CBSE Class 12 Question Paper 2024 Solution Biology

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior Secondary School Certificate Examination,2024
SUBJECT NAME BIOLOGY (Q.P. 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-XII, 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.

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

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

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 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).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:-
● 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 unassessed portion, non-carrying over of marks to the title page, or totaling 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 totaled 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.

XII_ 044 57/1/1 BIOLOGY # pg. 2

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

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

·
½

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