CBSE Class 12 Question Paper 2026 Solution Biology

FOR CBSE CLASS 12 EXAM PREPARATION

CBSE Class 12 2026
Question Paper
Solution · Biology
EXAM YEAR TYPE SUBJECT

CBSE Class 12 2026 Question Paper Solution Biology

Notes · Sample Papers · Previous Year Papers · Mock Tests

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Strictly Confidential
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(For Internal and Restricted use only)
Senior Secondary School Examination, 2026 (XIIth) a
SUBJECT NAME : Biology (Q.P. CODE 044/57-1-1)

General Instructions: -
1 The CBSE has decided to introduce On Screen Marking (OSM) for the evaluation of
Class XII answer Book with the 2026 Examination.
2 You are aware that evaluation is the most important process in the actual and correct
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assessment of the candidates. A small mistake in evaluation may lead to serious
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problems which may affect the future of the candidates, education system and

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teaching profession. To avoid mistakes, it is requested that before starting evaluation,
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you must read and understand the spot evaluation guidelines carefully.
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“Evaluation policy is a confidential policy as it is related to the confidentiality of
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the examinations conducted, evaluation done and several other aspects. Its
leakage to public in any manner could lead to derailment of the examination
system and affect the life and future of millions of candidates. Sharing this
policy/document to anyone, publishing in any magazine and printing in
Newspaper/Website, etc. may invite action under various rules of the Board and
IPC.”
4 Evaluation is to be done as per instructions provided in the Marking Scheme. It should
not be done according to one’s own interpretation or any other consideration.
Marking Scheme should be strictly adhered to and religiously followed. However,
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while evaluating, answers which are based on latest information or knowledge

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and/or are innovative, they may be assessed for their correctness otherwise and
due marks be awarded to them. In Class-XII, while evaluating two competency-
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based questions, please try to understand given answer and even if reply is not

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from marking scheme but correct competency is enumerated by the candidate,

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due marks should be awarded. a
The Marking scheme carries only suggested value points for the answers.
These are in the nature of Guidelines only and do not constitute the complete answer.
The students can have their own expression and if the expression is correct, the due
marks should be awarded accordingly.
6 The Head-Examiner must go through the first five answer books evaluated by each
evaluator on the first day, to ensure that evaluation has been carried out as per the
instructions given in the Marking Scheme. If there is any variation, the same should
be zero after deliberation and discussion. The remaining answer books meant for
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evaluation shall be given only after ensuring that there is no significant variation in

c. o7 the marking of individual evaluators.
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Evaluators will mark ( √ ) wherever answer is correct. For wrong answer CROSS ‘X’ e m
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be marked. Evaluators will not put right (✓) while evaluating which gives an
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impression that answer is correct and no marks are awarded. This is most common
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mistake which evaluators are committing.
If a question has parts, please award marks on the right-hand side for each part in the
OSM Portal. Marks awarded for different parts of the question will be totaled up by
the OSM System.
9 If a question does not have any parts, marks must be awarded in the left-hand margin
in the OSM Portal. This may also be followed strictly.

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10 No marks to be deducted for the cumulative effect of an error. It should be penalized
only once.
11 A full scale of marks 0 to 70 marks has to be used. Please do not hesitate to award
full marks if the answer deserves it.

12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8
hours every day and evaluate 20 answer books per day in main subjects and 25 answer
books per day in other subjects (Details are given in Spot Guidelines).This is in view
of the reduced syllabus and number of questions in question paper.
13 Ensure that you do not make the following common types of errors committed by the
Examiner in the past :-
● Answers marked as correct, but marks not awarded. (Ensure that the right tick
mark is correctly and clearly indicated. It should merely be a line. Same is with
the X for incorrect answer.)
● Half or a part of answer marked correct and the rest as wrong, but no marks
awarded.
14 While evaluating the answer books if the answer is found to be totally incorrect, it
should be marked as cross (X) and awarded zero (0) Marks.
15 The Examiners should acquaint themselves with the guidelines given in the
“Guidelines for Spot Evaluation” before starting the actual evaluation.
16 The candidates are entitled to obtain photocopy of the Answer Book on request on
payment of the prescribed processing fee. All Examiners/Additional Head
Examiners/Head Examiners are once again reminded that they must ensure that
evaluation is carried out strictly as per value points for each answer as given in the
Marking Scheme.
17 If a candidate attempts both alternatives/options in a question where only one
option/ alternative is required to be attempted, the Evaluator shall award marks
in both the options. The system will take the higher of two scores and disregard
the other response.
18 In a question having two options/alternatives, if a candidate has attempted only
one, then the evaluator shall mark “NA” (Not attempted) against the option that
has not been attempted by the candidate.

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MARKING SCHEME
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Biology (Subject Code-044)
(PAPER CODE : 57/1/1) (26-01-44N)

Q.No. EXPECTED OUTCOMES/VALUE POINTS Marks Total
Marks
SECTION – A
1. (D) / Papaya 1 1

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2. (C) / Red Blood Cells 1 1
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3. (B) /
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(i) and (iii) are correct 1
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4. (B) /
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Genetically different strains of rice in India – less than 1000
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5.
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(A) / Tomato 1 a 1
6. (A) a
/ Stratification 1 1
7. (A) / Jaintia hills 1 1
8. (B) / Independent assortment of genes 1 1
9. (C) / a – v, b – iv, c – ii, d – i 1 1
10. (D) / Lamarck 1 1
11.
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(B) / Their replication is controlled by chromosomal DNA 1 1
12. (B) / Ligaments
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mtrue and Reason is the correct
(A) / Both Assertion (A) and Reason (R) eare
explanation for Assertion (A). as
13. 1 1
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(C) / Assertion (A) is true, Reason(R) 1 1
15. (C)/ Assertion (A) is true, Reason(R) is false. 1 1
16. (D)/ Assertion (A) is false, Reason (R) is true. 1 1
SECTION – B
17. (a) An immune response where the B-lymphocytes produce an army of 1
proteins in response to pathogens into our blood to fight with them.
(b) Antibodies – IgA, IgE, IgM, IgG
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(Any two)

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(a)
(i) Dormancy is crucial for storage (dehydration) of seeds a
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raise the crops in next season / Seeds are used as source of food / for 1
commercial purpose
(ii)
Pea seed Castor seed
Non endospermic seed Endospermic seed

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Non-albuminous seed Albuminous seed
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No endosperm is left in the It contains some amount of
seed as it is fully consumed endosperm as it is not
during embryo development. consumed completely during
embryo development.
(Any one difference)
OR
(b)
• Tertiary follicle in the ovary 1
• Secondary oocyte and 1st polar body. ½+½ 2

19. (a)
• Palindromic sequences ½
• Restriction endonuclease / EcoRI. ½
(b)
Restriction endonucleases cut the strands of DNA a little away from
the centre of the palindrome sites to form sticky end on each strand /
1
Restriction endonucleases cut at specific recognition site between two
same bases in opposite strands to produce sticky ends / EcoRI cuts 2
the DNA between bases G and A in both the strand in the sequence
5’-GAATTC-3’
3’-CTTAAG-5’
20. (a)
Case 1 –Male heterogamety ½
Organism – Many insects / grasshopper / any correct example. ½
Case II–Female heterogamety ½
Organism – Birds / hen / any correct example. ½

OR
(b)
(i)
DNA nucleotide RNA nucleotide
1
Has deoxyribose sugar Has Ribose sugar
Types of nitrogen bases are Types of nitrogen bases are
A,G,C,T A,G,C,U
(Any one difference)
(ii) Bonds in a nucleotide
N – Glycosidic linkage (between Sugar and Base) ½
Phosphoester linkage – (Between Sugar and Phosphate) ½ 2

21. (a)

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(i)
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• Inverted Pyramid of Biomass
a ½

• Such pyramids are seen in-
Aquatic Conditions where a small standing crop of phytoplanktons
supports a large standing crop of zooplankton or fish / ½
In terrestrial ecosystem – when a large number of insects are feeding on a
single tree / or any other correct example

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(ii) -Ecological pyramids does not take into account the same species ½+½

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belonging to two or more trophic levels.
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-It does not accommodate a food web.
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-Saprophytes are not given any place in ecological pyramids even
though they play a vital role in the ecosystem.
(Any two)
OR
(b)
(i) -Decline in plant production.
-Lowered resistance to environmental perturbations as drought. ½+½
-Increased variability in certain ecosystem processes such as water
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use and pest and disease cycles
- Any other correct point
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(ii)
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-Habitat loss and fragmentation.
-Alien species invasion. ½+½
-Over exploitation
-Co-extinction 2
(Any two)
SECTION – C
22. (a)
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• No
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• -In cereals or rice or wheat pollen grains loose viability within 30
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minutes of their release ,
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maintain viability for months ,
a - or any other correct example
(b)
They are stored using Cryo-preservation techniques / In liquid nitrogen 3
½
at – 196oC

23. (a)
• Pedigree analysis –It is analysis of traits or inheritance of traits in 1
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several generations of a family.

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• Importance : To trace the inheritance of a trait , abnormality or a ½+½
disease.
(b)
(i) Autosomal- Recessive ½

½ 3
(ii) Sickle Cell Anaemia / Thalassemia / Phenylketonuria /Any other
correct example.

24. (a)
According to Hardy Weinberg principle –
Sum total of all the Allelic frequencies in a population = 1.
Let there be two alleles A and a in the population. Let their frequencies be
‘p’ and ‘q’ respectively.
p+q=1 ½

The frequency of AA will be p2 because - the probability that an allele A
with frequency ‘p’ would appear on both the chromosomes of a
½
diploid individual will be product of probabilities i.e. p2.

Similarly frequency of ‘a’ will be q2 , ½

and Aa is 2pq ½

Hence p² + 2pq + q² = 1

(b)
Gene migration or Gene flow, Genetic drift, Mutation, Genetic ½+½
3
Recombination, Natural selection
(Any two factors)

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½x6

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(Label any six parts) 3

26. (a)
Source plant - Papaver somniferum ½

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Part - latex of the plant ½
(b)
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Heroin is formed by acetylation of morphine. 1
•
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Morphine →It is effective sedative/ used as pain killer/ useful in
½
a
patient who have undergone surgery
Heroin is a depressant / slows down body function ½
3

27. (a)
Cry I Ab ½
(b)
The spores of Bt were mixed with water and sprayed in field A where these
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are eaten by the insect larvae, in the gut of the insect toxins are released, and
insect get killed.
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a Bt toxin
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The genetically modified plants (GM) express Bt gene /produce
a which when ingested will kill the pest. 1 3

28. (a)
• No bands will be seen in agarose gel. ½
• DNA fragments being negatively charged will not move towards
negative end or cathode / DNA being negatively charged will remain
½
positioned at anode or positive end.

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• Position of positive terminal or anode and negative terminal or 1
cathode has been interchanged. 3
• Band at X being the smallest fragment of DNA move farthest 1
towards the anode.

SECTION – D

29. (a)
-Competitive release. ½
-A species whose distribution is restricted to small geographical area due to
the presence of competitively superior species expands its distributional
½
range when competing superior species is removed.
OR
(a)
Gause’s competitive exclusion principle, states that two closely related
species cannot co-exist indefinitely and the competitively inferior one ½+½
will be eliminated eventually.
(b) In shallow south American lakes visiting flamingos and resident fishes
compete for common food Zooplankton / any other correct example 1

(c)
• By “Resource partitioning”
1
• They choose different times for feeding or different foraging patterns
e.g. Five closely related species of warblers avoid competition and
coexist due to behavioural differences in their foraging activities / any 1 4
other correct example.
30. (a)
Cross I – Genotypes TT and tt / TT and Tt / TT and TT ½
Cross II – Genotypes Tt and Tt. ½

OR
(a)
Monohybrid Cross 1
(b)
• Test Cross. 1
• It is done to find the genotype of an organism showing dominant 1
phenotype.
(c)
• It shows that tall trait is dominant over dwarf . ½

• Genetic principle – Law of Dominance. ½
4

SECTION – E
31. (a)
•
After the entry of sperm into female gamete, it undergoes meiotic
division II, to produce haploid ovum ie ootid, then the nucleus of
sperm and ootid fuse together to form zygote, zygote moves through
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the isthmus of oviduct towards uterus and it undergoes mitotic
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division or cleavage to form blastomeres, the embryo with 8 to 16
blastomere is called morula and it further divides and transforms into ½x8
Blastocyst, blastocyst have outer layer trophoblast and inner cell mass
and the trophoblast layer of blastocyst gets attached to endometrium,
Uterine wall divide rapidly and covers blastocyst and this it gets
embedded in the endometrium of uterus this is called implantation.

• The site of fertilization is-
ampullary region of fallopian tube /isthmus ampullary junction of
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fallopian tube. 1
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OR

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(b) (i)

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In vitroafertilization followed by zygote intra fallopian transfer or ZIFT / a
Inavitro fertilization followed by Intra uterine transfer or IUT / 1
Intra Uterine insemination or IUI / Artificial insemination or AI .

Case II -
Invitro fertilization followed by IUT
1

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Case III -
Intra Cytoplasmic Sperm Injection or ICSI / Artificial insemination or

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AI / Intra Uterine insemination or IUI. 1

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(ii)
a
• The copper releasing IUD’s are CUT, Cu7, Multiload 375
(Any two) ½+½
•
-Copper releasing IUD’s increases phagocytosis of sperms within
the uterus, 5
½+½
-Cu ions released suppress sperm motility and the fertilising capacity
of sperms.

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32. (a) (i)

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-Maurice Wilkins and Rosalind franklin, – X- ray diffraction studies of

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DNA.
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s e -Erwin Chargaff ,– For a double stranded DNA the ratios between
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a ½+½

a equal to one.
(ii) The salient features of double helix structure of DNA are as follows :
-It is made of two polynucleotide chains, in which backbone is
constituted by sugar phosphate and the bases project inside.
-The two chains have antiparallel polarity.
-The bases in two stands are paired through hydrogen bonds forming
base pairs(bp). Adenine form double hydrogen bond with thymine
and Guanine form triple hydrogen bond with cytosine.
1x3
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-The two chains are coiled in a right handed fashion.The pitch of
helix is 3.4 nm and 10b.p in each turn. Distance between base
pair in a helix is approximately 0.34 nm .
-The plane of one base pair stacks over the other in double helix.
(Any three points)
OR
(b)(i)
Template strand
3' – TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC 1
CGT AAT – 5'

m – RNA
5' – AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG 1
GCA UUA – 3'
(ii)
(1) In a bacterium
5' – AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG 1
GCA UUA – 3'
(2) In humans
5' – mGppp AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA 1
– poly A tail 3'

(iii) It will have 10 amino acids . 1 5

33.
(a)(i) Plasmodium falciparum
½
(ii)
• Female Anopheles Mosquito ½
•
-Mosquito picks up gametocytes from the infected person.
-Fertilization and development takes place in mosquito gut. ½x3
-Sporozoites from gut migrate to salivary glands of mosquito.

(iii) After entering the blood stream they enter liver, and then into red blood
cells and reproduce asexually, RBCs rupture and a toxic substance ½x3
haemozoin is released causing chills and high fever.
(iv) -Avoiding stagnation of water in and around residential areas.
-Regular cleaning of household coolers.
-Use of mosquito nets.
-Introduce fishes like Gambusia in ponds – they feed on mosquito ½+½
larval.
-Spraying insecticides in ditches of doors and windows to have
wire mesh.
-Any other appropriate preventive measures.
(Any two)

OR
(b)
(i) Bio fertilizers – Organisms that enrich the nutrient quality of soil. e.g. 1
bacteria fungi and cyanobacteria.
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(ii)
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-Bacteria / Rhizobium / Azotobacter / Azospirillum ½

Rhizobium forms symbiotic association with roots of leguminous plants
and fix atmospheric nitrogen into organic forms to be used by the plant, ½

Azotobacter or Azospirillum fix atmospheric nitrogen while free living
in the soil.
(Award one mark for any one organism with its correct contribution)

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-Fungi /genus Glomus . ½
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Fungi breakdown organic matter to release nutrients,
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Glomus form mycorrhiza with roots of the plants.This association helps to
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absorb phosphorous from soil and pass to the plant,
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plants show resistance to root-borne pathogens,
plants show tolerance to salinity and drought,
There is increase in plant growth.
(Any one)
-Cyanobacteria / Anabaena / Nostoc / Oscillatoria ½

½
Fix atmospheric Nitrogen , serve as important biofertilizers in paddy fields
(Any one )
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(iii)-Bio fertilizers do not pollute our environment (soil, ground water).
½+½
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-They conserve the beneficial soil microbes.

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-Do not cause contamination and harm human health. 5

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