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Physics - 50 Chemistry - 50 Biology - 100

Question Booklet Version Roll Number Question Booklet Sr. No.

11 (Write this number on

your Answer Sheet) 0
This is to certify that, the entries of RCC-2023 Roll No. and Answer Sheet No. have been correctly written and verified.

Question Paper
Candidate’s Signature With Solutions Invigilator’s Signature

“Stand up, be bold, be strong.

Take the whole responsibility on
your own shoulders and know
that you are the creator of your
own destiny.”
– Swami Vivekanand
• Today’s Test Syllabus •
Physics : Semiconductors + Dual Nature of Radiation and Matter + Ray Optics
Chemistry : Alcohols, Phenols, Ethers + Solutions + Biomolecules
Biology : Morphology + Plant Growth and Development + Respiration In Plants + Anatomy
+ Biotechnology (I+II)
• Next Test Syllabus • Date : 04/02/2024
Physics : Semiconductors + Mechanical Properties of Fluids + Work, Energy and Power
Chemistry : Alcohols, Phenols, Ethers + Solutions + Electrochemistry + Aldehydes, Ketones
and Carboxylic Acids
Biology : Morphology + Plant Growth and Development + Respiration In Plants
QUESTION BOOKLET VERSION : 11
PHYSICS Sol. (1) :
Section : A

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1. A p-n photo detector is fabricated from a 5. In pure semiconductor at zero kelvin
semiconductor with a band gap of 1.8 eV. It can temperature
detect a signal of wavelength 1) The valence band is completely filled and the
conduction band partially filled.
1) 6000 Å 2) 7200 Å 2) Valence band completely filled and
3) 7500 Å 4) 8000 Å conduction band completely empty
3) Both valence band and conduction band
Sol. (1) : empty
4) The valence band is partially empty and the
conduction band is partially filled.
Sol. (2) : Conceptual
2. Choose the only true statement from the 6. Depletion layer in the p-n junction consists of
following: 1) electrons
2) holes
1) The resistivity of a semiconductor , increase
3) positive and negative ions fixed in their
with increase in temperature
position
2) Substances with energy gap of 10 eV and 4) both electrons and holes
above are conductors. Sol. (3) : Depletion layer consists of immobile ions.
3) Conductivity in conductor increase with 7. The barrier potential of a p-n junction depends
temperature on
4) Conductivity of semiconductor decrease with a) type of semiconductor material
decrease in temperature b) amount of doping
c) temperature
Sol. (4) : As temperature is decreased, less electron
Which one of the following is correct?
get the energy to jump from conduction bond
1) (a) and (b) only 2) (b) only
to valence bond, and this decreases the
3) (b) and (c) only 4) (a), (b) and (c)
conductivity of semiconductor.
Sol. (4) : Conceptual
3. C and Si both have same lattice structure, having 8. Statement-1 : By doping silicon semiconductor
4 bonding electrons in each. However, C is with pentavalent material, the electrons density
insulator where as Si is intrinsic semiconductor. increases.
This is because Statement-2 : The n-type semiconductor has net
1) The four bonding electrons in the case of C lie negative charge.
in the second orbit, whereas in the case of Si 1) Both statement-1 and statement-2 are True
they lie in the third 2) Both statement-1 and statement-2 are False
2) The four bonding electrons in the case of C lie 3) Statement-1 is True but Statement-2 is False
in the third orbit, whereas for Si they lie in 4) Statement-1 is False but Statement-2 is True
the fourth orbit. Sol. (3) : Conceptual
9. The temperature (T) dependence of resistivity
3) In case of C the valance band is not
() of semiconductor is represented by
completely filled at absolute zero temperature
4) In case of C the conduction band is partly
filled even at absolute zero temperature
Sol. (1) : Conceptual 1) 2)
4. A semiconductor is known to have an electron
concentration of 8 × 10 13 per cm 3 and hole
concentration of 5 × 10 12 per cm 3 . The
semiconductor is
3) 4)
1) n-type 2) p-type
3) intrinsic 4) none of these

PCB TEST : 26 2 Date : 28/01/2024

QUESTION BOOKLET VERSION : 11
Sol. (3) : Conceptual Sol. (1) : Conceptual
10. Which energy band diagram represents the 13. Which statement is correct for p-type

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n-type of semiconductor semiconductor?
1) the number of electrons in conduction band
is more than the number of holes in valence
1) band at room temperature
2) the number of holes in valance band is more
than the number of electrons in conduction
band at room temperature
2)
3) there are no holes and electrons at room
temperature
4) number of holes and electrons is equal in
3) valence and conduction band
Sol. (2) : Conceptual
14. When light is incident on a surface, photoelectrons
4) are emitted. For photoelectrons.
1) the value of kinetic energy is same
Sol. (2) : Conceptual 2) kinetic energy does not depend on the wave
11. An intrinsic semiconductor is converted into length of incident light
n-type extrinsic semiconductor by doping it 3) the value of kinetic energy is equal to or less
with than a maximum kinetic energy
1) Germanium 2) Phosphorous
4) kinetic energy does not depend on frequency
3) Aluminium 4) Silver of incident light.
Sol. (2) : Doping with pentavalent impurities like Sol. (3) :
phosphorous, arsenic, antimony, bismuth
convert intrinsic s.c. to n-type extrinsic s.c. 15. The threshold wavelength for photoelectric
emission in tungsten is 400 nm The
12. The given graph represents V-I characteristic
wavelength of light that must be used in order
for a semiconductor device.
to eject electrons with a maximum energy of
0.9 eV is
1) 120 nm 2) 310 nm
3) 380 nm 4) 400 nm
Sol. (2) : Conceptual
16. Photoelectric emission occurs only when the
incident light has more than a certain
minimum
Which of the following statement is correct?
1) power 2) wavelength
1) It is V-I characteristic for solar cell where, point
A represents open circuit voltage and point B 3) intensity 4) frequency
short circuit current. Sol. (4) : Conceptual
2) It is a for a solar cell and point A and B represent 17. Kmax of photoelectron is independent of
open circuit voltage and current, respectively.
1) frequency of incidence light
3) It is for a photodiode and points A and B
2) threshold frequency of electron emitter
represent open circuit voltage and current,
respectively 3) number of photons incident on the electron
4) It is for a LED and points A and B represent emitter
open circuit voltage and short circuit current, 4) All of the above
respectively Sol. (3) : Conceptual

PCB TEST : 26 3 Date : 28/01/2024

QUESTION BOOKLET VERSION : 11
18. The work function of metal is 1 eV. Light of 22. Energy required to remove an electron from
wavelength 3000Å is incident on this metal aluminium surface is 4.2 eV. If light of

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surface. The maximum velocity of emitted wavelength 2000 Å falls on the surface. The
photoelectrons will be kinetic energy of the fastest electron ejected
1) 10 m/s 2) 103 m/s from the surface will be
3) 104 m/s 4) 106 m/s 1) 6.2 eV
2) 2.2 eV
Sol. (4) :
3) 2 eV
4) 1.2 eV

Sol. (3) :

19. When light of wavelength 300 nm (nanometer)

falls on a photoelectric emitter, photoelectrons 23. Light of wavelength 5000 Å is falling on a
are liberated of zero kinetic energy. For sensitive surface. If the surface has received
another emitter, however, light of 600 nm 10–7 J of energy , then the number of photons
wavelength is sufficient for creating falling on the surface will be
photoemission. What is the ratio of the work 1) 5 × 1011
function of the two emitter? 2) 2.5 × 1011
1) 1 : 2 2) 2 : 1 3) 3 × 1011
3) 4 : 1 4) 1 : 4 4) None of these

Sol. (2) : Sol. (2) :

20. A photosensitive metallic surface has work

function hv0 If photons of energy 2hv0 fall on this
surface the electrons come out with a maximum 24. A radiation of energy E falls normally on a
velocity of 4 × 106 m/s. When the photon energy perfectly reflecting surface. The momentum
is increases to 5hv0, then maximum velocity of transferred to the surface is (c = Velocity of
photoelectron will be light)
1) 2 × 106 m/s
2E 2E
2) 2 × 107 m/s 1) 2)
c c2
3) 8 × 105 m/s
4) 8 × 106 m/s E E
3) 4)
Sol. (4) : Conceptual c2 c
21. Assertion : In photoelectric effect, the minimum
Sol. (1) :
negative potential given to collector plate that can
stop photocurrent is stopping potential.
Reason : All the photoelectrons emitted from
metal surface do not have the same energy.
1) If both assertion and reason are true and the 25. If the de Broglie wavelengths for a proton and
reason is the correct explanation of assertion. for a -particle are equal, then the ratio of their
velocities will be
2) If both assertion and reason are true and the
reason is not the correct explanation of assertion. 1) 4 : 1 2) 2 : 1
3) If assertion is true but reason is false 3) 1 : 2 4) 1 : 4
4) If both assertion and reason are false.
Sol. (1) :
Sol. (2) : Conceptual

PCB TEST : 26 4 Date : 28/01/2024

QUESTION BOOKLET VERSION : 11
26. The critical angle for light going from medium Sol. (4) : Conceptual
x into medium y is . The speed of light 30. The graph between angle of deviation () and

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in medium x is v. The speed of light in medium angle of incidence (i) for a triangular prism is
y is represented by
1) v(1 – cos ) 2) v/sin
3) v/cos  4) v cos 

Sol. (2) : 1) 2)

27. A ray of light travels through a transparent

3) 4)
slab of refractive index 1.5. The speed of light
ray in the slab will be
1) 1.0 × 1010 cm s–1 Sol. (3) : For the prism as the angle of incidence (i)
2) 1.5 × 1010 cm s–1 increases, the angle of deviation () first decreases
goes to minimum value and then increases.
3) 2.0 × 1010 cm s–1
31. There are two plane
4) 1.5 × 1010 m s–1
mirror inclined at
40°, as shown. A
Sol. (3) : ray of light is
incident of mirror
M 1. What whould
28. An equiconvex lens of refractive index 1.5 is be the value of angle of incidence ‘i’ so that
the light ray retraces its path after striking the
 4 mirror M2?
put in water     . If its focal length in
 3 1) 60° 2) 50°
water is 1.20 m, then its radius of curvature 3) 30° 4) 40°
will be
1) 0.1 m 2) 0.2
Sol. (4) :
3) 0.3 m 4) 0.4 m

Sol. (3) :

32. A bubble in glass slab [µ = 1.5] when viewed

29. Assertion : The focal length of lens does not from one side appears at 5 cm and 2 cm from other
change when red light is replaced by blue light. side then thickness of slab is
Reason : The focal length of lens depends on the 1) 3.75 cm 2) 23 cm
color of light used. 3) 10.5 cm 4) 1.5 cm
1) If both assertion and reason are true and the Sol. (3) :
reason is the correct explanation of the
assertion
2) If both assertion and reason are true and the
reason is not the correct explanation of the
assertion
3) If assertion is true but reason is false
4) If assertion is false but reason is true

PCB TEST : 26 5 Date : 28/01/2024

QUESTION BOOKLET VERSION : 11
33. A lens of large focal length and large Section : B
aperture is best suited as an objective of an 36. What should be the value

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astronomical telescope since
of distance d so that
1) a large aperture contributes to the quality and final image is formed on
visibility of the images the object itself. (focal
2) a large area of the objective ensures better lengths of the lenses are
light gathering power written on the lenses).
3) a large aperture provides a better resolution 1) 10 cm 2) 20 cm
4) All of the above 3) 5 cm 4) None of these
Sol. (4) : Conceptual
Sol. (1) :
34. An object moving at a speed of 5 m/s towards
a concave mirror of focal length f = 1 m is at a
distance of 9 m. The average speed of the image
is
1 1
1) m/s 2) m/s
5 10
5 2
3) m/s 4) m/ s
9 5
Sol. (1) : Conceptual
35. A point object is placed at 37. Curved surfaces of a plano-

a distance of 0.3 m from a convex lens of refractive

convex lens of focal length index  1 and a plano-
0.2 m cut into two equal concave lens of refractive
halves, each of which is index 2 have equal radius
displaced by 0.0005 m, as of curvature as shown in
shown. If C1 and C2 be their figure. Find the ratio of radius of curvature to
optical centres then, the focal length of the combined lenses.
1) an image is formed at a distance of 0.6 m from
1
C1 or C2 along principal axis 1) 2) 1   2
 2  1
2) two images are formed, one at a distance of
0.6 m and other at a distance of 1.2 m from
1
C1 or C2 along principal axis 3) 4)  2  1
1   2
3) an image is formed at a distance of 0.12 m
from C1 or C2 along principal axis. Sol. (2) :
4) two images are formed at a distance of 0.6 m
from C 1 or C 2 along principal axis at a
separation of 0.003 m.

Sol. (4) :

PCB TEST : 26 6 Date : 28/01/2024

QUESTION BOOKLET VERSION : 11
38. A ray of light is incident on a prism as shown Sol. (3) :
in figure. The net deviation of the ray is

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1) 2.4° 2) 4.8°
3) 6.4° 4) 7.2°
Sol. (3) :

39. A point object O is 41. A proton a neutron, an electron, and an

-particle have same energies. Choose the
placed in front of correct option regarding their de Broglie
a glass rod having wavelengths.
spherical end
of radius of 1)  p  n   e    2)     p   ñ   e

curvature 30 cm. The image would be formed 3) e   p  n   4) e   p  n  

at (g = 3/2)
Sol. (2) :
1) 30 cm left
2) infinity 42. Light with an energy flux of 25 × 104 W m–2
falls on a perfectly reflecting surface at normal
3) 1 cm to the right
incidence. If the surface area is 15 cm 2, the
4) 18 cm to the left average force exerted on the surface is
1) 1.25 × 10–6 N 2) 2.50 × 10–6 N
Sol. (1) :
3) 1.20 × 10–6 N 4) 3.0 × 10–6 N

2 P 2  25  10 4  15  10 4
Sol. (2) : F   8
 2.5  10 6 N .
C 3  10
43. When a certain metallic surface is illuminated
with monochromatic light of wavelength ,
the stopping potential for photoelectric
40. When he rectangular metal tank is filled up current is 6 V 0. When the same surface is
to the top with an unknown liquid, an illuminated with light of wavelength 2 , the
observer with eyes level with the top of the stopping potential is 2V 0 . The threshold
tank can just see the corner E. Find the index wavelength of this surface for photoelectric
of refraction of the liquid.
effect is
1) 6  2) 4 /3
3) 4  4) 8 

Sol. (3) :

1) 1.5 2) 1.25
3) 1.67 4) 2.4

PCB TEST : 26 7 Date : 28/01/2024

QUESTION BOOKLET VERSION : 11
44. Figure is the plot of the stopping potential 46.
versus the frequency of the light used in an

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experiment on photoelectric effect.
Choose the correct option.

If in a p-n junction, a sinosudial input signal

is applied as shown, then output voltage
will be

1) 2)

1) The ratio h/e is 4 × 10–15 Vs and work function

is 0.8 eV 3) 4)
2) The ratio h/e is 2 × 10–15 Vs and work function
Sol. (2) :
is 0.4 eV
3) The ratio h/e is 2.5 × 10 –15 Vs and work
function is 0.6 eV
47. A sinosoidal voltage of peak value 200 volt is
4) The ratio h/e is 4 × 10–15 and work function is
connected to a diode and resistor R in the circuit
0.6 eV
shown. If the forward resistance of the diode is
Sol. (1) : negligible compared to R, the rms voltage across
R is approximately (in volt)

1) 200 2) 100

45. In the following diagram, if V2 > V1 then 100

3) 4) 283
2

Sol. (2) :
48. Two PN-junctions can be connected in series
by three different methods as shown in the
figure. If the potential difference in the
junction is the same, then the correct
1) 1   2 2)  1   2 connections will be

3)  1   2 4)  1   2
Sol. (4) :

1) In the circuit (A) and (B)

2) In the circuit (B) and (C)
3) In the circuit (A) and (C)
4) Only in the circuit (A)
PCB TEST : 26 8 Date : 28/01/2024
QUESTION BOOKLET VERSION : 11
Sol. (2) : Conceptual 50. What is the conductivity of a semiconductor
49. If the ratio of the concentration of electrons to if electron density = 5 × 10 12/cm 3 and hole

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7 density = 8 × 1013/cm3? (e = 2.3 V–1 s–1 m2 and
that of hole in a semiconductor is and the h = 0.01 V–1 s–1 m3?
5
7 1) 5.634
ratio of currents is , then what is the ratio
4 2) 1.968
of their drift velocities?
3) 3.421
5 4 5 4
1) 2) 3) 4)
8 5 4 7 4) 8.964
Sol. (3) : Sol. (2) :

PCB TEST : 26 9 Date : 28/01/2024

QUESTION BOOKLET VERSION : 11

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
Chemistry: Alcohols, Phenols and Ether + Biomolecules + Solutions
CHEMISTRY 52. The major product in the following reaction
SECTION ‘A’ is
51. Which of the following products is formed in O
Cl 1.CH 3 MgBr,dry ether, 0°C
 
good yeilds by the reduction of methyl 4-oxo- CH3 2. aq. acid

hexanoate (M) with NaBH4?

O OH
H3C
1) 2) CH3
CH3 H3C CH3

3) 4) CH3
1) Methyl 4-hydroxyhexanoate O CH2
O CH3
2) 6-Hydroxy-3-hexanone
3) 1, 4-Hexanediol
4) 4-Hydroxyhexanal Sol. (4) .

Sol. (1)

53. The final product (B) in the following reaction

sequence is
O (i)CH MgBr POCl 3
 3
(ii)H  /H 2 O
 A 
Pyridine,273K
 B

CH3 CH3
1) Cl 2) OPOCl 2

CH2 CH3
3) 4)

Sol. (4)
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PCB TEST - 26 10 Date: 28/01/2024

QUESTION BOOKLET VERSION : 11
57. Iso-propylbenzene on air oxidation in the

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COOH
H OH presence of dilute acid gives .....
1) C6H5COOH 2) C6H5COCH3
54. The absolute configuruation of H Cl
CH3 3) C6H5CHO 4) C6H5OH

is :
1) 2S, 3S 2) 2R, 3R Sol. (4)
3) 2R, 3S 4) 2S, 3R
Sol. (4) 58. Salicylic acid is prepared from phenol by ......
55. What is the product of the following reaction 1) Reimer-Tiemann reaction
?
2) Kolbe’s reaction
1.B H ·THF 3) Kolbe electrolysis reaction
2 6
2.NaOH,H 2 O 2
 ?
4) None of these
O Sol. (2)
OH
1) 2) 59. Which of the following is most acidic ?
1) p-Nitrophenol 2) o-Cresol
OH 3) Phenol 4) Anisole
3) 4) Sol. (1)
60. Victor Meyer’s test is not given by
Sol. (2) 1) (CH3 )3 COH
2) CH3 CH2 OH
3) (CH3 )2 CHOH
4) CH3 CH2 CH2 OH
56. Sol. (1)

61. Best method to prepare is

In the above given compound how many
functional groups are reduced by LAH the combination of
(Lithium aluminium hydride) and SBH
(Sodium borohydride) respectively?
1)
1) 4, 4 2) 4, 3
3) 3, 4 4) 4, 2
2)
Sol. (4)

3)
4) Both (1) & (2)
Sol. (4)
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PCB TEST - 26 11 Date: 28/01/2024

QUESTION BOOKLET VERSION : 11

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67. At higher altitudes, water boils at temperature
CH3 < 100°C because :
NBS/CCl 4 CH 3ONa 1) temperature of higher altitudes is low
62. 
 A   B
2) atmospheric pressure is low

The final product (B) formed in the above re- 3) the proportion of heavy water increases
action is 4) atmospheric pressure becomes more
Sol. (2)
1) CH2CH 3 2) OCH 3 68. Match the column I with column II and mark
the appropriate choice

3) CH 3O CH3 4) CH 2OCH3

Sol. (4)
63. Azeotropic mixture :
1) are ethose whihch can be fractionally distilled
2) have definite constant boiling point
3) have same definite composition at any
pressure 1) A-i, B-iii, C-ii, D-iv
4) are those which have different composition 2) A-iii, B-i, C-iv, D-ii
in liquid and vapour state
3) A-ii, B-iii, C-iv, D-i
Sol. (2)
4) A-iii, B-ii, C-i, D-iv
64. Which of the following liquid pairs shows a Sol. (2)
positive deviation from Raoult’s law ?
69. Colligative properties depend on
1) Acetone-chloroform
1) the nature of the solute particles dissolved in
2) Benzene-methanol solution
3) Water-nitric acid 2) the number of solute particles in solution
4) Water-hydrochloric acid 3) the physical properties of the solute particles
Sol. (2) dissolved in solution
65. Which of the following azeotropic solutions 4) the nature of solvent particles
has the b.p. less than b.p. of the constituents Sol. (2)
A and B ? 70. 3 moles of P and 2 moles of Q are mixed, what
1) CHCl3 and CH3COOH will be their total vapour pressure in the
2) CS2 and CH3COCH3 solution if their partial vapour pressures are
80 and 60 torr respectively?
3) CH3 CH2 OH and CH3COCH3
1) 80 torr 2) 140 torr
4) CH3CHO and CS2
3) 72 torr 4) 70 torr
Sol. (3)C 2 H 5 OH and CH 3 COCH 3 shows +ve
deviation. 3
Sol. (3)Mole fraction of P =  0.6
66. Which of the following is less than zero for 3 2
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ideal solutions ?
1) Hmix 2) Vmix
3) Gmix 4) Smix
Sol. (3)Mixing of solutions is spotaneous, hence,
Gmix < 0.

PCB TEST - 26 12 Date: 28/01/2024

QUESTION BOOKLET VERSION : 11

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71. What does A point signifies in the figure given 74. Which of the following is correct for a solution
below? showing positive deviations from Raoult's
law?
1) V = + ve, H = + ve
2) V = –ve, H = –ve
3) V = + ve, H = –ve
4) V = –ve, H = + ve
Sol. (1) For non–ideal solution showing positive dev.,
V = +ve, H = +ve.
1) Vapour pressure of solute
75. Which of the following solutions will have
2) Vapour pressure of pure solvent the lowest vapour pressure ?
3) Vapour pressure of solution 1) 0.1 M Na3PO4 2) 0.1 M Na2SO4
4) None of the above 3) 0.1 M BaCl2 4) 0.1 M Urea
Sol. (2) Sol. (1)
72. The given graph shows the vapour pressure 76. Select the No. of correct statement
temperature curves for some liquids.
i. Some common examples of carbohydrates
1 atm are cane sugar, glucose, starch.
A
B ii. Most of them have a general formula,
C Cx (H2 O) y.
D
V.P. iii. Carbohyrates Considered as hydrates of
carbon from where the name carbohydrate
T was derived. [Old NCERT Page-410]
which has highest B.P.
1) only one statement is correct
1) A 2) B
2) only two statement is correct
3) C 4) D
3) All three statements are incorrect
Sol. (4)
4) All three statements are correct
73. The diagram given below is a vapour pressure
composition diagram for a binary solution of Sol. (4)
A and B. 77. Match the sugars in column I with their types
given in column II and mark the appropriate
choice. [Old NCERT Page-412]
Column - I Column- II
a. Glucose i Ketohexose
b. Fructose ii Aldohexose
c. Ribose iii Aldotetrose
d. Erythrose iv Aldopentose
1) a-iv, b-i, c-iii, d-ii 2) a-iii, b-iv, c-i, d-ii
In the solution, A – B interactions are 3) a-i, b-ii, c-iii, d-iv 4) a-ii, b-i, c-iv, d-iii
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1) similar to A–A and B–B interactions Sol. (4)

2) greater than A–A and B–B interactions
3) smaller than A–A and B–B interactions
4) unpredictable.
Sol. (3) As solution shows positive deviations, A–B
interactions are smaller.
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QUESTION BOOKLET VERSION : 11

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
78. Assertion (A): Amylose is insoluble in water. 82. Which of the following statement is not
Reason (R): Amylose is a long linear molecule correct? [Old NCERT Page No. 420]
with more than 200 glucose units. 1) Only -amino acids are obtained on
In the light of the above statements, choose hydrolysis of proteins.
the correct answer from the options given 2) The amino acids which are synthesised in the
below: body are known as non-essential amino acids.
1) Both A and R are correct and R is the correct 3) There are 20 essential amino acids
explanation of A. 4) L-amino acid are represented by writing the
2) Both A and R are correct and R is NOT the –NH2 group on the left side.
correct explanation of A. Sol. (3)
3) A is correct but R is not correct. 83. Sulphur (S) containing amino acids from the
4) A is not correct but R is correct. following are:
Sol. (4) A. isoleucine B. cysteine
79. Which of the following statements is correct? C. lysine D. methionine
1) Gluconic acid can form cyclic (acetal/ E. glutamic acid
hemiacetal) structure 1) A, D 2) B, D
2) Gluconic acid is a dicarboxylic acid 3) B, C, E 4) A, B, C
3) Gluconic acid is a partial oxidation product of Sol. (2)
glucose 84. Mark the wrong statement about denaturation
4) Gluconic acid is obtained by oxidation of of proteins [Old NCERT Page No. 424]
glucose with HNO3 1) The primary structure of the protein does not
Sol. (3) change
80. Whcih of the following is not true about 2) Globular proteins are converted into fibrous
amino acids? proteins
1) They are constituents of all proteins 3) Fibrous proteins are converted into globular
proteins
2) Alanine having one amino and one carboxylic
group 4) The biological activity of the protein is lost
3) Most naturally occurring amino acids have D- Sol. (3)
configuration 85. Match the vitamins given in column I with
4) Glycine is the only naturally occurring amino the deficiency diseases caused by it given in
acid which is optically inactive column II and mark the appropriate choice.
Sol. (3)
81. Amino acids generally exist in the form of Column - I Column- II
zwitter ions. This means they contain a. Vitamin B1 i Convulsions
[Old NCERT Page No. 422] b. Vitamin B2 ii Pernicious anaemia
1) basic – NH2 group and acidic – COOH group c. Vitamin B12 iii Beri-beri
d. Vitamin B6 iv cheilosis

2) the basic – NH group and acidic – COO–
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3
[Old NCERT Page No. 426]
group
1) a-iv, b-iii, c-i, d-ii 2) a-i, b-iv, c-iii, d-ii
3) basic – NH2 and acidic – H+ group
3) a-ii, b-i, c-iv, d-iii 4) a-iii, b-iv, c-ii, d-i

–
4) basic – COO group and acidic – NH group Sol. (4)
3

Sol. (4)
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QUESTION BOOKLET VERSION : 11

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
SECTION ‘B’ 90. Match the column I with column II and mark
86. Match Column-I with Column-II the appropriate choice

Column-I Column-II Column - I Column- II

(Natural amino acid) (One letter code) a. Peptide linkage i Inversion
b. Nucleic acid ii Polysaccharide
a. Glutamic acid i. Q
Hydrolysis of
b. Glutamine ii. W c. iii Proteins
cane sugar
c. Tyrosine iii. E
d. Starch iv Nucleotides
d. Tryptophan iv. Y
1) a-ii, b-i, c-iii, d-iv 2) a-iv, b-i, c-ii, d-iii
1) a-ii, b-i, c-iv, d-iii
3) a-iii, b-iv, c-i, d-ii 4) a-i, b-iii, c-iv, d-ii
2) a-iv, b-iii, c-i, d-ii
Sol. (3)
3) a-iii, b-i, c-iv, d-ii
91. Match Column- I with Column- II
4) a-iii, b-iv, c-i, d-ii
Sol. (3) Column-I Column-II
87. Match the column I with column II and mark a. Glucose + HI i.
Gluconic acid
the appropriate choice Glucose
b. Glucose+Br 2 water ii.
pentacetate
Column - I Column- II
c. Glucose + acetic iii. Saccharic acid
sugar + base +
a. Nucleoside i d. Glucose + HNO3 iv. Hexane
phos-phoric aicd
Choose the correct answer from the options
b. Nucleotide ii Cytosine + uracil
given below:
c. DNA iii sugar + base
1) a-iv, b-i, c-ii, d-iii 2) a-iv, b-iii, c-ii, d-i
d. RNA iv Cytosine + thymine
3) a-iii, b-i, c-iv, d-ii 4) a-i, b-iii, c-iv, d-ii
[Old NCERT Page No. 428,429]
Sol. (1)
1) a-iii, b-i, c-iv, d-ii 2) a-i, b-iv, c-iii, d-ii 92. Which of the following statements is/are true
3) a-ii, b-iii, c-i, d-iv 4) a-iv, b-ii, c-i, d-iii for the diagram?
Sol. (1)
88. The linkage between the two monosaccharide
units in lactose is [Old NCERT Page No. 418]
1) C1 of -D-glucose and C4 of -D-galactose
2) C1 of -D-galactose and C4 of -D-glucose
3) C1 of -D-galactose and C4 of -D-glucose
4) C1 of -D-glucose and C4 of -D-galactose
Sol. (2) 1) The escaping tendency of molecule decreases
for each component
89. D - Glucose shows mutarotation because
2) Vapour pressure of the solution decreases
1) it is dextrorotatory
3) Solution shows negative deviation from
2) it undergoes inter conversion between its Raoult’s law
pyranose structure and furanose structure
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4) All of the above

3) it undergoes interconversion between its Sol. (4)
 and (+) Glucopyranose structures
4) it undergoes interconversion with
D(–) fructose
Sol. (3)

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QUESTION BOOKLET VERSION : 11

RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC** RCC**
93. The boiling point of an azeotropic mixture of
water and ethanol is less than that of water 97.
and ethanoL The mixture shows
1) negative deviations from Raoult's law H
|
2) positive deviations from Raoult's law 1) CH 3  C  OD 2)
|
3) no deviation from Raoult's law CH 3
4) deviations which cannot be predicted from
the given information. 3) 4)
Sol. (2) Boiling point of azeotropic mixture is less
when it shows positive deviations. Sol. (3)
94. If 0.1 M solutions of each electrolyte are taken 98. Which of the following reactions can be used
and if all electrolytes are completely to prepared 3-methyl-3-hexanol,
dissociated, then whose boiling point will be
highest ? OH
|
1) Glucose 2) KCl CH 3CH2CCH2CH 3?
|
3) BaCl2 4) K4 [Fe(CN)6 ] CH 3
Sol. (4)
95. True statement about non-ideal solutions is
1) do not obey Raoult’s law over the entire range
1)
of concentration
2) A – A or B–B type interactions > A–B type
interactions
3) A –A or B–B type interactions < A–B type
interactions 2)
4) All of the above
Sol. (4)
O 3)
96. The product obtained when is
OH
oxidised by HIO4, is 4) All of these
Sol. (4)
CHO COOH
1) CHO
2) CHO 99. The product of the following reaction is

COOH CHO
3) COOH
4) CH 2OH

Sol. (2)

1) 2)
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3) 4)

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Sol. (2)

100. In the following reaction,

the expected product(s) is/are

1) 2)

3) 4) All

Sol. (4)
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QUESTION BOOKLET VERSION : 11
Section-C-Biology 108. Primary growth in plants occur mainly by
elongation of main axis by the activity of -
Section A

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1) Cork cambium 2) Apical meristem
101. All cells in higher plants are descendents of-
3) Lateral meristem 4) Vascular cambium.
1) Zygote 2) Endosperm
Ans. (2)
3) Integument 4) Nucellus
109. Analysis of growth at cellular level is done by -
Ans. (1)
1) Increase in protoplasm
102. Growth occurs in plants when -
2) Increase in length
1) Anabolism > Catabolism
3) Increase in number
2) Anabolism < Metabolism
4) Increase in surface area.
3) Metabolism > Catabolism
Ans. (1)
4) Catabolism > Anabolism
110. Which of the plant part is not correctly related
Ans. (1) with the parameter of its growth?
103. Expansion of leaf is - 1) Pollen tube - Length
1) Not an example of growth 2) Dorsiventral leaf - Surface area
2) An example of unlimited growth 3) Meristem - Cell number
3) An example of unlimited differentiation 4) Watermelon cell - Dry weight.
4) An example of limited growth. Ans. (4)
Ans. (4) 111. Volume of a watermelon cell may increase upto-
104. Which set of plant parts tend to show limited 1) 10,000 times.
growth over a period of time?
2) 50,000 times.
1) Leaves and stem
3) 1,00,000 times.
2) Stem, root and flower
4) 3,50,000 times.
3) Flower, leaves and fruits
Ans. (4)
4) Differentiation and senescence
112. Usable enegy of respiration is-
Ans. (3)
1) used in charging biomolecule into activity
105. Growth in plants is -
2) Stored as heat
1) Always indeterminate
2) Generally indeterminate 3) Immediately consumed in cellular activities
3) Always restricted 4) Trapped in ATP molecules
4) Closed type Ans. (4)
Ans. (2) 113. Substrate phosphorylation occurs during
106. Incorrect about plant growth is - 1) Fumaric acid  Malic acid
1) Open type 2) Indeterminate 2) Oxalosuccinic acid  a- ketoglu- taric acid
3) Due to meristems 4) None. 3) Succinic acid  Fumaric acid
Ans. (4) 4) a- ketoglu- taric acid  Succinic acid
107. Select incorrect statement- Ans. (4)
1) Growth is a measurable phenomenon. 114. Respiration involves breaking of _____ bonds
2) Single root apical meristem can give rise to of complex compounds through oxidation
thousands of cells in a hour. within the cells, leading to release of
considerable amount of energy.
3) Plant growth is always indeterminate.
1) C–C 2) –O
4) Growth of pollen tube is measured in terms of
length. 3) C–H 4) Both (2) and (3)
Ans. (3) Ans. (1)

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QUESTION BOOKLET VERSION : 11
115. In electron transport chain, 1 molecule of NADH 121. Which type of phosphorylation occurs in
will regenerate - cytoplasm of Eukaryotic cell?

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1) Oxidative phosophorylation
1) 1 ATP 2) 2 ATP
2) Substrate level phosphorylation
3) 3 ATP 4) 6 ATP 3) Photophosphorylation
Ans. (3) 4) Both (1) and (2)
116. If one molecule of pyruvic acid enters the aerobic Ans. (2)
respiration, then how many NADH are 122. Select the correct statement regarding the
generated? cloning vectors -
1) None 2) 2 1) Plasmids can replicate within bacterial cell but
3) 4 4) 8 not bacteriophage
Ans. (3) 2) Bacteriophages can replicate within bacterial
cell but not plasmids
117. Which step of Kreb's cycle results in release of
CO2 3) Plasmids and bacteriophages, both can
replicate within bacterial cell
1) Isocitric acid  -ketoglutaric acid
4) Neither plasmids nor bacteriophages can
2) Succinyl CoA  Succinate replicate within bacterial cell.
3) Fumarate  Malate Ans. (3)
4) Both (1) and (2) 123. Which of the following is not a desirable feature
of a cloning vector? [NEET-2022]
Ans. (1)
1) Presence of origin of replication
118. Sequence of organic acids in Krebs' cycle is
2) Presence of a marker gene
1) Citric acid  oxalosuccinic acid  isocitric 3) Presence of single restriction enzyme site
acid
4) Presence of two or more rcognition sites of
2) Citric acid  isocitric acid  oxalosuccinic single restriction endo nuclease enzyme.
acid
Ans. (4)
3) Isocitric acid  oxalosuccinic acid  citric 124. In recombinant DNA technology, selection of
acid recombinants can be done by -
4) Oxalosuccinic acid  isocitric acid  citric 1) Inactivation of antibiotic resistant genes
acid
2) Insertional activation
Ans. (2)
3) Both (1) and (2)
119. Phospho-fructo-kinase enzyme is used in 4) Using Ethidium bromide.
converting –
Ans. (3)
1) Pyruvate to glucose
125. Recombinant DNA in host (E.coli) use ____
2) Glucose to Glucose-6-phosphate enzyme to form its multiple copies.
3) Glucose-6-phosphate to fructose-6-phosphate 1) Restriction enzyme 2) DNA ligase
4) Fructose-6-phosphate to fructose-1,6-biphosphate 3) DNA polymerase 4) RNA polymerase.
Ans. (4) Ans. (3)
120. ETS is present in 126. The sequence that controls the copy number of
1) stroma. the linked DNA in the vector, is termed
2) matrix of mitochondria. [NEET-2020]
3) inner membrane of mitochondria. 1) Selectable marker 2) Ori site
4) outer membrane of mitochondria. 3) Palindromic sequence4) Recognition site
Ans. (2)
Ans. (3)
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QUESTION BOOKLET VERSION : 11
127. Which category of products is usually produced 133. The separated bands of DNA are cut out from
by biotechnological applications? the agarose gel and extracted from the gel piece,

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1) Enzymes 2) Antibiotics is reffered as :
3) Vaccines 4) All of the above. 1) Gel electrophoresis
Ans. (4) 2) Elution
128. Number of copies of bacteriophages produced 3) Hybridization
in a bacteria is usually - 4) Spooling
1) Higher than number of plasmids Ans. (2)
2) Equal to the number of plasmids 134. An enzyme catalysing the removal of
3) Lesser than number of plasmids nucleotides from the ends of DNA is
4) Can't say. 1) Endonuclease
Ans. (1) 2) Exonuclease
129. Restriction endonucleases are enzymes which 3) DNA ligase
[AIPMT (Prelims)-2010] 4) Hind - II
1) Remove nucleotides from the ends of the DNA Ans. (2)
molecule 135. Competent host in recombinant DNA
2) Make cuts at specific positions within the DNA technology is
molecule 1) Any human cancer cell
3) Recognize a specific nucleotide sequence for
2) The Cell ready to uptake foreign DNA
binding of DNA ligase
3) A host cell without cell wall
4) Restrict the action of the enzyme DNA
polymerase 4) Agrobacterium cell
Ans. (2) Ans. (2)
130. Specific region or sequence of DNA which is Section B
responsible for initiating replication – 136. What is meant by the open type of growth in
1) Promoter region plants?
2) Operator region 1) New cells are always being added to
3) Origin of replication meristems.
4) Structural gene 2) Restricted growth which can open sometimes.
Ans. (3) 3) Each cell of plant body is capable of division.
131. The introduction of T-DNA into plants involves: 4) Meristematic cells divide continuously to add
1) exposing the plants to cold for a brief period cells to plant body.
2) allowing the plant roots stand in water Ans. (4)
137. Which of the given statement is correct in the
3) infection of the plant by Agrobacterium
context of observing DNA separated by agarose
tumefaciens
gel electrophoriesis?
4) altering the pH of the soil, then heat-shocking
1) DNA can be seen in visible light
the plants.
2) DNA can be seen without staining in visible
Ans. (3)
light.
132. Most important part of Ti- Plasmid at which 3) Ethidium bromide stained DNA can be seen
desired gene is put to target into plant cell is in visible light
1) Ti- gene 2) gene 4) Ethidium bromide stained DNA can be seen
3) ROP gene 4) Ori Site under exposure to UV light
Ans. (1) Ans. (4)

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QUESTION BOOKLET VERSION : 11
138. Read the following statements and select the 141. Read the statements (A-E) and answer the
incorrect ones :- question following them.

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A. In gel electrophoresis, DNA fragments A - BamHI site is located in tetR in pBR322.
separate according to their size through
B - Ti plasmid is obtained from E.coli.
sieving effect provided by the agarose gel.
C - Replication of bacteriophages in bacteria is
B. Larger the fragement size, the farther it moves
independent of chromosomal DNA
C. Unless one cuts the vector and the source
D - First artificial recombinant DNA was
DNA with the same restriction enzyme, the
constructed by Cohen and Boyer in 1972.
recombinant vector molecule cannot be
created. E - In animals, retroviruses can transform the
normal cells into cancerous cells.
D. Now a days the most commonly used matrix
is agrose which is a synthetic polymer How many of the above statement are not
extracted from sea weeds. incorrect?

1) A, B, C, D 1) All
2) One
2) A, C
3) Three
3) B, D
4) Four
4) C, D
Ans. (4)
Ans. (3)
142. Choose the incorrect option for why plants can
139. Analyse the given diagram as below get along without respiratory organs, unlike
animals that have specialised organs for gaseous
exchange.
1) Respiration rate is faster than animals in roots,
stems and leaves.
Which of the following option is not correct 2) O2 released during photosynthesis is utilised
for respiration.
1) Well are located toward negative electrode
3) Loose packing of parenchyma cells in leaves,
2) Toword cathod digested DNA fragement stems and roots facilitates respiration.
3) A stands for largest and B stands for smallest 4) There is very little transport of gases from one
DNA fragments plant part to another.
4) DNA fragments separate according to their size Ans. (1)
through sieving effect of agarose gel
143. Among the following, select the tools of
Ans. (2) recombinant DNA technology :
140. Which set of meristems appear later in life in A. Restriction enzyme
dicots and gymnosperms and increase the girth B. Polymerase enzyme
of organs?
C. Ligases
1) Apical meristem, Intercalary meristem.
D. Vectors
2) Lateral meristem, Vascular meristem, Apical E. Host organisms
meristem.
1) A, D, E
3) Intercalary meristem, Cork cambium. Lateral
2) A, D, C, E
meristem.
3) A, B, C, D, E
4) Lateral meristem Like Cork cambium and
Vascular meristem. 4) Only A
Ans. (3)
Ans. (4)
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QUESTION BOOKLET VERSION : 11
144. Indeterminate growth in plants is due to - 148. Plant body possess actively dividing cells at
1) Apical cell present in shoot apex only. certain locations which are giving rise to other

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2) Apical cell present in shoot and root apex both. cells. Cells formed later -
3) Meristems located throughout the plant body. 1) Have capacity of division and give rise to new
4) Meristems located at certain positions in plant meristems.
body. 2) Are known as meristems.
Ans. (4) 3) Lack capacity of division and are differentiated
145. Read the following passage carefully and select to perform specific function.
the option with all correct answer.
4) Are totipotent cells.
In glycolysis, glucose undergoes _(a) to form two
Ans. (3)
molecules of pyruvic acid. in plants, this glucose
is derived from _(b)_, which is converted into 149. Statement I : Starch is converted into glucose and
glucose and fructose by the enzyme, _(c)_, and fructose by the enzyme invertase.
these two monosaccharides readily enter the Statement II : Fructose-6-phosphate is split into
glycolytic pathway. _(d)_ is then the key product DHAp and PGAL.
of glycolysis. 1) Only statement I is correct
1) (a)-oxidation; (b)-galactose; (c)-hexokinase; 2) Only statement II is correct
(d)-oxygen 3) Both statements I and II are correct
2) (a)-reduction; (b)-sucrose; (c)-glucokinase; 4) Both statements I and II are incorrect
(d)-Acetyl co-A Ans. (4)
3) (a)-partial oxidation; (b)-sucrose; (c)-invertase;
150. Oxidative decarboxylation is summarised in the
(d)-Pyruvic acid
following reaction. A is an enzyme for which B
4) (a)-hydrolysis; (b)-maltose; (c)-zymas;
is a cofactor (mineral activator) and C is one of
(d)-Pyruvate
the product. Identify them correctly.
Ans. (3)
146. Match the Columns. B
Pyruvic acid + CoA + NAD 
A
Column I Column II
a. Complex I 1. Cytochrome bc1 complex Acetyl CoA + C + NADH + H+
b. Complex II 2. NADH dehydrogenase A B C
c. Complex III 3. Cytochrome c oxidase 1) Pyruvate synthase Mg 2+
CO2
d. Complex IV 4. ATP synthase 2) Pyruvate synthase Cu 2+
ATP
e. Complex V 5. FADH2 dehdrogenase 2+
3) Pyruvate dehydrogenase Cu ATP
1) a - 3, b - 4, c - 2, d - 5, e - 1
2+
2) a - 4, b - 5, c - 1, d - 2, e - 3 4) Pyruvate dehydrogenase Mg CO2
3) a - 2, b - 4, c - 3, d - 1, e - 5 Ans. (4)
4) a - 2, b - 5, c - 1, d - 3, e - 4 Section-D-Biology
Ans. (4)
147. Statement I: Restriction endonucleases recognise
Section A
specific sequence to cut DNA known as 151. If we added alien DNA at site Sal I in pBR322
palindromic nucleotide sequence. and E. Coli uptake recombinant from medium
Statement II: Hind II cut the DNA strand a little than
away from the centre of the palindromic site. 1) transformed colonies become sensitive for
In the light of the above statements, choose the tetracyclin
most appropriate answer from the options given 2) transformed colonies remains sensitive for
below : [NEET-2022] ampicilin
1) Both Statement I and Statement II are correct 3) transformed colonies remains resistant for both
2) Both Statement I and Statement II are incorrect tetracycline and ampicilin
3) Statement I is correct but Statement II is incorrect 4) proteins involved in the replication of the
4) Statement I is incorrect but Statement II is correct plasmid does not formed.
Ans. (3) Ans. (1)

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QUESTION BOOKLET VERSION : 11
152. In convention for naming restriction 158. Basmati is unique for its aroma and flavour,
endonuclease, second and third letter come from: whose A… documented varieties are cultivated

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1) Species of the prokaryotic cell from which they in B….
were isolated. Here, A and B refer to
2) Species of the eukaryotic cell on which they 1) A–27, B–America 2) A–30, B–America
shows action
3) A–27, B–India 4) A–30, B–India
3) Genus of the prokaryotic cell from which they
were isolated Ans. (3)
4) Genus of the eukaryotic cell on which they 159. Match the following columns.
shows action.
Column-I Column-II
Ans. (1) a Golden rice i Armyworm
153. In pBR322 ampR site have recognition sequence b Bt toxin ii Rich in vitamin-A
for :
c RNAi iii Cry protein
1) Pvu I and BamH I 2) Pvu I and Pst I
d Lepidopterans iv Gene silencing
3) BamH I and Sal I 4) BamH II and Hind III
1) a-ii, b-iii, c-iv, d-i 2) a-iii, b-iv, c-i, d-ii
Ans. (2)
3) a-iv, b-i, c-ii, d-iii 4) a-ii, b-i, c-iii, d-iv
154. Which site on vector must be present to make it
autonomously replicate in host cell? Ans. (1)
1) Selective marker 2) Recomgnition site 160. Choose the correct option regarding retrovirus.
3) Ori site 4) All of these 1) A RNA virus that can synthesise DNA during
Ans. (3) infection
155. Which one of the following molecular diagnostic 2) A DNA virus that can synthesise RNA during
techniques is used to detect the presence of a infection
pathogen in its early stage of infection? 3) A ssDNA virus
1) Angiography 4) A dsRNA virus
2) Radiography Ans. (1)
3) Enzyme replacement technique 161. In RNAi, genes are silenced using
4) Polymerase Chain Reaction (PCR) 1) ssDNA 2) dsDNA
Ans. (4) 3) dsRNA 4) ssRNA
156. Which of the following techniques is based on Ans. (3)
the principle of antigen-antibody interaction?
162. The site of production of ADA in the body is
1) PCR
1) erythrocytes 2) lymphocytes
2) ELISA
3) blood plasma 4) osteocytes
3) Recombinant DNA technology
Ans. (2)
4) Gene therapy
163. A probe which is a molecule used to locate
Ans. (2) homologous sequences in a mixture of DNA or
157. The first transgenic cow, Rosie produced RNA molecules could be
1) human protein enriched milk (2.4 g/L) 1) a single-stranded RNA
2) human protein enriched milk (2.8 g/L) 2) a single-stranded DNA
3) human calcium enriched milk (2.4 g/L) 3) either single-stranded RNA or DNA
4) human calcium enriched milk (2.8 g/L) 4) can be ssDNA but not ssRNA
Ans. (1) Ans. (3)

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QUESTION BOOKLET VERSION : 11
164. GEAC stands for 169. Phyllotaxy of sunflower is similar to that of -
1) Genome Engineering Action Committee 1) China rose 2) Calotropis

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2) Ground Environment Action Committee 3) Guava 4) Alstonia.
3) Genetic Engineering Approval Committee Ans. (1)
4) Genetic and Environment Approval Committee 170. Which of the following is incorrect ?
Ans. (3) 1) Root helps in water and mineral absorption
from soil
165. In floral meristem,
2) Roots provide a proper anchorage
1) Internodes do not elongate.
3) Roots store food material and synthesise plant
2) Axis gets condensed.
growth regulators
3) The apex produced different kinds of floral
4) Roots lack meristematic activity
appendages laterally at successive nodes
instead of leaves. Ans. (4)
4) All the above 171. Select the option with correct location of
bulliform cells?
Ans. (4)
1) Adaxial surface, monocot leaf
166. Pulvinus leaf base found in :
2) Abaxial surface, monocot leaf
1) Oil plant 2) Resinus plant
3) Adaxial surface, dicot leaf
3) Leguminous plant 4) Cereals plants
4) Abaxial surface, dicot leaf.
Ans. (3)
Ans. (1)
167. Select the correct option with respect to
Rhizophora. 172. Identify the true statements.
a. Grows in swampy areas i) Stem tendrils develop from axillary buds found
b. Pneumatophores are present for gaseous in gourds.
exchange ii) Cucumber, pumpkins, watermelon and grape
c. Many roots come out of the ground and grow have leaf tendrils.
vertical upwards iii)Axillary buds of stems may also get modified
d. Shows vivipary into thorns.
1) a and b only 2) a, b and c only iv) Citrus and Bougainvillea have leaf spine.
3) a and c only 4) all a, b, c and d v) Flattened chlorophyllus stems found in
Ans. (4) Euphorbia and fleshy cylindrical chlorophyllus
stems found in Opuntia.
168. Read the following statement about stem and
identify them as true(T) or false(F). 1) (i), (ii), (iii) and (iv)
A. Asceding part of plant axis 2) (i) and (iii)
B. Develops from radicle of the embryo of a 3) (i), (iii) and (iv)
germinating seed
4) (ii), (iv) and (v)
C. Generally green when older but brown when
Ans. (2)
young
D. It may preform function of vegetative propagation 173. Vascular bundles surrounded by bundle sheath
cells can be observed in :
A B C D
1) Monocot stem
1) T F F T
2) T T F T 2) Monocot leaf
3) T T F F 3) Dicot root
4) T F F F 4) Dicot leaf.
Ans. (1) Ans. (4)
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QUESTION BOOKLET VERSION : 11
174. Identify the type of vascular bundle as shown 178. The term Leptome refers to
in the figure and select the incorrect statement 1) conducting part of the phloem

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regarding it. 2) non-conducting part of phloem
3) conducting part of xylem
4) non-conducting part of xylem
Ans. (1)
179. Which of the following is incorrect? [NCERT Pg- 87,
91]
1) The root hairs are unicellular
1) Figure represents radial vascular bundles in 2) The first formed parimary phloem consists of
which xylem and phloem occur in the form of narrow sieve tubes and is called protophloem
separate bundles. whereas the later formed phloem has bigger
2) Xylem bundles and phloem bundles occur on sieve tube and is called metaphloem
different radii. 3) In stems, the protoxylem lie towards the pith
3) These are the characteristic of monocot and and metaxylem towards the periphery whereas
dicot leaves. in roots it is vice-versa
4) These are the characteristic of roots. 4) Trichomes in shoot systems are usually
unicellular
Ans. (3)
Ans. (4)
175. Select the mismatched pair out of the following. 180. In the T.S. of root :- [NCERT Pg- 91]
1) Radial vascular bundle – Xylem and phloem 1) Protoxylem and metaxylem are not present on
on different radii same radius
2) Bicollateral vascular – Phloem present on 2) Protoxylem is absent
bundle both sides of xylem 3) Protoxylem towards inside and metaxylem
3) Amphivasal vascular – Phloem surrounds towards outside
bundle xylem 4) Metaxylem is towards inside and protoxylem
towards outside
4) Conjoint vascular – Xylem and phloem on
bundle same radii Ans. (4)
181. Sieve tube differs from vessels in : [NCERT Pg- 87,
Ans. (3)
88]
176. The innermost layer of cortex in a dicot root :
1) Absence of nucleus
[NCERT Pg- 91]
2) Less deposition of lignin
1) Comprises a single layer of barrel shaped cells 3) Being dead
without any intercellular spaces 4) Lack of cytoplasm
2) Are composed of cells which have depositions Ans. (2)
of impermeable cutin and lignin mainly in their
182. Read the following and select out true
radial and tangential walls
statement:- [NCERT Pg- 86, 87]
3) Comprises a group of dead cells A) The cells of parenchyma are generally
4) Contain abundant intercellular spaces isodiametric
Ans. (1) B) Cell wall of xylem parenchyma are made up
177. Trichomes are of cellulose
1) epidermal hair of stem C) Xylem vessels and tracheids both are
multicellular
2) either soft or stiff
1) Only A 2) A and B
3) branched or unbranched
3) Only C 4) All A, B and C
4) all of these
Ans. (2)
Ans. (4)

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QUESTION BOOKLET VERSION : 11
183. In stem and root respectively the arrangement 188. Match the following columns.
of primary xylem is : [NCERT Pg- 87] Column-I Column-II

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1) Exarch and Endarch 2) Endarch and Exarch a Gene therapy i Effort to replace non-
3) Exarch and Exarch 4) Endarch and Endarch functional gene
Ans. (2) b Humulin ii A single-stranded DNA or
184. Pith or central part of ground tissue is made of RNA tagged with a
[NCERT Pg- 93] radioactive molecule
1) Collenchyma 2) Chlorenchyma c Probe iii Diagnostic test
3) Parenchyma 4) Sclerenchyma d ELISA iv Diabetes
Ans. (3) 1) a-i, b-iv, c-ii, d-iii 2) a-iv, b-ii, c-iii, d-i
185. Stele: [NCERT Pg- 91] 3) a-ii, b-iii, c-i, d-iv 4) a-iii, b-i, c-iv, d-ii
1) Constitute all the tissue inner side of pellogen Ans. (1)
2) Constitute all the tissues inner side of cork 189. Read the following four statements (A-D) :-
3) Constitute all the tissue inner side of vascular [NCERT Pg- 88]
cambium A) Bast fibre are made up of sclerenchymatous
4) Constitute all the tissue inner side of cells.
endodermis B) Bast fibre are absent in primary phloem.
Ans. (4) C) At maturity sieve tube element have
Section B developed nucleus.
186. Read the following statement and choose D) At maturity bast fibre lose their protoplasm
incorrect one :- and become dead.
1) DNA is hydrophobic molecule, it cannot pass How many of the above statements are correct?
through cell membrane. 1) Four 2) One
2) By micro-injection, recombinant DNA is 3) Two 4) Three
directly injected into the nucleus of an animals Ans. (4)
cell 190. In a dorsiventral leaf : [NCERT Pg- 93]
3) During DNA isolation we used chilled ethanol 1) The abaxial epidermis generally bears less
in final step number of stomata than the adaxial epidermis
4) Retoviruses have been use to deliver desirable 2) The upper epidermis may lack stomata
gene into animal cells after disarming. 3) Spongy parenchyma extends to the upper
Ans. (1) epidermis
4) Spongy parenchyma contain compactly
187. A diagram of T.S. of dicot root is given. Select the
arranged cells without intercellular spaces
option which correctly labels A, B, C, D, and E.
Ans. (2)
191. The transverse section of a plant shows
following anatomical features: [NEET-2020]
a) Large number of scattered vascular bundles
surrounded by bundle sheath.
b) Large conspicuous parenchymatous ground
1) A - Protoxylem, B - Metaxylem, C - Phloem, D tissue.
- Pericyde, E - Endodermis c) Vascular bundles conjoint and closed.
2) A - Metaxylem, B - Protoxylem, C - Phloem, D d) Phloem parenchyma absent.
- Pericyde, E - Endodermis Identify the category of plant and its part:
3) A - Protoxylem, B - Metaxylem, C - Phloem, D 1) Monocotyledonous stem
- Endodermis, E - Pericyde 2) Monocotyledonous root
4) A - Metaxylem, B - Protoxylem, C - Phloem, D 3) Dicotyledonous stem
- Endodermis, E - Pericyde 4) Dicotyledonous root
Ans. (2) Ans. (1)

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QUESTION BOOKLET VERSION : 11
192. Match the columns [NCERT Pg- 93]
196. Match the columns :
Column-I Column-II

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A. Bean shaped guard i. Dicot stem Column-I Column-II
cell a Prop root i Ficus benghalensis
B. Dumb-bell shaped ii. Monocot leaf b Stilt root ii Zea mays
guard cell c Respiratory root iii Pea
C. Trichome iii. Dicot leaf d Nodulated root iv Rhizophora
D. Exarch xylem iv. Dicot and monocot 1) a-i, b-ii, c-iii, d-iv 2) a-iv, b-iii, c-ii, d-i
root 3) a-ii, b-i, c-iii, d-iv 4) a-i, b-ii, c-iv, d-iii
1) A-ii, B-iii, C-i, D-iv 2) A-iii, B-ii, C-i, D-iv
Ans. (4)
3) A-iv, B-iii, C-i, D-ii 4) A-iii, B-iv, C-ii, D-i
197. The source of complementary RNA for RNA
Ans. (2) interference as a means of cellular defence in
193. Statement I : Canna flower is an asymmetric eukaryotic organism is
flower.
1) an infection by bacteria having RNA genome
Statement II : Canna flower cannot be cut into
2) an infection by viruses having RNA genome
two similar halves in any vertical plane.
3) mobile genetic elements that replicate via an
1) Statements I & II are true.
RNA intermediate
2) Statement I & II are false.
3) Statement I is true & Statement II is false. 4) Both (2) and (3)
4) Statement I is false & Statement II is true. Ans. (4)
Ans. (1) 198. Consider the following statements.
194. Which of the following statement true about I. Earlier, insulin was extracted from pancreas
intercalary meristem. [NCERT Pg- 85] of slaughtered cattle and pigs which was more
a) They occure between mature tissue efficient than the genetically engineered
b) They occure in grasses and regenerate parts insulin.
removed by the grazing herbivores. II. PCRtechnique is being used for the detection
c) They appear early in life of a plant and of HIV in suspected AIDS patients and
contribute to the formation of the primary genetic mutations in suspected cancer
plant body patients.
d) They are responsible for producing the III. ADA defeciency is treated by gene therapy.
secondary tissue Which of the above statements are correct?
1) Only a, b & c are correct 1) I and II 2) I and III
2) Only a & b are correct 3) II and III 4) None of these
3) Only b & d are correct
Ans. (3)
4) Only a & c are correct
199. Adenosine Deaminase (ADA) deficiency can be
Ans. (1) cured by …A… and …B… but it is not fully
195. Name the petals A, B and C in vexillary curative.
aestivation shown in the above figure - Here, A and B can be
1) A–gene therapy, B–radiation therapy
2) A–bone marrow transplantation, B–enzyme
replacement therapy
3) A–organ transplantation, B–hormone
1) A-Standard, B-Wing, Perianth
replacement therapy
2) A-Standard, B-Keel, C-Wing
4) A–radiation therapy, B–enzyme replacement
3) A-Wing, B-Keel, C-Wing
therapy
4) A-Standard, B-Wing, C-Keel
Ans. (2)
Ans. (4)

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QUESTION BOOKLET VERSION : 11
200. Choose the correct statements for the uses of PCR
technique in diagnosis.

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I. It is used to detect HIV in suspected AIDS
patients.
II. It is used to detect mutations in the genes in
suspected cancer patients.
III. It is used to detect different common diseases
in pigs, sheep and cows.
IV. It is a good technique to identify many other
genetic disorders.
1) I and II 2) III and IV
3) I, II and IV 4) II, III and IV
Ans. (3)

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