CBSE Class 12 Question Paper 2026 Solution Biotechnology

FOR CBSE CLASS 12 EXAM PREPARATION

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

CBSE Class 12 2026 Question Paper Solution Biotechnology

Notes · Sample Papers · Previous Year Papers · Mock Tests

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Marking Scheme

Strictly Confidential

(For Internal and Restricted use only)

Senior Secondary School Examination, 2026 (XIIth)

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SUBJECT NAME: - - (Q.P. CODE

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General

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 assessment of the candidates. A small mistake in evaluation may lead to
serious problems which may affect the future of the candidates, education system
and teaching profession. To avoid mistakes, it is requested that before starting

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evaluation, you must read and understand the spot evaluation guidelines carefully.

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of the examinations conducted, evaluation done and several other aspects.
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Its leakage to public in any manner could lead to derailment of the
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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

4 Evaluation is to be done as per instructions provided in the Marking Scheme. It

consideration. Marking Scheme should be strictly adhered to and religiously
followed. However, while evaluating, answers which are based on latest
information or knowledge and/or are innovative, they may be assessed for
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while evaluating two competency-based questions, please try to understand

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given answer and even if reply is not from marking scheme but correct
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las competency is enumerated by the candidate, due marks should be awarded.
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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.

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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
evaluation shall be given only after ensuring that there is no significant variation in
the marking of individual evaluators.

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be marked. Evaluators will not put right ( ) while evaluating which gives an
impression that answer is correct and no marks are awarded. This is most
common mistake which evaluators are committing.

8 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 If a student has attempted an extra question, answer of the question deserving
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more marks should be retained and the other answer scored out with a note

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a 11 No marks to be deducted for the cumulative effect of an error. It should be
penalized only once.

12 A full scale of marks __________ (example 0 to 80/70/60/50/40/30 marks as given
in Question Paper) has to be used. Please do not hesitate to award full marks if the
answer deserves it.

13 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
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answer books per day in other subjects (Details are given in Spot Guidelines).This
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is in view of the reduced syllabus and number of questions in question paper.
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Ensure that you do not make the following common types of errors committed by
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l asthe Examiner in the past :-
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a mark is correctly and clearly indicated. It should merely be a line. Same is with
Answers marked as correct, but marks not awarded. (Ensure that the right tick

the X for incorrect answer.)
Half or a part of answer marked correct and the rest as wrong, but no marks
awarded.
15 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.

16 The Examiners should acquaint themselves with the guidelines given in the
Guidelines for Spot Evaluation
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m.All Examiners/Additional
The candidates are entitled to obtain photocopy
payment of the prescribed processingefee.
of the Answer Book on request on

Examiners/Head Examiners are once s Head

la again reminded that they must ensure that
evaluation is carried out strictlygas per value points for each answer as given in the
Marking Scheme.
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MARKING SCHEME
BIOTECHNOLOGY (Subject Code 045)
(PAPER CODE: 99) (26-04-45N)

SECTION A
Q.No Expected Outcomes Book Reference Marks

1 (A) Incorrect folding and intracellular accumulation of recombinant Unit VI, Ch 1, p.102-103 1
proteins

2 (C) 5-10% CO 2 Unit VI, Ch 3, p.143 1

3 (A) cDNA Unit V, Ch 3, p.67 1

4 (C) Anther Unit VI, Ch 2, P.112 1

5 (B) 150 mg Unit VI, Ch 1, p.93 1

6 (C) Destruction of invading viral DNA Unit V, Ch1, p.04 1

7 (B) Rigby and Paul Berg in 1977 Unit V, Ch 3, p.65 1

8 (C) Provide information on official gene name Unit V, Ch 3, p.79 1

9 (A) Providing strength to our bones Unit V, Ch 2, p.29 1

10 (A) Monoclonal antibodies are preferred over polyclonal antibodies Unit VI, Ch 3, p.149 1

11 (C) In both coding and non-coding regions of a genome Unit V, Ch 3, p.63 1

12 (A) Adenosine deaminase Unit V, Ch 2, p.30 1

13 (B) Both Assertion (A) and Reason (R) are true, but the Reason (R) is Unit VI, Ch 3, p.153 1
not the correct explanation of the Assertion (A)

14 (C) Assertion (A) is true, but Reason (R) is false Unit VI, Ch 1, p.102 1

15 (A) Both Assertion (A) and Reason (R) are true, and the Reason (R) is Unit V, Ch 1, p.10-11 1
the correct explanation of the Assertion (A)

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a 16 (A) Both Assertion (A) and Reason (R) are true, and the Reason (R) isa Unit VI, Ch 2, p.128 1
the correct explanation of the Assertion (A)

SECTION B
Q.No Expected Outcomes Book Reference Marks
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17 (a) Fertility (F) plasmid/ plasmid from E.coli /bacterial plasmid (1) Unit V, Ch 1, p.12 1

o m genes/selectable markers/cloning (0.5+0.
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sites. (any TWO).(0.5+0.5)
Replication and maintenance 5)=2
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17 (b) EcoRI a
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Thegtwo generated fragments in double stranded forms
a Unit V, Ch 1, p.5-6
(0.5+0.
5)=2

(0.5+0.5) For making two fragments

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18 Antibiotics control the growth of bacteria/ fungi in culture Unit VI, Ch 3, p.142 2
medium (1)
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Penicillin and Streptomycin (any 2 antibiotics) (0.5+0.5) +0.5)

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Usefulness: (any one) (1) ag cells. (1)
Selectively removing a gene from mouse ES Unit VI, Ch 3, p.153-154 2
(1+1)
1) To understand the genetic basis of a disease
2) To search new diagnostic techniques
3) To search for therapeutic modalities

Alternative Question for Visually impaired
Unit VI, Ch 3, p.144 2
Animal cells at the bottom can be visualized, (1) (1+1)
because optical system is at the bottom with light source on the
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top (1)
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20 A proteolytic cut / cut by Trypsin in chymotrypsinogen results in 2
conformational change exposing the active site (1)
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Unit V, Ch 2, p.34-35 (1+1)

em In-situ activation (1)
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ag Due to existence of overlapping genes and splice variants 2
(1+1) Unit V, Ch 3, p.61 (1+1)

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SECTION C
Q.No Expected Outcomes Book Reference Marks
22 Substitution of Valine for Glutamic acid at 6th position of Beta Unit V, Ch 2, p.36-37 3
chain of haemoglobin (1) (1+1+1)
It results in the formation of fibres within RBCs causing sickling (1)
Sickle RBCs get stuck in the capillaries/ ScHb has impaired oxygen
carrying capacity. (1)

23 Culturable and non culturable bacteria/ utilises large number of Unit VI, Ch 1, p.103 3
genome (0.5) (0.5+0.5+
0.5+0.5+0
Isolation of DNA from environmental sample collection DNA .5+0.5)
manipulation Ligation Cloning / Library construction
Protein expression (0.5x5=2.5)
24 For longer shelf life/ transport to longer distances without Unit VI, Ch 2, p.128 3
spoilage (1) (1+1+1)

Ripening is slowed down by blocking/reducing ethylene
production/by introducing ethylene producing genes in a way that
suppresses its own production in the crop plant. (any two) (1+1)

25 Ser 221, His 64, and Asp 32 (1) Unit V, Ch 2, p.51-52 3

Subtilisin native enzyme is inactivated by bleach (oxidation of (1+1+1)
Met222) (1)

Met 222 was substituted by various amino acids/ alanine to retain
enzyme activity and stability in the presence of bleach./ SDM Site
Directed Mutagenesis/protein engineering (1)

Alternative Question for Visually impaired

Whey proteins elevate glutathione (1) Unit V, Ch 2, p.50 3

which detoxifies xenobiotics (1) (1+1+1)

and protects cellular components from oxygen intermediates and
free radicals.(1)

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Antibiotic resistance/
ag Unit V, Ch 1, p.15 3
-galactosidase /
(1+1+1)
protein such as GFP /

dependence or independence of a nutritional requirement such
as amino acid leucine.(any 3) (1mk each)

Transfection: Mixing of foreign DNA with charged substance
27 (a) calcium phosphate / cationic liposomes/DEAE dextran / Unit V, Ch 1, p.15 3
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overlaying on recipient/ host cells (1)
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Electroporation: An electric current is used to create transient (1+1+1)

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microscopic pores in the recipient host cell to allow rDNA(1)
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Biolistics: A gene / particle gun is used to bombard microscopic
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gold/ tungsten particles coated with DNA into the host cells (1)
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27 (b)
a OR
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Insertional inactivation of lac Z gene coding for B galactosidase Unit V, Ch 1, p.17
enzyme in pUC 19 plasmid; (1) (1+1+1)

plating of E.coli host cells after transformation on solid media
containing X-gal; (1)

white colonies à with rDNA; Blue colonies à without rDNA.(1)

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Differentiate on the basis of any 2 properties of the.cells from the Unit VI, Ch 3, p.138-140
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following: (2mk)

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(i)
(ii)
Limited life span
Contact inhibition ag
(iii) Density limitations
(iv) Anchorage dependence
(v) Growth rate
(vi) Growth form
(vii) Doubling time
(viii) Change in ploidy
(ix) Change in shape of cell

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Examples: CHO cell line and HeLa cell line, cos-1 cell lines (any
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other suitable example) (1mk)

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Book Reference Marks
a 29 (i) abl and bcr genes (0.5+0.5) Unit V, Ch 3, p.65-67 (0.5+0.5)=1

29 (ii) No; a part of chr 9 along with abl gene is translocated to chr 22/ 1
absence of fluorescent gene (0.5+0.5)
(0.5+0.5)

29 (iii) a
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Nicks created by DNase I, (0.5)
(0.5+0.5+1)
DNA polymerase I (0.5)

synthesizes new DNA ,incorporation of labelled nucleotides (1)

29 (iii) b OR 2

CML lymphocytes smear cells hybridized with two probes In- (0.5+0.5+0.
situ,(0.5) 5+0.5)

yellow fluorescence under fluorescent microscope.(0.5)

Fluorescent DNA probes for abl and bcr genes (0.5)

bind to exhibit merged fluorescence colour under fluorescent
microscope (0.5)

30 (i) Nutrient depletion/ done in small scale/havesting and addition of Unit VI, Ch 1, p.91-92 1
medium cannot be done at the same time/ waste accumulation/
limited growth/ cells exposed to continually changing (0.5+0.5)
environment (any two) (0.5+0.5)

30 (ii) continuously and sequentially fed with fresh medium (1)
1

30 (iii) a Chemostat maintains constant chemical environment (1) 2
(1+1)
Turbidostat maintains turbidity of culture medium/constant cell
concentration (1)

OR 2
30 (iii) b
Labelled graphs (I mk each) of batch and continuous cultures as (1+1)
on Fig. 5 and Fig. 7 (p.91-92)

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Where [X]à cell density

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[S] à Concentration of substrate
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[QS] à Cell specific substrate
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g of molecular ions according to mass to charge/ m/z a
31 (a)
a
Separation
ratio. (1) Unit V, Ch 2, p.45-46 5
Procedure of MALDI (all steps)
Ionisation,(1) (1+1+1+1+
Ion separation,(1) 1)
Detection/Analysis in a mass spectrum. (1)

Matrix absorbs laser light energy and vaporises/ Sample is
transferred from condensed phase to gas phase/ used to
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volatilise & protonate peptide and proteins ,(1)

OR
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31 (b)
l asacids in cereals and
Due to deficiencies of certain essential amino

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Unit VI, Ch 2, p.129 5
legumes. ,(1)
Unit V.Ch 2.p.53-54 (1+2+2)
Two genetic engineering approaches used to improve protein
quality in seeds, are:

(I) To engineer genes that encode seed storage proteins with
more nutritionally desirable amino acids by inserting genes of
additional amino acids and by substituting existing amino acids
with new ones. ,(2)

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2 (II) By introducing genes of entirely novel proteins that are highly
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s e32m(a) Unitg
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a VI, Ch 2, p.117 5

a Axillary meristem (any one) ,(1)
(1+0.5+0.5
Increased yield / quality / to obtain virus free plants from infected +1+2)
crop plant (any two) (0.5+0.5)

(ii) Artificial seeds are produced either by encapsulating
somatic embryos in a protective coating i.e Calcium alginate
beads or by desiccating the somatic embryos with or without
coating,(1)

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(iii) Diagram drawing (0.5)

labelling (0.5x3)
The artificial seed coat [A]
Somatic embryo at Torpedo stage [B]
Artificial endosperm [C]

OR
Three genes involved in the biosynthetic pathway (1)
-carotenoid production (1) 5
32 (b) Unit VI, Ch 2, p.129
were introduced under the control of endosperm specific (1+1+1+1+
promoter, so that gene products (enzymes) are synthesized in the 1)
rice endosperm. (1)
Advantages:
Provitamin A is produced in the entire grain/ prevents night
blindness (any one) (1)

Further engineered golden rice by introducing three more genes
from different organisms for iron source and its absorption. (1)

33 (a) (i) Brief description of Unit V, Ch 1, p.17-18 5
temperature (1x3)
(1+1+1+1+
n
(ii) Formula, 2 (where n = no of cycles) (1mk) 1)

Total no of double stranded DNA molecules = 3 X 2 10 = 3072 (1mk)

OR
33 (b) Unit V, Ch 1, p.6-7
Restriction Fragment Length Polymorphism (1)

Principle: Variation in size (length) of the restriction enzyme 5
generated fragments between individuals within a given species.
(1) (1+1+1+2)

Difference in movement of DNA fragments on electrophoresis
after digestion with restriction enzymes. (1)

Applications (Any Two (1mk each) ):

Forensic sciences /Identify individuals/ criminals
Relate individuals
Paternity/ parentage testing
DNA fingerprinting

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