of the plasmid and the gene of interest from the plant.                           [0.5+0.5=1] 
c) PUC18 as it has a higher copyrate.                                                                  [0.5+0.5=1] 
4 
30 a) P. aurelia species is competitively superior P. aurelia grows in numbers more quickly than P. 4 
Page 5


1 
 
MARKING SCHEME (2023-24 
Class XII  
Biology (Subject Code-044) 
Q. 
No. 
Answer 
Marks 
Section - A 
1 d) black pepper 1 
2 d) tapetum 1 
3 d)  7 1 
4 a) males and females, respectively 1 
5 a) 0.32 1 
6 a) random and directionless 1 
7  d) Nucleotide 1 
8 d) aa 1 
9 c) Cyclosporin A produced from Trichoderma polysporum 1 
10 d) 1 
11 b) Reduce pesticide accumulation in food chain 1 
12 d) Soil Sample C 1 
13 d) A is false but R is true 1 
14 c) A is true but R is false  1 
15 a) Both A and R are true and R is the correct explanation of A. 1 
16 a) Both A and R are true and R is the correct explanation of A. 1 
Section – B 
17 Spermatogenesis starts at the age of puberty due to significant increase in the secretion of 
gonadotropin releasing hormone (GnRH). This is a hypothalamic hormone.                           [0.5]                                                                                                                                 
The increased levels of GnRH then act at the anterior pituitary gland and stimulate secretion of 
two gonadotropins – luteinising hormone (LH) and follicle stimulating hormone (FSH).          [0.5]                                                                                                                                 
LH acts at the Leydig cells and stimulates synthesis and secretion of androgens. Androgens, in 
turn, stimulate the process of spermatogenesis.                                                                   [0.5]                                                                                                                                 
FSH acts on the Sertoli cells and stimulates secretion of some factors which help in the process of 
spermiogenesis.                                                                                                                 [0.5]                                                                                                                                 
2 
18 a) CTT would become CAT which codes for valine. Thus, valine would replace glutamic acid at 
that point.                                                                                                                     [0.5]                                                                                                                                 
b) Sickle cell anaemia [0.5], the mutant haemoglobin molecule undergoes polymerization [0.5] 
leading to the change in the shape of the RBC from biconcave disc to elongated sickle like 
structure.                                                                                                                      [0.5] 
2 
2 
 
19 On administration of the first dose of  the vaccine (L), the body shows a response of low 
intensity (X) as the immune system comes in contact with the  antigenic protein of the 
weakened/inactivated pathogen for the first time. This is called primary immune response. [1] 
On subsequent encounter with the same antigenic protein in the second dose (M), the body 
elicits a highly intensified secondary response (Y). Because of the memory of the first contact 
with the antigen, the secondary immune response is faster and stronger, leading to more 
effective pathogen elimination in comparison to the primary immune response.                        [1] 
2 
20 a) Plate 1, b-galactosidase enzyme is responsible for blue colour. Gene is inserted in the 
b-galactosidase site of the plasmid thereby causing insertional inactivation of the enzyme, 
so no blue colour is made.                                                                                               [1] 
b) Plate II - Gene of interest not inserted in the plasmid                                                      [0.5] 
     Plate III - No plasmid                                                                                                     [0.5] 
2 
21 
 
OR 
a) Gross Primary Productivity is  45000 + 40367 = 85367 KJm
-2
y
-1                                                              
[1] 
b) Net production is gradually reducing as we move from producers to consumers due to heat 
loss/respiration /10% law.                                                                                               [1] 
2 
Section – C 
22 a) Sperm A                                                                                                                [0.5] 
b) In the figure given, Sperm ‘A’has come in contact with the zona pellucida layer (P) of the 
ovum (Q), it will induce changes in the membrane that will block the entry of additional 
sperms (B and C). Thus, it ensures that only one sperm can fertilise the ovum.          [0.5] 
? The secretions of the acrosome of sperm A will help it to enter into the cytoplasm of the 
ovum (Q) through the zona pellucid (P) and the plasma membrane, this will induce the 
completion of the meiotic division of the secondary oocyte (Q).                                  [1] 
? The second meiotic division in Q being unequal will result in the formation of a second 
polar body and a haploid ovum. Then, the haploid nucleus of the sperm ‘A’ and that of the 
ovum (Q) will fuse together to form a diploid zygote.                                                  [1] 
3 
23 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 covers the blastocyst. 
? As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is 
called implantation and it leads to pregnancy.                                                         [0.5X6=3] 
3 
3 
 
 
Fig : Fertilisation and passage of growing embryo in fallopian tube 
24 a) The embryo has Turner’s Syndrome [0.5] due to aneuploidy of the sex chromosome. Such a 
disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with XO.      [0.5] 
 
b)   She was advised MTP as the child will have the following problems: 
· rudimentary ovaries 
· poorly developed breasts 
· lack of other secondary sexual characters 
· delayed or no onset of the menstrual cycle and infertile.                           [Any 2; 2 marks]   
3 
25 a) A -stabilising; B - directional; C - disruptive;                                                         [1.5] 
b) Graph A – Stabilising 
 Graph B – Directional  
 Graph C – Disruptive  
                            [1.5] 
3 
26 
? It will adversely affect the secondary treatment or biological treatment of sewage. 
? When the aeration tank is not functional, the air will not be pumped into it. 
? This will not allow the vigorous growth of useful aerobic microbes into flocs (masses 
of bacteria associated with fungal filaments to form mesh like structures). 
? Thus, the major part of the organic matter in the effluent will not be consumed by 
these bacteria. 
? The BOD (biochemical oxygen demand) of the effluent will not be reduced. BOD 
refers to the amount of the oxygen that would be consumed if all the organic matter 
in one liter of water were oxidised by bacteria. 
3 
4 
 
? The greater the BOD of waste water, more is its polluting potential. Thus, the effluent 
will remain polluted with high amount of organic matter and high BOD.        [0.5X6=3] 
27 a) Cry I Ab [0.5] 
b)  The spores of Bt contain crystalline toxin which is inactive [0.5]; for this crystalline toxin 
protein to become active it needs alkaline pH, which is present in insect  gut [0.5] The gut 
lining is broken down/mid gut epithelial cells become porous/swollen/cell lysis.                 [0.5] 
c) The Bt-toxin gene is cloned and inserted into the plant genome by recombinant 
DNA technology. These genetically modified (GM) plants express the Bt-toxin genes and 
become pest-resistant.                                                                                                    [1] 
OR 
a)    (i) Functional enzyme lipase is given to the patient by injection.                                       [0.5] 
(ii) This procedure is not completely curative.                                                                [0.5] 
b)  
? The disease can be treated by using Gene therapy.                                                   [0.5] 
? Gene therapy is a collection of methods that allows correction of a gene defect that has 
been diagnosed in a child/embryo.                                                                           [0.5] 
? 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.     [1] 
3 
28 Prokaryotic organisms’ diversity is not given any figures by ecologist because of following reasons. 
? Classification and identification of vast diversity of microbes is very difficult and cannot be 
efficiently done with use of currently available methods. 
? For many microorganisms, it is difficult to culture them under laboratory condition. 
? According to current biochemical and molecular techniques, it is estimated that microbes 
diversity can range in billions with microbes inhabiting diverse habitat on earth, with 
enormous diversity present in air, water and soil. Hence, more advanced molecular and 
biochemical techniques are needed to classify and identify this enormous diversity of 
microbes. 
3 
Section – D 
29 a) Plasmids which can be used to insert the geneof interest from a desired organism into a 
host/ they act as vectors to transfer gene of interest into the host.                             [1] 
OR  
Ori- Origin of replication (ori) - No replication will take place resulting in no copies of 
linked DNA.  
b)   i)   5’... ATC GTA/AAG CTT /CAT…3’ 
           3’... TAG CAT/TTC GAA /GTA…5’                                          [1 mark for both strand ] 
OR 
       5’... AAG CTT  …3’ 
           3’... TTC GAA …5’ ’                                                                 [1 mark for both strand ] 
 
ii)   No, as the restriction enzymes need to be the same which cut the     
           DNA of the plasmid and the gene of interest from the plant.                           [0.5+0.5=1] 
c) PUC18 as it has a higher copyrate.                                                                  [0.5+0.5=1] 
4 
30 a) P. aurelia species is competitively superior P. aurelia grows in numbers more quickly than P. 4 
5 
 
caudatum and shows more individuals in the same volume of culture/ 100 Paramecia aurelia in 6 
days whereas 60 P. caudatum in 8 days.                                                                              [2] 
b) Competitive Exclusion Principle’ which states that two closely related species competing for the 
same resources cannot co-exist indefinitely and the competitively inferior one will be 
eliminated.G.F. Gause,                                                                                                      [1] 
c) One such mechanism is ‘resource partitioning’. If two species compete for the same resource, 
they could avoid competition by choosing different times for feeding or different foraging 
patterns, to avoid competition and co-exist due to behavioural differences in their foraging 
activities.                                                                                                                         [1] 
OR 
    Graph A - As both species grow simultaneously. 
Section-E 
31 Couple 1: Normal reports of female, Normal sperms in testes, Missing connection in epididymis 
and Vas deferens in male. 
Assisted Reproductive Technology: 
Semen will be devoid of sperms in this case. So, In-vitro fertilization (IVF) by collecting the sperms 
from epididymis, followed by ZIFT or IUT (Test Tube Baby) is suggested. ZIFT is transfer of 
zygote or early embryo up to 8 blastomeres in fallopian tube and IUT refers to transfer of 
embryos with more than 8 blastomeres in uterus.                                                                  [1] 
Couple 2: Blockage in the fallopian tube in the female, Normal reports of male. 
Assisted reproductive Technology: 
Blockage of Fallopian Tube will not allow transfer of sperms to the site of fertilisation.  In-vitro 
fertilization (IVF) followed by IUT (Test Tube Baby). It would involve transfer of embryo with 
more than 8 blastomeres in uterus.                                                                                        [1] 
Couple 3: Normal reports of female, Poor semen parameters in terms of count, motility and 
morphology in male partner 
Assisted Reproductive Technology: 
Intracytoplasmic sperm injection (ICSI) in which sperm is directly injected into the ovum. Artificial 
insemination procedure is used mainly when sperms have poor characteristic or low sperm count. 
                                                                                                                                           [1] 
Couple 4: Low ovarian reserve in female, Normal reports in male 
Assisted Reproductive Technology: 
In-vitro-fertilization (IVF) by selection of normal blastocysts from ovary followed by Zygote intra-
fallopian transfer involving transfer of zygote or early embryos up to 8 blastomeres (ZIFT) or 
transfer of embryo with more than 8 blastomeres in the uterus (IUT).                                     [1] 
Couple 5: Poor ovarian reserve in female, morphologically abnormal sperms in male partner. 
Assisted Reproductive Technology: 
ICSI intracytoplasmic sperm injection in which selected normal sperms will be injected into the 
selected blastocyst. Intracytoplasmic sperm injection (ICSI) procedure is used mainly when 
sperms have poor characteristic or low sperm count.                                                             [1]  
OR 
Situatio
n No. 
Requirement of 
contraceptive for- 
Name of 
contraceptive device 
Mode of action 
5