General Instructions :

Read the following instructions carefully and follow them :

(i) This question paper contains 33 questions. All questions are compulsory.
(ii) Question paper is divided into five sections Sections A, B, C, D and E.
(iii) Section A questions number 1 to 16 are multiple choice type questions. Each
question carries 1 mark.
(iv) Section B questions number 17 to 21 are very short answer type questions.
Each question carries 2 marks.
(v) Section C questions number 22 to 28 are short answer type questions. Each
question carries 3 marks.
(vi) Section D questions number 29 and 30 are case-based questions. Each
question carries 4 marks. Each question has subparts with internal choice in
one of the subparts.
(vii) Section E questions number 31 to 33 are long answer type questions. Each
question carries 5 marks.
(viii) There is no overall choice. However, an internal choice has been provided in
Sections B, C and D of the question paper. A candidate has to write answer for
only one of the alternatives in such questions.
(ix) Kindly note that there is a separate question paper for Visually Impaired
candidates.
(x) Wherever necessary, neat and properly labelled diagrams should be drawn.

SECTION A

Questions no. 1 to 16 are Multiple Choice Type Questions, carrying 1 mark

each. 16 1=16

1. In which of the following plants are both male and female flowers born on
the same plant and the mode of pollination can be geitonogamy or
xenogamy ?
(A) Papaya (B) Date Palm
(C) Maize (D) Spinach

2. Which one of the following hormones is secreted by the human placenta

that helps in the maintenance of pregnancy ?
(A) Relaxin (B) Human Chorionic Gonadotropin
(C) Oxytocin (D) Human Placental Lactogen

57/1/1-11 3 P.T.O.
3. The periodic abstinence by a couple for family planning should be from :
(A) Day 5 to 10 of menstrual cycle
(B) Day 13 to 15 of menstrual cycle
(C) Day 10 to 17 of menstrual cycle
(D) Day 16 to 20 of menstrual cycle

4. Select the incorrect match from the following :

Human Karyotype Characters

(A) 45 + XX Broad palm with characteristic palm crease
(B) 44 + XXY Overall feminine development

(C) 44 + XO Sterile females as ovaries are rudimentary

(D) 44 + XY Normal male

5. You know that there are twenty different types of naturally occurring
amino acids and four different types of bases in the DNA. A combination
of 3 such bases code for a specific amino acid. If instead there are
96 different amino acids and 12 different bases in the DNA, then the
minimum number of combination of bases required to form a codon is :
(A) 6 (B) 8
(C) 2 (D) 4

6. The type of bond represented by the dotted line in a schematic

polynucleotide chain is :

(A) Hydrogen bond (B) Peptide bond

(C) N-glycosidic linkage (D) Phosphodiester bond

57/1/1-11 5 P.T.O.
7. In which of the following conditions/diseases is there a substantial
increase in the activity of mast cells observed in the human body ?
(A) Typhoid (B) Allergy
(C) Ascariasis (D) AIDS

8. Lactobacillus that sets milk into curd is categorised as :

(A) Cyanobacteria (B) Archaebacteria
(C) Chemosynthetic bacteria (D) Heterotrophic bacteria

9. Which one of the following transgenic animals is being used to test the
safety of the polio vaccine ?
(A) Sheep (B) Goat
(C) Pig (D) Mice

10. Restriction Endonuclease Hind II always cuts DNA molecules at a

particular point by recognising a specific sequence of :
(A) Six base pairs (B) Four base pairs
(C) Seven base pairs (D) Three base pairs

11. The improved trait that is found in the genetically modified transgenic
crop Golden rice is :
(A) High lysine content
(B) Insect resistant
(C) High protein content
(D) High vitamin-A content

12. The rate of formation of new organic matter by consumers, and the
biomass available for consumption of herbivores as well as decomposers
are referred to as :
(A) Gross primary productivity and Net primary productivity
(B) Net primary productivity and Secondary productivity
(C) Secondary productivity and Net primary productivity
(D) Secondary productivity and Gross primary productivity
57/1/1-11 7 P.T.O.
For Questions number 13 to 16, two statements are given one labelled as
Assertion (A) and the other labelled as Reason (R). Select the correct answer to these
questions from the codes (A), (B), (C) and (D) as given below.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the
correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not
the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

13. Assertion (A) : The laws of our country permit legal adoption and it is as
yet, one of the best methods for childless couples looking
for parenthood.
Reason (R) : Emotional, religious and social factors are no deterrents to
the legal adoption of orphaned and destitute children in
India.

14. Assertion (A) : Linked genes do not show dihybrid F2 ratio 9 : 3 : 3 : 1.

Reason (R) : Linked genes do not undergo independent assortment.

15. Assertion (A) : Agrobacterium tumefaciens is a pathogen of several

monocot plants.
Reason (R) : It is able to deliver a piece of DNA known as T-DNA to
transform normal plant cells into a tumor.

16. Assertion (A) : Indian Government has set up an organisation known as

GEAC to decide the validity of GM research.
Reason (R) : Genetic modification of organisms has no effect when such
organisms are introduced in the ecosystem.

57/1/1-11 9 P.T.O.
 SECTION B

1
17. Identify A, B, C and D in the table given below : 4 =2
2
Terms Part of the plant it represents
Pericarp
Cotyledon in seed of grass family
Embryonal axis
Remains of nucellus in a seed

18. Observe the picture given below. Name the naturalist and write the
explanation given by him that evolution of life forms had occurred on the
basis of this example. 2

19. Write the basic steps followed in the Assisted Reproductive Technologies
(ART) programme to help childless couples. Why is it also known as test
tube baby programme ? 2

20. (a) A farmer while working on his farm was bitten by a poisonous
snake. He was rushed to a nearby health centre where the doctor
gave him an injection to save his life.
1
(i) What did the doctor inject and why ? 1
2
1
(ii) Name the kind of immunity provided by this injection.
2
OR
(b) Why do organic farmers not recommend complete eradication of
insect pests ? Explain giving reason. 2

57/1/1-11 11 P.T.O.
21. Study the diagram of a pyramid of biomass given below. Name the two
Name
this type of pyramid and the ecosystem in which it is found. 2

SECTION C

22. Explain the mode of action of contraceptive pills taken by human

females. Mention the schedule to be followed for effective outcome. 3

23. (a) Name and write two characteristics of the type of DNA that forms

the basis of DNA fingerprinting technique. 1+1

1 1
(b) Mention any two applications of this technique. +
2 2

24. (a) Explain the significance of the experiment carried out by

S.L. Miller. Name the scientists whose hypothesis prompted him to

1
carry out this experiment. 2
2

1
(b) How does meteorite analysis favour this hypothesis ?
2

57/1/1-11 13 P.T.O.
25. Identify A, B, C, D, E and F in the table given below : 3

Name of Human Disease Causative Organism Symptoms

Pneumonia Streptococcus
Typhoid High fever, weakness,
headache, stomach
pain
Common Cold Rhino virus
Ringworm Dry scaly lesions on
body parts, redness,
itching
Ascariasis Ascaris
Entamoeba Constipation, cramps,
histolytica stools with mucous
and blood clots

26. In a family, the father, the daughter and the son are colour blind,
whereas the mother has normal vision. Do you think the son and the
daughter have inherited the disease from their father ? Work out a cross
to justify your answer. 3

27. (a) (i) What are transgenic animals ? 1

(ii) Name the first transgenic cow and state its importance. 2
OR
(b) (i) Explain the convention for naming EcoRI. 2

(ii) With the help of an illustration only, show the action of

EcoRI on a DNA Polynucleotide. 1

28. (a) Explain how it is ensured that the orchid Ophrys is pollinated by a
specific species of bee. 2

(b) Describe co-evolution with the help of this example. 1

57/1/1-11 15 P.T.O.
 SECTION D

Questions No. 29 and 30 are case-based questions. Each question has

3 sub-questions with internal choice in one sub-question.

29. Read the following passage and answer the questions that follow. 4

total land area whereas its share of the global species diversity is an
impressive 8·1 per cent ! However, in these estimates of species,
prokaryotes do not figure anywhere.

Biologists are always keen on collecting data with respect to species

diversity observed in different regions of the world. The data collected
based on the survey conducted for species richness of groups of
mammals in three different regions of the world is shown in the bar
graph given below :

(a) Why is the species richness maximum in Region III in the bar
graph ? 1
OR
(a) Why is the species richness minimum in Region I in the bar graph ? 1
(b) Plants and animals do not have uniform diversity in the world but
show rather uneven distribution. Mention what this kind of diversity
is referred to as. 1
(c) Why is it that prokaryotes do not have an estimated number of their
species diversity as seen in plants and animals ? Explain. 2
57/1/1-11 17 P.T.O.
30. Study the diagram given below that shows the steps involved in the
procedure of selecting transformed bacteria and answer the questions
that follow :

1 1
(a) Identify the colony that has got transformed. Justify your answer. +
2 2

(b) What are the sites in a plasmid called where ampicillin and
tetracycline resistance genes are inserted ? State their role in genetic
engineering. 2

(c) Name two enzymes playing an important role in genetic

engineering. 1

OR

(c) State the role of -galactosidase in insertional inactivation. 1

57/1/1-11 19 P.T.O.
 SECTION E

31. (a) (i) Explain the development of male gametophyte in an

1
angiosperm. 2
2
1
(ii) Draw a labelled diagram of a three-celled male gametophyte. 2
2

OR

(b) (i) Draw a diagrammatic sectional view of ovary of human

female and label the following : 2
(1) Blood vessels
(2) Primary follicle
(3) Tertiary follicle
(4) Ovum
(ii) At which stage of life are primary follicles formed in a human
1
female ?
2
(iii) Explain the events (both hormonal and structural) that occur
at the time of ovulation till the onset of the next menstrual
1
cycle. 2
2

32. (a) Stability, as one of the properties of genetic material, was very
evident in one of the very early experiments in genetics. Name the
scientist and describe his experiment. State the conclusion he arrived
at. 5

OR

(b) A tall pea plant bearing violet flowers with unknown genotype is
given. Find the genotype by working out different crosses only by
selfing the plants. Write the genotypic and phenotypic ratios of each
cross shown by you. 5
57/1/1-11 21 P.T.O.
33. (a) (i) Name and explain the property present in normal cells but is
1
lost in cancer cells. 1
2

(ii) All normal human cells have genes that may become
cancerous under certain conditions. Name them and explain
how. 1
(iii) State the role of the following techniques in detection and
diagnosis of cancer :

(1) Biopsy and Histopathology 1

1
(2) Magnetic Resonance Imaging 1
2

OR
(b) Large quantities of sewage are generated every day in cities as well
as in towns and are treated in Sewage Treatment Plants (STPs) to
make them less polluting. Given below is the flow diagram of stages
of STP.
Study the flow diagram and answer the questions that follow :

Primary effluent passed into large aeration tanks

Effluent passed into settling tanks to form sediment

(i) (1) Why is primary effluent passed into large aeration

tanks ? 1
(2) What is the sediment formed, referred to as ? Mention
its significance. 1
(3) Explain the final step in the settling tank before the
treated effluent is released into water bodies. 1
(ii) Name any two organisms commonly used as biofertilisers,
belonging to different kingdoms. Write how each one acts as
a biofertiliser. 2

57/1/1-11 23 P.T.O.
 MARKING SCHEME
Senior Secondary School Examination, 2024
BIOLOGY (Subject Code–044)
[ Paper Code: 57/1/1]
1 (C) / Maize 1 1
2. (B) / Human Chorionic Gonadotropin 1 1
3. (C) / Day 10 to 17 of menstrual cycle. 1 1
4. (B) / 44  XXY– Overall feminine development 1 1
5. (C) / 2 1 1
6. (C) / N–glycosidic linkage 1 1
7. (B) / Allergy 1 1
8. (D) / Heterotrophic bacteria 1 1
9. (D) / Mice 1 1
10. (A) / Six base pairs 1 1
11. (D) / High vitamin – A content 1 1
12. (C) / Secondary productivity and Net primary productivity 1 1
13. (C) / Assertion (A) is true, but Reason (R) is false. 1 1
14. (A) / Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation 1 1
of Assertion (A).
15. (D) / Assertion (A) is false, but Reason (R) is true 1 1
16. (C) / Assertion (A) is true, but reason (R) is false. 1 1
SECTION – B
17. A – Wall of fruit / Wall of ripened ovary ,

B – Scutellum,
C – Radicle / Plumule / epicotyl / hypocotyl, ½ x4
D –Perisperm,
2
18 Name of the Naturalist –Lamarck ½

-Evolution of life forms driven by use and disuse of organs. ½

-He said Giraffes in an attempt to forage leaves on tall trees had to adapt by
½
elongation of their necks.
-They passed on this acquired character of elongated neck to their succeeding generations
½
and slowly over the years all the future generations had long necks.

2
19.
• Ova from wife or female donor and sperm from husband or male donor are
collected , induced to form a zygote under simulated conditions in the ½x3
laboratory (In vitro) (outside body) , Zygote or embryo are transferred into
the female body for development. `
• Test tube baby programme – because initial process is carried out in the laboratory /
in vitro ½ 2

XII_ 044 57/1/1 BIOLOGY # pg. 3

20.
(a)

(i)

-Readymade or preformed antibodies or antitoxins against the snake venom , ½

-A quick immune response is required in this case. 1

(ii) Passive immunity ½

OR

O
(b)The organic farmers hold the view that complete eradication of pests is not only possible
but also undesirable because without them beneficial predatory and parasitic insects which
depend upon them as food or hosts will not survive /

One of the key belief of organic farmers is that biodiversity furthers health. Greater 2
biodiversity leads to more sustainable ecosystem. 2
21.
A – Zooplankton ,
B – Phytoplankton ,

-Inverted pyramid of biomass , ½x4

-Sea Ecosystem/ Aquatic ecosystem 2

SECTION – C

22.
• -Pills contain progestogens or progestogen – estrogen combination. ½

-They inhibit ovulation , and implantation as well as, alter the quality of ½x 3
cervical mucus to prevent or retard the entry of sperms.

• -Pills have to be taken daily for a period of 21 days starting within first five
days of menstrual cycle. ½
+

-After a gap of 7 days it has to be repeated in the same pattern till the female
½ 3
desires to prevent conception.
23.
(a)

-Satellite DNA / Repetitive DNA/ VNTR 1

XII_ 044 57/1/1 BIOLOGY # pg. 4

 -Do not code for any protein , form a large portion of human genome, show high ½+½
degree of polymorphism

(any two points) (ii)Form a large po

(b)

useful in forensic applications , helps in determining population and genetic

diversities ,forms the basis of paternity testing , to study evolution, to trace path ½+½
of hereditary diseases. 3
(any two applications)
24. (a)

• Miller experimentally showed formation of amino acids ,

½
and this proved theory of chemical evolution of life / formation of organic
1
molecules from inorganic molecules.
½+½
• Oparin , Haldane

(b)
Analysis of meteorite content also revealed similar compounds indicating that similar ½
processes are occurring in space. 3
25. A – Fever / chills / cough / headache / greyish blue lips and nails / severe problem in

Respiration ,
B – Salmonella typhi ,
C– Nasal congestion / discharge / sore throat / cough / hoarseness /tiredness,
D – Microsporum / Trichophyton / Epidermophyton ,
E– Internal bleeding / fever / muscular pain / anaemia / blockage of intestinal passage, ½x6
F – Amoebiasis / Amoebic dysentery.
3
26. No, ½

Son inherited disease from the mother and daughter inherited disease from both mother and ½
father.

Mother Father
XCX (1/2 Mark) X XCY (1/2 Mark) ½+½
Normal (Carrier) Colour blind

XII_ 044 57/1/1 BIOLOGY # pg. 5

 1

3
27. (a) (i) Transgenic animals are those which have their DNA manipulated to possess and express a
1
foreign gene.

(ii) First transgenic Cow – Rosie, 1

-Produced human protein enriched milk (24 g/litre) / cow milk containing human alpha
lactalbumin protein is nutritionally more balanced product for human babies than natural 1
cow milk.
OR
(b) (i) In EcoRI (comes from Escherichia coli RY13)
-E represent Genus Escherichia ,
-co represent species coli,
-R represent RY 13 strain,
½x4
-I represent order in which the enzyme were isolated from that strain of bacteria.
(ii)

½ 3

XII_ 044 57/1/1 BIOLOGY # pg. 6

28. (a)

-Orchid Ophrys employs ‘Sexual Deceit’to get pollinated by a species of bee

-one petal of flower resembles female of bee in size, colour and markings
-Male bee attracted and pseudocopulates the flower and gets dusted with pollens.
½ x4
-When same bee ‘pseudocopulates’ with other flower, it transfers the pollens to it.

(b)
If female bee pattern changes during evolution the flower needs to co-evolve to resemble the
female bee to get pollinated.
1 3
SECTION – D
29. (a)

This region is less seasonal with constant and more predictable environment / More solar energy
so higher productivity and higher diversity / it represent tropical lattitudes which remain 1
relatively undisturbed for millions of years and had a long evolutionary time for species
diversification .
OR
(a)
Region I represent temperate region subjected to frequent glaciation and get lesser evolutionary
time for species diversification / has more seasonal with less constant and less predictable
environment which lead to lower specie diversification / have lower solar energy available which 1
reduces productivity and inturn contributes to lesser diversity.
(b) Latitudinal gradient in diversity
1
(c) Conventional taxonomic methods are not suitable for identifying microbial species, and
many species are not culturable under laboratory conditions. 1+1
4
30. (a) Colony 4 is transformed with plasmid containing recombinant DNA, as they will not show ½ +½

resistance towards tetracycline.

(b) Award 2 marks to each student . 2

(c) Restriction endonuclease / ligase / Taq DNA Polymerase 1

OR
(c) Insertional inactivation of gene encoding for ß- galactosidase will lead to colorless 1 4
bacterial colonies (recombinant)

XII_ 044 57/1/1 BIOLOGY # pg. 7

 SECTION – E
31. (a)

(i) Each microspore mother cell in sporogenous tissue undergoes meiosis , forming
microspores tetrad by the process called microsporogenesis , microspores form cluster of
½x5
four cells called microspore tetrad , as the anther matures and dehydrates, each
microspore undergo assymetric division to produce a vegetative cell and a generative
cell.

(ii)

½x5

OR
(b)
(i)

½x4

(ii) At embryonic stage ½

XII_ 044 57/1/1 BIOLOGY # pg. 8

 (iii) Both LH and FSH attain a peak level in mid of menstrual cycle rapid secretion of
½x5
LH leading to maximum level is called LH surge , which induces rupture of graffian
follicle and cause ovulation , ovulatory phase is followed by luteal phase during which
remaining parts of graffian follicle transform as corpus luteum , it produces large
amount of progesterone needed for maintaining endometrium , in absence of fertilization
corpus luteum degenerates which causes disintegration of the endometrium leading to
5
menstruation.
32.
(a)

-Frederick Griffith ½
Took two strains of Streptococcus pneumoniae bacteria and inject them into mice

- R strain – Rough and Non–virulent ½

-S strain – Smooth and virulent (with mucous coat) ½

·
½

·
Conclusion :

• -R–strain bacteria had been transformed by heat killed S – strain. ½

• -‘Transforming principle’ transferred from heat killed S-Strain and enabled R–
strain to synthesise a smooth polysaccharide coat. This must be due to the ½
transfer of the genetic material.

• -Heat which killed bacteria did not destroy some of the properties of genetic
material which shows stability of genetic material. ½

XII_ 044 57/1/1 BIOLOGY # pg. 9

 OR

(b) Tall Pea plant with violet flowers can have 4 types of genotypes

TTVV , TtVV , TtVv , TTVv ½x4

Case I- TTVV  TTVV

↓

Genotype TTVV
½
Phenotype All will be tall and violet. ½
Case II-
Tt VV  Tt VV
↓

Gametes TV tV
TV TTVV Tall Violet TtVV Tall Violet
tV TtVV Tall Violet ttVV Dwarf Violet

Phenotypic ratio Tall Violet : Dwarf Violet

3: 1 ½
Genotypic ratio TTVV:TtVV : ttVV
1 : 2 : 1
½

XII_ 044 57/1/1 BIOLOGY # pg. 10

 Case III-
Tt Vv  Tt Vv
↓
Gametes TV Tv tV tv
TV TTVV TTVv TtVV TtVv
Tall violet Tall Violet Tall Violet Tall Violet
Tv TTVv TTvv TtVv Ttvv
Tall Violet Tall white Tall Violet Tall white
tV TtVV TtVv ttVV ttVv
Tall violet Tall Violet dwarf violet dwarf Violet
Tv TtVv Ttvv ttVv ttvv
Tall violet Tall white dwarf violet dwarf white
Phenotypic ratio - Tall Violet : Dwarf violet:Tall white: dwarf white
½
9 : 3 : 3: 1
Genotypic ratio- TTVV: TTVv:TtVV:TtVv:TTvv:ttVV: ttVv: Ttvv: ttvv
1: 2 : 2 : 4 : 1 :1 :2 :2 :1 ½
Case IV
TTVv X TTVv
↓

Gametes TV Tv
TV TTVV TTVv
Tall Violet Tall Violet
Tv TTVv TTvv
Tall Violet Tall White

Phenotypic ratio- Tall violet Tall white

3 : 1 ½
Genotypic ratio- TTVV : TTVv : TTvv
½
1: 2 : 1

• (Any three self cross shown and explained with genotypic and phenotypic ratio) 5

XII_ 044 57/1/1 BIOLOGY # pg. 11

33. (a)

(i)
-Contact Inhibition is present in normal cells but not in cancer cells, ½
1
-When normal cells come in contact with other cells it inhibits their uncontrolled growth.

(ii)
Cellular oncogenes / Proto-oncogenes , when activated under certain conditions could lead to ½ +½
oncogenic transformation of the cells.
(iii)
(1) Biopsy and histopathology– A piece of suspected tissue cut into thin sections is stained , and ½ +½
examined under microscope by pathologist for increased cell counts.

(2) MRI – detects cancer of internal organs , ½

uses strong magnetic fields and non–ionising radiations to detect pathological and physiological
1
changes in living tissue
OR
(b)
(i)
(1) In aeration tanks there is growth of aerobic microbes and fungi (flocs) that consume major 1
part of organic matter in effluent thus reducing BOD
(2)

• -Activated sludge ½

• -Used as inoculum in aeration tanks. ½

(3) Bacterial flocs are allowed to sediment. ( Activated sludge)

1

(ii)
-Rhizobium (Bacteria), live symbiotically in nodules of roots of leguminous plants and fix
½ +½
atmospheric nitrogen into organic form and provide nitrogen to the plant.

XII_ 044 57/1/1 BIOLOGY # pg. 12

-Glomus (fungi), live in symbiotic association with roots of higher plants and absorb phosphorus
½ +½
from the soil and passes it to plants.
-Cyanobacteria (Anabaena, Nostoc, Oscillatoria),
½ +½
Add organic matter to the soil and increase fertility (Paddy fields)
(Any two can be explained)
5

XII_ 044 57/1/1 BIOLOGY # pg. 13