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---

SET-1

Series SQR1P/1
àíZ-nÌ H$moS>

Q
...
Code

amob Z§
...


Candidates must write the Q
...
Code
on the title page of the answer-book
...
15 ~Oo {H$`m OmEJm &
10
...
30 ~Oo VH$ N>mÌ Ho$db
àíZdo CÎma-nwpñVH$m na H$moB© CÎma Zht {bI|Jo &

(V)

(I)

NOTE
Please check that this question paper
contains 23 printed pages
...

(III) Q
...
Code given on the right hand
side of the question paper should be
written on the title page of the
answer-book by the candidate
...

(V) 15 minute time has been allotted to
read this question paper
...
15 a
...
From 10
...
m
...
30 a
...
, the students will
read the question paper only and
will not write any answer on the
answer-book during this period
...
T
...


:
:
33

(i)
(ii)
(iii)

1

(iv)

17

21

(v)

22

28

(vi)

29
31

(vii)

16

1
2
3

30

4

33

5

(viii)
(ix)
(x)

IÊS> H$
1

16

1

16 1=16

1
...


{ZåZ{b{IV _| go H$m¡Z-gm hm°_m}Z _mZd Anam (ßb¡g|Q>m) Ûmam òm{dV hmoVm h¡ Omo gJ^©Vm
(J^m©dñWm) H$mo ~ZmE aIZo _| ghm`H$ h¡ ?
(A)
(C)

57/1/1-11

[ab¡

(B)

Q>mo{gZ

(D)
2

_mZd Oam`w JmoZ¡S>moQ´>m°{nZ
_mZd Anam b¡ oOZ

General Instructions :
Read the following instructions carefully and follow them :
(i)
This question paper contains 33 questions
...

(ii)
Question paper is divided into five sections Sections A, B, C, D and E
...
Each
question carries 1 mark
...

Each question carries 2 marks
...
Each
question carries 3 marks
...
Each
question carries 4 marks
...

(vii) Section E questions number 31 to 33 are long answer type questions
...

(viii) There is no overall choice
...
A candidate has to write answer for
only one of the alternatives in such questions
...

(x)
Wherever necessary, neat and properly labelled diagrams should be drawn
...
1 to 16 are Multiple Choice Type Questions, carrying 1 mark
each
...


2
...
T
...


3
...


{ZåZ{b{IV go

5
...


(A)

45 + XX

(B)

44 + XXY

(C)

44 + XO

(D)

44 + XY

$

g_J« ê$n go _mXm bjU H$m {dH$mg
Aënd{Y©V A§S>me` Ho$ H$maU Zmar ~m±P hmoVr h¡
gm_mÝ` Za (nwéf)

Amn OmZVo hr h¢ {H$ àmH¥${VH$ ê$n go ~rg {^Þ àH$ma Ho$ Eo_rZmo Aåb nmE OmVo h¢ VWm
S>rEZE _| Mma {^Þ àH$ma Ho$ jma nmE OmVo h¢ & Eogo 3 jmam| H$m g_yh EH$ {d{eîQ> Eo_rZmo
Aåb H$m Hy$Q> boIZ H$aVm h¡ & `{X Eogm Z hmoH$a 96 {^Þ àH$ma Ho$ Eo_rZmo Aåb hmoVo VWm
S>rEZE _| 12 {^Þ àH$ma Ho$ jma hmoVo, Vmo EH$ àHy$Q> Ho$ g§`moOZ hoVw Amdí`H$ Ý`yZV_ jmam|
H$s g§»`m hmoJr :
(A)

6

(B)

8

(C)

2

(D)

4

EH$ `moOZmË_H$ nm°br
{Zê$nU Xem©Vr h¡ :

(A)
(C)
57/1/1-11

ûm¥§Ibm _|

hmBS´>moOZ ~§Y
N-½bmBH$mo{g{S>H$ ~§Y

(B)
(D)
4

bmBZ BgHo$ àH$ma ~§Y H$m

noßQ>mBS> ~§Y
\$m°ñ\$moS>mBEñQ>a ~§Y

3
...


5
...


The periodic abstinence by a couple for family planning should be from :
(A)

Day 5 to 10 of menstrual cycle

(B)

Day 13 to 15 of menstrual cycle

(C)

Day 10 to 17 of menstrual cycle

(D)

Day 16 to 20 of menstrual cycle

Select the incorrect match from the following :
Human Karyotype

Characters

(A)

45 + XX

Broad palm with characteristic palm crease

(B)

44 + XXY

Overall feminine development

(C)

44 + XO

Sterile females as ovaries are rudimentary

(D)

44 + XY

Normal male

You know that there are twenty different types of naturally occurring
amino acids and four different types of bases in the DNA
...
If instead there are
96 different amino acids and 12 different bases in the DNA, then the
minimum number of combination of bases required to form a codon is :
(A)

6

(B)

8

(C)

2

(D)

4

The type of bond represented by the dotted line
polynucleotide chain is :

in a schematic

(A)

Hydrogen bond

(B)

Peptide bond

(C)

N-glycosidic linkage

(D)

Phosphodiester bond

57/1/1-11

5

P
...
O
...


{ZåZ{b{IV _| go {H$g àH$ma H$s n[apñW{V`m|/amoJ _| _mZd eara _| _mñQ> H$mo{eH$mAm| H$s
A{V{H«$`merbVm (_| d¥{Õ) ào{jV hmoVr h¡ ?
Q>mB\$m°BS>
EobOu
(A)
(B)
(C)
(D)
EoñHo$[aE{gg
ES²>g

8
...


{ZåZ{b{IV nmaOrdr O§VA
w m| (Q´>m§gOo{ZH$ E{Z_ëg) _| go {H$go nmo{b`mo Ho$ Q>rHo$
H$s gwajm narjU hoVw Cn`moJ {H$`m J`m h¡ ?
~H$ar
(A)
(B)
gyAa
Myhm
(C)
(D)

10
...


dh g_wÞV {deofH$ Omo AmZwd§{eH$V: ê$nm§V[aV nmaOrdr \$gb
OmVm h¡ :
(A)
_mÌm

JmoëS>Z amBg _|o nm`m

(B)
(C)
(D)
12
...


8
...


10
...


12
...
T
...


13

16

(A)
(A), (B), (C)

(R)
(D)
(A)

A{^H$WZ (A) Am¡a H$maU
ghr ì¶m»¶m H$aVm h¡ &

XmoZm| ghr h¢ Am¡a H$maU

(R),

A{^H$WZ

(A)

H$s

(B)

A{^H$WZ (A) Am¡a H$maU (R) XmoZm| ghr h¢, naÝVw H$maU
ghr ì¶m»¶m
H$aVm h¡ &

(R),

A{^H$WZ

(A)

H$s

(C)

A{^H$WZ

(A)

ghr h¡, naÝVw H$maU

(D)

A{^H$WZ

(A)

µJbV h¡, naÝVw H$maU

13
...


(A) :

(R) :
15
...


(A) :
(R) :

57/1/1-11

nm¡Yo _| àgm_mÝ` H$mo{eH$mAm| H$mo Q²>`y_a _| ê$nm§V[aV H$aZo hoVw `h S>rEZE
Ho$ EH$ A§e -S>rEZE H$mo hñVm§V[aV H$aZo _| g_W© h¡ &
^maV gaH$ma Zo OrE_ AZwg§YmZ g§~§Yr H$m`m] H$s d¡Ym{ZH$Vm {ZYm©aU hoVw
OrB©Egr
Zm_H$ g§JR>Z H$s ñWmnZm H$s h¡ &
O~ AmZwd§{eH$V: ê$nm§V[aV Ordm| H$mo nm[aV§Ì _| à{dîQ> H$am`m OmVm h¡
Vmo CZ na
8

For Questions number 13 to 16, two statements are given

one labelled as

Assertion (A) and the other labelled as Reason (R)
...

(A)

Both Assertion (A) and Reason (R) are true and Reason (R) is the
correct explanation of the Assertion (A)
...


13
...


(D)

Assertion (A) is false, but Reason (R) is true
...

Reason (R) :

Emotional, religious and social factors are no deterrents to
the legal adoption of orphaned and destitute children in
India
...


Assertion (A) : Linked genes do not show dihybrid F2 ratio 9 : 3 : 3 : 1
...


Linked genes do not undergo independent assortment
...

Reason (R) :

It is able to deliver a piece of DNA known as T-DNA to
transform normal plant cells into a tumor
...


Assertion (A) : Indian Government has set up an organisation known as
GEAC to decide the validity of GM research
...


57/1/1-11

9

P
...
O
...


18
...


ZrMo Xr JB© Vm{bH$m _|

A, B, C VWm D

H$mo nhMm{ZE :

\$b{^{Îm

A

B

Kmg Hw$b Ho$ ~rO H$m ~rOnÌ

^«yU Aj

C

D

~rO _| ~rOm§S>H$m` H$m Adeof

4

1
2

=2

ZrMo {XE JE {MÌ H$m àojU H$s{OE & Cg àH¥${V-d¡km{ZH$ H$m Zm_ {b{IE {OÝhmo§Zo {MÌ _|
{XE JE Ord Ho$ AmYma na Ordm| Ho$ {dH$mg H$m {gÕmÝV {X`m & {dH$mg Ho$ {bE CZH$s Xr
JB© ì`m»`m ^r {b{IE &

2

g§VmZhrZ `wJbm| Ho$ ghm`VmW© ghm`H$ OZZ àm¡Úmo{JH$s (EAmaQ>r) H$m`©H«$_ _| AnZmE OmZo
dmbo _yb MaUm| H$m CëboI H$s{OE & Bgo naIZbr {eew (Q>oñQ> Q²>`y~ ~o~r) H$m`©H«$_ ^r
?

20
...


Identify A, B, C and D in the table given below :
Terms
Pericarp

1
=2
2

4

Part of the plant it represents
Cotyledon in seed of grass family

Embryonal axis
Remains of nucellus in a seed
18
...


20
...
Name the naturalist and write the
explanation given by him that evolution of life forms had occurred on the
basis of this example
...
Why is it also known as test
tube baby programme ?

2

(a)

A farmer while working on his farm was bitten by a poisonous
snake
...

(i)

What did the doctor inject and why ?

(ii)

Name the kind of immunity provided by this injection
...

11

1
2
1
2

2

P
...
O
...


ZrMo {XE JE O¡d_mÌm Ho$ {nam{_S> H$m AÜ``Z H$s{OE & Eogr Xmo pñWV
Ho$ Zm_ {b{IE Omo ñVa

VWm ñVa

\$gb)

_| nmB© Om gH$Vr h¢ & Bg àH$ma Ho$ {nam{_S> H$m

Zm_ {bIH$a Cg nm[aV§Ì H$m Zm_ {b{IE {Og_| `h nm`m OmVm h¡ &

2

IÊS> J
22
...


3

(H$) Cg àH$ma Ho$ S>rEZE H$m Zm_ VWm Xmo A{^bjU {b{IE Omo S>rEZE A§Jw{bN>mnr
(S>rEZE q\$JaqàqQ>J) VH$ZrH$ H$m AmYma ~ZmVo h¢ &

1+1
1

(I) Bg VH$ZrH$ Ho$ H$moB© Xmo AZwà`moJm| H$m CëboI H$s{OE &

24
...
Eb
...


Study the diagram of a pyramid of biomass given below
...


2

SECTION C

22
...
Mention the schedule to be followed for effective outcome
...


(a)

Name and write two characteristics of the type of DNA that forms
the basis of DNA fingerprinting technique
...


3

(b)

Mention any two applications of this technique
...
L
...
Name the scientists whose hypothesis prompted him to
carry out this experiment
...
T
...


25
...


27
...


B©H$mo Ama
H$s{OE &

I (EcoRI)

Ho$ Zm_H$aU hoVw AnZmB© JB© na§nam H$s ì`m»`m
2

Ho$db AmaoI H$s ghm`Vm go B©H$mo Ama I H$s S>rEZE nm°br
{H«$`m H$mo àX{e©V H$s{OE &

(H$) `h H¡$go gw{ZpíMV hmoVm h¡ {H$ Am°{H©$S>
_{jH$m Ûmam hr hmo ? ì`m»`m H$s{OE &

14

1

H$m namJU EH$ {d{eîQ Om{V H$s>

(I) Bg CXmhaU H$s ghm`Vm go gh-{dH$mg H$m dU©Z H$s{OE &
57/1/1-11

na

2
1

25
...


27
...
Do you think the son and the
daughter have inherited the disease from their father ? Work out a cross
to justify your answer
...


2

OR
(b)

(i)

Explain the convention for naming EcoRI
...

28
...


(b)
57/1/1-11

1

2

Describe co-evolution with the help of this example
...
T
...


IÊS> K
29

30

3

ZrMo {XE JE àíZm| Ho$ CÎma Xr{OE &

29
...
29 and 30 are case-based questions
...

29
...


4

total land area whereas its share of the global species diversity is an
impressive 8·1 per cent ! However, in these estimates of species,
prokaryotes do not figure anywhere
...
The data collected
based on the survey conducted for species richness of groups of
mammals in three different regions of the world is shown in the bar
graph given below :

(a)

Why is the species richness maximum in Region III in the bar
graph ?

1

OR
(a)

Why is the species richness minimum in Region I in the bar graph ?

1

(b)

Plants and animals do not have uniform diversity in the world but
show rather uneven distribution
...


1

Why is it that prokaryotes do not have an estimated number of their
species diversity as seen in plants and animals ? Explain
...
T
...


30
...


Study the diagram given below that shows the steps involved in the
procedure of selecting transformed bacteria and answer the questions
that follow :

1 1
+
2 2

(a)

Identify the colony that has got transformed
...


(b)

What are the sites in a plasmid called where ampicillin and
tetracycline resistance genes are inserted ? State their role in genetic
engineering
...


1
OR

(c)

57/1/1-11

State the role of -galactosidase in insertional inactivation
...
T
...


IÊS> L>
31
...


(H$) AmZwd§{eH$s Ho$ àmapå^H$ à`moJm| _| go EH$ à`moJ Ûmam gwñnîQ> hmo J`m Wm {H$
AmZwd§{eH$ nXmW© H$m ñWm`r hmoZm CgH$m EH$ _hÎdnyU© A{^bjU h¡ & Cg
d¡km{ZH$ H$m Zm_ {b{IE {OZHo$ à`moJ Ûmam `h {gÕ hmo gH$m & à`moJ H$m dU©Z
H$s{OE VWm CgHo$ {ZîH$f© H$m dU©Z H$s{OE &

2

1
2

5

AWdm
(I) AmnH$mo ~¢JZr \y$bm| dmbm _Q>a H$m EH$ b§~m nm¡Ym {X`m J`m {OgH$m OrZràê$n
(OrZmoQ>mBn) kmV Zht h¡ & nm¡Ym| _| Ho$db ñdnamJU Ho$ Ûmam {H$E JE {d{^ÝZ H«$m°gm|
H$s ghm`Vm go nm¡Yo H$m OrZràê$n (OrZmoQ>mBn) kmV H$s{OE & AnZo Ûmam ~ZmE JE
àË`oH$ H«$m°g H$m OrZràê$nr (OrZmoQ>mBn) VWm Ñí`àê$nr (\$sZmoQ>mBn) AZwnmV
{b{IE &
57/1/1-11

20

5

SECTION E
31
...
2
2

angiosperm
...


32
...
Name the
scientist and describe his experiment
...


5
OR

(b)

A tall pea plant bearing violet flowers with unknown genotype is
given
...
Write the genotypic and phenotypic ratios of each
cross shown by you
...
T
...


33
...


(a)

(i)

Name and explain the property present in normal cells but is
lost in cancer cells
...
Name them and explain
how
...
Given below is the flow diagram of stages
of STP
...


(3)

1

Explain the final step in the settling tank before the
treated effluent is released into water bodies
...
Write how each one acts as
a biofertiliser
...
T
...


Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior Secondary School Certificate Examination,2024
SUBJECT NAME BIOLOGY (Q
...
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
...
To avoid mistakes, it is
requested that before starting evaluation, you must read and understand the spot evaluation
guidelines carefully
...
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
...
”

3

Evaluation is to be done as per instructions provided in the Marking Scheme
...
Marking Scheme should be
strictly adhered to and religiously followed
...
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
...
The students
can have their own expression and if the expression is correct, the due marks should be awarded
accordingly
...
If there is any variation, the same should be zero after delibration and discussion
...


6

Evaluators will mark( √ ) wherever answer is correct
...

Evaluators will not put right (✓)while evaluating which gives an impression that answer is correct
and no marks are awarded
...


7

If a question has parts, please award marks on the right-hand side for each part
...
This may be followed strictly
...

This may also be followed strictly
...
1

be retained and the other answer scored out with a note “Extra Question”
...
It should be penalized only once
...
Please do not hesitate to award full marks if the answer
deserves it
...
e
...
This is in view of the reduced syllabus and number
of questions in question paper
...

Giving more marks for an answer than assigned to it
...

Wrong transfer of marks from the inside pages of the answer book to the title page
...

Wrong totaling of marks of the two columns on the title page
...

Marks in words and figures not tallying/not same
...

Answers marked as correct, but marks not awarded
...
It should merely be a line
...
)
● Half or a part of answer marked correct and the rest as wrong, but no marks awarded
...

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
...

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


18

The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the
prescribed processing fee
...


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

3
...

5
...

7
...

9
...

11
...

13
...


(C) / Maize
(B) / Human Chorionic Gonadotropin
(C) / Day 10 to 17 of menstrual cycle
...

(A) / Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation
of Assertion (A)
...
(D) / Assertion (A) is false, but Reason (R) is true
16
...

SECTION – B
17
...


½

-He said Giraffes in an attempt to forage leaves on tall trees had to adapt by

½

elongation of their necks
...


½
2

19
...

• Test tube baby programme – because initial process is carried out in the laboratory /
in vitro

½x3
`
½

2

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

(a)
(i)
-Readymade or preformed antibodies or antitoxins against the snake venom ,

½

-A quick immune response is required in this case
...
Greater
biodiversity leads to more sustainable ecosystem
...

A – Zooplankton ,
B – Phytoplankton ,

½x4

-Inverted pyramid of biomass ,

2

-Sea Ecosystem/ Aquatic ecosystem
SECTION – C
22
...


½

-They inhibit ovulation , and implantation as well as, alter the quality of
cervical mucus to prevent or retard the entry of sperms
...

-After a gap of 7 days it has to be repeated in the same pattern till the female
desires to prevent conception
...

(a)
-Satellite DNA / Repetitive DNA/ VNTR

1

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

(any two applications)

3

24
...


1
½+½

• Oparin , Haldane
(b)
Analysis of meteorite content also revealed similar compounds indicating that similar
processes are occurring in space
...
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

No,

½

Son inherited disease from the mother and daughter inherited disease from both mother and

½

father
...
5

1

3
27
...


(ii) First transgenic Cow – Rosie,
-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
cow milk
...


½x4

(ii)

½

½

3

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

(b)
If female bee pattern changes during evolution the flower needs to co-evolve to resemble the

1

female bee to get pollinated
...
(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
...

(b)

Latitudinal gradient in diversity

(c)

Conventional taxonomic methods are not suitable for identifying microbial species, and

1
1+1

many species are not culturable under laboratory conditions
...
(a) Colony 4 is transformed with plasmid containing recombinant DNA, as they will not show

½ +½

resistance towards tetracycline
...


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
...
(a)
(i) Each microspore mother cell in sporogenous tissue undergoes meiosis , forming
microspores tetrad by the process called microsporogenesis , microspores form cluster of
four cells called microspore tetrad , as the anther matures and dehydrates, each

½x5

microspore undergo assymetric division to produce a vegetative cell and a generative
cell
...
8

(iii) Both LH and FSH attain a peak level in mid of menstrual cycle rapid secretion of
LH leading to maximum level is called LH surge , which induces rupture of graffian

½x5

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
...

(a)
½

-Frederick Griffith
Took two strains of Streptococcus pneumoniae bacteria and inject them into mice

½

- R strain – Rough and Non–virulent
-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
...
This must be due to the
transfer of the genetic material
...


½

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


Case IITt 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
...
11

33
...


½
1

(ii)
Cellular oncogenes / Proto-oncogenes , when activated under certain conditions could lead to

½ +½

oncogenic transformation of the cells
...

(2) MRI – detects cancer of internal organs ,

½

uses strong magnetic fields and non–ionising radiations to detect pathological and physiological
changes in living tissue

1

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
...
( Activated sludge)
(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
...
12

-Glomus (fungi), live in symbiotic association with roots of higher plants and absorb phosphorus
from the soil and passes it to plants
...
13



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