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

 Page 1


for-B-16 - 57/2/1, 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


for-B-16 - 57/2/1, 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
for-B-16 - 57/2/1, 2, 3  DPSVK/4
Question Paper Code 57/2/1
SECTION – A
Q. Nos. 1 - 5 are of one marks each
1. Name two animals that exhibit Oestrus cycle.
Ans. cow / sheep / rat / deer / dog / tiger / anyother (correct example) = ½ × 2
[1Mark]
2. What is point mutation? Give one example.
Ans. Arising due to change in a single base pair of DNA , sickle cell anemia = ½ × 2
[1 Mark]
3. Mention one difference to distinguish an exon from an intron.
Ans. Exon : coded / expressed sequence of nucleotides in mRNA , = ½
Intron : Intervening sequence of nucleotides not appearing in processed mRNA = ½
[1 Mark]
4. Suggest a molecular diagnostic procedure that detects HIV in a suspected AIDS patient.
Ans. PCR / ELISA = 1
[1 Mark]
5. What does nature’s carrying capacity for a species indicate ?
Ans. (In nature) a given habitat has enough (limited) resources to support a maximum possible number ,
no further growth in population is possible = ½ + ½
[1 Mark]
SECTION –B
Q. Nos. 6-10 are of two marks each
6. Write the location and functions of Myometrium and Endometrium.
Ans. Myometrium : middle layer of uterus , contractions of the uterus during delivery / child birth / parturition
= ½ + ½
Endometrium : Inner layer of uterus , cyclic changes during menstruation / implantation of embryo
= ½ + ½
[2 Marks]
7. How does a test cross help to determine the genotype of an individual ?
Ans. Individual of unknown genotype crossed with recessive parent , = 1
All dominant in progeny - Homozygosity , dominant to recessive ratio  1:1   in progeny - Heterozygosity
= ½ + ½
[2 Marks]
Page 3


for-B-16 - 57/2/1, 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
for-B-16 - 57/2/1, 2, 3  DPSVK/4
Question Paper Code 57/2/1
SECTION – A
Q. Nos. 1 - 5 are of one marks each
1. Name two animals that exhibit Oestrus cycle.
Ans. cow / sheep / rat / deer / dog / tiger / anyother (correct example) = ½ × 2
[1Mark]
2. What is point mutation? Give one example.
Ans. Arising due to change in a single base pair of DNA , sickle cell anemia = ½ × 2
[1 Mark]
3. Mention one difference to distinguish an exon from an intron.
Ans. Exon : coded / expressed sequence of nucleotides in mRNA , = ½
Intron : Intervening sequence of nucleotides not appearing in processed mRNA = ½
[1 Mark]
4. Suggest a molecular diagnostic procedure that detects HIV in a suspected AIDS patient.
Ans. PCR / ELISA = 1
[1 Mark]
5. What does nature’s carrying capacity for a species indicate ?
Ans. (In nature) a given habitat has enough (limited) resources to support a maximum possible number ,
no further growth in population is possible = ½ + ½
[1 Mark]
SECTION –B
Q. Nos. 6-10 are of two marks each
6. Write the location and functions of Myometrium and Endometrium.
Ans. Myometrium : middle layer of uterus , contractions of the uterus during delivery / child birth / parturition
= ½ + ½
Endometrium : Inner layer of uterus , cyclic changes during menstruation / implantation of embryo
= ½ + ½
[2 Marks]
7. How does a test cross help to determine the genotype of an individual ?
Ans. Individual of unknown genotype crossed with recessive parent , = 1
All dominant in progeny - Homozygosity , dominant to recessive ratio  1:1   in progeny - Heterozygosity
= ½ + ½
[2 Marks]
for-B-16 - 57/2/1, 2, 3  DPSVK/5
OR
Mention two applications of DNA polymorphism.
Ans. Genetic mapping , DNA finger printing = 1 + 1
[2 Marks]
8. What kind of areas are suitable for practicing apiculture? Write the scientific name of  the
variety commonly reared for the purpose.
Ans. (Bee pastures of) wild shrub , fruit orchards , cultivated crop (any two) = ½ + ½
Apis indica = 1
[2 Marks]
9. Suggest four advanced ex-situ methods to conserve threatened biodiversity.
Ans. Cryopreservation , in vitro fertilisation , tissue culture , seed banks = ½ × 4
[2 Marks]
10. Lower BOD of a water body helps reappearance of clean-water organisms. Explain.
Ans. Lowering of BOD results in decreased biodegradable material  reduced microbial decomposition
 oxygen utilisation reduced  more Dissolved Oxygen (DO) available (clean water - organisms
reappear) = ½ × 4
[2 Marks]
SECTION –C
Q. Nos. 11-22 are of three marks each
11.  “Post-industrialization, the population of melanised moth increased in England at the expense
of white-winged moths.” Provide explanations.
Ans. Pre Industrialisation had more white winged moth against grey lichens on tree trunk, industrialisation
led to deposition of soot & smoke on tree bark , making bark of trees dark , against the dark
background white moth could easily be preyed upon , melanised moth could camouflage against
dark bark , increased in number (through reproduction) / natural selection  = ½ × 6
[3 Marks]
12. Why does the ‘insertional   inactivation’ method to detect recombinant DNA is preferred to
‘antibiotic resistance’ procedure?
Ans. The presence of a chromogenic substrate gives blue coloured colonies , in absence  of an insert / in
non-transformants , presence of an insert (in the enzyme site) , results into (insertional inactivation of
the ß-galactosidase) colonies which do not produce colour = ½ × 4
Antibiotic resistance method requires duplicate plating / cumbersome procedure  = 1
[3 Marks]
13. Explain the role of the enzyme EcoRI in recombinant DNA technology.
Ans. EcoRI inspects length of DNA and recognises specific palindromic nucleotide sequence , binds with
DNA , cuts each of the two strands of double helix at specific points = 1 × 3
[3 Marks]
Page 4


for-B-16 - 57/2/1, 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
for-B-16 - 57/2/1, 2, 3  DPSVK/4
Question Paper Code 57/2/1
SECTION – A
Q. Nos. 1 - 5 are of one marks each
1. Name two animals that exhibit Oestrus cycle.
Ans. cow / sheep / rat / deer / dog / tiger / anyother (correct example) = ½ × 2
[1Mark]
2. What is point mutation? Give one example.
Ans. Arising due to change in a single base pair of DNA , sickle cell anemia = ½ × 2
[1 Mark]
3. Mention one difference to distinguish an exon from an intron.
Ans. Exon : coded / expressed sequence of nucleotides in mRNA , = ½
Intron : Intervening sequence of nucleotides not appearing in processed mRNA = ½
[1 Mark]
4. Suggest a molecular diagnostic procedure that detects HIV in a suspected AIDS patient.
Ans. PCR / ELISA = 1
[1 Mark]
5. What does nature’s carrying capacity for a species indicate ?
Ans. (In nature) a given habitat has enough (limited) resources to support a maximum possible number ,
no further growth in population is possible = ½ + ½
[1 Mark]
SECTION –B
Q. Nos. 6-10 are of two marks each
6. Write the location and functions of Myometrium and Endometrium.
Ans. Myometrium : middle layer of uterus , contractions of the uterus during delivery / child birth / parturition
= ½ + ½
Endometrium : Inner layer of uterus , cyclic changes during menstruation / implantation of embryo
= ½ + ½
[2 Marks]
7. How does a test cross help to determine the genotype of an individual ?
Ans. Individual of unknown genotype crossed with recessive parent , = 1
All dominant in progeny - Homozygosity , dominant to recessive ratio  1:1   in progeny - Heterozygosity
= ½ + ½
[2 Marks]
for-B-16 - 57/2/1, 2, 3  DPSVK/5
OR
Mention two applications of DNA polymorphism.
Ans. Genetic mapping , DNA finger printing = 1 + 1
[2 Marks]
8. What kind of areas are suitable for practicing apiculture? Write the scientific name of  the
variety commonly reared for the purpose.
Ans. (Bee pastures of) wild shrub , fruit orchards , cultivated crop (any two) = ½ + ½
Apis indica = 1
[2 Marks]
9. Suggest four advanced ex-situ methods to conserve threatened biodiversity.
Ans. Cryopreservation , in vitro fertilisation , tissue culture , seed banks = ½ × 4
[2 Marks]
10. Lower BOD of a water body helps reappearance of clean-water organisms. Explain.
Ans. Lowering of BOD results in decreased biodegradable material  reduced microbial decomposition
 oxygen utilisation reduced  more Dissolved Oxygen (DO) available (clean water - organisms
reappear) = ½ × 4
[2 Marks]
SECTION –C
Q. Nos. 11-22 are of three marks each
11.  “Post-industrialization, the population of melanised moth increased in England at the expense
of white-winged moths.” Provide explanations.
Ans. Pre Industrialisation had more white winged moth against grey lichens on tree trunk, industrialisation
led to deposition of soot & smoke on tree bark , making bark of trees dark , against the dark
background white moth could easily be preyed upon , melanised moth could camouflage against
dark bark , increased in number (through reproduction) / natural selection  = ½ × 6
[3 Marks]
12. Why does the ‘insertional   inactivation’ method to detect recombinant DNA is preferred to
‘antibiotic resistance’ procedure?
Ans. The presence of a chromogenic substrate gives blue coloured colonies , in absence  of an insert / in
non-transformants , presence of an insert (in the enzyme site) , results into (insertional inactivation of
the ß-galactosidase) colonies which do not produce colour = ½ × 4
Antibiotic resistance method requires duplicate plating / cumbersome procedure  = 1
[3 Marks]
13. Explain the role of the enzyme EcoRI in recombinant DNA technology.
Ans. EcoRI inspects length of DNA and recognises specific palindromic nucleotide sequence , binds with
DNA , cuts each of the two strands of double helix at specific points = 1 × 3
[3 Marks]
for-B-16 - 57/2/1, 2, 3  DPSVK/6
14. Draw a labelled diagram of the embryonic stage that gets implanted in the human uterus.
State the functions of the two parts labelled.
Ans.
inner cell mass (½)
inner cell mass (½)
-Trophoblast - helps in implantation / attachment to endometrium / attachment to uterus = 1
-Inner cell mass - gets differentiated into an embryo = 1
[3 Marks]
15. (a) Draw a labelled sketch of a mature 7-celled,8-nucleate embryo-sac.
(b) Which one of the cell in an embryo-sac produce endosperm after double fertilization?
Ans. (a)
 = ½ × 5 = 2½
(b) Central cell  =  ½
 [3 Marks]
16. Narrowly utilitarian arguments are put forth in support of biodiversity conservation. Explain
the other two arguments that are put forth in support of the same cause.
Ans. - Broadly utilitarian = ½
Ecosystem services - Purify air , cycling of nutrients , habitat for wildlife , pollinating crops , aesthetic
pleasure (any two) = ½ × 2  = 1
Page 5


for-B-16 - 57/2/1, 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
for-B-16 - 57/2/1, 2, 3  DPSVK/4
Question Paper Code 57/2/1
SECTION – A
Q. Nos. 1 - 5 are of one marks each
1. Name two animals that exhibit Oestrus cycle.
Ans. cow / sheep / rat / deer / dog / tiger / anyother (correct example) = ½ × 2
[1Mark]
2. What is point mutation? Give one example.
Ans. Arising due to change in a single base pair of DNA , sickle cell anemia = ½ × 2
[1 Mark]
3. Mention one difference to distinguish an exon from an intron.
Ans. Exon : coded / expressed sequence of nucleotides in mRNA , = ½
Intron : Intervening sequence of nucleotides not appearing in processed mRNA = ½
[1 Mark]
4. Suggest a molecular diagnostic procedure that detects HIV in a suspected AIDS patient.
Ans. PCR / ELISA = 1
[1 Mark]
5. What does nature’s carrying capacity for a species indicate ?
Ans. (In nature) a given habitat has enough (limited) resources to support a maximum possible number ,
no further growth in population is possible = ½ + ½
[1 Mark]
SECTION –B
Q. Nos. 6-10 are of two marks each
6. Write the location and functions of Myometrium and Endometrium.
Ans. Myometrium : middle layer of uterus , contractions of the uterus during delivery / child birth / parturition
= ½ + ½
Endometrium : Inner layer of uterus , cyclic changes during menstruation / implantation of embryo
= ½ + ½
[2 Marks]
7. How does a test cross help to determine the genotype of an individual ?
Ans. Individual of unknown genotype crossed with recessive parent , = 1
All dominant in progeny - Homozygosity , dominant to recessive ratio  1:1   in progeny - Heterozygosity
= ½ + ½
[2 Marks]
for-B-16 - 57/2/1, 2, 3  DPSVK/5
OR
Mention two applications of DNA polymorphism.
Ans. Genetic mapping , DNA finger printing = 1 + 1
[2 Marks]
8. What kind of areas are suitable for practicing apiculture? Write the scientific name of  the
variety commonly reared for the purpose.
Ans. (Bee pastures of) wild shrub , fruit orchards , cultivated crop (any two) = ½ + ½
Apis indica = 1
[2 Marks]
9. Suggest four advanced ex-situ methods to conserve threatened biodiversity.
Ans. Cryopreservation , in vitro fertilisation , tissue culture , seed banks = ½ × 4
[2 Marks]
10. Lower BOD of a water body helps reappearance of clean-water organisms. Explain.
Ans. Lowering of BOD results in decreased biodegradable material  reduced microbial decomposition
 oxygen utilisation reduced  more Dissolved Oxygen (DO) available (clean water - organisms
reappear) = ½ × 4
[2 Marks]
SECTION –C
Q. Nos. 11-22 are of three marks each
11.  “Post-industrialization, the population of melanised moth increased in England at the expense
of white-winged moths.” Provide explanations.
Ans. Pre Industrialisation had more white winged moth against grey lichens on tree trunk, industrialisation
led to deposition of soot & smoke on tree bark , making bark of trees dark , against the dark
background white moth could easily be preyed upon , melanised moth could camouflage against
dark bark , increased in number (through reproduction) / natural selection  = ½ × 6
[3 Marks]
12. Why does the ‘insertional   inactivation’ method to detect recombinant DNA is preferred to
‘antibiotic resistance’ procedure?
Ans. The presence of a chromogenic substrate gives blue coloured colonies , in absence  of an insert / in
non-transformants , presence of an insert (in the enzyme site) , results into (insertional inactivation of
the ß-galactosidase) colonies which do not produce colour = ½ × 4
Antibiotic resistance method requires duplicate plating / cumbersome procedure  = 1
[3 Marks]
13. Explain the role of the enzyme EcoRI in recombinant DNA technology.
Ans. EcoRI inspects length of DNA and recognises specific palindromic nucleotide sequence , binds with
DNA , cuts each of the two strands of double helix at specific points = 1 × 3
[3 Marks]
for-B-16 - 57/2/1, 2, 3  DPSVK/6
14. Draw a labelled diagram of the embryonic stage that gets implanted in the human uterus.
State the functions of the two parts labelled.
Ans.
inner cell mass (½)
inner cell mass (½)
-Trophoblast - helps in implantation / attachment to endometrium / attachment to uterus = 1
-Inner cell mass - gets differentiated into an embryo = 1
[3 Marks]
15. (a) Draw a labelled sketch of a mature 7-celled,8-nucleate embryo-sac.
(b) Which one of the cell in an embryo-sac produce endosperm after double fertilization?
Ans. (a)
 = ½ × 5 = 2½
(b) Central cell  =  ½
 [3 Marks]
16. Narrowly utilitarian arguments are put forth in support of biodiversity conservation. Explain
the other two arguments that are put forth in support of the same cause.
Ans. - Broadly utilitarian = ½
Ecosystem services - Purify air , cycling of nutrients , habitat for wildlife , pollinating crops , aesthetic
pleasure (any two) = ½ × 2  = 1
for-B-16 - 57/2/1, 2, 3  DPSVK/7
- Ethical =  ½
Philosophical / spiritual / moral duty towards future generations = ½ × 2 = 1
(½ + 1 + ½ + 1) [3 Marks]
17. On a visit to a Hill station, one of your friend suddenly become unwell and felt uneasy.
(a) List two symptoms you would look for to term it to be due to allergy.
(b) Explain the response of the body to an allergen.
(c) Name two drugs that can be recommended for immediate relief.
Ans. (a) sneezing , watery eyes, running nose , difficulty in breathing (any two)= ½ + ½
(b) body releases antibodies , IgE type = ½ + ½
(c) Antihistamine , adrenalin , steroids (any two) = ½ +½
[3 Marks]
18. (a) Why did Hershey and Chase use radioactive sulfur and radioactive phosphorus in
their experiment ?
(b) Write the conclusion they arrived at and how.
Ans. (a) In order to label protein coat of virus with radioactive sulfur , label DNA with radioactive
phosphorus = ½ + ½
(b) Bacteria which were infected with viruses having radioactive DNA were found to contain
radioactive DNA later on = ½
Bacteria which were infected with viruses having radioactive protein coat were not found to
contain radioactivity = ½
Conclusion - DNA is the genetic material = 1
[3 Marks]
19. (a)   Explain any two defence mechanisms plants evolved against their predators.
(b)   How does predation differ from parasitism?
Ans. (a) (i) Thorns are (morphological) means of defence = 1
(ii) produce / store chemicals which inhibit digestion / disrupts reproduction / kill //
Calotropis produces highly poisonous cardiac glycosides // plants may produce
chemcials such as nicotine / caffiene / quinine / strychnine / opium are produced as
defence  = 1
(b) Parasitism Predation
- Lives & feed on the host Only feeds on prey
- host specific prudent / not prey specific
- Co-evolve with the host Control / check prey population
(any one difference) = 1
[3 Marks]