Class 12 Biology: CBSE Marking Scheme (2024-25) | Sample Papers for Class 12 Medical and Non-Medical PDF Download

 Page 1


 
Marking Scheme   
Biology (044) 
Class XII (2024 – 25) 
Q. 
No. 
Answer Marks 
Section - A 
1 B. Both placenta as well as fully developed foetus.  1 
2 B. 2000  
Formation of one seed requires fertilsation between one pollen grain and 
one egg. To produce 1600 seeds, 1600 pollen grains and 1600 eggs will be 
required. Each microspore mother cell results in the formation of 4 pollen 
grains after one cycle of meiotic division. So, 400 meiotic divisions will result 
in the production of 1600 pollen grains. One megaspore mother cell after 
one cycle of meiotic division results in the formation of 1 egg; so, 1600 
meiotic divisions will take place to form 1600 eggs. Thus, total number of 
meiotic divisions required for the formation of 1600 seeds will be 400 + 1600 
=2000. 
1 
3 A. 23% 
According to Chargaff’s rules, in DNA, A =T and G=C; 
Thus, A +T+G+C =100 
Given T = 27% so A=T =27% 
Thus A+T = 27 +27 =54% 
Thus, G+C =100 – 54 = 46% 
Since G = C so G = 46/2 =23% 
1 
4 B. CGTA  
- - - - - - - - - - - - - - - - - - - 
For Visual Impaired Students 
B. 4000 bp/s  
 It completes replication process in 18 minutes i.e. 18x 60 seconds.  
Rate of polymerization = 4.6 x 10
6 
bp/ 18x 60 s  
= 460000/108  
= 4259.1bp/s or approximately 4000 bp/sec Thus, the correct option is B. 
1 
5 C. 
            Suresh                                             Rajesh 
Sickle Cell Anaemia – Autosomal 
linked Recessive trait  
Thalassemia – Autosomal 
Recessive blood disorder  
 
1 
6 A. present in the medium and it binds to the repressor.  1 
7 A. (i) and (ii)  1 
Page 2


 
Marking Scheme   
Biology (044) 
Class XII (2024 – 25) 
Q. 
No. 
Answer Marks 
Section - A 
1 B. Both placenta as well as fully developed foetus.  1 
2 B. 2000  
Formation of one seed requires fertilsation between one pollen grain and 
one egg. To produce 1600 seeds, 1600 pollen grains and 1600 eggs will be 
required. Each microspore mother cell results in the formation of 4 pollen 
grains after one cycle of meiotic division. So, 400 meiotic divisions will result 
in the production of 1600 pollen grains. One megaspore mother cell after 
one cycle of meiotic division results in the formation of 1 egg; so, 1600 
meiotic divisions will take place to form 1600 eggs. Thus, total number of 
meiotic divisions required for the formation of 1600 seeds will be 400 + 1600 
=2000. 
1 
3 A. 23% 
According to Chargaff’s rules, in DNA, A =T and G=C; 
Thus, A +T+G+C =100 
Given T = 27% so A=T =27% 
Thus A+T = 27 +27 =54% 
Thus, G+C =100 – 54 = 46% 
Since G = C so G = 46/2 =23% 
1 
4 B. CGTA  
- - - - - - - - - - - - - - - - - - - 
For Visual Impaired Students 
B. 4000 bp/s  
 It completes replication process in 18 minutes i.e. 18x 60 seconds.  
Rate of polymerization = 4.6 x 10
6 
bp/ 18x 60 s  
= 460000/108  
= 4259.1bp/s or approximately 4000 bp/sec Thus, the correct option is B. 
1 
5 C. 
            Suresh                                             Rajesh 
Sickle Cell Anaemia – Autosomal 
linked Recessive trait  
Thalassemia – Autosomal 
Recessive blood disorder  
 
1 
6 A. present in the medium and it binds to the repressor.  1 
7 A. (i) and (ii)  1 
8 C. 5´ UAACGG 3´ 1 
9 B.  CO2  1 
10 D. Rapid divergence of traits among populations inhabiting a given 
geographical area. 
1 
11 A. 1& 5; 5 &1 1 
12 A. Reduction in BOD 1 
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). 
Answer these questions selecting the appropriate option given below:  
A. Both A and R are true and R is the correct explanation of A.  
B. Both A and R are true and R is not the correct explanation of A.  
C. A is true but R is false. 
D.  A is False but R is true. 
13 C. A is true but R is false.  1 
14 C. A is true but R is false.  1 
15 A. Both A and R are true and R is correct explanation of A. 1 
16 C. A is true but R is false. 1 
Section - B 
17 Student to attempt either option A or B. 
A.  
(i) Negative hCG implies no pregnancy (0.5); Placenta. (0.5) 
(ii) Human placental lactogen (hPL), estrogen, progestogens, relaxin  
(any two) (0.5 x 2 = 1) 
OR 
B.  
(i) A sperm induces changes in the zona pellucida membrane on 
contact, blocking entry of other sperms.                                         (1)  
(ii)  Ovum and sperms should be transported simultaneously to the 
ampullary region for fertilization.                                                     (1)  
2 
18 Student to attempt either option A or B. 
A.  
(i) I is point mutation; II is Frame shift                                                   (1) 
(ii) II as more codons are affected;                                                        (0.5) 
It is extremely likely to lead to large-scale changes to polypeptide length 
and chemical composition/ resulting in a non-functional protein that often 
disrupts the biochemical processes of a cell/Incorrect amino acids are 
inserted/ often premature termination occurs when a nonsense codon is 
read/ Frameshifts have very severe phenotypic effects.       (any one) (0.5) 
OR 
2 
Page 3


 
Marking Scheme   
Biology (044) 
Class XII (2024 – 25) 
Q. 
No. 
Answer Marks 
Section - A 
1 B. Both placenta as well as fully developed foetus.  1 
2 B. 2000  
Formation of one seed requires fertilsation between one pollen grain and 
one egg. To produce 1600 seeds, 1600 pollen grains and 1600 eggs will be 
required. Each microspore mother cell results in the formation of 4 pollen 
grains after one cycle of meiotic division. So, 400 meiotic divisions will result 
in the production of 1600 pollen grains. One megaspore mother cell after 
one cycle of meiotic division results in the formation of 1 egg; so, 1600 
meiotic divisions will take place to form 1600 eggs. Thus, total number of 
meiotic divisions required for the formation of 1600 seeds will be 400 + 1600 
=2000. 
1 
3 A. 23% 
According to Chargaff’s rules, in DNA, A =T and G=C; 
Thus, A +T+G+C =100 
Given T = 27% so A=T =27% 
Thus A+T = 27 +27 =54% 
Thus, G+C =100 – 54 = 46% 
Since G = C so G = 46/2 =23% 
1 
4 B. CGTA  
- - - - - - - - - - - - - - - - - - - 
For Visual Impaired Students 
B. 4000 bp/s  
 It completes replication process in 18 minutes i.e. 18x 60 seconds.  
Rate of polymerization = 4.6 x 10
6 
bp/ 18x 60 s  
= 460000/108  
= 4259.1bp/s or approximately 4000 bp/sec Thus, the correct option is B. 
1 
5 C. 
            Suresh                                             Rajesh 
Sickle Cell Anaemia – Autosomal 
linked Recessive trait  
Thalassemia – Autosomal 
Recessive blood disorder  
 
1 
6 A. present in the medium and it binds to the repressor.  1 
7 A. (i) and (ii)  1 
8 C. 5´ UAACGG 3´ 1 
9 B.  CO2  1 
10 D. Rapid divergence of traits among populations inhabiting a given 
geographical area. 
1 
11 A. 1& 5; 5 &1 1 
12 A. Reduction in BOD 1 
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). 
Answer these questions selecting the appropriate option given below:  
A. Both A and R are true and R is the correct explanation of A.  
B. Both A and R are true and R is not the correct explanation of A.  
C. A is true but R is false. 
D.  A is False but R is true. 
13 C. A is true but R is false.  1 
14 C. A is true but R is false.  1 
15 A. Both A and R are true and R is correct explanation of A. 1 
16 C. A is true but R is false. 1 
Section - B 
17 Student to attempt either option A or B. 
A.  
(i) Negative hCG implies no pregnancy (0.5); Placenta. (0.5) 
(ii) Human placental lactogen (hPL), estrogen, progestogens, relaxin  
(any two) (0.5 x 2 = 1) 
OR 
B.  
(i) A sperm induces changes in the zona pellucida membrane on 
contact, blocking entry of other sperms.                                         (1)  
(ii)  Ovum and sperms should be transported simultaneously to the 
ampullary region for fertilization.                                                     (1)  
2 
18 Student to attempt either option A or B. 
A.  
(i) I is point mutation; II is Frame shift                                                   (1) 
(ii) II as more codons are affected;                                                        (0.5) 
It is extremely likely to lead to large-scale changes to polypeptide length 
and chemical composition/ resulting in a non-functional protein that often 
disrupts the biochemical processes of a cell/Incorrect amino acids are 
inserted/ often premature termination occurs when a nonsense codon is 
read/ Frameshifts have very severe phenotypic effects.       (any one) (0.5) 
OR 
2 
B.  
(i) Translational unit in mRNA is the sequence of RNA that is flanked 
by the start codon (AUG) and the stop codon (UAA) and codes for a 
polypeptide/ AUG AUC UCG UAA.                                                (1) 
(ii) Untranslated regions (UTR). The UTRs are present at both 5' -end 
(before start codon) and at 3' -end (after stop codon). They are 
required for an efficient translation process.                                  (1) 
19 A. As the adaptive immune response gears up, there is a reciprocal 
relationship between virus levels in the blood and helper T 
lymphocytes levels. As the level of helper T levels rises, the virus 
levels decline.                                                                                   (1) 
B. Several years later, if untreated, HIV patient will lose the adaptive 
immune response, including the ability to make antibodies, as 
gradually the HIV enters the helper T lymphocytes leading to a 
progressive decrease in the number of helper T lymphocytes.         (1) 
- - - - - - - - - - - - - - - - - - - 
For visually impaired students.  
After getting into the body of the person, the virus enters into 
macrophages where the RNA genome of the virus replicates to form viral 
DNA with the help of the enzyme reverse transcriptase. The viral DNA 
gets incorporated into the host cell’s DNA and directs the infected cells to 
produce virus particles. Macrophages continue to produce virus particles; 
in this way they act like HIV factory.                                                                         (1)  
Simultaneously, HIV enters into helper T-lymphocytes (TH), replicates and 
produces progeny viruses. The progeny virus released in blood attack 
other T lymphocytes leading to a progressive decrease in the number of 
helper T-lymphocytes in the body of the infected person. Due to decrease 
in the number of helper T lymphocytes, the person becomes immuno-
deficient.                                                                                                   (1) 
2 
20 ? The variation in colour of colonies is due to the principle of 
insertional inactivation.                                                              (0.5)                                                                 
? In this, a recombinant DNA is inserted within the coding sequence 
of an enzyme, ß-galactosidase. This results into inactivation of the 
enzyme, which is referred to as insertional inactivation.           (0.5)                                                                           
? The presence of a chromogenic substrate gives blue-coloured 
colonies if the plasmid in the bacteria does not have an insert.   (0.5) 
? Presence of insert results into insertional inactivation of the ß -
galactosidase and the colonies do not produce any colour, these 
are identified as recombinant colonies.                                      (0.5) 
 
2 
Page 4


 
Marking Scheme   
Biology (044) 
Class XII (2024 – 25) 
Q. 
No. 
Answer Marks 
Section - A 
1 B. Both placenta as well as fully developed foetus.  1 
2 B. 2000  
Formation of one seed requires fertilsation between one pollen grain and 
one egg. To produce 1600 seeds, 1600 pollen grains and 1600 eggs will be 
required. Each microspore mother cell results in the formation of 4 pollen 
grains after one cycle of meiotic division. So, 400 meiotic divisions will result 
in the production of 1600 pollen grains. One megaspore mother cell after 
one cycle of meiotic division results in the formation of 1 egg; so, 1600 
meiotic divisions will take place to form 1600 eggs. Thus, total number of 
meiotic divisions required for the formation of 1600 seeds will be 400 + 1600 
=2000. 
1 
3 A. 23% 
According to Chargaff’s rules, in DNA, A =T and G=C; 
Thus, A +T+G+C =100 
Given T = 27% so A=T =27% 
Thus A+T = 27 +27 =54% 
Thus, G+C =100 – 54 = 46% 
Since G = C so G = 46/2 =23% 
1 
4 B. CGTA  
- - - - - - - - - - - - - - - - - - - 
For Visual Impaired Students 
B. 4000 bp/s  
 It completes replication process in 18 minutes i.e. 18x 60 seconds.  
Rate of polymerization = 4.6 x 10
6 
bp/ 18x 60 s  
= 460000/108  
= 4259.1bp/s or approximately 4000 bp/sec Thus, the correct option is B. 
1 
5 C. 
            Suresh                                             Rajesh 
Sickle Cell Anaemia – Autosomal 
linked Recessive trait  
Thalassemia – Autosomal 
Recessive blood disorder  
 
1 
6 A. present in the medium and it binds to the repressor.  1 
7 A. (i) and (ii)  1 
8 C. 5´ UAACGG 3´ 1 
9 B.  CO2  1 
10 D. Rapid divergence of traits among populations inhabiting a given 
geographical area. 
1 
11 A. 1& 5; 5 &1 1 
12 A. Reduction in BOD 1 
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). 
Answer these questions selecting the appropriate option given below:  
A. Both A and R are true and R is the correct explanation of A.  
B. Both A and R are true and R is not the correct explanation of A.  
C. A is true but R is false. 
D.  A is False but R is true. 
13 C. A is true but R is false.  1 
14 C. A is true but R is false.  1 
15 A. Both A and R are true and R is correct explanation of A. 1 
16 C. A is true but R is false. 1 
Section - B 
17 Student to attempt either option A or B. 
A.  
(i) Negative hCG implies no pregnancy (0.5); Placenta. (0.5) 
(ii) Human placental lactogen (hPL), estrogen, progestogens, relaxin  
(any two) (0.5 x 2 = 1) 
OR 
B.  
(i) A sperm induces changes in the zona pellucida membrane on 
contact, blocking entry of other sperms.                                         (1)  
(ii)  Ovum and sperms should be transported simultaneously to the 
ampullary region for fertilization.                                                     (1)  
2 
18 Student to attempt either option A or B. 
A.  
(i) I is point mutation; II is Frame shift                                                   (1) 
(ii) II as more codons are affected;                                                        (0.5) 
It is extremely likely to lead to large-scale changes to polypeptide length 
and chemical composition/ resulting in a non-functional protein that often 
disrupts the biochemical processes of a cell/Incorrect amino acids are 
inserted/ often premature termination occurs when a nonsense codon is 
read/ Frameshifts have very severe phenotypic effects.       (any one) (0.5) 
OR 
2 
B.  
(i) Translational unit in mRNA is the sequence of RNA that is flanked 
by the start codon (AUG) and the stop codon (UAA) and codes for a 
polypeptide/ AUG AUC UCG UAA.                                                (1) 
(ii) Untranslated regions (UTR). The UTRs are present at both 5' -end 
(before start codon) and at 3' -end (after stop codon). They are 
required for an efficient translation process.                                  (1) 
19 A. As the adaptive immune response gears up, there is a reciprocal 
relationship between virus levels in the blood and helper T 
lymphocytes levels. As the level of helper T levels rises, the virus 
levels decline.                                                                                   (1) 
B. Several years later, if untreated, HIV patient will lose the adaptive 
immune response, including the ability to make antibodies, as 
gradually the HIV enters the helper T lymphocytes leading to a 
progressive decrease in the number of helper T lymphocytes.         (1) 
- - - - - - - - - - - - - - - - - - - 
For visually impaired students.  
After getting into the body of the person, the virus enters into 
macrophages where the RNA genome of the virus replicates to form viral 
DNA with the help of the enzyme reverse transcriptase. The viral DNA 
gets incorporated into the host cell’s DNA and directs the infected cells to 
produce virus particles. Macrophages continue to produce virus particles; 
in this way they act like HIV factory.                                                                         (1)  
Simultaneously, HIV enters into helper T-lymphocytes (TH), replicates and 
produces progeny viruses. The progeny virus released in blood attack 
other T lymphocytes leading to a progressive decrease in the number of 
helper T-lymphocytes in the body of the infected person. Due to decrease 
in the number of helper T lymphocytes, the person becomes immuno-
deficient.                                                                                                   (1) 
2 
20 ? The variation in colour of colonies is due to the principle of 
insertional inactivation.                                                              (0.5)                                                                 
? In this, a recombinant DNA is inserted within the coding sequence 
of an enzyme, ß-galactosidase. This results into inactivation of the 
enzyme, which is referred to as insertional inactivation.           (0.5)                                                                           
? The presence of a chromogenic substrate gives blue-coloured 
colonies if the plasmid in the bacteria does not have an insert.   (0.5) 
? Presence of insert results into insertional inactivation of the ß -
galactosidase and the colonies do not produce any colour, these 
are identified as recombinant colonies.                                      (0.5) 
 
2 
21 Student to attempt either option A or B. 
A.  
(i) NPP = GPP – R;  
Given GPP = 400 J/m
2
/day 
R = 150 J/m
2
/day 
       NPP = 400 J/m
2
/day – 150 J/m
2
/day = 250 J/m
2
/day                       (1) 
(ii) Pyramid of energy is always upright. As energy flows from one trophic 
level to the next trophic level some amount of energy is lost in each 
trophic level in the form of heat. Therefore, the pyramid of energy is 
always upright and can never be inverted.                                          (1) 
OR 
B.  
(i) If GPP is equal, then we can manipulate the NPP equation and 
solve.  
? NPP = GPP – Respiration of plants;  
? Respiration of Plants = GPP – NPP.  
? This means that the smallest NPP corresponds to the largest 
respiration. That is forest C.                                                              (1) 
(ii)   
(a) 
 
(0.5) 
(b)  
 (0.5) 
 
2 
Section – C 
22 
A. Seed X- 3 embryos; 1embryo sac; 1ovule;                        (0.5 x 3=1.5) 
B. The nucellar cells grow mitotically and develop into the embryos by 
asexual reproduction.                                                                       (0.5) 
C. The plants growing from seed X will have to share the 
resources/endosperm so there is a possibility of some plant being 
undernourished/; only one plant in seed Y will use the entire endosperm 
for its growth or as the plants of seed X are clones they will not show 
variation and may succumb to environmental stress;/ plants from seed 
Y will have genetic variation and so can show greater adaptability.    (1) 
3 
Page 5


 
Marking Scheme   
Biology (044) 
Class XII (2024 – 25) 
Q. 
No. 
Answer Marks 
Section - A 
1 B. Both placenta as well as fully developed foetus.  1 
2 B. 2000  
Formation of one seed requires fertilsation between one pollen grain and 
one egg. To produce 1600 seeds, 1600 pollen grains and 1600 eggs will be 
required. Each microspore mother cell results in the formation of 4 pollen 
grains after one cycle of meiotic division. So, 400 meiotic divisions will result 
in the production of 1600 pollen grains. One megaspore mother cell after 
one cycle of meiotic division results in the formation of 1 egg; so, 1600 
meiotic divisions will take place to form 1600 eggs. Thus, total number of 
meiotic divisions required for the formation of 1600 seeds will be 400 + 1600 
=2000. 
1 
3 A. 23% 
According to Chargaff’s rules, in DNA, A =T and G=C; 
Thus, A +T+G+C =100 
Given T = 27% so A=T =27% 
Thus A+T = 27 +27 =54% 
Thus, G+C =100 – 54 = 46% 
Since G = C so G = 46/2 =23% 
1 
4 B. CGTA  
- - - - - - - - - - - - - - - - - - - 
For Visual Impaired Students 
B. 4000 bp/s  
 It completes replication process in 18 minutes i.e. 18x 60 seconds.  
Rate of polymerization = 4.6 x 10
6 
bp/ 18x 60 s  
= 460000/108  
= 4259.1bp/s or approximately 4000 bp/sec Thus, the correct option is B. 
1 
5 C. 
            Suresh                                             Rajesh 
Sickle Cell Anaemia – Autosomal 
linked Recessive trait  
Thalassemia – Autosomal 
Recessive blood disorder  
 
1 
6 A. present in the medium and it binds to the repressor.  1 
7 A. (i) and (ii)  1 
8 C. 5´ UAACGG 3´ 1 
9 B.  CO2  1 
10 D. Rapid divergence of traits among populations inhabiting a given 
geographical area. 
1 
11 A. 1& 5; 5 &1 1 
12 A. Reduction in BOD 1 
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). 
Answer these questions selecting the appropriate option given below:  
A. Both A and R are true and R is the correct explanation of A.  
B. Both A and R are true and R is not the correct explanation of A.  
C. A is true but R is false. 
D.  A is False but R is true. 
13 C. A is true but R is false.  1 
14 C. A is true but R is false.  1 
15 A. Both A and R are true and R is correct explanation of A. 1 
16 C. A is true but R is false. 1 
Section - B 
17 Student to attempt either option A or B. 
A.  
(i) Negative hCG implies no pregnancy (0.5); Placenta. (0.5) 
(ii) Human placental lactogen (hPL), estrogen, progestogens, relaxin  
(any two) (0.5 x 2 = 1) 
OR 
B.  
(i) A sperm induces changes in the zona pellucida membrane on 
contact, blocking entry of other sperms.                                         (1)  
(ii)  Ovum and sperms should be transported simultaneously to the 
ampullary region for fertilization.                                                     (1)  
2 
18 Student to attempt either option A or B. 
A.  
(i) I is point mutation; II is Frame shift                                                   (1) 
(ii) II as more codons are affected;                                                        (0.5) 
It is extremely likely to lead to large-scale changes to polypeptide length 
and chemical composition/ resulting in a non-functional protein that often 
disrupts the biochemical processes of a cell/Incorrect amino acids are 
inserted/ often premature termination occurs when a nonsense codon is 
read/ Frameshifts have very severe phenotypic effects.       (any one) (0.5) 
OR 
2 
B.  
(i) Translational unit in mRNA is the sequence of RNA that is flanked 
by the start codon (AUG) and the stop codon (UAA) and codes for a 
polypeptide/ AUG AUC UCG UAA.                                                (1) 
(ii) Untranslated regions (UTR). The UTRs are present at both 5' -end 
(before start codon) and at 3' -end (after stop codon). They are 
required for an efficient translation process.                                  (1) 
19 A. As the adaptive immune response gears up, there is a reciprocal 
relationship between virus levels in the blood and helper T 
lymphocytes levels. As the level of helper T levels rises, the virus 
levels decline.                                                                                   (1) 
B. Several years later, if untreated, HIV patient will lose the adaptive 
immune response, including the ability to make antibodies, as 
gradually the HIV enters the helper T lymphocytes leading to a 
progressive decrease in the number of helper T lymphocytes.         (1) 
- - - - - - - - - - - - - - - - - - - 
For visually impaired students.  
After getting into the body of the person, the virus enters into 
macrophages where the RNA genome of the virus replicates to form viral 
DNA with the help of the enzyme reverse transcriptase. The viral DNA 
gets incorporated into the host cell’s DNA and directs the infected cells to 
produce virus particles. Macrophages continue to produce virus particles; 
in this way they act like HIV factory.                                                                         (1)  
Simultaneously, HIV enters into helper T-lymphocytes (TH), replicates and 
produces progeny viruses. The progeny virus released in blood attack 
other T lymphocytes leading to a progressive decrease in the number of 
helper T-lymphocytes in the body of the infected person. Due to decrease 
in the number of helper T lymphocytes, the person becomes immuno-
deficient.                                                                                                   (1) 
2 
20 ? The variation in colour of colonies is due to the principle of 
insertional inactivation.                                                              (0.5)                                                                 
? In this, a recombinant DNA is inserted within the coding sequence 
of an enzyme, ß-galactosidase. This results into inactivation of the 
enzyme, which is referred to as insertional inactivation.           (0.5)                                                                           
? The presence of a chromogenic substrate gives blue-coloured 
colonies if the plasmid in the bacteria does not have an insert.   (0.5) 
? Presence of insert results into insertional inactivation of the ß -
galactosidase and the colonies do not produce any colour, these 
are identified as recombinant colonies.                                      (0.5) 
 
2 
21 Student to attempt either option A or B. 
A.  
(i) NPP = GPP – R;  
Given GPP = 400 J/m
2
/day 
R = 150 J/m
2
/day 
       NPP = 400 J/m
2
/day – 150 J/m
2
/day = 250 J/m
2
/day                       (1) 
(ii) Pyramid of energy is always upright. As energy flows from one trophic 
level to the next trophic level some amount of energy is lost in each 
trophic level in the form of heat. Therefore, the pyramid of energy is 
always upright and can never be inverted.                                          (1) 
OR 
B.  
(i) If GPP is equal, then we can manipulate the NPP equation and 
solve.  
? NPP = GPP – Respiration of plants;  
? Respiration of Plants = GPP – NPP.  
? This means that the smallest NPP corresponds to the largest 
respiration. That is forest C.                                                              (1) 
(ii)   
(a) 
 
(0.5) 
(b)  
 (0.5) 
 
2 
Section – C 
22 
A. Seed X- 3 embryos; 1embryo sac; 1ovule;                        (0.5 x 3=1.5) 
B. The nucellar cells grow mitotically and develop into the embryos by 
asexual reproduction.                                                                       (0.5) 
C. The plants growing from seed X will have to share the 
resources/endosperm so there is a possibility of some plant being 
undernourished/; only one plant in seed Y will use the entire endosperm 
for its growth or as the plants of seed X are clones they will not show 
variation and may succumb to environmental stress;/ plants from seed 
Y will have genetic variation and so can show greater adaptability.    (1) 
3 
- - - - - - - - - - - - - - - - - - - 
For visually impaired students 
A. Seed X- 3 embryos; 1embryo sac; 1ovule;                        (0.5 x 3=1.5) 
B. The nucellar cells grow mitotically and develop into the embryos by 
asexual reproduction.                                                                       (0.5) 
C. The plants growing from seed X will have to share the 
resources/endosperm so there is a possibility of some plant being 
undernourished/; only one plant in seed Y will use the entire endosperm 
for its growth or as the plants of seed X are clones they will not show 
variation and may succumb to environmental stress;/ plants from seed 
Y will have genetic variation and so can show greater adaptability.      (1) 
23 
? The first meiotic division is completed in the primary oocyte during 
oogenesis.                                                                                         (1) 
? Then primary oocyte undergoes first meiotic division to form a large 
haploid secondary oocyte and a tiny first polar body.                             (1) 
? The primary oocyte comprises of 46 chromosomes, whereas 
secondary oocyte and first polar body have 23 chromosomes each. 
 (1) 
3 
24 
A. During replication, Adenine pairs with thymine in DNA; during 
transcription, adenine pairs with uracil in RNA.                          (0.5+0.5) 
B.  In retrovirus the nucleic acid is RNA and it is used to synthesize DNA; 
the process is called reverse transcription.                              (0.5+0.5) 
C. It is a highly energy-rich process/ or as per the need only the gene 
coding for a specific protein is transcribed.                                         (1) 
- - - - - - - - - - - - - - - - - - - 
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A. During replication, Adenine pairs with thymine in DNA; during 
transcription, adenine pairs with uracil in RNA.                       (0.5+0.5) 
B. In retrovirus the nucleic acid is RNA and it is used to synthesize DNA; 
the process is called reverse transcription.                              (0.5+0.5) 
C. It is a highly energy-rich process/ or as per the need only the gene 
coding for a specific protein is transcribed.                                        (1) 
3 
25 ? isolation of DNA, 
? digestion of DNA by restriction endonucleases, 
? separation of DNA fragments by electrophoresis, 
? transferring (blotting) of separated DNA fragments to synthetic 
membranes, such as nitrocellulose or nylon, 
? hybridisation using labelled VNTR probe, and 
? detection of hybridised DNA fragments by autoradiography.                    
(0.5 x 6 =3) 
3