CBSE Class 12 Question Paper 2023 Solution Biology

e s t i o n P a p er
Qu
Solu t i o n
2023

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior School Certificate Examination, 2023
SUBJECT NAME BIOLOGY (SUBJECT CODE 044) (PAPER CODE 57/1/1)

General Instructions: -
1 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 evaluation, you must read and
understand the spot evaluation guidelines carefully.

2 “Evaluation policy is a confidential policy as it is related to the confidentiality of 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 News Paper/Website etc may
invite action under various rules of the Board and IPC.”

3 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, while
evaluating, answers which are based on latest information or knowledge and/or are
innovative, they may be assessed for their correctness otherwise and due marks be
awarded to them. In class-X, while evaluating two competency-based questions,
please try to understand given answer and even if reply is not from marking scheme
but correct competency is enumerated by the candidate, due marks should be
awarded.

4 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.

5 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 delibration 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.

6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer CROSS ‘X” 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.

7 If a question has parts, please award marks on the right-hand side for each part. Marks
awarded for different parts of the question should then be totaled up and written in the left-
hand margin and encircled. This may be followed strictly.

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8 If a question does not have any parts, marks must be awarded in the left-hand margin and
encircled. This may also be followed strictly.

9 If a student has attempted an extra question, answer of the question deserving more
marks should be retained and the other answer scored out with a note “Extra Question”.

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 0-70 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).

13 Ensure that you do not make the following common types of errors committed by the
Examiner in the past:-
● Leaving answer or part thereof unassessed in an answer book.
● Giving more marks for an answer than assigned to it.
● Wrong totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● Wrong transfer of marks from the answer book to online award list.
● 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 Any un assessed portion, non-carrying over of marks to the title page, or totalling error
detected by the candidate shall damage the prestige of all the personnel engaged in the
evaluation work as also of the Board. Hence, in order to uphold the prestige of all
concerned, it is again reiterated that the instructions be followed meticulously and
judiciously.

16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines
for spot Evaluation” before starting the actual evaluation.

17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to
the title page, correctly totalled and written in figures and words.

18 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.

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MARKING SCHEME
Senior Secondary School Examination, 2023
BIOLOGY (Subject Code– 044)
[ Paper Code:57/1/1]
Maximum Marks: 70
Q. Marks Total
EXPECTED ANSWER / VALUE POINTS Marks
No.
SECTION A
1. (b) / (ii) and (iv) 1 1
2. (c) / X = Promoter, Y = Sigma factor, Z = RNA polymerase. 1 1
3. (a) / Declining population. 1 1
4. (a) / Point (i) 1 1
5. (d) / P - (iii), Q - (i), R- (ii), S- (iv) 1 1
6. (c) / 1 and 3 1 1
7. (c) / Directional selection as giraffes with longer neck lengths are selected 1
// //
(d)/Stabilizing selection as giraffe with medium neck lengths are selected. 1 1
8. (c) / 1300 1 1
9. (a) / P– (ii), Q – (iv), R – (iii), S – (i) 1 1
10. (c) / Leydig cells – Androgen – Initiate the production of sperms. 1 1
11. (b) / Salmonella typhimurium. 1 1
12. (d) / Dr. R.A. Mashelkar. 1 1
13. (c) / A is true, but R is false. 1 1
14. (a) / A and R are true and R is the correct explanation of A. 1 1
15. (c) / A is true , but R is false. 1 1
16. (a) / A and R are true and R is correct explanation of A. 1 1
SECTION B
17. (a) Blastocyst ½
(b) Uterine wall/endometrium/innermost layer of uterine wall. ½
(c) (Outer layer/trophoblast) ‘X’- helps in implantation in uterus/attachment to ½
endometrium.
(Inner cell mass) ‘Y’- gets differentiated into embryo. ½ 2

18. (a) From micropylar end, through the synergids (filiform apparatus)/filiform (within ½×2
synergids) apparatus guides the entry of pollen tube
(b) One male nucleus fuses with two polar nuclei to form Primary endosperm nucleus 2
and termed triple fusion, other male nucleus fuses with egg cell nucleus to form ½×2
zygote i.e. undergoes Syngamy
19. (a)
(i) ‘A; Circular DNA/Plasmid ½
‘B’ Bacteriophage ½
(ii)(Plasmid)-Can carry foreign gene into the host cell/acts as cloning vector/has
selectable marker/ independent of the control of chromosomal DNA/ high ½
copy number
(Bacteriophage) -Cloning vector have the ability to replicate in bacterial cells /
independent of the control of chromosomal DNA / high copy number per cell. ½
OR
(b) Treating bacteria with specific concentration of calcium (ions)which increases the
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efficiency with which DNA enters the bacteria through pores in its cell
wall ,recombinant DNA can then be forced into such cells by incubating the cells
with recombinant DNA on ice, followed by placing them briefly at 420C (heat shock), ½×4 2
then putting them back on ice.
20. • Doesn’t take into account same species belonging to two or more trophic levels.

• Assumes a simple food chain which never exists in nature/does not accommodate a
food web. 1×2
• Saprophytes are not given any place though they play an important ecological role.

(Any two points) 2
21. 2·4 percent 1

8·1 percent share of the global species diversity 1 2
SECTION C
22.

½×6

3

23. (a) Recessive trait, both the parents in generation I do not express the trait yet it ½×2
appears in the progeny.

(b) Autosomal trait, both male and females have equal chances of getting the trait. ½×2

(c) Child ‘1’ : Aa/AA , Child ‘3’ : Aa ½×2
// //
(a) Recessive trait, both the parents in generation I do not express the trait, yet it
appears in the progeny. ½×2

(b) Sex linked trait, comparatively more male are getting affected. ½×2

(c) Child ‘1’ : XY, Child ‘3’: X’ X(carrier) ½×2
3
24. (a) They have the ability of self-renewal / to divide, and differentiate into any ½×2
kind of cell/tissue/organ.

(b) – Inner cell mass of blastocyst / umbilical cord / Bone marrow
(or any other correct source) 1

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(Any one)
(c) Diabetes treatment via forming islets of Langerhans,
Restoration of vision by injecting stem cells, to treat rheumatoid arthritis, reduces ½×2
pancreatic cancer, to treat genetic disorder like cystic fibrosis, spinal cord injurie,
heart disease, any other correct application 3
(Any two)
25. (a)

S. Malignant tumor Benign tumor
No. 1
1 Cells grow very rapidly and Comparatively slow growth and remain
invade and damage the confined to their original location and do
surrounding normal tissue. not spread to other parts of the body
2 Show metastasis Do not show metastasis
(Any one difference)

(b)• Metastasis
1
• Cells from these tumors slough off and reach distant sites through blood,
wherever they get lodged in the body they start a new tumor there. ½×2 3
26. (a) Primary Sludge: All the solids that settle down, during the primary treatment of ½×2
sewage water.

(b) Activated Sludge: Produced during the secondary treatment or biological ½×2
treatment of sewage, primary effluent + aerobic microbes flocs (bacteria and
fungus) – get converted to a sediment whose BOD has reduced significantly.
½×2
(c) Anaerobic sludge digesters: Large tanks where activated sludge is treated with
anaerobic bacteria which digest the bacteria and fungi, and produce a mixture of CH4, 3
H2S and CO2/ Biogas
27. (a) Dryopithecus, Ramapithecus ½+½

(b) Time period : 2 million years ago ½
Place : East African grasslands ½

(c)

Homo habilis Homo erectus
Brain capacity Brain capacity 900 cc
between 650 – 800 cc ½
probably did not eat probably ate meat
meat. ½
3
28. (a) (i) (1) ZIFT : Zygote intrafallopian transfer.
(2) ICSI : Intracytoplasmic sperm injection.
(3) IUT : Intra uterine transfer.
(4) GIFT : Gamete intrafallopian transfer. ½×4

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(ii) •GIFT ½
•GIFT allows the eggs to fertilize and develop in the fallopian tube/ IVF places ½
a directly fertilized egg (zygote) into the uterus/ in vivo fertilisation is involved in
GIFT.

1
OR
(b) (i)
Perisperm Pericarp:
Persistent nucellus in some The wall of ovary develops into
seeds wall of fruit.

1

(ii)
Syncarpous Apocarpous
fused pistils. free pistils.

iii)

Plumule : Radicle :
Future stem/ terminal part Future root/ terminal 1
of epicotyl / shoot tip of part of hypocotyl /
embryonal axis root tip of embryonal
axis 3

SECTION D
29. (a) In presence of lactose repressor protein dose not bind to the operator region (O) 1
and allow RNA polymerase to transcribe the operon.
// //
In absence of lactose repressor protein bind to the operator region (O) and prevent
RNA polymerase from transcribing the operon. 1

(b) Presence of Permease enzyme coded by gene ‘y’ is required that allows lactose to
enter the cell for switching on the operon / so that lactose enter inside the cell. ½

(c) ‘i’ stands for ‘inhibitor/ this gene transcribes repressor protein which binds to
the ‘operator’ site and switch off the operon. ½

(d)

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½×4

OR

(d)

½×4

4
30. (a) Species diversity decreases as we move from region A to region B. 1

Reasons : less Constant mean annual temperature, lesser habitable land 1+1
area, availability of lesser solar energy, lesser productivity, any other correct
reason in ‘B’ region.
(Any two)
(b) More the 1200 species of birds, Indian land mass being largely in the tropical ½+½
latitudes.

OR
(b) Amazonian rainforest (in South America), mainly being in tropical region. ½+½ 4
SECTION-D
31. (a)(i) Normal nitrogen/ N, heavy nitrogen/ 15N
14
½+½
to produce two types of DNA / light and heavy DNA respectively. 1

(ii) E. coli has generation time of 20 minutes so the samples taken at intervals of 20 1
minutes, to understand the mode of replication when E.coli with 15N DNA was
cultured in medium 14N (normal) nitrogen.

(iii) To distinguish or separate heavy DNA from light DNA on the basis of density. 1

(iv) Mode of DNA replication is semiconservative. 1

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OR

(b) (i) TTRR ttrr ½
Tall round seeds Dwarf wrinkled seeds
Parents
Gametes TR tr ½

F1 generation TrRr (All tall round) ½
Selfing
TtRr X TtRr ½

2

(Male and Female gametes along with Punnett square are to be awarded two marks)

F gen. Tall Round Dwarf Tall Dwarf
2
Round wrinkled Wrinkled
½
9 : 3 : 3 : 1

(ii) ‘when two pairs of traits are combined in a hybrid, segregation of one pair
of characters is independent of the other pair of characters’(law of independent ½
assortment ).
5
32. (a) (i) The vaccine contains the antigen, which stimulates or activates immune cells to ½+1
produce antibodies (by B lymphocytes) / which generates primary response or
humoral immune response.

(ii) Memory cells generate, amnestic response/secondary response ½+½

(iii) P = Yes
Q = Catching an infection/getting infected
R = No ½×5
S = Yes
T = No
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OR

(b) (i) Diacetylmorphine 1
 as it is highly addictive, and being a depressant it slows down body functions. ½+½

(ii) (1) Cannabis sativa, affects the cardiovascular system of the body. ½+½

(2) Erythroxylum coca /coca plant , interferes with the transport of ½+½
neurotransmitter dopamine / produces sense of euphoria / increased energy.
(3) Papaver somniferum, acts as depressant/ slows down body function/ reduces ½+½
pain/sedative 5
.
33. (a) (i) (1) Autogamy
(2) Geitonogamy ½×3
(3) Xenogamy

(ii) (1) Water lily: pollinated by insects/wind. ½

(2) Vallisneria : Female flowers on long stalks reach water surface male flowers or
pollen released on water and carried by water current to female flowers to achieve 1
pollination.

(iii) Genetic : Self-incompatibility / prevents self-pollen (same flower or other flowers
of same plant) from fertilizing the ovules by inhibiting pollen germination, pollen ½×2
tube growth in pistil.

Physiological : Pollen release and stigma receptivity are not synchronized, either
pollen matures earlier and stigma later or pollen matures later than stigma. ½×2

OR

(b) (i) (1) Menstrual period
(2) Follicular phase/proliferative phase ½×4
(3) Luteal phase/secretory phase
(4) Ovulatory phase

(ii)

½×6

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Days Ovarian Pituitary hormones
hormones
1 8-12 Follicular Simulates follicular
growth / Development/ secretion of
proliferation estrogen by growing follicles
of
endometrial
cells.
2 13-15 Maturation of Rupture of graafian follicle to
ovarian release ovum.
follicles/
formation of
graafian
follicles /
thickening of
endometrium. 5
3 16-18 Maintenance Secretion of progesterone
of from corpus luteum.
endometrium

**********************

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