PREBOARD EXAMINATION 2024-2025

Class-12
(Biology-044/001)
Roll No.: Maximum Marks: 70
Date: Time allowed: 3 hours

(GENERAL INSTRUCTIONS)
Please check that this question paper contains
11printed pages.
[1]. Please check that this question paper contains 33 questions.
[2]. Reading time of 15 minutes is given to read the question paper alone. No writing during this
time.
[3]. All questions are compulsory.
[4]. The question paper has five sections: Section A, Section B, Section C, Section D and Section
E. There are 33 questions in the question paper.
[5]. Section-A has 12 questions of MCQ and 04 Assertion & Reason questions of each 1 mark.
Section-B has 5 questions of 2 marks each. Section-C has 7 questions of 3 marks each and
Section-D has 2 questions of 4 marks each, Section-E has 3 questions of 5 marks each.
[6]. There is no overall choice. However, internal choices have been provided in some
questions. A student has to attempt only one of the alternatives in such questions.
[7]. Wherever necessary, neat and properly labelled diagrams should be drawn.

SECTION-A
Q.No. Question Marks
1 Name the complex mechanism and its hormones involved in 1
parturition.
a. Neuroendocrine, cortisol, estrogen and oxytocin
b. Immunoendocrine, colostrum, oxytoxin and cortisol
c. Cardioendocrine, oxytoxin, relaxin and progestogen
d. Gynoendocrine, estrogen, progestin and relaxin
2 How many meiotic divisions are needed for forming 120 pollen 1
grains in wheat?
a. 30
b. 60
c. 90
d. 120

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3 A DNA molecule is 160 base pairs long. It has 30% Guanine. How 1
many adenine bases are present in this DNA molecule?
a. 32
b. 48
c. 64
d. 96
4 Identify the genetic information flow from the image depicted 1
below and also the scientist who had proposed the same.

a. Central dogma in PCR and Maurice Wilkins

b. Central dogma in Molecular Biology & Francis Crick
c. Central dogma in biotechnology & Rosalind Franklin
d. Central dogma in Ecology & Odum
5 Match List I, II with List III 1

List I List II List III

A Monohybrid i. Snapdragon a. Pleiotropy
cross
B Incomplete ii. Pisum sativum b. Multiple alleles
dominance
C Starch synthesis iii. ABO blood c. 3 :1
group
D Co-dominance iv. Pea seed d. No complete
dominance

a. A-ii-c, B-i-d, C-iv-a, D-iii-b

b. A-iii-c, B-i-d, C-iv-a, D-ii-b
c. A-i-c, B-ii-d, C-iv-a, D-iii-b
d. A-ii-c, B-i-d, C-iii-a, D-iv-b
6 How many nucleotides had reported in bacteriophage ϕ x 174? 1
a. 5386
b.6386
c. 7386
d. 8386

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7 In the following human pedigree, the filled symbols represent the 1
affected individuals. Identify the type of given pedigree

a. Autosomal Dominant
b. Autosomal recessive
c. X-linked dominant
d. Y-linked dominant
8 If the sequence of nitrogen bases of the coding strand of DNA in a 1
transcription unit is: 5’ – ATGAATG – 3’, the sequence of bases in
its RNA transcript would be ____________

a. 5’ – AUGAAUG – 3’
b. 5’ – UACUU AC – 3’
c. 5’ – CAUUCAU – 3’
d. 5’ – GUAAGUA – 3’
9 Which of the following is not a SCP? 1
a. Spirullina sp.
b. Chlorella sp.
c. Baculoviral sp.
d. Saccharomyces sp.
10 In which era the following Gymnosperms such as Cycads, 1
Conifers, Gnetales and Ginkgos were dominated in geological time
scale?

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 a. Paleozoic
b. Mesozoic
c. Cenozoic
d. Proteozoic
11 Identify the type of Restriction enzyme to cut the following 1
nucleotide sequence.

a. EcoRI
b. BamHI
c. HaeIII
d. HindII
12 Select the mismatched pair from the following :- 1
a. Glomus sp. – biofertilizer
b. Penicillium notatum – antibiotic
c. Clostridium butylicum – lactic acid
d. Aspergillus niger – citric acid
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R).
Answerthese 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 Assertion (A): The synergids have special cellular thickenings at 1
the micropylar tip called filiform apparatus.
Reason (R):it play an important role in guiding the pollen tubes
into the synergids
14 Assertion (A): There are at least three RNA polymerases (I-III) in 1
the eukaryotic nucleus.
Reason (R): The single DNA-dependent RNA polymerase that
catalyses all types of bacterial transcription.

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15 Assertion(A): Cyclosporin-A is a bioactive molecule and used for 1
organ-transplant patients.
Reason (R):It is produced by Monascus purpureus.
16 Assertion(A): rop is a gene found on Ti-plasmid. 1
Reason (R):rop codes for the proteins involved in the replication
of the plasmid.
SECTION-B
17 Attempt either option A or B 2
A.
Study the image given below and answer the following:-

i. Find the labelled parts of A to D.

ii. . Name the hormone which acts on C and stimulates secretion of
some factors which help in the process of spermiogenesis.
OR
B.
The process of formation of male gametes is called
spermatogenesis.
i. Name the hormones involved in it.
ii. What do you mean by spermiation?
18 Attempt either option A or B 2
A.
How can you determine genotype from individual dominate
phenotype? Vv or vv (Flower colour in Pisum sativum).
OR

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 B.
If E.coli was allowed to grow for 80 minutes then what would be
the proportions of light and hybrid densities DNA molecule?
19 Identify the chemical structure of A and B in the following and 2
mention any one ill effect of both identified structures:-
A B

20 Agrobacterium tumefaciens is referred to as Natural Genetic 2

Engineer of plant. Why?
21 Attempt either option A or B 2
A.
i. Construct a grazing food chain and detritus food chain.
ii. Give any one difference between them.
OR
B.Construct an ideal pyramid of energy when 1,000,000 Joules of
sunlight is available. Mention it for four trophic levels.
SECTION-C
22 This is an image of Commelina sp. bearing flowers. 3

i. What is the significance of the flowers borne on this plant?

ii. Find out Flower 1 and 2.
iii. Write one advantage and disadvantage of Flower 2

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23 i. Diagrammatic presentation of ovarian events during a menstrual 3
cycle is given below. Identify A,B,C and D.

ii. What are ovarian and pituitary hormones?

24 List in tabular column, any three differences between prokaryotic 3
and eukaryotic transcription.
25 Outline the sex determination in birds OR honeybees. Support your 3
answer with flow-chart.
26 What does IPM stand for? How does the programme help in 3
controlling pest in the cultivated field?
27 Expand PCR. Point any two of its applications in medicinal field. 3
28 i. What do you mean by mutualism? Give an example. 3
ii. Give an significance of the Mediterranean orchid Ophry sp.
SECTION-D
29 Study the figures given below and answer the questions. 4

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 A. Find out the types of natural selection (A, B and C).
B. Under the influence of which type of natural selection would
graph become like graph C?
Attempt either option C or D
C. What could be the likely reason of new variation arising in
the population?
OR
D. Who was the founder of the modern synthetic theory of
evolution? Why?
30 The next 90 days: India’s ambitious attempt to vaccinate all adults 4
(Health Express Report, OCT 05, 2021). Almost three-fourths of
the elderly population in India has already gotten at least one dose,
and the coverage in the 45-59 age group was even higher. Overall,
more than one-third of the 45+ aged citizens in India are fully
vaccinated.
From the early days of the Liberalized Pricing and Accelerated
National COVID-19 Vaccination Strategy in May 2021, when the
daily vaccine doses administered were around 2 million, India’s
vaccination drive has come a long way with the month of
September witnessing an average of over 7.5 million doses a day.
An analysis of monthly vaccination numbers from the start of the
vaccination drive reveals a substantial scale up.
India’s Covid-19 Vaccination Drive
(Average daily doses administered – 100,000s)

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 Answer the following questions based on your concept on
vaccination and the above mentioned graph.

A. Identify the second highest month and year successful

vaccination in India?
B. Mention the challenges faced worldwide while in COVID-19
Vaccine preparation. Support your answer with 2 reasons.
Attempt either option C or D
C. Give one point of difference between active and passive
immunity
OR
D. Why antitoxins are called performed antibodies?
SECTION-E
31 Attempt either option A or B 5
A.
The diagram below shows sectional view of the female
reproductive system. Observe carefully and answer the following:-

i. Identify the labelled parts 1 to 5.

ii. When and where are primary oocytes formed in human female?
iii. Trace the development of these oocytes till ovulation
iv. How do gonadotropins influence this development?
v. Draw a labelled diagram of Graffian follicle.
OR

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 B.
i. A matured typical angiosperm embryo sac at maturity, though 8-
nucleate is 7-celled. Justify.
ii. Draw a diagram of an angiospermic embryo sac where
fertilisation is just completed and label the following parts:-
a. micropylar end of the embryo sac
b. the part that develop into an embryo
c. the part that develop into an endosperm
d. the degenerating cells at the chalazal end
32 Attempt either option A or B 5

A.
i. Identify the technology from the depicted image given below and
briefly explain about its differentiated cells:

ii. Discuss about its various kinds of cell used in this technology
iii. List down any 4 applications of this technology.
OR
B.
i. What is biopiracy? State the initiative taken by the Indian
parliament towards it.
ii. Briefly explain any 2 controversies in India regarding Patent and
Biopiracy.
iii. Expand: WTO and TRIPS.

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33 Attempt either option A or B 5

A
i. What are the two types of desirable approaches to conserve
biodiversity? Explain with examples bringing out the difference
between the two types.
ii. India has more than 50000 strains of rice. Mention the level of
biodiversity it represents.
iii. Define biopropecting.
OR
B
i. Explain Species-Area relationship with suitable graphical
representation.
ii. Give the normal range of Z-value and also mention for
frugivorous birds and mammals in the tropical forest on different
continents.

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 PREBOARD EXAMINATION 2024-2025
Class-12
(Biology-044/001)
MARKING SCHEME

GENERAL INSTRUCTIONS:

1. The Marking scheme carries only suggested value points for the answers, the due marks
to be awarded to the relevant answers.
2. ‘/’ - indicate the alternate point. The evaluator should consider accordingly.
3. If a student has attempted an extra question, answer of the question deserving more
marksshould be retained and the other answer scored out with a note “Extra Question”
4. Evaluators are asked to follow the BOARD EVALUATION POLICYfor an effective
evaluation.
5. A full scale of marks 0 to 70 has to be used. Please do not hesitate to award full
marks if the answer deserves it.

Q.No. Expected Answer/ Value Points Marks Total

marks
SECTION-A
1 a. / Neuroendocrine, cortisol, estrogen and oxytocin 1 1
2 a. /30 1 1
3 a. /32 1 1
4 b./ Central dogma in Molecular Biology & Francis Crick 1 1
5 a. /A-ii-c, B-i-d, C-iv-a, D-iii-b 1 1
6 a./ 5386 1 1
7 b./ Autosomal recessive 1 1
8 a. /5’ – AUGAAUG – 3’ 1 1
9 c. /Baculoviral sp. 1 1
10 b./ Mesozoic 1 1
11 a./ EcoRI 1 1
12 c./ Clostridium butylicum – lactic acid 1 1
13 A. /Both A and R are true and R is the correct explanation 1 1
of A
14 B./ Both A and R are true and R is not the correct 1 1
explanation of A
15 C. /A is true but R is false 1 1
16 D. /A is False but R is true 1 1

1
 SECTION-B
17 A 1+1 2
i.
A- Spermatid
B- primary spermatocyte
C- Sertoli cell
D- Spermatogonia ……………………. 4 x ¼ = 1 Mark
ii. FSH / Follicle stimulating hormone (any one) …..1
Mark

OR OR OR
B
i. 1+1 2
GnRH / Gonadotropin Releasing Hormone (any one) – 1
Mark
//
LH acts on Leydig cell (androgen) & FSH on Sertoli cell
…. ½ + ½ = 1 Mark
ii.
After spermiogenesis, sperms heads become embedded in
the Sertoli cells and are finally released from the
seminiferous tubules / relevant definition ……. 1 Mark
18 A 1+1 2
Test cross / F1 individual crossed with recessive parents
…… 1 mark
Used to predict the genotype of the test organism / genetic
cross with ratio (1:1) ….. 1 mark
OR OR OR
B
I generation …. 2 DNA strands ( All 2 hybrids)
II generation ….. 4 Strands (2 hybrid DNA strands and 2
Light DNA strands) 1+1 2
III generation ….. 8 Strands (2 hybrid DNA strands and 6
Light DNA strands)
IV generation ….. 16 Strands (2 hybrid DNA strands and
214Light DNA strands). Its proportion 12.5 % Hybrid
DNA and 87.5% Light DNA) / any related calculation with
appropriate proportion …. 1 + 1 = 2 Marks
19 A- Morphine ….. ½ Mark 4x½ 2
affects CNS / gastrointestinal tract / slow down body
2
 function (any one) ……. ½ mark
B- Cannabinoid……. ½ mark
Principally affects in the brain / damage to cardiovascular
system (any one)……. ½ mark
20  Their capability to transfer T-DNA of their plasmid 1+1 2
into desired plant genome at wound site.
 It leads of the formation of crown gall tumor i.e.
unorganized growth of cells.
 Able to deliver a piece of DNA known as T-DNA
 Tumor inducing (Ti) plasmid used as cloning vector
in many dicot plants
 Helps to delivery gene of interest into a variety of
plants. (any two relevant points)
21 A. 4x½ 2
i. Grazing food chain (GFC)

,…………. ½ mark
Detritus food chain (DFC)
Decaying
Plant matter Earthworm Bird Snake … ½ mark
ii. Any one corresponding difference …. ½ + ½ Mark
S.No. GFC DFC
1. It starts from the living It starts form the dead
green plant organic matter or
decomposers called
detritivores
2. Energy is derived from Energy comes from
sun organic matter or
detritus
3. It may be shorter or It is usually smaller as
longer food chain compared to GFC

OR OR OR
B 1+1 2
Sun
1000000 J

Trophic level/TL-1 TL-2 TL-3 TL-4

3
 (10000 J) ( 1000 J) (100 J) (10 J)
Food Chain construction ….. 1 Mark

1% Law: Producer receiving energy from ultimate sunlight

10% Law: Transfer of 10% of energy from one trophic
level to next and subsequent …. Explanation ….1 Mark
// relevant explanation with food chain construction ….1+ 1
SECTION-C
22 i. produces two different types of flowers ( regular flower 1 3
born at shoot system and another flower at root)
ii. Flower-1 Chasmogamous flower and Flower -2 ½+½
Cleistogamous flower
iii. Advantage – Produce assure seedset even in the absent ½+½
of pollinator and disadvantage- flowers are invariably
autogamous / no chance of cross-pollen landing on the
stigma / no variation (any one point in each)
23 i. 4x½ 3
A- developing follicle
B- mature follicle
C-developing corpus luteum
D- regressing corpus luteum
ii. ½+½
Ovarian hormone: Estrogen and progesterone
Pituitary Hormone: FSH and LS
24 Any three corresponding differences 6x½ 3

S.No. Prokaryotic Eukaryotic

transcription transcription
1. It occurs in cytoplasm It occur is nucleus
2. A single RNA RNA polymerase I, II
polymerase involved and III involved
3. No post-transcriptional Additional
process occurs complexities occurs
in post –
transcriptional
process
4. Transcriptional unit Transcriptional unit
(structural gene) could (structural gene)
be said as polycistronic could be said as

4
 monocistronic
5. Sigma and rho factors Complex initiation,
are involved elongation and
termination factors
are involved
25 a. Sex determination in birds – ZZ-ZW mechanism 1½ +1½ 3

b. Sex termination in Honey bee – haplodiploid

mechanism

Two relevant flowchart or suitable explanation 1½ +1½

26  Integrated pest management 1+1+1 3
 Beneficial insects are being conserved to aid

5
  Ecological sensitive area is being treated
 Baculovirus / Nucleopolyhedrovirus used as
biological control agents / any relevant points

27 PCR- Polymerase chain reaction 1+2 3

Any two applications
 Pathogen deduction at low concentration
 Powerful technique to identify many other genetic
disorders
 Routinely used to detect HIV in suspected AIDS
patients
 To deduct mutations in genes in suspected cancer
patients
28 i.
The interaction confers benefits on both the interacting 1+1 3
species.
Example: Mutualistic relationship between a fungus and
photosynthetic algae or cyanobacteria / mycorrhizae are
association between fungi and the roots of higher roots (any
one example)
ii. 1
employs sexual deceit to get pollination done by a species
of bee / one petal of its orchid flower resemblance of the
female bee in size, colour and marking/pseudocopulation
will occur by male be and pollen transfer takes place (any
one relevant point)
SECTION-D
29 A. 2 4
A- Stabilizing Natural selection
B-Directional Natural selection
C- Disruptive Natural selection
B. More individuals acquire peripheral character value at 1
both ends of the distribution curve / both extreme traits on
either end of the spectrum are favoured over intermediate
traits / graph shows bimodal distribution (any one point /
relevant point)
C-Mutation/genetic recombination/natural selection/genetic ½+½
drift/ gene migration / gene flow (any two points)
contribute to new variation arising in the population

6
 OR OR
D- Charles Darwin and to explain genetic basis of
evolution / explain how evolution can lead to the
emergency of new species (any one point) ½+½
30 A- 79.08 % and September 2021 ½+½ 4
B- Emergency of novel sub-type/antigenic property 1+1
change/ time factor/mutated strain/ issues of vaccine
distribution/ conducting preclinical trials (any two points /
relevant points)
C- Any one corresponding difference ½+½
S.No. Active immunity Passive immunity
1 It is produced due to It is produced due to
the contact with antibodies obtained
pathogen from outside the body
2 It is a time taking It develops immediately
immunity with no
immediate effect
3 It sustains for the It may be only for few
entire life of an days to months
organism
4 No side effects are There are few side
shown by the body effects that prevail up
of an organism to the destruction of
pathogens

OR OR
D- They are pre-formed antibodies that are injected into a 1
person / to provide temporary projection against a toxin
(any one point)

SECTION-E
31 A
i.
1-Fallopian tube / Ampulla 1 5
2-Ovary
3-Cervix
4-Uterine cavity
d- Vagina
ii. Embryonic development of foetus & primary oocytes are ½+½

7
 formed within each fetal ovary, these cells start division
and enter into prophase-I of meiotic division and get
temporarily arrested at the stage / relevant points
iii. oogonia to primary oocyte(by mitosis) to secondary 1
oocyte (by meiosis) / flow chart
iv. Gonadotropins hormone (LH and FSH) increases 1
gradually during follicular phase / FSH stimulate oocyte
growth / LH promotes oocyte maturation and ovulation
(any one point)
iii. Labelled (one label of Antrum) diagram of Graffian ½+½
follicle – Refer NCERT textbook Page.No.33, Fig. No. 2.7-
only Graffian follicle diagram(not necessary to draw entire
sectional view of ovary)

OR OR OR

B.
i. Six of the eight nuclei are enclosed by cell walls, 2 5
whereas, the remaining two nuclei (polar nuclei) are
located in the central cell. Hence, a typical angiosperm
embryo sac at maturity is 8-nucleate and 7-celled.

ii. Diagram (Refer NCERT textbook, Fig. No.1.8-c-A 2

diagrammatic representation of the mature embryo sac) – 2
marks and 4 labellings …1 mark 4x¼
32 A.
i. Stem cell technology / process of stem cells & A cell has ½ + ½ 5
ability of self renewal and are specialized cells / used in
developing different human tissues.
ii. three kinds of stem cells-Totipotent, Pluripotent, 2
Multipotent with explanation and examples
iii. any four applications (useful in treatment disease / cures 4 x ½
genetics diseases / tissue repairing/ adult bone marrow are
injected in heart arteries to improve cardiac functions/
injection of stem cells reduces pancreatic cancer/any
relevant 4 points)
OR OR OR
B. 5
i. 1
Biopiracy is the term used to refer to the use of bio-
8
 resources by multinational companies and other
organisations without proper authorization from the
countries and people concerned without compensatory
payment.
& 1
Indian parliament has cleared the second amendment of the
Indian patent bills that takes such issues into consideration,
including patent terms emergency provisions and research
and development initiative. 1+1
ii. controversies in India regarding patent and biopiracy-
Neem/Turmeric/Basmati Rice to be explained ½+½
iii. WTO-World Trade organisation & TRIPS – Trade
related aspects of Intellectual Property Rights)
33 A.
i. in-situ conservation and ex-site conservation ½+½ 5
any two differences
S.No. In-situ conservation Ex-situ conservation 1+1
1. On-site / conservation Off-site /conservation
and protection of of selected rare plants
biodiversity in its animals outside their
natural habitat. habitat.
2. It conserve in natural Man-made habitat
habitat
3. It is not suitable in all High risk of extinction
event of a rapid or endangered species
decline in the number (desirable approach)
of species.
( any two corresponding differences- 4 x ½ = 2 Marks )
ii. Genetic diversity 1
iii. Exploring molecular, genetic and species-level diversity 1
for products of economic importance
OR OR OR
B. 2+2 5
i. Species-Area relationship
 It was observed by Alexander von Humboldt that
within a region species richness increased with
increasing explored area, but only up to a limit.
 On a logarithmic scale, the relationship between
species richness and area for a wide variety of taxa

9
 (angiosperm plants, birds, bats, freshwater fishes) is
a straight line described by the following
equation:log S = log C + Z log A (where,S = Species
richness A=Area, Z = slope of the line (regression
coefficient)&C = Y-intercept)
 In this equation, the slope of a regression line (Z)
represents the rate of change in species richness as
area changes.
 It is very important for determining species-area
relationship.
 On analyzing species-area relationship, regardless of
the taxonomic group or the region, the value of Z lies
in the range of 0.1 to 0.2 for smaller area whereas Z
ranges from 0.6 to 1.2 for large areas.
 It signifies that species richness increases with
increasing area of observation.
 In larger areas, the slope of regression coefficient
becomes steeper.
 Steeper slopes signifies that number of species found
increases faster than the area explored in very large
areas like the entire continents

½
Formula and Graph - 2 (1 + 1) Marks and any 4 ½
Explanation points- 4 x ½ = 2 Marks
ii. normal range of Z-value : 0.1 and 0.2
Z- Value of frugivorous birds and mammals in the
tropical forest on different continents

10