Past Year Paper - Solutions, Biology (Set - 1, 2 and 3), Delhi, 2016, Class 12, Biology | Additional Study Material for NEET PDF Download

 Page 1


Out-B-16 - 57/1, 2, 3  DPSVK/3
Senior School Certificate Examination
March 2016
Marking Scheme - Biology (Theory)
Expected Answers/Value Points
General Instructions :
The Marking Scheme and mechanics of marking
1 In the marking scheme the marking points are separated by commas, one oblique line
(/) indicates acceptable alternative, two obliques (//) indicate complete acceptable
alternative set of marking points.
2. Any words/phrases given within brackets do not have marks.
3. Allow spelling mistakes unless the misspelt word has another biological meaning.  Ignore
plurals unless otherwise stated in the marking scheme.
4. In any question exclusively on diagram no marks on any description.  But in questions on
descriptions, same value points may be marked on the diagrams as a subsitute.
5. All awarded marks are to be written in the left hand margin at the end of the question or
its part.
6. Place a tick ( ?) in red directly on the key/operative term or idea provided it is in correct
context. Place “Half-tick” ½ wherever there is ½ mark in the marking scheme.  (Do not
place tick indiscriminately just to show that you have read the answer).
7. If no marks are awarded to any part or question put a cross (×) at incorrect value portion
and mark it zero (in words only).
8. Add up ticks or the half ticks for a part of the question, do the calculation if any, and write
the part total or the question total in the left hand margin.
9. Add part totals of the question and write the question total at the end.  Count all the ticks
for the entire question as a recheck and draw a circle around the question total to confirm
correct addition.
10. If parts have been attempted at different places do the totalling at the end of the part
attempted last.
11. If any extra part is attempted or any question is reattempted, score out the last one and
write “extra”.
12. In questions where only a certain number of items are asked evaluate only that many
numbers in sequence as is asked ignoring all the extra ones even if otherwise correct.
13. Transcribe the marks on the cover page.  Add up question totals.  Recheck the script
total by adding up circled marks in the script.
14. Points/answer given in brackets in marking scheme are not so important and may be
ignored for marking.
Confidential
(For Restricted circulation)
28.03.2016
Page 2


Out-B-16 - 57/1, 2, 3  DPSVK/3
Senior School Certificate Examination
March 2016
Marking Scheme - Biology (Theory)
Expected Answers/Value Points
General Instructions :
The Marking Scheme and mechanics of marking
1 In the marking scheme the marking points are separated by commas, one oblique line
(/) indicates acceptable alternative, two obliques (//) indicate complete acceptable
alternative set of marking points.
2. Any words/phrases given within brackets do not have marks.
3. Allow spelling mistakes unless the misspelt word has another biological meaning.  Ignore
plurals unless otherwise stated in the marking scheme.
4. In any question exclusively on diagram no marks on any description.  But in questions on
descriptions, same value points may be marked on the diagrams as a subsitute.
5. All awarded marks are to be written in the left hand margin at the end of the question or
its part.
6. Place a tick ( ?) in red directly on the key/operative term or idea provided it is in correct
context. Place “Half-tick” ½ wherever there is ½ mark in the marking scheme.  (Do not
place tick indiscriminately just to show that you have read the answer).
7. If no marks are awarded to any part or question put a cross (×) at incorrect value portion
and mark it zero (in words only).
8. Add up ticks or the half ticks for a part of the question, do the calculation if any, and write
the part total or the question total in the left hand margin.
9. Add part totals of the question and write the question total at the end.  Count all the ticks
for the entire question as a recheck and draw a circle around the question total to confirm
correct addition.
10. If parts have been attempted at different places do the totalling at the end of the part
attempted last.
11. If any extra part is attempted or any question is reattempted, score out the last one and
write “extra”.
12. In questions where only a certain number of items are asked evaluate only that many
numbers in sequence as is asked ignoring all the extra ones even if otherwise correct.
13. Transcribe the marks on the cover page.  Add up question totals.  Recheck the script
total by adding up circled marks in the script.
14. Points/answer given in brackets in marking scheme are not so important and may be
ignored for marking.
Confidential
(For Restricted circulation)
28.03.2016
Out-B-16 - 57/1, 2, 3  DPSVK/4
Question Paper Code 57/1
SECTION – A
Q. Nos. 1 - 5 are of one marks each
1. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one
reason.
Ans Male Honey bee develops from unfertilized female gamete / unfertilised egg / Parthenogenesis of
female gamete (16 chromosomes), female develops by fertilization / fertilised egg (32
chromosomes) = ½ + ½
[1 Mark]
2. Mention the role of 'genetic mother' in MOET.
Ans Genetic mother is used to produce many eggs / for superovulation // 6-8 eggs (under the influence of
FSH)
[1 Mark]
3. What is biopiracy ?
Ans Biopiracy is the use of bioresources by multinational companies and other organization without
proper authorization/compensation payment to the concern country /organisation.
[1 Mark]
4. Mention two advantages for preferring CNG over diesel as an automobile fuel.
Ans Advantages of CNG-
i) burns efficiently / less unburnt residues.
ii) Cheaper than petrol / diesel.
iii) Causes less pollution.
iv) cannot be adulterated.
v) cannot be siphoned by thieves. (any two) = ½ × 2
[1 Mark]
5. Write the probable differences in eating habits of Homo habilis and Homo erectus.
Ans Homo habilis did not eat meat / vegetarian
Homo erectus ate meat (meat eater ) = ½ × 2
[1 Mark]
SECTION - B
Q Nos. 6-10 are of two marks each
6. A single pea plant in your kitchen garden produces pods with viable seeds, but the
individual papaya plant does not. Explain.
Ans Pea- flowers of pea plants are bisexual , monoecious /  self pollinated (to produce pods with viable
seeds) = ½ + ½
Papaya-Dioecious plant / unisexual plant bearing male and female flowers on seperate plants, unable
to produce viable seeds as there is no cross pollination / it could be a male plant which is unable to
produce fruit and seeds = ½ + ½
[1+1=2 Marks]
Page 3


Out-B-16 - 57/1, 2, 3  DPSVK/3
Senior School Certificate Examination
March 2016
Marking Scheme - Biology (Theory)
Expected Answers/Value Points
General Instructions :
The Marking Scheme and mechanics of marking
1 In the marking scheme the marking points are separated by commas, one oblique line
(/) indicates acceptable alternative, two obliques (//) indicate complete acceptable
alternative set of marking points.
2. Any words/phrases given within brackets do not have marks.
3. Allow spelling mistakes unless the misspelt word has another biological meaning.  Ignore
plurals unless otherwise stated in the marking scheme.
4. In any question exclusively on diagram no marks on any description.  But in questions on
descriptions, same value points may be marked on the diagrams as a subsitute.
5. All awarded marks are to be written in the left hand margin at the end of the question or
its part.
6. Place a tick ( ?) in red directly on the key/operative term or idea provided it is in correct
context. Place “Half-tick” ½ wherever there is ½ mark in the marking scheme.  (Do not
place tick indiscriminately just to show that you have read the answer).
7. If no marks are awarded to any part or question put a cross (×) at incorrect value portion
and mark it zero (in words only).
8. Add up ticks or the half ticks for a part of the question, do the calculation if any, and write
the part total or the question total in the left hand margin.
9. Add part totals of the question and write the question total at the end.  Count all the ticks
for the entire question as a recheck and draw a circle around the question total to confirm
correct addition.
10. If parts have been attempted at different places do the totalling at the end of the part
attempted last.
11. If any extra part is attempted or any question is reattempted, score out the last one and
write “extra”.
12. In questions where only a certain number of items are asked evaluate only that many
numbers in sequence as is asked ignoring all the extra ones even if otherwise correct.
13. Transcribe the marks on the cover page.  Add up question totals.  Recheck the script
total by adding up circled marks in the script.
14. Points/answer given in brackets in marking scheme are not so important and may be
ignored for marking.
Confidential
(For Restricted circulation)
28.03.2016
Out-B-16 - 57/1, 2, 3  DPSVK/4
Question Paper Code 57/1
SECTION – A
Q. Nos. 1 - 5 are of one marks each
1. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one
reason.
Ans Male Honey bee develops from unfertilized female gamete / unfertilised egg / Parthenogenesis of
female gamete (16 chromosomes), female develops by fertilization / fertilised egg (32
chromosomes) = ½ + ½
[1 Mark]
2. Mention the role of 'genetic mother' in MOET.
Ans Genetic mother is used to produce many eggs / for superovulation // 6-8 eggs (under the influence of
FSH)
[1 Mark]
3. What is biopiracy ?
Ans Biopiracy is the use of bioresources by multinational companies and other organization without
proper authorization/compensation payment to the concern country /organisation.
[1 Mark]
4. Mention two advantages for preferring CNG over diesel as an automobile fuel.
Ans Advantages of CNG-
i) burns efficiently / less unburnt residues.
ii) Cheaper than petrol / diesel.
iii) Causes less pollution.
iv) cannot be adulterated.
v) cannot be siphoned by thieves. (any two) = ½ × 2
[1 Mark]
5. Write the probable differences in eating habits of Homo habilis and Homo erectus.
Ans Homo habilis did not eat meat / vegetarian
Homo erectus ate meat (meat eater ) = ½ × 2
[1 Mark]
SECTION - B
Q Nos. 6-10 are of two marks each
6. A single pea plant in your kitchen garden produces pods with viable seeds, but the
individual papaya plant does not. Explain.
Ans Pea- flowers of pea plants are bisexual , monoecious /  self pollinated (to produce pods with viable
seeds) = ½ + ½
Papaya-Dioecious plant / unisexual plant bearing male and female flowers on seperate plants, unable
to produce viable seeds as there is no cross pollination / it could be a male plant which is unable to
produce fruit and seeds = ½ + ½
[1+1=2 Marks]
Out-B-16 - 57/1, 2, 3  DPSVK/5
7. Following are the features of genetic codes. What does each one indicate ?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon.
Ans Stop codon - does not code for any amino acid / terminates the synthesis of polypeptide chain
Unambiguous codon - one codon codes for one amino acid only
Degenerate codon - some amino acid are coded by more than one codon
Universal codon - genetic code is same for all organisms (bacteria to humans) = ½ ×4
[2 Marks]
8. Suggest four important steps to produce a disease resistant plant through conventional
plant breeding technology.
Ans Steps for producing disease resistant plants-
i) Screening of germplasm ( for resistance sources)
ii) Hybridization of selected parents
iii) Selection and evaluation of hybrids
iv) Testing and release of new varieties = ½ ×4
[2 Marks]
9. Name a genus of baculovirus. Why are they considered good biocontrol agents ?
Ans Nucleopolyhedrovirus  = ½
Species specific, narrow spectrum insecticidal application , no negative impact on non target
organisms = ½ ×3
[½ + 1½ = 2 Marks]
10. Explain the relationship between CFC's and Ozone in the stratosphere.
Ans UV rays act on CFC's , release Cl  atom ,which act on ozone to release O
2 
, resulting in ozone layer
depletion / causing ozone  hole = ½ ×4
[2 Marks]
OR
Why are sacred groves highly protected ?
Ans Sacred groves are highly protected - because of religious and cultural traditions , refuges for large
number of rare and threatened plants / ecologically unique and biodiversity rich regions =1+1
[2 Marks]
SECTION - C
Q Nos. 11-22 are of three marks each
11. (a) Name the organic material exine of the pollen grain is made up of. How is this
material advantageous to pollen grain ?
(b) Still it is observed that it does not form a continuous layer around the pollen grain.
Give reason.
(c) How are ‘pollen banks’ useful ?
Page 4


Out-B-16 - 57/1, 2, 3  DPSVK/3
Senior School Certificate Examination
March 2016
Marking Scheme - Biology (Theory)
Expected Answers/Value Points
General Instructions :
The Marking Scheme and mechanics of marking
1 In the marking scheme the marking points are separated by commas, one oblique line
(/) indicates acceptable alternative, two obliques (//) indicate complete acceptable
alternative set of marking points.
2. Any words/phrases given within brackets do not have marks.
3. Allow spelling mistakes unless the misspelt word has another biological meaning.  Ignore
plurals unless otherwise stated in the marking scheme.
4. In any question exclusively on diagram no marks on any description.  But in questions on
descriptions, same value points may be marked on the diagrams as a subsitute.
5. All awarded marks are to be written in the left hand margin at the end of the question or
its part.
6. Place a tick ( ?) in red directly on the key/operative term or idea provided it is in correct
context. Place “Half-tick” ½ wherever there is ½ mark in the marking scheme.  (Do not
place tick indiscriminately just to show that you have read the answer).
7. If no marks are awarded to any part or question put a cross (×) at incorrect value portion
and mark it zero (in words only).
8. Add up ticks or the half ticks for a part of the question, do the calculation if any, and write
the part total or the question total in the left hand margin.
9. Add part totals of the question and write the question total at the end.  Count all the ticks
for the entire question as a recheck and draw a circle around the question total to confirm
correct addition.
10. If parts have been attempted at different places do the totalling at the end of the part
attempted last.
11. If any extra part is attempted or any question is reattempted, score out the last one and
write “extra”.
12. In questions where only a certain number of items are asked evaluate only that many
numbers in sequence as is asked ignoring all the extra ones even if otherwise correct.
13. Transcribe the marks on the cover page.  Add up question totals.  Recheck the script
total by adding up circled marks in the script.
14. Points/answer given in brackets in marking scheme are not so important and may be
ignored for marking.
Confidential
(For Restricted circulation)
28.03.2016
Out-B-16 - 57/1, 2, 3  DPSVK/4
Question Paper Code 57/1
SECTION – A
Q. Nos. 1 - 5 are of one marks each
1. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one
reason.
Ans Male Honey bee develops from unfertilized female gamete / unfertilised egg / Parthenogenesis of
female gamete (16 chromosomes), female develops by fertilization / fertilised egg (32
chromosomes) = ½ + ½
[1 Mark]
2. Mention the role of 'genetic mother' in MOET.
Ans Genetic mother is used to produce many eggs / for superovulation // 6-8 eggs (under the influence of
FSH)
[1 Mark]
3. What is biopiracy ?
Ans Biopiracy is the use of bioresources by multinational companies and other organization without
proper authorization/compensation payment to the concern country /organisation.
[1 Mark]
4. Mention two advantages for preferring CNG over diesel as an automobile fuel.
Ans Advantages of CNG-
i) burns efficiently / less unburnt residues.
ii) Cheaper than petrol / diesel.
iii) Causes less pollution.
iv) cannot be adulterated.
v) cannot be siphoned by thieves. (any two) = ½ × 2
[1 Mark]
5. Write the probable differences in eating habits of Homo habilis and Homo erectus.
Ans Homo habilis did not eat meat / vegetarian
Homo erectus ate meat (meat eater ) = ½ × 2
[1 Mark]
SECTION - B
Q Nos. 6-10 are of two marks each
6. A single pea plant in your kitchen garden produces pods with viable seeds, but the
individual papaya plant does not. Explain.
Ans Pea- flowers of pea plants are bisexual , monoecious /  self pollinated (to produce pods with viable
seeds) = ½ + ½
Papaya-Dioecious plant / unisexual plant bearing male and female flowers on seperate plants, unable
to produce viable seeds as there is no cross pollination / it could be a male plant which is unable to
produce fruit and seeds = ½ + ½
[1+1=2 Marks]
Out-B-16 - 57/1, 2, 3  DPSVK/5
7. Following are the features of genetic codes. What does each one indicate ?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon.
Ans Stop codon - does not code for any amino acid / terminates the synthesis of polypeptide chain
Unambiguous codon - one codon codes for one amino acid only
Degenerate codon - some amino acid are coded by more than one codon
Universal codon - genetic code is same for all organisms (bacteria to humans) = ½ ×4
[2 Marks]
8. Suggest four important steps to produce a disease resistant plant through conventional
plant breeding technology.
Ans Steps for producing disease resistant plants-
i) Screening of germplasm ( for resistance sources)
ii) Hybridization of selected parents
iii) Selection and evaluation of hybrids
iv) Testing and release of new varieties = ½ ×4
[2 Marks]
9. Name a genus of baculovirus. Why are they considered good biocontrol agents ?
Ans Nucleopolyhedrovirus  = ½
Species specific, narrow spectrum insecticidal application , no negative impact on non target
organisms = ½ ×3
[½ + 1½ = 2 Marks]
10. Explain the relationship between CFC's and Ozone in the stratosphere.
Ans UV rays act on CFC's , release Cl  atom ,which act on ozone to release O
2 
, resulting in ozone layer
depletion / causing ozone  hole = ½ ×4
[2 Marks]
OR
Why are sacred groves highly protected ?
Ans Sacred groves are highly protected - because of religious and cultural traditions , refuges for large
number of rare and threatened plants / ecologically unique and biodiversity rich regions =1+1
[2 Marks]
SECTION - C
Q Nos. 11-22 are of three marks each
11. (a) Name the organic material exine of the pollen grain is made up of. How is this
material advantageous to pollen grain ?
(b) Still it is observed that it does not form a continuous layer around the pollen grain.
Give reason.
(c) How are ‘pollen banks’ useful ?
Out-B-16 - 57/1, 2, 3  DPSVK/6
Ans (a) Sporopollenin = ½
Most resistant to high temperature / strong acids / alkali / no enzymes can degrade it (any
one) =  ½
(b) (Germs pores)  to allow pollen tube to emerge out / pollen germination = 1
(c) Helps in storing pollen grains for years / for crop breeding programmes = 1
[3 Marks]
OR
(a) Mention the problems that are taken care of by Reproduction and Child Health Care
programme.
(b) What is amniocentesis and why there is a statutory ban on it ?
Ans (a) Uncontrolled population growth / social evil like sex abuse / sex related crime / STDs (any
two) = ½ × 2
(b) Foetal sex determination tests based on chromosomal pattern in the amniotic fluid / to study
chromosomal abnormalities in the foetus = 1
Banned to legally check  female foeticide =1
[3 Marks]
12. What is a test cross ? How can it decipher the heterozygosity of a plant ?
Ans. - A cross to analyse whether genotype of dominant individual is homozygous or heterozygous
=1
- On crossing with a recessive parent  ,  if   50% of progeny have dominant trait and 50% have
recessive trait then the plant  is said to be heterozygous = 1+1
//
The above value points can be considered with the help of a test cross  = 1 + 1
[3 Marks]
13. (a) What do ‘Y and ‘B’ stand for in ‘YAC’ and ‘BAC’ used in Human Genome Project
(HGP). Mention their role in the project.
(b) Write the percentage of the total human genome that codes for proteins and the
percentage of discovered genes whose functions are known as observed during HGP.
(c) Expand ‘SNPs’ identified by scientists in HGP.
Ans. (a) - Y = Yeast = ½
B = Bacterial = ½
- Used as vector for cloning foreign DNA = ½
(b) (<) 2% , (<) 50% = ½ + ½
(c) Single Nucleotide Polymorphism = ½
[3 Marks]
Page 5


Out-B-16 - 57/1, 2, 3  DPSVK/3
Senior School Certificate Examination
March 2016
Marking Scheme - Biology (Theory)
Expected Answers/Value Points
General Instructions :
The Marking Scheme and mechanics of marking
1 In the marking scheme the marking points are separated by commas, one oblique line
(/) indicates acceptable alternative, two obliques (//) indicate complete acceptable
alternative set of marking points.
2. Any words/phrases given within brackets do not have marks.
3. Allow spelling mistakes unless the misspelt word has another biological meaning.  Ignore
plurals unless otherwise stated in the marking scheme.
4. In any question exclusively on diagram no marks on any description.  But in questions on
descriptions, same value points may be marked on the diagrams as a subsitute.
5. All awarded marks are to be written in the left hand margin at the end of the question or
its part.
6. Place a tick ( ?) in red directly on the key/operative term or idea provided it is in correct
context. Place “Half-tick” ½ wherever there is ½ mark in the marking scheme.  (Do not
place tick indiscriminately just to show that you have read the answer).
7. If no marks are awarded to any part or question put a cross (×) at incorrect value portion
and mark it zero (in words only).
8. Add up ticks or the half ticks for a part of the question, do the calculation if any, and write
the part total or the question total in the left hand margin.
9. Add part totals of the question and write the question total at the end.  Count all the ticks
for the entire question as a recheck and draw a circle around the question total to confirm
correct addition.
10. If parts have been attempted at different places do the totalling at the end of the part
attempted last.
11. If any extra part is attempted or any question is reattempted, score out the last one and
write “extra”.
12. In questions where only a certain number of items are asked evaluate only that many
numbers in sequence as is asked ignoring all the extra ones even if otherwise correct.
13. Transcribe the marks on the cover page.  Add up question totals.  Recheck the script
total by adding up circled marks in the script.
14. Points/answer given in brackets in marking scheme are not so important and may be
ignored for marking.
Confidential
(For Restricted circulation)
28.03.2016
Out-B-16 - 57/1, 2, 3  DPSVK/4
Question Paper Code 57/1
SECTION – A
Q. Nos. 1 - 5 are of one marks each
1. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one
reason.
Ans Male Honey bee develops from unfertilized female gamete / unfertilised egg / Parthenogenesis of
female gamete (16 chromosomes), female develops by fertilization / fertilised egg (32
chromosomes) = ½ + ½
[1 Mark]
2. Mention the role of 'genetic mother' in MOET.
Ans Genetic mother is used to produce many eggs / for superovulation // 6-8 eggs (under the influence of
FSH)
[1 Mark]
3. What is biopiracy ?
Ans Biopiracy is the use of bioresources by multinational companies and other organization without
proper authorization/compensation payment to the concern country /organisation.
[1 Mark]
4. Mention two advantages for preferring CNG over diesel as an automobile fuel.
Ans Advantages of CNG-
i) burns efficiently / less unburnt residues.
ii) Cheaper than petrol / diesel.
iii) Causes less pollution.
iv) cannot be adulterated.
v) cannot be siphoned by thieves. (any two) = ½ × 2
[1 Mark]
5. Write the probable differences in eating habits of Homo habilis and Homo erectus.
Ans Homo habilis did not eat meat / vegetarian
Homo erectus ate meat (meat eater ) = ½ × 2
[1 Mark]
SECTION - B
Q Nos. 6-10 are of two marks each
6. A single pea plant in your kitchen garden produces pods with viable seeds, but the
individual papaya plant does not. Explain.
Ans Pea- flowers of pea plants are bisexual , monoecious /  self pollinated (to produce pods with viable
seeds) = ½ + ½
Papaya-Dioecious plant / unisexual plant bearing male and female flowers on seperate plants, unable
to produce viable seeds as there is no cross pollination / it could be a male plant which is unable to
produce fruit and seeds = ½ + ½
[1+1=2 Marks]
Out-B-16 - 57/1, 2, 3  DPSVK/5
7. Following are the features of genetic codes. What does each one indicate ?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon.
Ans Stop codon - does not code for any amino acid / terminates the synthesis of polypeptide chain
Unambiguous codon - one codon codes for one amino acid only
Degenerate codon - some amino acid are coded by more than one codon
Universal codon - genetic code is same for all organisms (bacteria to humans) = ½ ×4
[2 Marks]
8. Suggest four important steps to produce a disease resistant plant through conventional
plant breeding technology.
Ans Steps for producing disease resistant plants-
i) Screening of germplasm ( for resistance sources)
ii) Hybridization of selected parents
iii) Selection and evaluation of hybrids
iv) Testing and release of new varieties = ½ ×4
[2 Marks]
9. Name a genus of baculovirus. Why are they considered good biocontrol agents ?
Ans Nucleopolyhedrovirus  = ½
Species specific, narrow spectrum insecticidal application , no negative impact on non target
organisms = ½ ×3
[½ + 1½ = 2 Marks]
10. Explain the relationship between CFC's and Ozone in the stratosphere.
Ans UV rays act on CFC's , release Cl  atom ,which act on ozone to release O
2 
, resulting in ozone layer
depletion / causing ozone  hole = ½ ×4
[2 Marks]
OR
Why are sacred groves highly protected ?
Ans Sacred groves are highly protected - because of religious and cultural traditions , refuges for large
number of rare and threatened plants / ecologically unique and biodiversity rich regions =1+1
[2 Marks]
SECTION - C
Q Nos. 11-22 are of three marks each
11. (a) Name the organic material exine of the pollen grain is made up of. How is this
material advantageous to pollen grain ?
(b) Still it is observed that it does not form a continuous layer around the pollen grain.
Give reason.
(c) How are ‘pollen banks’ useful ?
Out-B-16 - 57/1, 2, 3  DPSVK/6
Ans (a) Sporopollenin = ½
Most resistant to high temperature / strong acids / alkali / no enzymes can degrade it (any
one) =  ½
(b) (Germs pores)  to allow pollen tube to emerge out / pollen germination = 1
(c) Helps in storing pollen grains for years / for crop breeding programmes = 1
[3 Marks]
OR
(a) Mention the problems that are taken care of by Reproduction and Child Health Care
programme.
(b) What is amniocentesis and why there is a statutory ban on it ?
Ans (a) Uncontrolled population growth / social evil like sex abuse / sex related crime / STDs (any
two) = ½ × 2
(b) Foetal sex determination tests based on chromosomal pattern in the amniotic fluid / to study
chromosomal abnormalities in the foetus = 1
Banned to legally check  female foeticide =1
[3 Marks]
12. What is a test cross ? How can it decipher the heterozygosity of a plant ?
Ans. - A cross to analyse whether genotype of dominant individual is homozygous or heterozygous
=1
- On crossing with a recessive parent  ,  if   50% of progeny have dominant trait and 50% have
recessive trait then the plant  is said to be heterozygous = 1+1
//
The above value points can be considered with the help of a test cross  = 1 + 1
[3 Marks]
13. (a) What do ‘Y and ‘B’ stand for in ‘YAC’ and ‘BAC’ used in Human Genome Project
(HGP). Mention their role in the project.
(b) Write the percentage of the total human genome that codes for proteins and the
percentage of discovered genes whose functions are known as observed during HGP.
(c) Expand ‘SNPs’ identified by scientists in HGP.
Ans. (a) - Y = Yeast = ½
B = Bacterial = ½
- Used as vector for cloning foreign DNA = ½
(b) (<) 2% , (<) 50% = ½ + ½
(c) Single Nucleotide Polymorphism = ½
[3 Marks]
Out-B-16 - 57/1, 2, 3  DPSVK/7
14. Differentiate between homology and analogy. Give one example of each.
Ans. Homology Analogy
- Organisms having the same structure - Different structures having the same
developed along different directions function ( in different organisms)
due to adaptations / different functions
- Result of divergent evolution - Result of convergent evolution
- Indicates common ancestry - Does not indicate common ancestry
- Anatomically same structures - Anatomically different structures
(Any two difference)  = 1 + 1
Example Example
Forelimbs of whale - bats - cheetah - human // Wings of butterfly and birds //
Thorns of Bougainvillea - tendrils of cucurbits Sweet potato and potato
( Any other correct example) = ½ × 2
[3 Marks]
15. (a) It is generally observed that the children who had suffered from chicken - pox in their
childhood may not contract the same disease in their adulthood. Explain giving
reasons the basis of such an immunity in an individual. Name this kind of immunity.
(b) What are interferons ? Mention their role.
Ans. (a) The first infection of chicken pox produce a primary response and antibodies are generated
against chicken pox virus , subsequent encounter with the same virus elicits a highly intensified
secondary response , due to the memory cells formed during the first encounter , active
immunity = ½ × 4
(b) Proteins secreted by viral infected cells , which protects non infected cells from viral infection
/ when a - interferon is given to cancer patient (it activates immune system) , destroys tumour
= ½  × 2
[3 Marks]
16. (a) Write the two limitations of traditional breeding technique that led to promotion of
micro propagation.
(b) Mention two advantages of micro propagation.
(c) Give two examples where it is commercially adopted.
Ans. (a) Failed to keep pace with demand , failed to provide fast and efficient system of crop
improvement = ½ × 2
(b) Large number of plants can be developed in a short duration / production of genetically
identical plants / somaclones  / healthy plants can be recovered from diseased plants
(Any two) = ½ × 2
(c) Tomato / banana / apple (Any two) = ½ × 2
[3 Marks]
17. (a) How do organic farmers control pests ? Give two examples.
(b) State the difference in their approach from that of conventional pest control methods.
Ans. (a) Natural predation / biological control = 1